diff --git "a/ap_stat_inference_data.csv" "b/ap_stat_inference_data.csv"
deleted file mode 100644--- "a/ap_stat_inference_data.csv"
+++ /dev/null
@@ -1,1015 +0,0 @@
-Resource,Chapter,Number,Question,Answer,Type,,,,
-TPS,8.1,1,"Identify the point estimator you would use to estimate the parameter and calculate the value of the point estimate: How many pairs of shoes, on average, do female teens have? To find out, an AP ® Statistics class selected an SRS of 20 female students from their school. Then they recorded the number of pairs of shoes that each student reported having. Here are the data: 50, 26, 26, 31, 57, 19, 24, 22, 23, 38, 13, 50, 13,  34, 23, 30, 49, 13, 15, 51","Sample mean, x_bar = 30.35",Free Response,,,,
-TPS,8.1,2,"Identify the point estimator you would use to estimate the parameter and calculate the value of the point estimate: What is the variability in the number of pairs of shoes that female students have? To find out, an AP ® Statistics class selected an SRS of 20 female students from their school. Then they recorded the number of pairs of shoes that each student reported having. Here are the data: 50, 26, 26, 31, 57, 19, 24, 22, 23, 38, 13, 50, 13, 34, 23, 30, 49, 13, 15, 51","Sample standard deviation, sx = 14.24",Free Response,,,,
-TPS,8.1,3,Identify the point estimator you would use to estimate the parameter and calculate the value of the point estimate: Tonya wants to estimate the pro-portion of seniors in her school who plan to attend the prom. She interviews an SRS of 50 of the 750 seniors in her school and finds that 36 plan to go to the prom.,"Sample Proportion, p^ = 36/50 = 0.72",Free Response,,,,
-TPS,8.1,4,Identify the point estimator you would use to estimate the parameter and calculate the value of the point estimate: What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172 undergraduates at a large university: “You witness two students cheating on a quiz. Do you go to the professor?” Only 19 answered “Yes.”,"Sample porportion, p^ = 19/172 = 0.11",Free Response,,,,928
-TPS,8.1,5a,"A New York Times /CBS News Poll asked a random sample of U.S. adults the question “Do you favor an amendment to the Constitution that would permit organized prayer in public schools?” Based on this poll, the 95% confidence interval for the population proportion who favor such an amendment is (0.63, 0.69). Interpret the confidence interval.",We are 95% confident that the interval from 0.63 to 0.69 captures the true proportion of all U.S. adults who favor an amendment to the Constitution that would permit organized prayer in public schools.,Free Response,,,,
-TPS,8.1,5b,"A New York Times /CBS News Poll asked a random sample of U.S. adults the question “Do you favor an amendment to the Constitution that would permit organized prayer in public schools?” Based on this poll, the 95% confidence interval for the population proportion who favor such an amendment is (0.63, 0.69). What is the point estimate that was used to create the interval? What is the margin of error?",The point estimate = p^ = (0.63 + 0.69) / 2 = 0.66; the margin of error is = 0.69-0.66 = 0.03,Free Response,,,,
-TPS,8.1,5c,"A New York Times /CBS News Poll asked a random sample of U.S. adults the question “Do you favor an amendment to the Constitution that would permit organized prayer in public schools?” Based on this poll, the 95% confidence interval for the population proportion who favor such an amendment is (0.63, 0.69). Based on this poll, a reporter claims that more than two-thirds of U.S. adults favor such and amendment. Use the confidence interval to evaluate this claim.","Because the value 2/3 5 0.667 (and values less than 2/3) are in the interval, it is plausible that two-thirds or less of the population favor such an amendment. There is not convincing evidence that more than two-thirds of U.S. adults favor such an amendment.",Free Response,,,,
-TPS,8.1,6a,"A Gallup poll asked a random sample of U.S. adults, “Would you like to lose weight?” Based on this poll, the 95% confidence interval for the population proportion who want to lose weight is (0.56, 0.62). Interpret the confidence interval.",We are 95% confident that the interval from 0.56 to 0.62 captures the true proportion of all U.S. adults who would like to lose weight.,Free Response,,,,
-TPS,8.1,6b,"A Gallup poll asked a random sample of U.S. adults, “Would you like to lose weight?” Based on this poll, the 95% confidence interval for the population proportion who want to lose weight is (0.56, 0.62). What is the point estimate that was used to create the interval? What is the margin of error?",The point estimate = p^ = (0.56 + 0.62) / 2 = 0.59; the margin of error is = 0.62-0.59 = 0.03,Free Response,,,,
-TPS,8.1,6c,"A Gallup poll asked a random sample of U.S. adults, “Would you like to lose weight?” Based on this poll, the 95% confidence interval for the population proportion who want to lose weight is (0.56, 0.62). Based on this pool, Gallop claims that more than half of U.S. adults want to lose weight. Use the confidence interval to evaluate this claim.","Because all the plausible values in the interval are greater than 0.5, there is convincing evidence that more than half of U.S. adults want to lose weight.",Free Response,,,,
-TPS,8.1,7a,"A particular type of diet cola advertises that each can contains 12 ounces of the beverage. Each hour, a supervisor selects 10 cans at random, measures their contents, and computes a 95% confidence interval for the true mean volume. For one particular hour, the 95% confidence interval is 11.97 ounces to 12.05 ounces. Does the confidence interval provide convincing evidence that the true mean volume is different than 12 ounces? Explain your answer.","Because 12 is one of the plausible values in the 95% confidence interval, there is not convincing evidence that the true mean volume is different from 12 ounces.",Free Response,,,,
-TPS,8.1,7b,"A particular type of diet cola advertises that each can contains 12 ounces of the beverage. Each hour, a supervisor selects 10 cans at random, measures their contents, and computes a 95% confidence interval for the true mean volume. For one particular hour, the 95% confidence interval is 11.97 ounces to 12.05 ounces. Does the confidence interval provide convincing evidence that the true mean volume is 12 ounces? Explain your answer.","No; although 12 is a plausible value for the true mean volume, there are many other plausible values in the confidence interval. Because any of these values could be the true mean, there is not convincing evidence that the true mean volume is 12 ounces.",Free Response,,,,
-TPS,8.1,8a,A candy bar manufacturer sells a “fun size” version that is advertised to weigh 17 grams. A hungry teacher selected a random sample of 44 fun size bars and found a 95% confidence interval for the true mean weight to be 16.945 grams to 17.395 grams. Does the confidence interval provide convincing evidence that the true mean weight is different than 17 grams? Explain your answer.,"Because 17 is one of the plausible values in the 95% confidence interval, there is not convincing evidence that the true mean weight is different than 17 grams.",Free Response,,,,
-TPS,8.1,8b,A candy bar manufacturer sells a “fun size” version that is advertised to weigh 17 grams. A hungry teacher selected a random sample of 44 fun size bars and found a 95% confidence interval for the true mean weight to be 16.945 grams to 17.395 grams. Does the confidence interval provide convincing evidence that the true mean weight is 17 grams? Explain your answer.,"No; although 17 is a plausible value for the true mean weight, there are many other plausible values in the confidence interval. Because any of these values could be the true mean, there is not convincing evidence that the true mean weight is 17 grams.",Free Response,,,,
-TPS,8.1,9a,The AP ® Statistics class in Exercise 1 also asked an SRS of 20 boys at their school how many pairs of shoes they have. A 95% confidence interval for μG – μB = the true difference in the mean number of pairs of shoes for girls and boys is 10.9 to 26.5. Interpret the confidence interval.,We are 95% confident that the interval from 10.9 to 26.5 captures the true difference (Girls – Boys) in the mean number of pairs of shoes at this school.,Free Response,,,,
-TPS,8.1,9b,The AP ® Statistics class in Exercise 1 also asked an SRS of 20 boys at their school how many pairs of shoes they have. A 95% confidence interval for μG – μB = the true difference in the mean number of pairs of shoes for girls and boys is 10.9 to 26.5. Does the confidence interval give convincing evidence of a difference in the true mean number of pairs of shoes for boys and girls at the school? Explain your answer.,"Yes; because the 95% confidence interval does not include 0 as a plausible value for the difference in means, there is convincing evidence of a difference in the mean number of shoes for boys and girls.",Free Response,,,,
-TPS,8.1,10a,"Many teens have posted profiles on sites such as Facebook. A sample survey asked random samples of teens with online profiles if they included false information in their profiles. Of 170 younger teens (ages 12 to 14) polled, 117 said “Yes.” Of 317 older teens (ages 15 to 17) polled, 152 said “Yes.” A 95% confidence interval for pY – pO = the true difference in the proportions of younger teens and older teens who include false information in their profile is 0.120 to 0.297. Interpret the confidence interval.",We are 95% confident that the interval from 0.120 to 0.297 captures the true difference (Younger – Older) in the proportions of all teens who include false information on their profiles. ,Free Response,,,,
-TPS,8.1,10b,"Many teens have posted profiles on sites such as Facebook. A sample survey asked random samples of teens with online profiles if they included false information in their profiles. Of 170 younger teens (ages 12 to 14) polled, 117 said “Yes.” Of 317 older teens (ages 15 to 17) polled, 152 said “Yes.” A 95% confidence interval for pY – pO = the true difference in the proportions of younger teens and older teens who include false information in their profile is 0.120 to 0.297. Does the confidence interval give convincing evidence of a difference in the true proportions of younger and older teens who include false information in their profiles? Explain your answer.","Yes; because the 95% confidence interval does not include 0 as a plausible value for the difference in proportions, there is convincing evidence of a difference in the proportion of younger teens and older teens who include false information on their profiles.",Free Response,,,,
-TPS,8.1,11,"A New York Times /CBS News Poll asked a random sample of U.S. adults the question “Do you favor an amendment to the Constitution that would permit organized prayer in public schools?” Based on this poll, the 95% confidence interval for the population proportion who favor such an amendment is (0.63, 0.69). Interpret the confidence level.","If we were to select many random samples of the same size from the population of U.S. adults and construct a 95% confidence interval using each sample, about 95% of the intervals from each random sample would capture the true proportion of all U.S. adults who would favor an amendment to the Constitution that would permit organized prayer in public schools. ",Free Response,,,,
-TPS,8.1,12,"A Gallup poll asked a random sample of U.S. adults, “Would you like to lose weight?” Based on this poll, the 95% confidence interval for the population proportion who want to lose weight is (0.56, 0.62). Interpret the confidence level.","If we were to select many random samples of the same size from the population of U.S. adults and construct a 95% confidence interval using each sample, about 95% of the intervals would capture the true proportion of all U.S. adults who want to lose weight. ",Free Response,,,,
-TPS,8.1,13a,"The 2015 American Community Survey estimated the median household income for each state. According to ACS, the 90% confidence interval for the 2015 median household income in Arizona is $51,492 ± $431. Interpret the confidence interval.","The confidence interval is $51,492 - $431 = $51,061 to $51,492 - $431 = $51,923. We are 90% confident that the interval from $51,061 to $51,923 captures the true median household income for all households in Arizona in 2015. ",Free Response,,,,
-TPS,8.1,13b,"The 2015 American Community Survey estimated the median household income for each state. According to ACS, the 90% confidence interval for the 2015 median household income in Arizona is $51,492 ± $431. Interpret the confidence level.",About 90% of the intervals from the random samples would capture the true median household income for all households in Arizona in 2015.,Free Response,,,,
-TPS,8.1,14a,"The 2015 American Community Survey estimated the median household income for each state. According to ACS, the 90% confidence interval for the 2015 median household income in New Jersey is $72,222 ± $610. Interpret the confidence interval.","The confidence interval is $71,612 to $72,832. We are 90% confident that the interval from $71,612 to $72,832 captures the true median household income for all households in New Jersey in 2015. ",Free Response,,,,
-TPS,8.1,14b,"The 2015 American Community Survey estimated the median household income for each state. According to ACS, the 90% confidence interval for the 2015 median household income in New Jersey is $72,222 ± $610. Interpret the confidence level.",About 90% of the intervals from the random samples would capture the true median household income for all households in New Jersey in 2015.,Free Response,,,,
-TPS,8.1,17a,"A 95% confidence interval for the mean body mass index (BMI) of young American women is 26.8 ± 0.6. Discuss whether the following explanation is correct, based on that information: We are confident that 95% of all young women have BMI between 26.2 and 27.4.","Incorrect; the interval provides plausible values for the mean BMI of all women, not plausible values for individual BMI measurements, which will be much more variable.",Free Response,,,,
-TPS,8.1,17b,"A 95% confidence interval for the mean body mass index (BMI) of young American women is 26.8 ± 0.6. Discuss whether the following explanation is correct, based on that information: We are 95% confident that future samples of young women will have mean BMI between 26.2 and 27.4.",Incorrect; we shouldn’t use the results of one sample to predict the results for future samples.,Free Response,,,,
-TPS,8.1,17c,"A 95% confidence interval for the mean body mass index (BMI) of young American women is 26.8 ± 0.6. Discuss whether the following explanation is correct, based on that information: Any value from 26.2 to 27.4 is believable as the true mean BMI of young American women.",Correct; a confidence interval provides an interval of plausible values for a parameter. ,Free Response,,,,
-TPS,8.1,17d,"A 95% confidence interval for the mean body mass index (BMI) of young American women is 26.8 ± 0.6. Discuss whether the following explanation is correct, based on that information: If we take many samples, the population mean BMI will be between 26.2 and 27.4 in about 95% of those samples.","Incorrect; the population mean always stays the same, regardless of the number of samples taken. ",Free Response,,,,
-TPS,8.1,17e,"A 95% confidence interval for the mean body mass index (BMI) of young American women is 26.8 ± 0.6. Discuss whether the following explanation is correct, based on that information: The mean BMI of young American women cannot be 28.","Incorrect; we are 95% confident that the population mean is between 26.2 and 27.4, but that doesn’t rule out any other possibilities. ",Free Response,,,,
-TPS,8.1,18a,"The admissions director for a university found that (107.8, 116.2) is a 95% confidence interval for the mean IQ score of all freshmen. Discuss whether the following explanation is correct, based on that information: There is a 95% probability that the interval from 107.8 to 116.2 contains μ.","Incorrect; the population mean is always the same, so the probability that m is in a particular interval is either 0 or 1 (but we don’t know which). ",Free Response,,,,
-TPS,8.1,18b,"The admissions director for a university found that (107.8, 116.2) is a 95% confidence interval for the mean IQ score of all freshmen. Discuss whether the following explanation is correct, based on that information: There is a 95% chance that the interval (107.8, 116.2) contains x_bar.","Incorrect; the point estimate x will always be in the center of the confidence interval, so there is a 100% chance that x will be in the interval. ",Free Response,,,,
-TPS,8.1,18c,"The admissions director for a university found that (107.8, 116.2) is a 95% confidence interval for the mean IQ score of all freshmen. Discuss whether the following explanation is correct, based on that information: This interval was constructed using a method that produces intervals that capture the true mean in 95% of all possible samples.",Correct; this is the meaning of 95% confidence.,Free Response,,,,
-TPS,8.1,18d,"The admissions director for a university found that (107.8, 116.2) is a 95% confidence interval for the mean IQ score of all freshmen. Discuss whether the following explanation is correct, based on that information: If we take many samples, about 95% of them will contain the interval (107.8, 116.2).",Incorrect; it doesn’t make sense to say that a sample contains an interval.,Free Response,,,,
-TPS,8.1,18e,"The admissions director for a university found that (107.8, 116.2) is a 95% confidence interval for the mean IQ score of all freshmen. Discuss whether the following explanation is correct, based on that information: The probability that the interval (107.8, 116.2) captures μ is either 0 or 1, but we don’t know which.","Correct; the value of m is always the same, so it is either always in a particular interval or always not in a particular interval.",Free Response,,,,
-TPS,8.1,19a,"A New York Times /CBS News Poll asked a random sample of U.S. adults the question “Do you favor an amendment to the Constitution that would permit organized prayer in public schools?” Based on this poll, the 95% confidence interval for the population proportion who favor such an amendment is (0.63, 0.69). Explain what would happen to the length of the interval if the confidence level were increased to 99%.",The length would increase. ,Free Response,,,,
-TPS,8.1,19b,"A New York Times /CBS News Poll asked a random sample of U.S. adults the question “Do you favor an amendment to the Constitution that would permit organized prayer in public schools?” Based on this poll, the 95% confidence interval for the population proportion who favor such an amendment is (0.63, 0.69). How would the width of a 95% confidence interval based on a sample size 4 times as large compare to the original 95% interval, assuming the sample proportion remained the same?",The confidence interval would be half as wide. ,Free Response,,,,
-TPS,8.1,19c,"A New York Times /CBS News Poll asked a random sample of U.S. adults the question “Do you favor an amendment to the Constitution that would permit organized prayer in public schools?” Based on this poll, the 95% confidence interval for the population proportion who favor such an amendment is (0.63, 0.69). The news article goes on to say: “The theoretical errors do not take into account additional errors resulting from the various practical difficulties in taking any survey of public opinion.” List some of the “practical difficulties” that may cause errors which are not included in the 3± percentage point margin of error.","One of the practical difficulties would include nonresponse. For example, if people selected but not responding have different views from those responding, the estimated proportion may be off by more than 3 percentage points. ",Free Response,,,,
-TPS,8.1,20a,"A Gallup poll asked a random sample of U.S. adults, “Would you like to lose weight?” Based on this poll, the 95% confidence interval for the population proportion who want to lose weight is (0.56, 0.62). Interpret the confidence interval. Explain what would happen to the length of the interval if the confidence level was decreased to 90%.",The length would decrease. ,Free Response,,,,
-TPS,8.1,20b,"A Gallup poll asked a random sample of U.S. adults, “Would you like to lose weight?” Based on this poll, the 95% confidence interval for the population proportion who want to lose weight is (0.56, 0.62). Interpret the confidence interval. How would the width of a 95% confidence interval based on a sample size 4 times as large compare to the original 95% interval, assuming the sample proportion remained the same?",The confidence interval would be half as wide. ,Free Response,,,,
-TPS,8.1,20c,"A Gallup poll asked a random sample of U.S. adults, “Would you like to lose weight?” Based on this poll, the 95% confidence interval for the population proportion who want to lose weight is (0.56, 0.62). Interpret the confidence interval. As Gallup indicates, the 3 percentage point margin of error for this poll includes only sampling variability (what they call “sampling error”). What other potential sources of error (Gallup calls these “nonsampling errors”) could affect the accuracy of the 95% confidence interval?","One of the practical difficulties would include nonresponse. For example, if people selected but not responding have different views from those responding, the estimated proportion may be off by more than 3 percentage points. ",Free Response,,,,
-TPS,8.1,21a,"People love living in California for many reasons, but traffic isn’t one of them. Based on a random sample of 572 employed California adults, a 90% confidence interval for the average travel time to work for all employed California adults is 23 minutes to 26 minutes. Interpret the confidence level.","If we constructed a 90% confidence interval using each of the many random samples taken, about 90% of the intervals would capture the true average travel time to work for all employed California adults. ",Free Response,,,,
-TPS,8.1,21b,"People love living in California for many reasons, but traffic isn’t one of them. Based on a random sample of 572 employed California adults, a 90% confidence interval for the average travel time to work for all employed California adults is 23 minutes to 26 minutes. Name two things you could do to reduce the margin of error. What drawbacks do these actions have?",Decrease confidence level. Drawback: less confidence that our interval captures the true average. Increase sample size. Drawback: larger samples cost more time and money. ,Free Response,,,,
-TPS,8.1,21c,"People love living in California for many reasons, but traffic isn’t one of them. Based on a random sample of 572 employed California adults, a 90% confidence interval for the average travel time to work for all employed California adults is 23 minutes to 26 minutes. Describe how nonresponse might lead to bias in this survey. Does the stated margin of error account for this possible bias?", People who have longer travel times to work may have less time to respond to a survey. This would cause our estimate from the sample to be less than the true mean travel time to work. The bias due to nonresponse is not accounted for by the margin of error because the margin of error accounts for only the variability we expect from random sampling. ,Free Response,,,,
-TPS,8.1,22a,"Employment in California Each month the government releases unemployment statistics. The stated unemployment rate doesn’t include people who choose not to be employed, such as retirees. Based on a random sample of 1000 California adults, a 99% confidence interval for the proportion of all California adults employed in the workforce is 0.532 to 0.612. Interpret the confidence level.","If the government took many random samples and constructed a 99% confidence interval for each sample, about 99% of these intervals would capture the true proportion of all California adults in the workforce.",Free Response,,,,
-TPS,8.1,22b,"Employment in California Each month the government releases unemployment statistics. The stated unemployment rate doesn’t include people who choose not to be employed, such as retirees. Based on a random sample of 1000 California adults, a 99% confidence interval for the proportion of all California adults employed in the workforce is 0.532 to 0.612. Name two things you could do to reduce the margin of error. What drawbacks do these actions have?",Decrease the confidence level. Drawback: less confidence that our interval will capture the true proportion. Increase the sample size. Drawback: larger samples cost more time and money. ,Free Response,,,,
-TPS,8.1,22c,"Employment in California Each month the government releases unemployment statistics. The stated unemployment rate doesn’t include people who choose not to be employed, such as retirees. Based on a random sample of 1000 California adults, a 99% confidence interval for the proportion of all California adults employed in the workforce is 0.532 to 0.612. Describe how untruthful answers might lead to bias in this survey. Does the stated margin of error account for this possible bias?",People ashamed of being unemployed may give untruthful answers on the survey by claiming to be currently employed. This would cause the estimate from the sample to be greater than the true proportion of California adults who are in the workforce. The bias due to untruthful answers is not accounted for by the margin of error because the margin of error accounts for only the variability we expect from random sampling.,Free Response,,,,
-TPS,8.1,23,"A researcher plans to use a random sample of houses to estimate the mean size (in square feet) of the houses in a large population. The researcher is deciding between a 95% confidence level and a 99% confidence level. Compared with a 95% confidence interval, a 99% confidence interval will be (a)narrower and would involve a larger risk of being incorrect. (b) wider and would involve a smaller risk of being incorrect. (c) narrower and would involve a smaller risk of being incorrect.  (d) wider and would involve a larger risk of being incorrect. (e) wider and would have the same risk of being incorrect.",b,Multiple Choice,,,,
-TPS,8.1,24,"A researcher plans to use a random sample of houses to estimate the mean size (in square feet) of the houses in a large population. After deciding on a 95% confidence level, the researcher is deciding between a sample of size n=500 and a sample of size n=1000. Compared with using a sample size of =500, a confidence interval based on a sample size of n=1000 will be (a) narrower and would involve a larger risk of being incorrect. (b) wider and would involve a smaller risk of being incorrect. (c) narrower and would involve a smaller risk of being incorrect. (d) wider and would involve a larger risk of being incorrect. (e) narrower and would have the same risk of being incorrect.",e,Multiple Choice,,,,
-TPS,8.1,25,"In a poll conducted by phone, I) Some people refused to answer questions. II) People without telephones could not be in the sample. III) Some people never answered the phone in several calls. Which of these possible sources of bias is included in the ± 2% margin of error announced for the poll? (a) I only (b) II only (c) III only (d) I, II, and III (e) None of these",e,Multiple Choice,,,,
-TPS,8.1,26,"You have measured the systolic blood pressure of an SRS of 25 company employees. Based on the sample, a 95% confidence interval for the mean systolic blood pressure for the employees of this company is (122, 138). Which of the following statements is true? (a) 95% of the sample of employees have a systolic blood pressure between 122 and 138. (b) 95% of the population of employees have a systolic blood pressure between 122 and 138. (c) If the procedure were repeated many times, 95% of the resulting confidence intervals would contain the population mean systolic blood pressure. (d) If the procedure were repeated many times, 95% of the time the population mean systolic blood pressure would be between 122 and 138. (e) If the procedure were repeated many times, 95% of the time the sample mean systolic blood pressure would be between 122 and 138.",c,Multiple Choice,,,,
-TPS,8.2,29,Check whether each of the conditions is met for calculating a confidence interval for the population proportion p: Latoya wants to estimate the pro-portion of the seniors at her boarding school who like the cafeteria food. She interviews an SRS of 50 of the 175 seniors and finds that 14 think the cafeteria food is good.,Random: Met—SRS. 10%: Not met because 50 > 10% of seniors in the dormitory. Large Counts: Met because 14 ≥ 10 and 36 ≥ 10. ,Free Response,,,,
-TPS,8.2,30,Check whether each of the conditions is met for calculating a confidence interval for the population proportion p: Glenn wonders what proportion of the students at his college believe that tuition is too high. He interviews an SRS of 50 of the 2400 students and finds 38 of those interviewed think tuition is too high.,Random: Met—SRS. 10%: Met because 50 < 10% of students at his college. Large Counts: Met because 38 ≥ 10 and 12 ≥ 10. ,Free Response,,,,
-TPS,8.2,31,"Check whether each of the conditions is met for calculating a confidence interval for the population proportion p: A quality control inspector takes a random sam-ple of 25 bags of potato chips from the thousands of bags filled in an hour. Of the bags selected, 3 had too much salt.",Random: Met—SRS. 10%: Met because 25 < 10% of the thousands of bags filled in an hour. Large Counts: Not met because 3 bags with too much salt < 10. ,Free Response,,,,
-TPS,8.2,32,"Check whether each of the conditions is met for calculating a confidence interval for the population proportion p: The small round holes you often see in seashells were drilled by other sea creatures, who ate the former dwellers of the shells. Whelks often drill into mussels, but this behavior appears to be more or less common in different locations. Researchers collected whelk eggs from the coast of Oregon, raised the whelks in the laboratory, then put each whelk in a container with some delicious mussels. Only 9 of 98 whelks drilled into a mussel. ",Random: May not be met because we do not know if the whelk eggs were a random sample. 10%: Met because 98 < 10% of all whelk eggs. Large Counts: Not met because np^ = 9 < 10. ,Free Response,,,,
-TPS,8.2,33a,"When constructing a confidence interval for a population proportion, we check that the sample size is less than 10% of the population size. Why is it necessary to check this condition?","To ensure that the observations can be viewed as independent. Otherwise, the formula for the standard error of p^ will overestimate the standard deviation of the sampling distribution of p^. ",Free Response,,,,
-TPS,8.2,33b,"When constructing a confidence interval for a population proportion, we check that the sample size is less than 10% of the population size. What happens to the capture rate if this condition is violated?",The capture rate will usually be greater than the specified confidence level. ,Free Response,,,,
-TPS,8.2,34a,"When constructing a confidence interval for a population proportion, we check that both np^ and n(1-p^) are at least 10. Why is it necessary to check this condition?","So the shape of the sampling distribution of p^ will be approximately Normal, which allows use of a Normal distribution to calculate the critical value z*. ",Free Response,,,,
-TPS,8.2,34b,"When constructing a confidence interval for a population proportion, we check that both np^ and n(1-p^) are at least 10. What happens to the capture rate if this condition is violated?",The capture rate will almost always be less than the specified confidence level. ,Free Response,,,,
-TPS,8.2,35a,"According to a recent Pew Research Center report, many American adults have made money by selling something online. In a random sample of 4579 American adults, 914 reported that they earned money by selling something online in the previous year. Assume the conditions for inference are met. Determine the critical value z * for a 98% confidence interval for a proportion.",Table A: z* 5 2.33; Tech: z* 5 2.326,Free Response,,,,
-TPS,8.2,35b,"According to a recent Pew Research Center report, many American adults have made money by selling something online. In a random sample of 4579 American adults, 914 reported that they earned money by selling something online in the previous year. Assume the conditions for inference are met. Construct a 98% confidence interval for the proportion of all American adults who would report having earned money by selling something online in the previous year.","0.1996±2.326 sqrt((0.1996(1-0.1996))/4579) → (0.1859, 0.2133)",Free Response,,,,
-TPS,8.2,35c,"According to a recent Pew Research Center report, many American adults have made money by selling something online. In a random sample of 4579 American adults, 914 reported that they earned money by selling something online in the previous year. Assume the conditions for inference are met. A 98% confidence interval was found to be (0.1859, 0.2133). Interpret this interval.",We are 95% confident that the interval from 0.1859 to 0.2133 captures p5 the true proportion of American adults who have earned money by selling something online in the previous year. ,Free Response,,,,
-TPS,8.2,36a,What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172 undergraduates at a large university: “You witness two students cheating on a quiz. Do you go to the professor?” Only 19 answered “Yes.”  Assume the conditions for inference are met. Determine the critical value z * for a 96% confidence interval for a proportion.,Table A: z* 5 2.05; Tech: z* 5 2.054,Free Response,,,,
-TPS,8.2,36b,What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172 undergraduates at a large university: “You witness two students cheating on a quiz. Do you go to the professor?” Only 19 answered “Yes.”  Assume the conditions for inference are met. Construct a 96% confidence interval for the proportion of all undergraduates at this university who would go to the professor.,"0.1105±2.054 sqrt((0.1105(1-0.1105))/172) → (0.0614, 0.1596)",Free Response,,,,
-TPS,8.2,36c,"What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172 undergraduates at a large university: “You witness two students cheating on a quiz. Do you go to the professor?” Only 19 answered “Yes.”  Assume the conditions for inference are met. A 96% confidence interval was found to be (0.0614, 0.1596). Interpret this interval.",We are 98% confident that the interval from 0.0614 to 0.1596 captures p5 the true proportion of undergraduates at the large university who would go to the professor if they witnessed two students cheating on a quiz.,Free Response,,,,
-TPS,8.2,37,"According to a recent Pew Research Center report, many American adults have made money by selling something online. In a random sample of 4579 American adults, 914 reported that they earned money by selling something online in the previous year. Assume the conditions for inference are met. Determine the critical value z * for a 98% confidence interval for a proportion. Calculate and interpret the standard error of p^ for these data.","SEp^ = sqrt((0.1996(1-0.1996))/4579) = 0.0059; in repeated SRSs of size 4579, the sample proportion of American adults who have earned money by selling something online in the previous year typically varies from the population proportion by about 0.0059.",Free Response,,,,
-TPS,8.2,38,What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172 undergraduates at a large university: “You witness two students cheating on a quiz. Do you go to the professor?” Only 19 answered “Yes.”  Assume the conditions for inference are met. Calculate and interpret the standard error of p^ for these data.,"SEp^ = sqrt((0.1105(1-0.1105))/172) = 0.0239;in repeated SRSs of size 172, the sample proportion of undergraduate students at a large university who would go to the professor if they witness two students cheating on a quiz typically varies from the population proportion by about 0.0239. ",Free Response,,,,
-TPS,8.2,39a,Tonya wants to estimate what proportion of her school’s seniors plan to attend the prom. She interviews an SRS of 50 of the 750 seniors in her school and finds that 36 plan to go to the prom. Identify the population and parameter of interest.,All seniors at Tonya’s high school; the true proportion of all seniors who plan to attend the prom. ,Free Response,,,,
-TPS,8.2,39b,Tonya wants to estimate what proportion of her school’s seniors plan to attend the prom. She interviews an SRS of 50 of the 750 seniors in her school and finds that 36 plan to go to the prom. Check conditions for constructing a confidence interval for the parameter.,"Random: Random sample. 10%: 50<, 10% of the population. Large Counts: 36 ≥ 10 and 14 ≥ 10.",Free Response,,,,
-TPS,8.2,39c,Tonya wants to estimate what proportion of her school’s seniors plan to attend the prom. She interviews an SRS of 50 of the 750 seniors in her school and finds that 36 plan to go to the prom. Construct a 90% confidence interval for p^.,"0.72±1.645 sqrt((0.72(1-0.72))/50) = (0.616, 0.824)",Free Response,,,,
-TPS,8.2,39d,"Tonya wants to estimate what proportion of her school’s seniors plan to attend the prom. She interviews an SRS of 50 of the 750 seniors in her school and finds that 36 plan to go to the prom. A 90% confidence was found to be (0.616, 0.824). Interpret this interval in context. ",We are 90% confident that the interval from 0.616 to 0.824 captures p = the true proportion of all seniors at Tonya’s high school who plan to attend the prom. ,Free Response,,,,
-TPS,8.2,40a,"The student body president of a high school claims to know the names of at least 1000 of the 1800 students who attend the school. To test this claim, the student government advisor randomly selects 100 students and asks the president to identify each by name. The president successfully names only 46 of the students. Identify the population and parameter of interest.",All students at this high school; the true proportion of all students that the student body president knows by name. ,Free Response,,,,
-TPS,8.2,40b,"The student body president of a high school claims to know the names of at least 1000 of the 1800 students who attend the school. To test this claim, the student government advisor randomly selects 100 students and asks the president to identify each by name. The president successfully names only 46 of the students. Check conditions for constructing a confidence interval for the parameter.",Random: Random sample. 10%: 100 < 10% of the population. Large Counts: 46 ≥ 10 and 54 ≥ 10.,Free Response,,,,
-TPS,8.2,40c,"The student body president of a high school claims to know the names of at least 1000 of the 1800 students who attend the school. To test this claim, the student government advisor randomly selects 100 students and asks the president to identify each by name. The president successfully names only 46 of the students. Construct a 99% confidence interval for p^.","0.46±2.576 sqrt((0.46(1-0.46))/100) = (0.332, 0.588)",Free Response,,,,
-TPS,8.2,40d,"The student body president of a high school claims to know the names of at least 1000 of the 1800 students who attend the school. To test this claim, the student government advisor randomly selects 100 students and asks the president to identify each by name. The president successfully names only 46 of the students. A 99% confidence was found to be (0.332, 0.588). Interpret this interval in context",We are 99% confident that the interval from 0.332 to 0.588 captures p = the true proportion of all students at this high school whom the student body president knows by name.,Free Response,,,,
-TPS,8.2,41,"A Pew Research Center report on gamers and gaming estimated that 49% of U.S. adults play video games on a computer, TV, game console, or portable device such as a cell phone. This estimate was based on a random sample of 2001 U.S. adults. Construct and interpret a 95% confidence interval for the proportion of all U.S. adults who play video games.","S: p = the true proportion of all U.S. adults who play video games. P: One-sample z interval for p. Random: Random sample. 10%: 2001 < 10% of all U.S. adults. Large Counts: 980 ≥ 10 and 1021 ≥ 10. D: (0.468, 0.512). C: We are 95% confident that the interval from 0.468 to 0.512 captures p5 the true proportion of U.S. adults who play video games. ",Free Response,,,,
-TPS,8.2,42,"A recent study asked U.S. adults to name 10 historic events that occurred in their lifetime that have had the greatest impact on the country. The most frequently chosen answer was the September 11, 2001, terrorist attacks, which was included by 76% of the 2025 randomly selected U.S. adults. Construct and interpret a 95% confidence interval for the proportion of all U.S. adults who would include the 9/11 attacks on their list of 10 historic events.","S: p = the true proportion of all U.S. adults who would include the 9/11 attacks on their list of 10 historic events. P: One-sample z interval for p. Random: Random sample. 10%: 2025 < 10% of all U.S. adults. Large Counts: 1539 ≥ 10 and 486 ≥ 10. D: (0.741, 0.779). C: We are 95% confident that the interval from 0.741 to 0.779 captures p5 the true proportion of U.S. adults who would include the 9/11 attacks on their list of 10 historic events.",Free Response,,,,
-TPS,8.2,43a,"A Pew Research Center report on gamers and gaming estimated that 49% of U.S. adults play video games on a computer, TV, game console, or portable device such as a cell phone. This estimate was based on a random sample of 2001 U.S. adults. The study also estimated that 67% of adults aged 18–29 play video games, but only 25% of adults aged 65 and older play video games. Explain why you do not have enough information to give confidence intervals for these two age groups separately.",We do not know the size of the sample that was taken from each age group. ,Free Response,,,,
-TPS,8.2,43b,"A Pew Research Center report on gamers and gaming estimated that 49% of U.S. adults play video games on a computer, TV, game console, or portable device such as a cell phone. This estimate was based on a random sample of 2001 U.S. adults. The study also estimated that 67% of adults aged 18–29 play video games, but only 25% of adults aged 65 and older play video games. Do you think a 95% confidence interval for adults aged 18–29 would have a larger or smaller margin of error than the estimate from the interval (0.468, 0.512)? Explain your answer.","Larger, because this is a smaller sample size than the original group.",Free Response,,,,
-TPS,8.2,44a,"A recent study asked U.S. adults to name 10 historic events that occurred in their lifetime that have had the greatest impact on the country. The most frequently chosen answer was the September 11, 2001, terrorist attacks, which was included by 76% of the 2025 randomly selected U.S. adults. The study also reported that 86% of millennials included 9/11 in their top-10 list and 70% of baby boomers included 9/11. Explain why you do not have enough information to give confidence intervals for millennials and baby boomers separately.",We do not know the size of the sample that was taken from each group. ,Free Response,,,,
-TPS,8.2,44b,"A recent study asked U.S. adults to name 10 historic events that occurred in their lifetime that have had the greatest impact on the country. The most frequently chosen answer was the September 11, 2001, terrorist attacks, which was included by 76% of the 2025 randomly selected U.S. adults. The study also reported that 86% of millennials included 9/11 in their top-10 list and 70% of baby boomers included 9/11. Do you think a 95% confidence interval for baby boomers would have a larger or smaller margin of error than the estimate than (0.741, 0.779)? Explain your answer.","Larger, because this is a smaller sample size than the original group. ",Free Response,,,,
-TPS,8.2,45a,A 2016 survey of 1480 randomly selected U.S. adults found that 55% of respondents agreed with the following statement: “Organic produce is better for health than conventionally grown produce.” Construct and interpret a 99% confidence interval for the proportion of all U.S. adults who think that organic produce is better for health than conventionally grown produce.,"S: p = the true proportion of all U.S. adults who think that organic produce is better for health than conventionally grown produce. P: One-sample z interval for p. Random: Random sample. 10%: 1480 < 10% of all U.S. adults. Large Counts: 814 ≥ 10 and 666 ≥ 10. D: (0.517, 0.583). C: We are 99% confident that the interval from 0.517 to 0.583 captures p5 the true proportion of U.S. adults who would agree with the statement: “Organic produce s better for health than conventionally grown produce.” ",Free Response,,,,
-TPS,8.2,45b,A 2016 survey of 1480 randomly selected U.S. adults found that 55% of respondents agreed with the following statement: “Organic produce is better for health than conventionally grown produce.” Does the interval from part (a) provide convincing evidence that a majority of all U.S. adults think that organic produce is better for health? Explain your answer.,"Yes; all the plausible values in the interval are greater than 0.5, which provides convincing evidence that a majority of all U.S. adults think that organic produce is better for health than conventionally grown produce. ",Free Response,,,,
-TPS,8.2,46a,"According to a recent study by the Annenberg Foundation, only 36% of adults in the United States could name all three branches of government. This was based on a survey given to a random sample of 1416 U.S. adults. Construct and interpret a 90% confidence interval for the proportion of all U.S. adults who could name all three branches of government.","S: p = the true proportion of all U.S. adults who could name all three branches of government. P: One-sample z interval for p. Random: Random sample. 10%: 1416 < 10% of all U.S. adults. Large Counts: 510 ≥ 10 and 906 ≥ 10. D: (0.339, 0.381). C: We are 90% confident that the interval from 0.339 to 0.381 captures p = the true proportion of U.S. adults who could name all three branches of government. ",Free Response,,,,
-TPS,8.2,46b,"According to a recent study by the Annenberg Foundation, only 36% of adults in the United States could name all three branches of government. This was based on a survey given to a random sample of 1416 U.S. adults. Does the interval from part (a) provide convincing evidence that less than half of all U.S. adults could name all three branches of government? Explain your answer.","Yes; all the plausible values in the interval are less than 0.5, which provides convincing evidence that less than half of all U.S. adults could name all three branches of government.",Free Response,,,,
-TPS,8.2,49a,A small pilot study estimated that 44% of all American adults agree that parents should be given vouchers that are good for education at any public or private school of their choice. How large a random sample is required to obtain a margin of error of at most 0.03 with 99% confidence? Answer this question using the pilot study’s result as the guessed value for p^.,Solving 2.576 sqrt((0.44(0.56))/n) ≤ 0.03 gives n ≥ 1817.,Free Response,,,,
-TPS,8.2,49b,A small pilot study estimated that 44% of all American adults agree that parents should be given vouchers that are good for education at any public or private school of their choice. How large a random sample is required to obtain a margin of error of at most 0.03 with 99% confidence? Answer the question using the conservative guess p^ = 0.5. By how much do using this sample size and the pilot study’s result as a sample size differ? ,Solving 2.576 sqrt((0.50(0.50))/n) ≤ 0.03 gives n ≥ 1844. The conservative approach requires 27 more adults.,Free Response,,,,
-TPS,8.2,50a,"PTC is a substance that has a strong bitter taste for some people and is tasteless for others. The ability to taste PTC is inherited. About 75% of Italians can taste PTC, for example. You want to estimate the proportion of Americans who have at least one Italian grandparent and who can taste PTC. How large a sample must you test to estimate the proportion of PTC tasters within 0.04 with 90% confidence? Answer this question using the 75% estimate as the guessed value for p^.",Solving 1.645 sqrt((0.75(0.25))/n) ≤ 0.04 gives n ≥ 318.,Free Response,,,,
-TPS,8.2,50b,"PTC is a substance that has a strong bitter taste for some people and is tasteless for others. The ability to taste PTC is inherited. About 75% of Italians can taste PTC, for example. You want to estimate the proportion of Americans who have at least one Italian grandparent and who can taste PTC. How large a sample must you test to estimate the proportion of PTC tasters within 0.04 with 90% confidence? Answer the question in part (a) again, but this time use the conservative guess p^ = 0.5. By how much do using this estimate and the 75% estimate differ?",Solving 1.645 sqrt((0.50(0.50))/n) ≤ 0.04 gives n ≥ 423. The conservative approach requires 105 more people.,Free Response,,,,
-TPS,8.2,51,"A college student organization wants to start a nightclub for students under the age of 21. To assess support for this proposal, they will select an SRS of students and ask each respondent if he or she would patronize this type of establishment. What sample size is required to obtain a 90% confidence interval with a margin of error of at most 0.04?",Solving 1.645 sqrt((0.50(0.50))/n) ≤ 0.04 gives n ≥ 423.,Free Response,,,,
-TPS,8.2,52,Gloria Chavez and Ronald Flynn are the candidates for mayor in a large city. We want to estimate the proportion p of all registered voters in the city who plan to vote for Chavez with 95% confidence and a margin of error no greater than 0.03. How large a random sample do we need?,Solving 1.96 sqrt((0.50(0.50))/n) ≤ 0.03 gives n ≥ 1068.,Free Response,,,,
-TPS,8.2,53a,"According to a Gallup Poll report, 64% of teens aged 13 to 17 have TVs in their rooms. Here is part of the footnote to this report: These results are based on telephone interviews with a randomly selected national sample of 1028 teenagers in the Gallup Poll Panel of households, aged 13 to 17 For results based on this sample, one can say . . . that the maximum error attributable to sampling and other random effects is ±3 percentage points. In addition to sampling error, question wording and practical difficulties in conducting surveys can introduce error or bias into the findings of public opinion polls. We omitted the confidence level from the footnote. Use what you have learned to estimate the confidence level, assuming that Gallup took an SRS.",Solving 0.03 = z* sqrt((0.64(0.36))/1028) gives z* = 2.00. The confidence level is likely 95%. ,Free Response,,,,
-TPS,8.2,53b,"According to a Gallup Poll report, 64% of teens aged 13 to 17 have TVs in their rooms. Here is part of the footnote to this report: These results are based on telephone interviews with a randomly selected national sample of 1028 teenagers in the Gallup Poll Panel of households, aged 13 to 17 For results based on this sample, one can say . . . that the maximum error attributable to sampling and other random effects is ±3 percentage points. In addition to sampling error, question wording and practical difficulties in conducting surveys can introduce error or bias into the findings of public opinion polls. Give an example of a “practical difficulty” that could lead to bias in this survey.","Teens are hard to reach and often unwilling to participate in surveys, so nonresponse bias is a “practical difficulty.” If teens with TVs in their rooms are less likely to answer the poll, the estimate from the poll would likely be too small.",Free Response,,,,
-TPS,8.2,54a,"Gambling is an issue of great concern to those involved in college athletics. Because of this concern, the National Collegiate Athletic Association (NCAA) surveyed randomly selected student athletes concerning their gambling-related behaviors.21 Of the 5594 Division I male athletes who responded to the survey, 3547 reported participation in some gambling behavior. This includes playing cards, betting on games of skill, buying lottery tickets, betting on sports, and similar activities. A report of this study cited a 1% margin of error. The confidence level was not stated in the report. Use what you have learned to estimate the confidence level, assuming that the NCAA took an SRS.",Solving 0.01 = z* sqrt((0.6341(0.3659))/1028) gives z* = 1.55. The area between 21.55 and 1.55 under the standard Normal curve is 0.8788. The confidence level is likely 88%. ,Free Response,,,,
-TPS,8.2,54b,"Gambling is an issue of great concern to those involved in college athletics. Because of this concern, the National Collegiate Athletic Association (NCAA) surveyed randomly selected student athletes concerning their gambling-related behaviors.21 Of the 5594 Division I male athletes who responded to the survey, 3547 reported participation in some gambling behavior. This includes playing cards, betting on games of skill, buying lottery tickets, betting on sports, and similar activities. A report of this study cited a 1% margin of error. The study was designed to protect the anonymity of the student athletes who responded. As a result, it was not possible to calculate the number of students who were asked to respond but did not. How does this fact affect the way that you interpret the results?","We do not know if those who did respond can reliably represent those who did not. Because the student athletes might worry about being identified if they have gambled, the estimate from the poll would likely be too small.",Free Response,,,,
-TPS,8.2,55,"A Gallup poll found that only 28% of American adults expect to inherit money or valuable possessions from a relative. The poll’s margin of error was ±3 percentage points at a 95% confidence level. This means that (a) the poll used a method that gets an answer within 3% of the truth about the population 95% of the time. (b) the percent of all adults who expect an inheritance must be between 25% and 31%. (c) if Gallup takes another poll on this issue, the results of the second poll will lie between 25% and 31%. (d) there’s a 95% chance that the percent of all adults who expect an inheritance is between 25% and 31%. (e) Gallup can be 95% confident that between 25% and 31% of the sample expect an inheritance.",a,Multiple Choice,,,,
-TPS,8.2,56,A Gallup poll found that only 28% of American adults expect to inherit money or valuable possessions from a relative. The poll’s margin of error was ±3 percentage points at a 95% confidence level. Suppose that Gallup wanted to cut the margin of error in half from 3 percentage points to 1.5 percentage points. How should they adjust their sample size? (a) Multiply the sample size by 4. (b) Multiply the sample size by 2. (c) Multiply the sample size by 1/2. (d) Multiply the sample size by 1/4. (e) There is not enough information to answer this question.,a,Multiple Choice,,,,
-TPS,8.2,57,"Most people can roll their tongues, but many can’t. The ability to roll the tongue is genetically determined. Suppose we are interested in determining what proportion of students can roll their tongues. We test a simple random sample of 400 students and find that 317 can roll their tongues. The margin of error for a 95% confidence interval for the true proportion of tongue rollers among students is closest to which of the following? (a) 0.0008 (b) 0.02 (c) 0.03 (d) 0.04 (e) 0.05",d,Multiple Choice,,,,
-TPS,8.2,58,"A newspaper reporter asked an SRS of 100 residents in a large city for their opinion about the mayor’s job performance. Using the results from the sample, the C% confidence interval for the proportion of all residents in the city who approve of the mayor’s job performance is 0.565 to 0.695. What is the value of C? (a) 82 (b) 86 (c) 90 (d) 95 (e) 99",a,Multiple Choice,,,,
-TPS,8.3,61,"The movie A Civil Action (1998) tells the story of a major legal battle that took place in the small town of Woburn, Massachusetts. A town well that supplied water to east Woburn residents was contaminated by industrial chemicals. During the period that residents drank water from this well, 16 of 414 babies born had birth defects. On the west side of Woburn, 3 of 228 babies born during the same time period had birth defects. Let p1 = true the proportion of all babies born with birth defects in west Woburn and p2 = the true proportion of all babies born with birth defects in east Woburn. Check if the conditions for calculating a confidence interval for p1 – p2 are met","Random: Not met because these data do not come from independent random samples or from two groups in a randomized experiment. 10%: Since no sampling took place, the 10% condition does not apply. Large Counts: Not met because there were fewer than 10 successes (3) in the group from the west side of Woburn. ",Multiple Choice,,,,
-TPS,8.3,62,"We don’t like to find broken crackers when we open the package. How can makers reduce breaking? One idea is to microwave the crackers for 30 seconds right after baking them. Randomly assign 65 newly baked crackers to the microwave and another 65 to a control group that is not microwaved. After 1 day, none of the microwave group were broken and 16 of the control group were broken.  Let p1 = the true proportion of crackers like these that would break if baked in the microwave and p2 = the true proportion of crackers like these that would break if not microwaved. Check if the conditions for calculating a confidence interval for p1 – p2 are met.","Random: Crackers were randomly assigned to one of the two treatment groups. 10%: Since no sampling took place, the 10% condition does not apply. Large Counts: Not met because there were fewer than 10 successes (0) in the microwave group. ",Multiple Choice,,,,
-TPS,8.3,63,"The pesticide diazinon is commonly used to treat infestations of the German cockroach, Blattella germanica. A study investigated the persistence of this pesticide on various types of surfaces. Researchers applied a 0.5% emulsion of diazinon to glass and plasterboard. After 14 days, they randomly assigned 72 cockroaches to two groups of 36, placed one group on each surface, and recorded the number that died within 48 hours. On glass, 18 cockroaches died, while on plasterboard, 25 died. If p1 and p2 are the true proportions of cockroaches like these that would die within 48 hours on glass treated with diazinon and on plasterboard treated with diazinon, respectively, check if the conditions for calculating a confidence interval for p1 – p2 are met.","Random: Random assignment. Large Counts: 25, 11, 18, 18 ≥ 10. ",Multiple Choice,,,,
-TPS,8.3,64,"A company that records and sells rewritable DVDs wants to compare the reliability of DVD fabricating machines produced by two different manufacturers. They randomly select 500 DVDs produced by each fabricator and find that 484 of the disks produced by the first machine are acceptable and 480 of the disks produced by the second machine are acceptable. If p1 and p2 are the proportions of acceptable DVDs produced by the first and second machines, respectively, check if the conditions for calculating a confidence interval for p1 – p2 are met.","Random: Independent random samples. 10%: 500 < 10% of all DVDs produced by each machine. Large Counts: 480, 20, 484, 16 ≥ 10. ",Multiple Choice,,,,
-TPS,8.3,65a,A surprising number of young adults (ages 19 to 25) still live in their parents’ homes. The National Institutes of Health surveyed independent random samples of 2253 men and 2629 women in this age group. The survey found that 986 of the men and 923 of the women lived with their parents. Construct and interpret a 99% confidence interval for the difference in the true proportions of men and women aged 19 to 25 who live in their parents’ homes.,"S: p1 = true proportion of young men who live in their parents’ home and p2 = that of young women . . . P: Two-sample z interval for p1 - p2. Random: Independent random samples. 10%: 2253 < 10% of all young men and 2629 < 10% of all young women. Large Counts: 986, 1267, 923, 1706 ≥ 10. D: (0.051, 0.123) C: We are 99% confident that the interval from 0.051 to 0.123 captures p1 - p2 = the true difference in the proportions of young men and young women who live in their parents’ home. ",Multiple Choice,,,,
-TPS,8.3,65b,A surprising number of young adults (ages 19 to 25) still live in their parents’ homes. The National Institutes of Health surveyed independent random samples of 2253 men and 2629 women in this age group. The survey found that 986 of the men and 923 of the women lived with their parents. Does your interval from part (a) give convincing evidence of a difference between the population proportions? Justify your answer.,"Because the interval does not contain 0, there is convincing evidence that the true proportion of young men who live at their parents’ home is different from the true proportion of young women who do.",Multiple Choice,,,,
-TPS,8.3,66a,"In a Pew Research poll, 287 out of 522 randomly selected U.S. men were able to identify Egypt when it was highlighted on a map of the Middle East. When 520 randomly selected U.S. women were asked, 233 were able to do so. Construct and interpret a 95% confidence interval for the difference in the true proportions of U.S. men and U.S. women who can identify Egypt on a map.","S: pM= true proportion of men who can identify Egypt on a map and pW = that of women . . . P: Two-sample z-interval for pM - pW. Random: Independent random samples. 10%: 522 ≥ 10% of all men and 520 < 10% of all women. Large Counts: 287, 235, 233, 287 ≥ 10. D: (0.041, 0.162) C: We are 95% confident that the interval from 0.041 to 0.162 captures p1 - p2 5 the true difference in the proportions of men and women who can identify Egypt on a map. (b) Because the interval does not contain 0, there is convincing evidence that the true proportion of men who can identify Egypt on a map is different from the true proportion of women who can identify Egypt on a map. ",Free Response,,,,
-TPS,8.3,66b,"In a Pew Research poll, 287 out of 522 randomly selected U.S. men were able to identify Egypt when it was highlighted on a map of the Middle East. When 520 randomly selected U.S. women were asked, 233 were able to do so. Based on your interval, is there convincing evidence of a difference in the true proportions of U.S. men and women who can identify Egypt on a map? Justify your answer.","Because the interval does not contain 0, there is convincing evidence that the true proportion of men who can identify Egypt on a map is different from the true proportion of women who can identify Egypt on a map. ",Free Response,,,,
-TPS,8.3,67,"A surprising number of young adults (ages 19 to 25) still live in their parents’ homes. The National Institutes of Health surveyed independent random samples of 2253 men and 2629 women in this age group. The survey found that 986 of the men and 923 of the women lived with their parents. The 99% confidence interval was found to be (0.051, 0.123).  Interpret the confidence level for the interval.","If we were to select many independent random samples of 2253 men and 2629 women from the population of all young adults (ages 19 to 25) and construct a 99% confidence interval for the difference in the true proportions each time, about 99% of the intervals would capture the difference in the true proportions of men and women aged 19 to 25 who live in their parents’ homes.",Free Response,,,,
-TPS,8.3,68,"In a Pew Research poll, 287 out of 522 randomly selected U.S. men were able to identify Egypt when it was highlighted on a map of the Middle East. When 520 randomly selected U.S. women were asked, 233 were able to do so. The 95% confidence interval was found to be (0.041, 0.162). Interpret the confidence level for the interval.","If we were to select many independent random samples of 522 men and 520 women from the population of all men and women in the U.S. and construct a 95% confidence interval for the difference in the true proportions each time, about 95% of the intervals would capture the difference in the true proportions of U.S. men and U.S. women who can identify Egypt on a map.",Free Response,,,,
-TPS,8.3,69,"Does the appearance of the interviewer influence how people respond to a survey question? Ken (white, with blond hair) and Hassan (darker, with Middle Eastern features) conducted an experiment to address this question. They took turns (in a random order) walking up to people on the main street of a small town, identifying themselves as students from a local high school, and asking them, “Do you support President Obama’s decision to launch airstrikes in Iraq?” Of the 50 people Hassan spoke to, 11 said “Yes,” while 21 of the 44 people Ken spoke to said “Yes.” Construct and interpret a 90% confidence interval for the difference in the proportions of people like these who would say they support President Obama’s decision when asked by Hassan versus when asked by Ken.","S: p1 = true proportion of people like the ones in this study who would say they support President Obama’s decision when asked by Hassan and p2 = true proportion . . . when asked by Ken. P: Two-sample z-interval for p1 - p2. Random: Random assignment. Large Counts: 11, 39, 21, 23 are ≥ 10. D: (20.414, 20.100) C: We are 90% confident that the interval from 20.414 to 20.100 captures p1 - p2 = the true difference in the proportions of people like the ones in this study who would say they support President Obama’s decision when asked by Hassan or by Ken. ",Free Response,,,,
-TPS,8.3,70,"Nicotine patches are often used to help smokers quit. Does giving medicine to fight depression help? A randomized double-blind experiment assigned 244 smokers to receive nicotine patches and another 245 to receive both a patch and the antidepressant drug bupropion. After a year, 40 subjects in the nicotine patch group had abstained from smoking, as had 87 in the patch-plus-drug group. Construct and interpret a 99% confidence interval for the difference in the true proportion of smokers like these who would abstain when using bupropion and a nicotine patch and the proportion who would abstain when using only a patch.","S: p1 = true proportion of smokers like these who would abstain when using only a patch and p2 = true proportion . . . when using both. P: Two-sample z-interval for p1 - p2. Random: Random assignment. Large Counts: 40, 204, 87, 158 are ≥ 10. D: (20.291, 20.092) C: We are 99% confident that the interval from 120.291, 20.0922 captures p1 - p2 = the true difference in the proportion of smokers like these who would abstain when using only a patch and the proportion of smokers like these who would abstain when using bupropion and a nicotine patch. ",Free Response,,,,
-TPS,8.3,71a,"A CBS News poll asked 606 randomly selected women and 442 randomly selected men, “Do you think putting a special tax on junk food would encourage more people to lose weight?” 170 of the women and 102 of the men said “Yes.”26 A 99% confidence interval for the difference (Women – Men) in the true proportion of people in each population who would say “Yes” is -0.020 to 0.120. Does the confidence interval provide convincing evidence that the two population proportions are different? Explain your answer.","Because the interval includes 0 as a plausible value for the difference (Women – Men) in the true proportion of people in each population who would say “Yes,” we do not have convincing evidence that the two population proportions are different. ",Free Response,,,,
-TPS,8.3,71b,"A CBS News poll asked 606 randomly selected women and 442 randomly selected men, “Do you think putting a special tax on junk food would encourage more people to lose weight?” 170 of the women and 102 of the men said “Yes.”26 A 99% confidence interval for the difference (Women – Men) in the true proportion of people in each population who would say “Yes” is -0.020 to 0.120. Does the confidence interval provide convincing evidence that the two population proportions are equal? Explain your answer.","No; because 0 is captured in the interval, it is plausible that the population proportions are equal, but this does not provide convincing evidence that the two proportions are equal. ",Free Response,,,,
-TPS,8.3,72a,"An association of Christmas tree growers in Indiana wants to know if there is a difference in preference for natural trees between urban and rural households. So the association sponsored a survey of Indiana households that had a Christmas tree last year to find out. In a random sample of 160 rural households, 64 had a natural tree. In a separate random sample of 261 urban households, 89 had a natural tree. A 95% confidence interval for the difference (Rural – Urban) in the true proportion of households in each population that had a natural tree is -0.036 to 0.154. Does the confidence interval provide convincing evidence that the two population proportions are different? Explain your answer.","Because the interval includes 0 as a plausible value for the difference (Rural – Urban) in the true proportion of households that have a natural tree, we do not have convincing evidence that the two population proportions are different. ",Free Response,,,,
-TPS,8.3,72b,"An association of Christmas tree growers in Indiana wants to know if there is a difference in preference for natural trees between urban and rural households. So the association sponsored a survey of Indiana households that had a Christmas tree last year to find out. In a random sample of 160 rural households, 64 had a natural tree. In a separate random sample of 261 urban households, 89 had a natural tree. A 95% confidence interval for the difference (Rural – Urban) in the true proportion of households in each population that had a natural tree is -0.036 to 0.154. Does the confidence interval provide convincing evidence that the two population proportions are equal? Explain your answer.","No; because 0 is captured in the interval, it is plausible that the population proportions are equal, but this does not provide convincing evidence that the two proportions are equal.",Free Response,,,,
-TPS,8.3,73,"Earlier in this section, you read about an experiment comparing surgery and observation as treatments for men with prostate cancer. After 20 years, p^s = 141/364 = 0.387 of the men who were assigned to surgery were still alive and p^o = 122/367 = 0.332 of the men who were assigned to observation were still alive. Which of the following is the 95% confidence interval for ps-po? (a) (141 – 122) ± 1.96 sqrt(((141*223)/364) + ((122*245)/367)) (b) (141 – 122) ± 1.96 (sqrt(((141*223)/364)) + sqrt(((122*245)/367))) (c) (0.387 – 0.332) ± 1.96 sqrt(((0.387*0.613)/364) + ((0.332*0.668)/367)) (d) (0.387 – 0.332) ± 1.96 (sqrt(((0.387*0.613)/364)) + sqrt(((0.332*0.668)/367))) (e) (0.387 – 0.332) ± 1.96 sqrt(((0.387*0.613)/364) –  ((0.332*0.668)/367))",c,Multiple Choice,,,,
-TPS,8.3,74,"When constructing a confidence interval for a difference between two population proportions, why is it important to check that the number of successes and the number of failures in each sample is at least 10? (a) So we can generalize the results to the populations from which the samples were selected. (b) So we can assume that the two samples are independent. (c) So we can assume that the observations within each sample are independent. (d) So we can assume the sampling distribution of p^1 – p^2 is approximately Normal. (e) So we can assume that population 1 and population 2 are approximately Normal.",d,Multiple Choice,,,,
-TPS,8.3,75,"To estimate the difference in the proportion of students at high school A and high school B who drive themselves to school, a district administrator selected a random sample of 100 students from each school. At school A, 23 of the students said they drive themselves; at school B, 29 of the students said they drive themselves. A 90% confidence interval for pA – pB is -0.16 to 0.04. Based on this interval, which conclusion is best? (a) Because -0.06 is in the interval, there is convincing evidence of a difference in the population proportions. (b) Because 0 is in the interval, there is convincing evidence of a difference in the population proportions. (c) Because -0.06 is in the interval, there is not convincing evidence of a difference in the population proportions. (d) Because 0 is in the interval, there is not convincing evidence of a difference in the population proportions. (e) Because most of the interval is negative, there is convincing evidence that a greater proportion of students at high school B drive themselves to school.",d,Multiple Choice,,,,
-TPS,R8,1a,"A Gallup poll conducted telephone interviews with a random sample of adults aged 18 and older. Data were obtained for 1000 people. Of these, 370 said that football is their favorite sport to watch on television. Define the parameter p in this setting. ",p= the proportion of all adults aged 18 and older who would say that football is their favorite sport to watch on television. ,Free Response,,,,
-TPS,R8,1b,"A Gallup poll conducted telephone interviews with a random sample of adults aged 18 and older. Data were obtained for 1000 people. Of these, 370 said that football is their favorite sport to watch on television. What point estimator will you use to estimate p? What is the value of the point estimate?",Point estimator = p^; p^ = 0.37 ,Free Response,,,,
-TPS,R8,1c,"A Gallup poll conducted telephone interviews with a random sample of adults aged 18 and older. Data were obtained for 1000 people. Of these, 370 said that football is their favorite sport to watch on television. Do you believe that the value of the point estimate is equal to the value of p? Explain your answer.","Because of sampling variability, I doubt that the value of the point estimate is exactly equal to the value of p. ",Free Response,,,,
-TPS,R8,2a,Are you a sports fan? That’s the question the Gallup polling organization asked a random sample of 1527 U.S. adults.  Gallup reported that a 95% confidence interval for the proportion of all U.S. adults who are sports fans is 0.565 to 0.615. Calculate the point estimate and the margin of error. ,The point estimate is (0.565 + 0.615)/2 = 0.59. The margin of error is 0.615 - 0.59 = 0.025. ,Free Response,,,,
-TPS,R8,2b,Are you a sports fan? That’s the question the Gallup polling organization asked a random sample of 1527 U.S. adults.  Gallup reported that a 95% confidence interval for the proportion of all U.S. adults who are sports fans is 0.565 to 0.615. Interpret the confidence level. ,"If we were to select many random samples of size 1527 from the population of U.S. adults and construct a 95% confidence interval using each sample, about 95% of the intervals would capture the true proportion of all U.S. adults who are sports fans. ",Free Response,,,,
-TPS,R8,2c,"Are you a sports fan? That’s the question the Gallup polling organization asked a random sample of 1527 U.S. adults.  Gallup reported that a 95% confidence interval for the proportion of all U.S. adults who are sports fans is 0.565 to 0.615. Based on the interval, is there convincing evidence that a majority of U.S. adults are sports fans? Explain your answer.","Yes; because the confidence interval lies entirely above 0.5, there is convincing evidence that a majority of U.S. adults are sports fans.",Free Response,,,,
-TPS,R8,3a,"Explain how increasing the confidence level would affect the margin of error of a confidence interval, if all other things remained the same.",The margin of error must get larger to increase the capture rate of the intervals. ,Free Response,,,,
-TPS,R8,3b,"Explain how quadrupling the sample size would affect the margin of error of a confidence interval, if all other things remained the same.",The margin of error will decrease by a factor of 2. ,Free Response,,,,
-TPS,R8,4a,"A random digit dialing telephone survey of 880 drivers asked, “Recalling the last 10 traffic lights you drove through, how many of them were red when you entered the intersections?” Of the 880 respondents, 171 admitted that at least one light had been red. Construct and interpret a 95% confidence interval for the population proportion.","S: p5 the true proportion of all drivers who have run at least one red light in the last 10 intersections they have entered. P: One-sample z interval. Random: Random sample. 10%: 880 , 10% of all drivers. Large Counts: 171 $ 10 and 709 $ 10. D: (0.168, 0.220). C: We are 95% confident that the interval from 0.168 to 0.220 captures the true proportion of all drivers who have run at least one red light in the last 10 intersections they have entered. ",Free Response,,,,
-TPS,R8,4b,"A random digit dialing telephone survey of 880 drivers asked, “Recalling the last 10 traffic lights you drove through, how many of them were red when you entered the intersections?” Of the 880 respondents, 171 admitted that at least one light had been red. Nonresponse is a practical problem for this survey—only 21.6% of calls that reached a live person were completed. Another practical problem is that people may not give truthful answers. What is the likely direction of the bias: Do you think more or fewer than 171 of the 880 respondents really ran a red light? Why? Are these sources of bias included in the margin of error?","It is likely that more than 171 respondents have run red lights. We would not expect very many people to claim they have run red lights when they have not, but some people will deny running red lights when they have. The margin of error does not account for these sources of bias, only sampling variability.",Free Response,,,,
-TPS,R8,5,The Gallup Poll plans to ask a random sample of adults whether they attended a religious service in the past 7 days. How large a sample would be required to obtain a margin of error of at most 0.01 in a 99% confidence interval for the population proportion who would say that they attended a religious service?,"Solving 2.576 sqrt((0.5(0.5))/ n) ≤ 0.01gives n = 16,590 adults.",Free Response,,,,
-TPS,R8,6a,"As part of the Pew Internet and American Life Project, researchers conducted two surveys. The first survey asked a random sample of 1060 U.S. teens about their use of social media. A second survey posed similar questions to a random sample of 2003 U.S. adults. In these two studies, 71.0% of teens and 58.0% of adults used Facebook. 30 Let pT = the true proportion of all U.S. teens who use Facebook and pA = the true proportion of all U.S. adults who use Facebook. Calculate and interpret a 99% confidence interval for the difference in the true proportions of U.S. teens and adults who use Facebook.","S: 99% CI for pT - pA, where pT5 the true proportion of U.S. teens who use Facebook and pA 5 the true proportion of U.S. adults who use Facebook. P: Two-sample z-interval for pT = pA. Random: The data come from independent random samples of 1060 U.S. teens and 2003 U.S. adults. Large Counts: 1060(0.71) < 753, 1060(1 - 0.71) < 297, 2003(0.58) < 1162, and 2003(1 - 0.58) < 841 are all $ 10. D: p^T= 0.71 and p^A = 0.58(0.084, 0.176) C: We are 99% confident that the interval from 0.084 to 0.176 captures pT - pA = the difference in the true proportions of teens and adults who use Facebook. ",Free Response,,,,
-TPS,R8,6b,"As part of the Pew Internet and American Life Project, researchers conducted two surveys. The first survey asked a random sample of 1060 U.S. teens about their use of social media. A second survey posed similar questions to a random sample of 2003 U.S. adults. In these two studies, 71.0% of teens and 58.0% of adults used Facebook. 30 Let pT = the true proportion of all U.S. teens who use Facebook and pA = the true proportion of all U.S. adults who use Facebook. A 99% confidence interval was found to be (0.084, 0.176). Based on this confidence interval, is there convincing evidence of a difference in the population proportions? Explain your answer.","Because the interval does not include 0 as a plausible value for the difference in the true proportion of U.S. teens and U.S. adults who use Facebook, we do have convincing evidence that the two population proportions are different.",Free Response,,,,
-TPS,T8,1,"The Gallup Poll interviews 1600 people. Of these, 18% say that they jog regularly. The news report adds: “The poll had a margin of error of plus or minus 3 percentage points at a 95% confidence level.” You can safely conclude that (a) 95% of all Gallup Poll samples like this one give answers within ±3% of the true population value. (b) the percent of the population who jog is certain to be between 15% and 21%. (c) 95% of the population jog between 15% and 21% of the time. (d) we can be 95% confident that the sample proportion is captured by the confidence interval. (e) if Gallup took many samples, 95% of them would find that 18% of the people in the sample jog.",a,Multiple Choice,,,,
-TPS,T8,2,"A confidence interval for a difference in proportions is −0.077 to 0.013. What are the point estimate and the margin of error for this interval? (a) −0.032, 0.045 (b) −0.032, 0.090 (c) −0.032, 0.180 (d) −0.045, 0.032 (e) −0.045, 0.090 ",a,Multiple Choice,,,,
-TPS,T8,3,"In a random sample of 100 students from a large high school, 37 regularly bring a reusable water bottle from home. Which of the following gives the correct value and interpretation of the standard error of the sample proportion? (a) In samples of size 100 from this school, the sample proportion of students who bring a reusable water bottle from home will be at most 0.095 from the true proportion. (b) In samples of size 100 from this school, the sample proportion of students who bring a reusable water bottle from home will be at most 0.048 from the true proportion. (c) In samples of size 100 from this school, the sample proportion of students who bring a reusable water bottle from home typically varies by about 0.095 from the true proportion. (d) In samples of size 100 from this school, the sample proportion of students who bring a reusable water bottle from home typically varies by about 0.048 from the true proportion. (e) There is not enough information to calculate the standard error.",d,Multiple Choice,,,,
-TPS,T8,4,"Many television viewers express doubts about the validity of certain commercials. In an attempt to answer their critics, Timex Group USA wishes to estimate the true proportion p of all consumers who believe what is shown in Timex television commercials. Which of the following is the smallest number of consumers that Timex can survey to guarantee a margin of error of 0.05 or less at a 99% confidence level? (a) 550 (b) 600 (c) 650 (d) 700 (e) 750 ",d,Multiple Choice,,,,
-TPS,T8,5,Which of the following is the critical value for calculating a 94% confidence interval for a population proportion? (a) 1.555 (b) 1.645 (c) 1.881 (d) 1.960 (e) 2.576 ,c,Multiple Choice,,,,
-TPS,T8,6,"A radio talk show host with a large audience is interested in the proportion p of adults in his listening area who think the drinking age should be lowered to 18. To find this out, he poses the following question to his listeners: “Do you think that the drinking age should be reduced to 18 in light of the fact that 18-year-olds are eligible for military service?” He asks listeners to go to his website and vote “Yes” if they agree the drinking age should be lowered and “No” if not. Of the 100 people who voted, 70 answered “Yes.” Which of the following conditions are violated? I) Random II) 10% III) Large Counts (a) I only (b) II only (c) III only (d) I and II only (e) I, II, and III",a,Multiple Choice,,,,
-TPS,T8,7,"A marketing assistant for a technology firm plans to randomly select 1000 customers to estimate the pro-portion who are satisfied with the firm’s performance. Based on the results of the survey, the assistant will construct a 95% confidence interval for the proportion of all customers who are satisfied. The marketing manager, however, says that the firm can afford to survey only 250 customers. How will this decrease in sample size affect the margin of error? (a) The margin of error will be about 4 times larger. (b) The margin of error will be about 2 times larger. (c) The margin of error will be about the same size. (d) The margin of error will be about half as large. (e) The margin of error will be about one-fourth as large.",b,Multiple Choice,,,,
-TPS,T8,8,Thirty-five people from a random sample of 125 workers from Company A admitted to using sick leave when they weren’t really ill. Seventeen employees from a random sample of 68 workers from Company B admitted that they had used sick leave when they weren’t ill. Which of the following is a 95% confidence interval for the difference in the proportions of workers at the two companies who would admit to using sick leave when they weren’t ill? (a) 0.03 ± sqrt(((0.28*0.72)/125) + ((0.25*0.75)/68)) (b) 0.03 ± 1.96 sqrt(((0.28*0.72)/125) + ((0.25*0.75)/68)) (c) 0.03 ± 1.96 sqrt(((0.28*0.72)/125) - ((0.25*0.75)/68)) (d) 57 ± 1.96 sqrt(((0.28*0.72)/125) + ((0.25*0.75)/68)) (e) 57 ± 1.96 sqrt(((0.28*0.72)/125) - ((0.25*0.75)/68)),b,Multiple Choice,,,,
-TPS,T8,9,A telephone poll of an SRS of 1234 adults found that 62% are generally satisfied with their lives. The announced margin of error for the poll was 3%. Does the margin of error account for the fact that some adults do not have telephones? (a) Yes; the margin of error accounts for all sources of error in the poll. (b) Yes; taking an SRS eliminates any possible bias in estimating the population proportion. (c) Yes; the margin of error accounts for undercoverage but not nonresponse. (d) No; the margin of error accounts for nonresponse but not undercoverage. (e) No; the margin of error only accounts for sampling variability.,e,Multiple Choice,,,,
-TPS,T8,10,"At a baseball game, 42 of 65 randomly selected people own an iPod. At a rock concert occurring at the same time across town, 34 of 52 randomly selected people own an iPod. A researcher wants to test the claim that the proportion of iPod owners at the two venues is different. A 90% confidence interval for the difference (Game – Concert) in population proportions is (- 0.154, 0.138). Which of the following gives the correct outcome of the researcher’s test of the claim? (a) Because the interval includes 0, the researcher can conclude that the proportion of iPod owners at the two venues is the same. (b) Because the center of the interval is −0.008, the researcher can conclude that a higher proportion of people at the rock concert own iPods than at the baseball game. (c) Because the interval includes 0, the researcher cannot conclude that the proportion of iPod owners at the two venues is different. (d) Because the interval includes −0.008, the researcher cannot conclude that the proportion of iPod owners at the two venues is different. (e) Because the interval includes more negative than positive values, the researcher can conclude that a higher proportion of people at the rock concert own iPods than at the baseball game.",c,Multiple Choice,,,,
-TPS,T8,11a,"The U.S. Forest Service is considering additional restrictions on the number of vehicles allowed to enter Yellowstone National Park. To assess public reaction, the service asks a random sample of 150 visitors if they favor the proposal. Of these, 89 say “Yes.” Construct and interpret a 99% confidence interval for the proportion of all visitors to Yellowstone who favor the restrictions.","STATE: p= the true proportion of all visitors to Yellowstone who would say they favor the restrictions. PLAN: One-sample z interval. Random: The visitors were selected randomly. 10%: n= 150 is less than 10% of all visitors to Yellowstone National Park. Large Counts: np^ = 89 ≥ 10 and n(1 - p^) = 61 ≥ 10. DO: p^ = 89/150 = 0.593; (0.490, 0.696). CONCLUDE: We are 99% confident that the interval from 0.490 to 0.696 captures p5 the true proportion of all visitors who would say that they favor the restrictions.",Free Response,,,,
-TPS,T8,11b,"The U.S. Forest Service is considering additional restrictions on the number of vehicles allowed to enter Yellowstone National Park. To assess public reaction, the service asks a random sample of 150 visitors if they favor the proposal. Of these, 89 say “Yes.” A 99% confidence was found to be (0.490, 0.696) based on this,  is there convincing evidence that more than half of all visitors to Yellowstone National Park favor the proposal? Justify your answer.","Because there are values less than 0.50 in the confidence interval, the U.S. Forest Service cannot conclude that more than half of visitors to Yellowstone National Park favor the proposal. It is plausible that only 49% favor the proposal. ",Free Response,,,,
-TPS,T8,12a,"For some people, mistletoe is a symbol of romance. Mesquite trees, however, have no love for the parasitic plant that attaches itself and steals nutrients from the tree. To estimate the proportion of mesquite trees in a desert park that are infested with mistletoe, a random sample of mesquite trees was randomly selected. After inspecting the trees in the sample, the park supervisor calculated a 95% confidence interval of 0.2247 to 0.5753. Interpret the confidence level. ","If we were to select many random samples of the size used in this study from the same population of mesquite trees and construct a 95% confidence interval using each sample, about 95% of the intervals would capture the true proportion of all mesquite trees in this park that are infested with mistletoe. ",Free Response,,,,
-TPS,T8,12b,"For some people, mistletoe is a symbol of romance. Mesquite trees, however, have no love for the parasitic plant that attaches itself and steals nutrients from the tree. To estimate the proportion of mesquite trees in a desert park that are infested with mistletoe, a random sample of mesquite trees was randomly selected. After inspecting the trees in the sample, the park supervisor calculated a 95% confidence interval of 0.2247 to 0.5753. Calculate the sample size used to create this interval.",Point estimate = 0.4 and margin of error = 0.1753; 0.1753 = 1.96 sqrt((0.4(0.62))/n); n = 30,Free Response,,,,
-TPS,T8,13,"Do “props” make a difference when researchers interact with their subjects? Emily and Madi asked 100 people if they thought buying coffee at Star-bucks was a waste of money. Half of the subjects were asked while Emily and Madi were holding cups from Starbucks, and the other half of the subjects were asked when the girls were empty handed. The choice of holding or not holding the cups was deter-mined at random for each subject. When holding the cups, 19 of 50 subjects agreed that buying cof-fee at Starbucks was a waste of money. When they weren’t holding the cups, 23 of 50 subjects said it was a waste of money. Calculate and interpret a 90% confidence interval for the difference in the proportion of people like the ones in this experiment who would say that buying coffee from Starbucks is a waste of money when asked by interviewers holding or not holding a cup from Starbucks.","STATE: 90% CI for p1 - p2, where p1 = the true proportion of people like these who would say that buying coffee at Starbucks is a waste of money when the girls hold cups from Starbucks and p2 = the true proportion of people like these who would say that buying coffee at Starbucks is a waste of money when the girls were empty handed. PLAN: Two-sample z-interval for p1 - p2. Random: Subjects were randomly assigned to be asked while the girls held Starbucks cups or were empty handed. Large Counts: 19, 50 - 19 = 31, 23, 50 - 23 = 27 are all ≥ 10. DO: p1 = 19/50 = 0.38 and p2 =23/50 =0.46; (20.242, 0.082). CONCLUDE: We are 90% confident that the interval from 20.242 to 0.082 captures p1 - p2 = the difference (Holding – Empty handed) in the true proportions of people like these who would say that buying coffee at Starbucks is a waste of money.",Free Response,,,,
-TPS,9.1,1,"For the following scenario, state appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest: Mr. Tabor believes that less than 75% of the students at his school completed their math homework last night. The math teachers inspect the homework assignments from a random sample of 50 students at the school.","H0: p=0.75; Ha: p< 0.75, where p= the true proportion of the students at Mr. Tabor’s school who completed their math homework last night. ",Free Response,,,,
-TPS,9.1,2,"For the following scenario, state appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest: A Gallup poll report revealed that 72% of teens said they seldom or never argue with their friends. 3 Yvonne wonders whether this result holds true in her large high school, so she surveys a random sample of 150 students at her school.","H0: p= 0.72; Ha: p≠ 0.72, where p= the true proportion of teens in Yvonne’s school who rarely or never argue with their friends.",Free Response,,,,
-TPS,9.1,3,"For the following scenario, state appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest: One company’s bottles of grape-fruit juice are filled by a machine that is set to dispense an average of 180 milliliters (ml) of liquid. A quality-control inspector must check that the machine is working properly. The inspector takes a random sample of 40 bottles and measures the volume of liquid in each bottle.","H0: µ = 180; Ha: µ ≠ 180, where m = the true mean volume of liquid dispensed by the machine. ",Free Response,,,,
-TPS,9.1,4,"For the following scenario, state appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest: The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students’ attitudes toward school and study habits. Scores range from 0 to 200. Higher scores indicate better attitudes and study habits. The mean score for U.S. college students is about 115. A teacher suspects that older students have better attitudes toward school, on average. She gives the SSHA to an SRS of 45 of the over 1000 students at her college who are at least 30 years of age.","H0: µ = 115; Ha: µ > 115, where m = the true mean score on the SSHA for all students at least 30 years of age at the teacher’s college. ",Free Response,,,,
-TPS,9.1,5,"For the following scenario, state appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest: During the winter months, the temperatures at the Starneses’ Colorado cabin can stay well below freezing (32°F or 0°C) for weeks at a time. To prevent the pipes from freezing, Mrs. Starnes sets the thermostat at 50°F. The manufacturer claims that the thermostat allows variation in home temperature of σ =°3F. Mrs. Starnes suspects that the manufacturer is overstating the consistency of the thermostat.","H0: σ = 3; Ha: σ > 3, where s = the true standard deviation of the temperature in the cabin. ",Free Response,,,,
-TPS,9.1,6,"For the following scenario, state appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest: When ski jumpers take off, the distance they fly varies considerably depending on their speed, skill, and wind conditions. Event organizers must position the landing area to allow for differences in the distances that the athletes fly. For a particular competition, the organizers estimate that the variation in distance flown by the athletes will be σ=10 meters. An experienced jumper thinks that the organizers are underestimating the variation.","H0: σ = 10; Ha: σ > 10, where s = the true standard deviation of the distance jumped by the ski jumpers. ",Free Response,,,,
-TPS,9.1,7a,"Explain what’s wrong with the stated hypotheses. Then give correct hypotheses. A change is made that should improve student satisfaction with the parking situation at a local high school. Before the change, 37% of students approve of the parking that’s provided. The null hypothesis Ho: p>0.37 is tested against the alternative: Ha: p=0.37.",The null hypothesis is always that there is “no difference” or “no change”; the alternative hypothesis is what we suspect is true. These ideas are reversed in the stated hypotheses. Correct: H0: p= 0.37; Ha: p> 0.37. ,Free Response,,,,
-TPS,9.1,7b,"Explain what’s wrong with the stated hypotheses. Then give correct hypotheses. A researcher suspects that the mean birth weights of babies whose mothers did not see a doctor before delivery is less than 3000 grams. The researcher states the hypotheses Ho: x_bar = 3000 grams, Ha: x_bar < 3000 grams","Hypotheses are always about population parameters. However, the stated hypotheses are in terms of the sample statistic. Correct: H0: µ = 3000 grams; Ha: µ < 3000 grams. ",Free Response,,,,
-TPS,9.1,8a,"Explain what’s wrong with the stated hypotheses. Then give correct hypotheses. A change is made that should improve student satisfaction with the parking situation at your school. Before the change, 37% of students approve of the parking that’s provided. The null hypothesis Ho: p^= 0.37 is tested against the alternative Ha: p^> 0.37.","Hypotheses are always about population parameters, but the stated hypotheses are about the sample statistic. Correct: H0: p= 0.37; Ha: p> 0.37. ",Free Response,,,,
-TPS,9.1,8b,"Explain what’s wrong with the stated hypotheses. Then give correct hypotheses. A researcher suspects that the mean birth weights of babies whose mothers did not see a doctor before delivery is less than 3000 grams. The researcher states the hypotheses as Ho: μ =3000 grams,  Ha: μ ≤ 2999 grams",Values in both hypotheses must be the same; the null hypothesis should have 5 and the alternative should have <. Correct: H0: µ=3000 grams; Ha: µ<3000 grams. ,Free Response,,,,
-TPS,9.1,9a,Mr. Tabor believes that less than 75% of the students at his school completed their math homework last night. The math teachers inspect the homework assignments from a random sample of 50 students at the school. The math teachers inspect the homework assignments from a random sample of 50 students at the school. Only 68% of the students completed their math homework. A significance test yields a P-value of 0.1265. Explain what it would mean for the null hypothesis to be true in this setting.,"If H0: p= 0.75 is true, then the proportion of all students at Mr. Tabor’s school who completed their homework last night is 0.75. ",Free Response,,,,
-TPS,9.1,9b,Mr. Tabor believes that less than 75% of the students at his school completed their math homework last night. The math teachers inspect the homework assignments from a random sample of 50 students at the school. The math teachers inspect the homework assignments from a random sample of 50 students at the school. Only 68% of the students completed their math homework. A significance test yields a P-value of 0.1265. Interpret the P -value. ,"Assuming the proportion is 0.75, there is a 0.1265 probability of getting a sample proportion of 0.68 or less by chance in a random sample of 50 students at the school. ",Free Response,,,,
-TPS,9.1,10a,"The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students’ attitudes toward school and study habits. Scores range from 0 to 200. Higher scores indicate better attitudes and study habits. The mean score for U.S. college students is about 115. A teacher suspects that older students have better attitudes toward school, on average. She gives the SSHA to an SRS of 45 of the over 1000 students at her college who are at least 30 years of age. Explain what it would mean for the null hypothesis to be true in this setting.","If H0: µ = 115 is true, then the true mean score is 115.",Free Response,,,,
-TPS,9.1,10b,"The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students’ attitudes toward school and study habits. Scores range from 0 to 200. Higher scores indicate better attitudes and study habits. The mean score for U.S. college students is about 115. A teacher suspects that older students have better attitudes toward school, on average. She gives the SSHA to an SRS of 45 of the over 1000 students at her college who are at least 30 years of age. Interpret the P -value. ","Assuming the true mean score is 115, there is a 0.0101 probability of getting a sample mean of 125.7 or greater just by chance in an SRS of 45 older students. ",Free Response,,,,
-TPS,9.1,11,One company’s bottles of grape-fruit juice are filled by a machine that is set to dispense an average of 180 milliliters (ml) of liquid. A quality-control inspector must check that the machine is working properly. The inspector takes a random sample of 40 bottles and measures the volume of liquid in each bottle. The mean amount of liquid in the bottles is 179.6 ml and the standard deviation is 1.3 ml. A significance test yields a P -value of 0.0589. Interpret the P -value.,"Assuming the true mean volume of liquid dispensed by the machine is 180 ml, there is a 0.0589 probability of getting a sample mean at least as far from 180 as 179.6 (in either direction) by chance in a random sample of 40 bottles filled by the machine. ",Free Response,,,,
-TPS,9.1,12,"A Gallup poll report revealed that 72% of teens said they seldom or never argue with their friends. 3 Yvonne wonders whether this result holds true in her large high school, so she surveys a random sample of 150 students at her school. Yvonne finds that 96 of the 150 students (64%) say they rarely or never argue with friends. A significance test yields a P-value of 0.0291. Interpret the P-value.","Assuming the true proportion of teens in Yvonne’s school who rarely or never argue with their friends is 0.72, there is a 0.0291 probability of getting a sample proportion at least as far from 0.72 as 0.64 (in either direction) by chance in a random sample of 150 students from her school.",Free Response,,,,
-TPS,9.1,13,"A student performs a test of Ho: μ = 100 versus Ha: μ > 100 and gets a P-value of 0.044. The student says, “There is a 0.044 probability of getting the sample result I did by chance alone.” Explain why the student’s explanation is wrong.","The student forgot to include the conditions and the direction in the interpretation. Assuming the null hypothesis is true, there is a 0.044 probability of getting the sample result I did or one even larger by chance alone. ",Free Response,,,,
-TPS,9.1,14,"A student performs a test of Ho: p=0.3 versus Ha: p<0.3 and gets a P-value of 0.22. The student says, “This means there is about a 22% chance that the null hypothesis is true.” Explain why the student’s explanation is wrong.","Either H0 is true or H0 is false. Assuming the null hypothesis is true, there is a 0.22 probability of getting a sample proportion as small as or smaller than the one observed just by chance.",Free Response,,,,
-TPS,9.1,15,Mr. Tabor believes that less than 75% of the students at his school completed their math homework last night. The math teachers inspect the homework assignments from a random sample of 50 students at the school. The math teachers inspect the homework assignments from a random sample of 50 students at the school. Only 68% of the students completed their math homework. A significance test yields a P-value of 0.1265. What conclusion would you make at the α=0.05 level?,"Because the P-value of 0.1265 is greater than α= 0.05, we fail to reject H0. We do not have convincing evidence that the true proportion of students at Mr. Tabor’s school who completed their math homework last night is less than 0.75.",Free Response,,,,
-TPS,9.1,16,"The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students’ attitudes toward school and study habits. Scores range from 0 to 200. Higher scores indicate better attitudes and study habits. The mean score for U.S. college students is about 115. A teacher suspects that older students have better attitudes toward school, on average. She gives the SSHA to an SRS of 45 of the over 1000 students at her college who are at least 30 years of age. In the study of older students’ attitudes, the sample mean SSHA score was 125.7 and the sample standard deviation was 29.8. A significance test yields a P -value of 0.0101. What conclusion would you make at the α=0.05 level?","Because the P-value of 0.0101 is less than α= 0.05, we reject H0. We have convincing evidence that the true mean score on the SSHA for all students at least 30 years of age at the teacher’s college is greater than 115.",Free Response,,,,
-TPS,9.1,17a,One company’s bottles of grape-fruit juice are filled by a machine that is set to dispense an average of 180 milliliters (ml) of liquid. A quality-control inspector must check that the machine is working properly. The inspector takes a random sample of 40 bottles and measures the volume of liquid in each bottle. The mean amount of liquid in the bottles is 179.6 ml and the standard deviation is 1.3 ml. A significance test yields a P -value of 0.0589. Interpret the P -value. What conclusion would you make at the α=0.10 level?,"Because the P-value of 0.0589 is less than α = 0.10, we reject H0. We have convincing evidence that the true mean volume of liquid dispensed differs from 180 ml. ",Free Response,,,,
-TPS,9.1,17b,One company’s bottles of grape-fruit juice are filled by a machine that is set to dispense an average of 180 milliliters (ml) of liquid. A quality-control inspector must check that the machine is working properly. The inspector takes a random sample of 40 bottles and measures the volume of liquid in each bottle. The mean amount of liquid in the bottles is 179.6 ml and the standard deviation is 1.3 ml. A significance test yields a P -value of 0.0589. Interpret the P -value. Would your conclusion change if a 5% significance level was used instead of a 10% significance level? Explain your reasoning.,"Yes; because the P-value of 0.0589 > α = 0.05, we would fail to reject H0. We lack convincing evidence that the true mean volume differs from 180 ml.",Free Response,,,,
-TPS,9.1,18a,"A Gallup poll report revealed that 72% of teens said they seldom or never argue with their friends. 3 Yvonne wonders whether this result holds true in her large high school, so she surveys a random sample of 150 students at her school. Yvonne finds that 96 of the 150 students (64%) say they rarely or never argue with friends. A significance test yields a P-value of 0.0291. Interpret the P-value. What conclusion would you make at the α=0.01 level?","Because the P-value of 0.0291 > α=0.01, we fail to reject H0. We lack convincing evidence that the true proportion of teens in Yvonne’s school who rarely or never argue differs from 0.72. ",Free Response,,,,
-TPS,9.1,18b,"A Gallup poll report revealed that 72% of teens said they seldom or never argue with their friends. 3 Yvonne wonders whether this result holds true in her large high school, so she surveys a random sample of 150 students at her school. Yvonne finds that 96 of the 150 students (64%) say they rarely or never argue with friends. A significance test yields a P-value of 0.0291. Interpret the P-value. Would your conclusion change if a 5% significance level was used instead of a 1% significance level? Explain your reasoning.","Yes; because the P-value of 0.0291< α = 0.05, we would reject H0. We have convincing evidence that the true proportion differs from 0.72.",Free Response,,,,
-TPS,9.1,19,"A student performs a test of H0: p= 0.75 versus Ha: p< 0.75 at the α=0.05 significance level and gets a P-value of 0.22. The student writes: “Because the P-value is large, we accept H0. The data provide convincing evidence that the null hypothesis is true.” Explain what is wrong with this conclusion.","It is never correct to “accept the null hypothesis.” If the P-value is large, the data do not provide convincing evidence that the alternative hypothesis is true. Lacking evidence for the alternative hypothesis does not provide convincing evidence that the null hypothesis is true. ",Free Response,,,,
-TPS,9.1,20,"A student performs a test of H0: μ =12 versus Ha μ ≠ 12 at the α=0.05 significance level and gets a P-value of 0.01. The student writes: “Because the P -value is small, we reject H0. The data prove that Ha is true.” Explain what is wrong with this conclusion.","The data never “prove” a hypothesis true, no matter how large or small the P-value. There can always be error. ",Free Response,,,,
-TPS,9.1,21a,"The mean weight of loaves of bread produced at the bakery where you work is supposed to be 1 pound. You are the supervisor of quality control at the bakery, and you are concerned that new employees are producing loaves that are too light. Suppose you weigh an SRS of bread loaves and find that the mean weight is 0.975 pound. State appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest.","H0: µ = 1, Ha: µ < 1, where µ = the true mean weight (in pounds) of bread loaves produced. ",Free Response,,,,
-TPS,9.1,21b,"The mean weight of loaves of bread produced at the bakery where you work is supposed to be 1 pound. You are the supervisor of quality control at the bakery, and you are concerned that new employees are producing loaves that are too light. Suppose you weigh an SRS of bread loaves and find that the mean weight is 0.975 pound. Explain why there is some evidence for the alternative hypothesis.",Some evidence for the alternative hypothesis exists because the mean weight of an SRS of bread loaves is only 0.975 pound—less than the suspected mean (1 pound).,Free Response,,,,
-TPS,9.1,21c,"The mean weight of loaves of bread produced at the bakery where you work is supposed to be 1 pound. You are the supervisor of quality control at the bakery, and you are concerned that new employees are producing loaves that are too light. Suppose you weigh an SRS of bread loaves and find that the mean weight is 0.975 pound. The P -value for the test is 0.0806. Interpret the P -value.","Assuming the true mean weight of bread loaves is 1 pound, a 0.0806 probability exists of getting a sample mean of 0.975 pound or less by chance in a random sample of loaves. ",Free Response,,,,
-TPS,9.1,21d,"The mean weight of loaves of bread produced at the bakery where you work is supposed to be 1 pound. You are the supervisor of quality control at the bakery, and you are concerned that new employees are producing loaves that are too light. Suppose you weigh an SRS of bread loaves and find that the mean weight is 0.975 pound. What conclusion would you make at the α 0.01= significance level with a P-value of 0.0806?","Because the P-value of 0.0806 > α = 0.01, we fail to reject H0. We do not have convincing evidence that the true mean weight for all loaves of bread produced is less than 1 pound. ",Free Response,,,,
-TPS,9.1,22a,"Nationally, the proportion of red cars on the road is 0.12. A statistically minded fan of the Philadelphia Phillies (whose team color is red) wonders if Phillies fans are more likely to drive red cars. One day during a home game, he takes a random sample of 210 cars parked at Citizens Bank Park (the Phillies’ home field), and counts 35 red cars. State appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest.","H0: p = 0.12, Ha: p > 0.12, where p= the true proportion of all cars parked at Phillies home field that are red. ",Free Response,,,,
-TPS,9.1,22b,"Nationally, the proportion of red cars on the road is 0.12. A statistically minded fan of the Philadelphia Phillies (whose team color is red) wonders if Phillies fans are more likely to drive red cars. One day during a home game, he takes a random sample of 210 cars parked at Citizens Bank Park (the Phillies’ home field), and counts 35 red cars. Explain why there is some evidence for the alternative hypothesis.","Some evidence for the alternative hypothesis exists because the sample proportion of red parked cars is 35/210 = 0.167, which is greater than the national proportion of red cars (0.12). ",Free Response,,,,
-TPS,9.1,22c,"Nationally, the proportion of red cars on the road is 0.12. A statistically minded fan of the Philadelphia Phillies (whose team color is red) wonders if Phillies fans are more likely to drive red cars. One day during a home game, he takes a random sample of 210 cars parked at Citizens Bank Park (the Phillies’ home field), and counts 35 red cars. The P -value for the test is 0.0187. Interpret the P -value.","Assuming the true proportion of all red cars parked at Phillies home field is 0.12, a 0.0187 probability exists of getting a sample proportion of 35/210 = 0.167 or greater by chance in a random sample of 210 cars. ",Free Response,,,,
-TPS,9.1,22d,"Nationally, the proportion of red cars on the road is 0.12. A statistically minded fan of the Philadelphia Phillies (whose team color is red) wonders if Phillies fans are more likely to drive red cars. One day during a home game, he takes a random sample of 210 cars parked at Citizens Bank Park (the Phillies’ home field), and counts 35 red cars. What conclusion would you make at the α 0.05= significance level for test with a p-value of 0.0187?","Because the P-value of 0.0187 < α = 0.05, we reject H0. We have convincing evidence that the true proportion of all cars parked at Phillies home field that are red is greater than 0.12.",Free Response,,,,
-TPS,9.1,23,"You are thinking about opening a restaurant and are searching for a good location. From research you have done, you know that the mean income of those living near the restaurant must be over $85,000 to support the type of upscale restaurant you wish to open. You decide to take a simple random sample of 50 people living near one potential location. Based on the mean income of this sample, you will perform a test of H0: μ =  $85, 000, Ha: μ > $85, 000 where μ is the true mean income in the population of people who live near the restaurant.  Describe a Type I error and a Type II error in this setting, and give a possible consequence of each.","Type I: You find convincing evidence that the mean income of all residents near the restaurant exceeds $85,000 when in reality it does not. Consequence: You open in an undesirable location, so your restaurant may go out of business. Type II: You do not find convincing evidence that the mean income of all residents near the restaurant exceeds $85,000 when in reality it does. Consequence: You do not open in a desirable location and lose potential income. ",Free Response,,,,
-TPS,9.1,24,"Television networks rely heavily on ratings of TV shows when deciding whether to renew a show for another season. Suppose a network has decided that “Miniature Golf with the Stars” will only be renewed if it can be established that more than 12% of U.S. adults watch the show. A polling company asks a random sample of 2000 U.S. adults if they watch “Miniature Golf with the Stars.” The network uses the data to perform a test of H0: p=0.12, Ha p>0.12 where p is the true proportion of all U.S. adults who watch the show. Describe a Type I error and a Type II error in this setting, and give a possible consequence of each.","Type I: The network finds convincing evidence that the true proportion of U.S. adults who watch “Miniature Golf with the Stars” is greater than 0.12, when it really equals 0.12. Consequence: The show will be renewed when interest does not support it. Type II: The network doesn’t find convincing evidence that the true proportion who watch “Miniature Golf with the Stars” is greater than 0.12, when it really is greater than 0.12. Consequence: The show will not renew when there was a market for it.",Free Response,,,,
-TPS,9.1,25a,"Slow response times by paramedics, firefighters, and policemen can have serious consequences for accident victims. In the case of life-threatening injuries, victims generally need medical attention within 8 minutes of the accident. Several cities have begun to monitor emergency response times. In one such city, emergency personnel took more than 8 minutes to arrive on 22% of all calls involving life-threatening injuries last year. The city manager shares this information and encourages these first responders to “do better.” After 6 months, the city manager selects an SRS of 400 calls involving life-threatening injuries and examines the response times. She then performs a test at the α=0.05 level of H0 p= 0.22 versus Ha: p< 0.22, where p is the true proportion of calls involving life-threatening injuries during this 6-month period for which emergency personnel took more than 8 minutes to arrive. Describe a Type I error and a Type II error in this setting.",A Type I error would be finding convincing evidence that the proportion of all calls in which first responders took more than 8 minutes to arrive had decreased when it really hadn’t. A Type II error would be not finding convincing evidence that the proportion of all calls in which first responders took more than 8 minutes to arrive decreased when it really had. ,Free Response,,,,
-TPS,9.1,25b,"Slow response times by paramedics, firefighters, and policemen can have serious consequences for accident victims. In the case of life-threatening injuries, victims generally need medical attention within 8 minutes of the accident. Several cities have begun to monitor emergency response times. In one such city, emergency personnel took more than 8 minutes to arrive on 22% of all calls involving life-threatening injuries last year. The city manager shares this information and encourages these first responders to “do better.” After 6 months, the city manager selects an SRS of 400 calls involving life-threatening injuries and examines the response times. She then performs a test at the α=0.05 level of H0 p= 0.22 versus Ha: p< 0.22, where p is the true proportion of calls involving life-threatening injuries during this 6-month period for which emergency personnel took more than 8 minutes to arrive. Which type of error is more serious in this case? Justify your answer.",A Type I error would be worse because the city would overestimate the ability of the emergency personnel to get to the scene quickly and people may end up dying. ,Free Response,,,,
-TPS,9.1,25c,"Slow response times by paramedics, firefighters, and policemen can have serious consequences for accident victims. In the case of life-threatening injuries, victims generally need medical attention within 8 minutes of the accident. Several cities have begun to monitor emergency response times. In one such city, emergency personnel took more than 8 minutes to arrive on 22% of all calls involving life-threatening injuries last year. The city manager shares this information and encourages these first responders to “do better.” After 6 months, the city manager selects an SRS of 400 calls involving life-threatening injuries and examines the response times. She then performs a test at the α=0.05 level of H0 p= 0.22 versus Ha: p< 0.22, where p is the true proportion of calls involving life-threatening injuries during this 6-month period for which emergency personnel took more than 8 minutes to arrive. Do you agree with the manager’s choice of α=0.05? Why or why not?","The probability of a Type I error is α = 0.05; because the consequence may be the difference between life and death, the manager should use a value for a that is lower, such as α = 0.01. ",Free Response,,,,
-TPS,9.1,26a,"The Environmental Protection Agency (EPA) has determined that safe drinking water should contain at most 1.3 mg/liter of copper, on average. A water supply company is testing water from a new source and collects water in small bottles at each of 30 randomly selected locations. The company performs a test at the α=0.05 significance level of H0: μ =1.3 versus Ha: μ >1.3, where μ is the true mean copper content of the water from the new source. Describe a Type I error and a Type II error in this setting.",A Type I error would be finding convincing evidence that the true mean copper content of the water from the new source is greater than 1.3 mg/liter when it really isn’t. A Type II error would be not finding convincing evidence that the true mean copper content of the water from the new source is greater than 1.3 mg/l when it really is. ,Free Response,,,,
-TPS,9.1,26b,"The Environmental Protection Agency (EPA) has determined that safe drinking water should contain at most 1.3 mg/liter of copper, on average. A water supply company is testing water from a new source and collects water in small bottles at each of 30 randomly selected locations. The company performs a test at the α=0.05 significance level of H0: μ =1.3 versus Ha: μ >1.3, where μ is the true mean copper content of the water from the new source. Which type of error is more serious in this case? Justify your answer.","A Type II error would be worse because the water would not be safe for drinking, yet would seem safe. ",Free Response,,,,
-TPS,9.1,26c,"The Environmental Protection Agency (EPA) has determined that safe drinking water should contain at most 1.3 mg/liter of copper, on average. A water supply company is testing water from a new source and collects water in small bottles at each of 30 randomly selected locations. The company performs a test at the α=0.05 significance level of H0: μ =1.3 versus Ha: μ >1.3, where μ is the true mean copper content of the water from the new source. Do you agree with the company’s choice of α=0.05? Why or why not?","Here, we seek to minimize the probability of making a Type II error. As the probability of a Type II error increases, the probability of Type I error decreases; so I would recommend that the company use a value of a greater than 0.05. ",Free Response,,,,
-TPS,9.1,29,Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. The mean time is 18 seconds for one particular maze. A researcher thinks that a loud noise will cause the mice to complete the maze faster. She measures how long each of 10 mice takes with a loud noise as stimulus. The appropriate hypotheses for the significance test are  (a) H0: μ = 18; Ha: µ ≠ 18  (b) H0: μ = 18; Ha: µ > 18  (c) H0: μ > 18; Ha: µ = 18 (d) H0: μ = 18; Ha: µ < 18 (e) H0: x_bar = 18; Ha: x_bar < 18,d,Multiple Choice,,,,
-TPS,9.1,30,"Members of the city council want to know if a majority of city residents supports a 1% increase in the sales tax to fund road repairs. To investigate, they survey a random sample of 300 city residents and use the results to test the following hypotheses: H0: p =0.50; Ha: p > 0.50 where p is the proportion of all city residents who support a 1% increase in the sales tax to fund road repairs. A Type I error in the context of this study occurs if the city council (a) finds convincing evidence that a majority of residents supports the tax increase, when in reality there isn’t convincing evidence that a majority supports the increase. (b) finds convincing evidence that a majority of residents supports the tax increase, when in reality at most 50% of city residents support the increase. (c) finds convincing evidence that a majority of residents supports the tax increase, when in reality more than 50% of city residents do support the increase. (d) does not find convincing evidence that a majority of residents supports the tax increase, when in reality more than 50% of city residents do support the increase. (e) does not find convincing evidence that a majority of residents supports the tax increase, when in reality at most 50% of city residents do support the increase.",b,Multiple Choice,,,,
-TPS,9.1,31,"Members of the city council want to know if a majority of city residents supports a 1% increase in the sales tax to fund road repairs. To investigate, they survey a random sample of 300 city residents and use the results to test the following hypotheses: H0: p = 0.50; Ha: p > 0.50 where p is the proportion of all city residents who support a 1% increase in the sales tax to fund road repairs. In the sample, p^= 158 / 300 = 0.527. The resulting P-value is 0.18. What is the correct interpretation of this P-value? (a) Only 18% of the city residents support the tax increase. (b) There is an 18% chance that the majority of residents supports the tax increase. (c) Assuming that 50% of residents support the tax increase, there is an 18% probability that the sample proportion would be 0.527 or greater by chance alone. (d) Assuming that more than 50% of residents support the tax increase, there is an 18% probability that the sample proportion would be 0.527 or greater by chance alone. (e) Assuming that 50% of residents support the tax increase, there is an 18% chance that the null hypothesis is true by chance alone.",c,Multiple Choice,,,,
-TPS,9.1,32,"Members of the city council want to know if a majority of city residents supports a 1% increase in the sales tax to fund road repairs. To investigate, they survey a random sample of 300 city residents and use the results to test the following hypotheses: H0: p =0.50; Ha: p > 0.50 where p is the proportion of all city residents who support a 1% increase in the sales tax to fund road repairs. Based on the P-value of 0.18, which of the following would be the most appropriate conclusion? (a) Because the P-value is large, we reject H0. There is sufficient evidence that more than 50% of city residents support the tax increase. (b) Because the P-value is large, we fail to reject H0. There is sufficient evidence that more than 50% of city residents support the tax increase. (c) Because the P-value is large, we reject H0. There is sufficient evidence that at most 50% of city residents support the tax increase. (d) Because the P-value is large, we fail to reject H0. There is sufficient evidence that at most 50% of city residents support the tax increase. (e) Because the P-value is large, we fail to reject H0. There is not sufficient evidence that more than 50% of city residents support the tax increase.",e,Multiple Choice,,,,
-TPS,9.2,35,"Jason reads a report that says 80% of U.S. high school students have a computer at home. He believes the proportion is smaller than 0.80 at his large rural high school. Jason chooses an SRS of 60 students and finds that 41 have a computer at home. He would like to carry out a test at the α= 0.05 significance level of H0: p = 0.80 versus Ha: p < 0.80, where p = the true proportion of all students at Jason’s high school who have a computer at home. Check if the conditions for performing the significance test are met.",Random: We have an SRS of 60 students from a large rural high school. 10%: The sample size (60) is less than 10% of all students at this large high school. Large Counts: np0 = 60(0.80) = 48 ≥ 10 and n(1 - p0) = 60(0.20) = 12 ≥ 10.,Free Response,,,,
-TPS,9.2,36,"A recent report claimed that 13% of students typically walk to school.  DeAnna thinks that the proportion is higher than 0.13 at her large elementary school. She surveys a random sample of 100 students and finds that 17 typically walk to school. DeAnna would like to carry out a test at the α= 0.05 significance level of H0: p = 0.13 versus Ha: p > 0.13, where p = the true proportion of all students at her elementary school who typically walk to school. Check if the conditions for performing the significance test are met.",Random: We have a random sample of 100 students from a large elementary school. 10%: The sample size (100) is less than 10% of the students at this large elementary school. Large Counts: np0 = 100(0.13) = 13 ≥ 10 and n(1 - p0) = 100(0.87) = 87 ≥ 10.,Free Response,,,,
-TPS,9.2,37a,"The chips project Zenon decided to investigate whether students at his school prefer name-brand potato chips to generic potato chips. He randomly selected 50 of the 400 students at his school and asked each student to try both types of chips, in a random order. Overall, 41 of the 50 students preferred the name-brand chips. Zenon wants to perform a test of H0: p = 0.5 versus Ha: p > 0.5, where p = the true proportion of students at his school who prefer name-brand chips. Can the observations in the sample be viewed as independent? Justify your answer.","No; the students (n = 50) are selected from a finite population (N = 400) without replacement. Because n = 50 is not less than 10% of the population size (400), the observations are not independent. ",Free Response,,,,
-TPS,9.2,37b,"The chips project Zenon decided to investigate whether students at his school prefer name-brand potato chips to generic potato chips. He randomly selected 50 of the 400 students at his school and asked each student to try both types of chips, in a random order. Overall, 41 of the 50 students preferred the name-brand chips. Zenon wants to perform a test of H0: p = 0.5 versus Ha: p > 0.5, where p = the true proportion of students at his school who prefer name-brand chips. Is the sampling distribution of p^ approximately Normal? Explain your reasoning.","Yes; assuming p = 0.5 is true, the sampling distribution of p^ will be approximately Normal because np0 = 50(0.5) = 25 and n(1 - p0) = 50(1 - 0.5) = 25 are both ≥ 10.",Free Response,,,,
-TPS,9.2,38a,"On TV shows that feature singing competitions, contestants often wonder if there is an advantage in performing last. To investigate, researchers selected a random sample of 100 students from their large university and showed each student the audition video of 12 different singers with similar vocal skills. Each student viewed the videos in a random order. We would expect approximately 1/12 of the students to prefer the last singer seen, assuming order doesn’t matter. In this study, 11 of the 100 students preferred the last singer they viewed. The researchers want to perform a test of H0: p =1/12 versus Ha: p > 1/12, where p = the true proportion of students at this university who prefer the singer they last see. Can the observations in the sample be viewed as independent? Justify your answer.","Yes; although the students (n = 100) are selected from a finite population (a large university) without replacement, n = 100, 10% of the population size (all students at the large university). ",Free Response,,,,
-TPS,9.2,38b,"On TV shows that feature singing competitions, contestants often wonder if there is an advantage in performing last. To investigate, researchers selected a random sample of 100 students from their large university and showed each student the audition video of 12 different singers with similar vocal skills. Each student viewed the videos in a random order. We would expect approximately 1/12 of the students to prefer the last singer seen, assuming order doesn’t matter. In this study, 11 of the 100 students preferred the last singer they viewed. The researchers want to perform a test of H0: p =1/12 versus Ha: p > 1/12, where p = the true proportion of students at this university who prefer the singer they last see. Is the sampling distribution of p^ approximately Normal? Explain your reasoning.","No; assuming that p = 1/12 is true, the sampling distribution of p^ will not be approximately Normal because np0 = 100(1/12) = 8.33 and n(1 - p0) = 100(1 - 1/12) = 91.67 are not both ≥ 10.",Free Response,,,,
-TPS,9.2,39a,"Jason reads a report that says 80% of U.S. high school students have a computer at home. He believes the proportion is smaller than 0.80 at his large rural high school. Jason chooses an SRS of 60 students and finds that 41 have a computer at home. He would like to carry out a test at the α= 0.05 significance level of H0: p = 0.80 versus Ha: p < 0.80, where p = the true proportion of all students at Jason’s high school who have a computer at home. Explain why the sample result gives some evidence for the alternative hypothesis.","The sample result gives some evidence for Ha: p < 0.80 because p^ = 41/60 = 0.683, which is less than 0.80.",Free Response,,,,
-TPS,9.2,39b,"Jason reads a report that says 80% of U.S. high school students have a computer at home. He believes the proportion is smaller than 0.80 at his large rural high school. Jason chooses an SRS of 60 students and finds that 41 have a computer at home. He would like to carry out a test at the α= 0.05 significance level of H0: p = 0.80 versus Ha: p < 0.80, where p = the true proportion of all students at Jason’s high school who have a computer at home.  Calculate the standardized test statistic and P -value. ",z = (0.683 - 0.80) / sqrt((0.80(0.20) / 60) = -2.27; P-value = 0.0116.,Free Response,,,,
-TPS,9.2,39c,"Jason reads a report that says 80% of U.S. high school students have a computer at home. He believes the proportion is smaller than 0.80 at his large rural high school. Jason chooses an SRS of 60 students and finds that 41 have a computer at home. He would like to carry out a test at the α= 0.05 significance level of H0: p = 0.80 versus Ha: p < 0.80, where p = the true proportion of all students at Jason’s high school who have a computer at home. Based on a test with a P-value of 0.0116. What conclusion would you make? ","Because the P-value of 0.0116 < α = 0.05, we reject H0. We have convincing evidence that the true proportion of all students at this large rural high school who have a computer at home is less than 0.80.",Free Response,,,,
-TPS,9.2,40a,"A recent report claimed that 13% of students typically walk to school.  DeAnna thinks that the proportion is higher than 0.13 at her large elementary school. She surveys a random sample of 100 students and finds that 17 typically walk to school. DeAnna would like to carry out a test at the α= 0.05 significance level of H0: p = 0.13 versus Ha: p > 0.13, where p = the true proportion of all students at her elementary school who typically walk to school.  Explain why the sample gives some evidence for the alternative hypothesis.","The sample result gives some evidence for Ha: p>0.13 because p^ = 17/100 = 0.17, which is greater than 0.13.",Free Response,,,,
-TPS,9.2,40b,"A recent report claimed that 13% of students typically walk to school.  DeAnna thinks that the proportion is higher than 0.13 at her large elementary school. She surveys a random sample of 100 students and finds that 17 typically walk to school. DeAnna would like to carry out a test at the α= 0.05 significance level of H0: p = 0.13 versus Ha: p > 0.13, where p = the true proportion of all students at her elementary school who typically walk to school.  Calculate the standardized test statistic and P -value. ",z = (0.17 - 0.13) / sqrt((0.13)(0.87)/100) = 1.19; P-value = 0.1170,Free Response,,,,
-TPS,9.2,40c,"A recent report claimed that 13% of students typically walk to school.  DeAnna thinks that the proportion is higher than 0.13 at her large elementary school. She surveys a random sample of 100 students and finds that 17 typically walk to school. DeAnna would like to carry out a test at the α= 0.05 significance level of H0: p = 0.13 versus Ha: p > 0.13, where p = the true proportion of all students at her elementary school who typically walk to school. Based on a test with a P-value of 0.1170, what conclusion would you make? ","Because the P-value of 0.1170 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true proportion of all students at this elementary school who typically walk to school is greater than 0.13. ",Free Response,,,,
-TPS,9.2,41a,A test of H0: p = 0.5 versus Ha: p > 0.5 based on a sample of size 200 yields the standardized test statistic z = 2.19. Assume that the conditions for performing inference are met. Find and interpret the P -value.,"The P-value is P(z≥ 2.19) = 0.0143. Assuming that the true population proportion is 0.5, there is a 0.0143 probability of getting a sample proportion as large as or larger than the one observed just by chance in a random sample of size n=200. ",Free Response,,,,
-TPS,9.2,41b,A test of H0: p = 0.5 versus Ha: p > 0.5 based on a sample of size 200 yields the standardized test statistic z = 2.19. Assume that the conditions for performing inference are met. What conclusion would you make at the α=0.01? Would your conclusion change if you used α=0.05 instead? Explain your reasoning.,"α = 0.01: Because the P-value of 0.0143 < α = 0.01, we fail to reject H0. There is not convincing evidence that p < 0.5. α = 0.05: Yes! The P-value of 0.0143 < α = 0.05, so this time we will reject H0. There is convincing evidence that p > 0.5. ",Free Response,,,,
-TPS,9.2,41c,A test of H0: p = 0.5 versus Ha: p > 0.5 based on a sample of size 200 yields the standardized test statistic z = 2.19. Assume that the conditions for performing inference are met. Determine the value of p^= the sample proportion of successes.,Solve for p^: 2.19 = (p^ - 0.5) sqrt((0.5(0.5)) / 200); p^ = 0.5774 ,Free Response,,,,
-TPS,9.2,42a,A test of H0: p = 0.65 versus Ha: p <0.65 based on a sample of size 400 yields the standardized test statistic z =− 1.78. Find and interpret the P -value.,"P-value = 0.0375; if the true population proportion is 0.65, a 0.0375 probability exists of getting a sample proportion as small as or smaller than the one observed by chance in a random sample of size n= 400. ",Free Response,,,,
-TPS,9.2,42b,A test of H0: p = 0.65 versus Ha: p <0.65 based on a sample of size 400 yields the standardized test statistic z =− 1.78. What conclusion would you make at the α=0.10? Would your conclusion change if you used α=0.05 instead? Explain your reasoning. ,"When α = 0.10 and α = 0.05, reject H0. ",Free Response,,,,
-TPS,9.2,42c,A test of H0: p = 0.65 versus Ha: p <0.65 based on a sample of size 400 yields the standardized test statistic z =− 1.78. Determine the value of p^= the sample proportion of successes. ,p^ = 0.6075 ,Free Response,,,,
-TPS,9.2,43,"A media report claims that more than 75% of middle school students engage in bullying behavior. A University of Illinois study on aggressive behavior surveyed a random sample of 558 middle school students. When asked to describe their significance level? Would your conclusion change if you used behavior in the last 30 days, 445 students admitted that they had engaged in physical aggression, social ridicule, teasing, name-calling, and issuing threats—all of which would be classified as bullying.  Do these data provide convincing evidence at the α=0.05 significance level that the media report’s claim is correct?","S: H0: p = 0.75, Ha: p > 0.75, where p= the true proportion of all middle school students who engage in bullying behavior using α = 0.05. P: One-sample z test for p. Random: Random sample. 10%: 558 < 10% of all middle school students. Large Counts: 418.5 ≥ 10 and 139.5 ≥ 10. D: z= 2.59; P-value = 0.0048. C: Because the P-value of 0.0048 < α = 0.05, we reject H0. We have convincing evidence that the true proportion of all middle school students who engage in bullying behavior is > 0.75. ",Free Response,,,,
-TPS,9.2,44,"The germination rate of seeds is defined as the proportion of seeds that sprout and grow when properly planted and watered. A certain variety of grass seed usually has a germination rate of 0.80. A company wants to see if spraying the seeds with a chemical that is known to increase germination rates in other species will increase the germination rate of this variety of grass. The company researchers spray a random sample of 400 grass seeds with the chemical, and 339 of the seeds germinate. Do these data provide convincing evidence at the α=0.05 significance level that the chemical is effective for this variety of grass? Explain why the sample result gives some evidence for the alternative hypothesis.","S: H0: p= 0.80, Ha: p> 0.80 P: Large Counts: 320 ≥ 10 and 80 ≥ 10. D: z= 2.38; P-value = 0.0087. C: Reject H0. ",Free Response,,,,
-TPS,9.2,45a,"A local high school makes a change that should improve student satisfaction with the parking situation. Before the change, 37% of the school’s students approved of the parking that was provided. After the change, the principal surveys an SRS of 200 from the more than 2500 students at the school. In all, 83 students say that they approve of the new parking arrangement. The principal cites this as evidence that the change was effective. Describe a Type I error and a Type II error in this setting, and give a possible consequence of each.","Type I: Finding convincing evidence that more than 37% of students were satisfied with the new arrangement, when in reality only 37% were satisfied. Consequence: The principal believes students are satisfied and takes no further action. Type II: Failing to find convincing evidence that more than 37% are satisfied with the new arrangement, when in reality more than 37% are satisfied. Consequence: The principal takes further action when none is needed. ",Free Response,,,,
-TPS,9.2,45b,"A local high school makes a change that should improve student satisfaction with the parking situation. Before the change, 37% of the school’s students approved of the parking that was provided. After the change, the principal surveys an SRS of 200 from the more than 2500 students at the school. In all, 83 students say that they approve of the new parking arrangement. The principal cites this as evidence that the change was effective. Is there convincing evidence that the principal’s claim is true?","S: H0: p= 0.37, Ha: p> 0.37, where p= the true proportion of all students who are satisfied with the parking situation after the change using α = 0.05. P: One-sample z test for p. Random: Random sample. 10%: 200 < 10% of 2500. Large Counts: 74 ≥ 10 and 126 ≥ 10. D: z= 1.32; P-value = 0.0934. C: Because the P-value of 0.0934 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true proportion of all students who are satisfied with the parking situation after the change is greater than 0.37.",Free Response,,,,
-TPS,9.2,46a,"Members of the city council want to know if a majority of city residents supports a 1% increase in the sales tax to fund road repairs. To investigate, council members survey a random sample of 300 city residents. In the sample, 158 residents say that they are in favor of the sales tax increase. Describe a Type I error and a Type II error in this setting, and give a possible consequence of each.","Type I error: Finding convincing evidence that a majority of city residents support a 1% increase in the sales tax to fund road repairs, when in reality at most 50% would. Consequence: The sales tax will be increased by 1%, potentially upsetting most city residents. Type II error: Failing to find convincing evidence that a majority of city residents support a 1% increase in the sales tax to fund road repairs, when in reality a majority would. Consequence: The sales tax will not increase and the road repairs will not take place, despite the support of the city residents. ",Free Response,,,,
-TPS,9.2,46b,"Members of the city council want to know if a majority of city residents supports a 1% increase in the sales tax to fund road repairs. To investigate, council members survey a random sample of 300 city residents. In the sample, 158 residents say that they are in favor of the sales tax increase. Do these data provide convincing evidence that a majority of city residents support the tax increase?","S: H0: p= 0.5, Ha: p> 0.5. P: Large Counts: 150 ≥ 10 and 150 ≥ 10. D: z= 0.92; P-value = 0.1788. C: Fail to reject H0. ",Free Response,,,,
-TPS,9.2,47,"Cell-phone passwords A consumer organization suspects that less than half of parents know their child’s cell-phone password. The Pew Research Center asked a random sample of parents if they knew their child’s cell-phone password. Of the 1060 parents surveyed, 551 reported that they knew the password. Explain why it isn’t necessary to carry out a significance test in this setting.","Because p^ > 0.5, there is no evidence for Ha: p< 0.50. ",Free Response,,,,
-TPS,9.2,48,"A political organization wants to determine if there is convincing evidence that a majority of registered voters in a large city favor Proposition X. In an SRS of 1000 registered voters, 482 favor the proposition. Explain why it isn’t necessary to carry out a significance test in this setting.","Because p^ < 0.5, there is no evidence for Ha: p> 0.50.",Free Response,,,,
-TPS,9.2,49a,"Gregor Mendel (1822–1884), an Austrian monk, is considered the father of genetics. Mendel studied the inheritance of various traits in pea plants. One such trait is whether the pea is smooth or wrinkled. Mendel predicted a ratio of 3 smooth peas for every 1 wrinkled pea. In one experiment, he observed 423 smooth and 133 wrinkled peas. Assume that the conditions for inference are met. State appropriate hypotheses for testing Mendel’s claim about the true proportion of smooth peas.","H0: p= 0.75, Ha: p≠ 0.75, where p5 the true proportion of peas that will be smooth using α = 0.05. ",Free Response,,,,
-TPS,9.2,49b,"Gregor Mendel (1822–1884), an Austrian monk, is considered the father of genetics. Mendel studied the inheritance of various traits in pea plants. One such trait is whether the pea is smooth or wrinkled. Mendel predicted a ratio of 3 smooth peas for every 1 wrinkled pea. In one experiment, he observed 423 smooth and 133 wrinkled peas. Assume that the conditions for inference are met. Calculate the standardized test statistic and P -value. ",z= 0.59; P-value = 0.5552 ,Free Response,,,,
-TPS,9.2,49c,"Gregor Mendel (1822–1884), an Austrian monk, is considered the father of genetics. Mendel studied the inheritance of various traits in pea plants. One such trait is whether the pea is smooth or wrinkled. Mendel predicted a ratio of 3 smooth peas for every 1 wrinkled pea. In one experiment, he observed 423 smooth and 133 wrinkled peas. Assume that the conditions for inference are met. A test was done and a P-value was found to be 0.5552. Interpret the P -value. What conclusion would you make?","Assuming the true proportion of smooth peas is 0.75, there is a 0.5552 probability of getting a sample proportion as different from 0.75 as 0.761 (in either direction) by chance in a random sample of 556 peas. Because the P-value of 0.5552 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true proportion of peas that are smooth differs from 0.75. ",Free Response,,,,
-TPS,9.2,50a,"When a fair coin is flipped, we all know that the probability the coin lands on heads is 0.50. However, what if a coin is spun? According to the article “Euro Coin Accused of Unfair Flipping” in the New Scientist, two Polish math professors and their students spun a Belgian euro coin 250 times. It landed heads 140 times. One of the professors concluded that the coin was minted asymmetrically. A representative from the Belgian mint indicated the result was just chance. Assume that the conditions for inference are met. State appropriate hypotheses for testing these competing claims about the true proportion of spins that will land on heads.","H0: p= 0.50, Ha: p≠ 0.50 ",Free Response,,,,
-TPS,9.2,50b,"When a fair coin is flipped, we all know that the probability the coin lands on heads is 0.50. However, what if a coin is spun? According to the article “Euro Coin Accused of Unfair Flipping” in the New Scientist, two Polish math professors and their students spun a Belgian euro coin 250 times. It landed heads 140 times. One of the professors concluded that the coin was minted asymmetrically. A representative from the Belgian mint indicated the result was just chance. Assume that the conditions for inference are met. Calculate the standardized test statistic and P -value.",z= 1.90; P-value = 0.0574 ,Free Response,,,,
-TPS,9.2,50c,"When a fair coin is flipped, we all know that the probability the coin lands on heads is 0.50. However, what if a coin is spun? According to the article “Euro Coin Accused of Unfair Flipping” in the New Scientist, two Polish math professors and their students spun a Belgian euro coin 250 times. It landed heads 140 times. One of the professors concluded that the coin was minted asymmetrically. A representative from the Belgian mint indicated the result was just chance. Assume that the conditions for inference are met. A test was done and a P-value was found to be 0.074. Interpret the P -value. What conclusion would you make?","If the true proportion of heads is 0.50, a 0.0574 probability exists of getting a sample proportion as different from 0.50 as 0.56 (in either direction) just by chance in a random sample of 250 spun Belgian euros. Fail to reject H0.",Free Response,,,,
-TPS,9.2,51,A state’s Division of Motor Vehicles (DMV) claims that 60% of all teens pass their driving test on the first attempt. An investigative reporter examines an SRS of the DMV records for 125 teens; 86 of them passed the test on their first try. Is there convincing evidence at the α =0.05 significance level that the DMV’s claim is incorrect?,"S: H0: p= 0.60, Ha: p≠ 0.60, where p= the true proportion of teens who pass on the first attempt using α = 0.05. P: One-sample z test for p. Random: Random sample. 10%: 125 < 10% of population. Large Counts: 75 ≥ 10 and 50 $≥ 10. D: z= 2.01; P-value = 0.0444. C: Reject H0. There is convincing evidence that the true proportion of teens who pass on their first attempt differs from 0.60. ",Free Response,,,,
-TPS,9.2,52,"In a recent year, 73% of first-year college students responding to a national survey identified “being very well-off financially” as an important personal goal. A state university finds that 132 of an SRS of 200 of its first-year students say that this goal is important. Is there convincing evidence at the α=0.05 significance level that the proportion of all first-year students at this university who think being very well-off is important differs from the national value of 73%?","S: H0: p= 0.73, Ha: p≠ 0.73 P: Large Counts: 146 ≥ 10 and 54 $≥ 10. D: z=-2.23; P-value = 0.0258. C: Reject H0. ",Free Response,,,,
-TPS,9.2,53a,A state’s Division of Motor Vehicles (DMV) claims that 60% of all teens pass their driving test on the first attempt. An investigative reporter examines an SRS of the DMV records for 125 teens; 86 of them passed the test on their first try.  Construct and interpret a 95% confidence interval for the true proportion p of all teens in the state who passed their driving test on the first attempt. Assume that the conditions for inference are met.,"S: p= the true proportion of teens who pass on the first attempt. P: One-sample z interval for p. D: (0.607,0.769) C: We are 95% confident that the interval from 0.607 to 0.769 captures the true proportion of teens who pass on the first attempt. ",Free Response,,,,
-TPS,9.2,53b,A state’s Division of Motor Vehicles (DMV) claims that 60% of all teens pass their driving test on the first attempt. An investigative reporter examines an SRS of the DMV records for 125 teens; 86 of them passed the test on their first try. Explain why calculating this as an interval provides more information than a significance test.,The confidence interval gives the values of p that are plausible based on the sample data and not just a reject or fail to reject decision about H0. ,Free Response,,,,
-TPS,9.2,54a,"In a recent year, 73% of first-year college students responding to a national survey identified “being very well-off financially” as an important personal goal. A state university finds that 132 of an SRS of 200 of its first-year students say that this goal is important.  Construct and interpret a 95% confidence interval for the true proportion p of all first-year students at the university who would identify being very well-off as an important personal goal. Assume that the conditions for inference are met.","S: p= the true proportion of first-year students who think being very well-off financially is an important personal goal. P: One-sample z interval for p. D: (0.594, 0.726) C: We are 95% confident that the interval from 0.594 to 0.726 captures the true proportion of first-year students who identify being very well-off as an important personal goal. ",Free Response,,,,
-TPS,9.2,54b,"In a recent year, 73% of first-year college students responding to a national survey identified “being very well-off financially” as an important personal goal. A state university finds that 132 of an SRS of 200 of its first-year students say that this goal is important. Explain why calculating this as an interval provides more information than a significance test.",The confidence interval gives the values of p that are plausible based on the sample data and not just a reject or fail to reject decision about H0. ,Free Response,,,,
-TPS,9.2,55,"The Pew Internet and American Life Project asked a random sample of U.S. adults, “Do you ever . . . use Twitter or another service to share updates about yourself or to see updates about others?” According to Pew, the resulting 95% confidence interval is (0.123, 0.177).  Based on the confidence interval, is there convincing evidence that the true proportion of U.S. adults who would say they use Twitter or another service to share updates differs from 0.17? Explain your reasoning.","No; because the value 0.17 is included in the confidence interval, it is a plausible value for the true proportion of U.S. adults who would say they use Twitter. ",Free Response,,,,
-TPS,9.2,56," A Gallup poll found that 59% of the people in its sample said “Yes” when asked, “Would you like to lose weight?” Gallup announced: “For results based on the total sample of national adults, one can say with 95% confidence that the margin of (sampling) error is 3± percentage points.”  Based on the confidence interval, is there convincing evidence that the true proportion of U.S. adults who would say they want to lose weight differs from 0.55? Explain your reasoning.","Yes; because 0.55 is not included in the confidence interval from 0.56 to 0.62, it is not a plausible value for the true proportion of U.S. adults who would say that they want to lose weight. ",Free Response,,,,
-TPS,9.2,59,"A company that makes potato chips requires each shipment of potatoes to meet certain quality standards. If the company finds convincing evidence that more than 8% of the potatoes in the shipment have “blemishes,” the truck will be sent back to the supplier to get another load of potatoes. Otherwise, the entire truckload will be used to make potato chips. To make the decision, a supervisor will inspect a random sample of potatoes from the shipment. He will then perform a test of H0 p =: 0.08 versus Ha: p > 0.08, where p is the true proportion of potatoes with blemishes in a given truckload. The power of the test to detect that p = 0.11, based on a random sample of 500 potatoes and significance level α= 0.05 is 0.764. Interpret this value.","If the true proportion of potatoes with blemishes in a given truckload is p = 0.11, there is a 0.764 probability that the company will find convincing evidence for Ha: p > 0.08. ",Free Response,,,,
-TPS,9.2,60,"You are thinking about opening a restaurant and are searching for a good location. From the research you have done, you know that the mean income of those living near the restaurant must be over $85,000 to support the type of upscale restaurant you wish to open. You decide to take a simple random sample of 50 people living near one potential site. Based on the mean income of this sample, you will perform a test at the α= 0.05 significance level of H0: μ= $85,000 versus Ha: μ > $85,000, where μ is the true mean income in the population of people who live near the restaurant. The power of the test to detect that μ= $85,000 is 0.64. Interpret this value.","If the true mean income in the population of people who live near the restaurant is µ = $86,000, there is a 0.64 probability that I will find convincing evidence for Ha: µ > $85,000. ",Free Response,,,,
-TPS,9.2,61a,"A company that makes potato chips requires each shipment of potatoes to meet certain quality standards. If the company finds convincing evidence that more than 8% of the potatoes in the shipment have “blemishes,” the truck will be sent back to the supplier to get another load of potatoes. Otherwise, the entire truckload will be used to make potato chips. To make the decision, a supervisor will inspect a random sample of potatoes from the shipment. He will then perform a test of H0 p =: 0.08 versus Ha: p > 0.08, where p is the true proportion of potatoes with blemishes in a given truckload. The power of the test to detect that p = 0.11, based on a random sample of 500 potatoes and significance level α= 0.05 is 0.764. Determine if the following changes would increase or decrease the power of the test. Explain your answers. Change the significance level to α= 0.10.",Increase; using a larger significance level makes it easier to reject H0 when Ha is true. ,Free Response,,,,
-TPS,9.2,61b,"A company that makes potato chips requires each shipment of potatoes to meet certain quality standards. If the company finds convincing evidence that more than 8% of the potatoes in the shipment have “blemishes,” the truck will be sent back to the supplier to get another load of potatoes. Otherwise, the entire truckload will be used to make potato chips. To make the decision, a supervisor will inspect a random sample of potatoes from the shipment. He will then perform a test of H0 p =: 0.08 versus Ha: p > 0.08, where p is the true proportion of potatoes with blemishes in a given truckload. The power of the test to detect that p = 0.11, based on a random sample of 500 potatoes and significance level α= 0.05 is 0.764. Determine if the following changes would increase or decrease the power of the test. Explain your answers. Take a random sample of 250 potatoes instead of 500 potatoes.",Decrease; a smaller sample size gives less information about the true proportion p. ,Free Response,,,,
-TPS,9.2,61c,"A company that makes potato chips requires each shipment of potatoes to meet certain quality standards. If the company finds convincing evidence that more than 8% of the potatoes in the shipment have “blemishes,” the truck will be sent back to the supplier to get another load of potatoes. Otherwise, the entire truckload will be used to make potato chips. To make the decision, a supervisor will inspect a random sample of potatoes from the shipment. He will then perform a test of H0 p =: 0.08 versus Ha: p > 0.08, where p is the true proportion of potatoes with blemishes in a given truckload. The power of the test to detect that p = 0.11, based on a random sample of 500 potatoes and significance level α= 0.05 is 0.764. Determine if the following changes would increase or decrease the power of the test. Explain your answers. The true proportion is p = 0.10 instead of p = 0.11.",Decrease; it is harder to detect a smaller difference between the null and alternative parameter value. ,Free Response,,,,
-TPS,9.2,62a,"You are thinking about opening a restaurant and are searching for a good location. From the research you have done, you know that the mean income of those living near the restaurant must be over $85,000 to support the type of upscale restaurant you wish to open. You decide to take a simple random sample of 50 people living near one potential site. Based on the mean income of this sample, you will perform a test at the α= 0.05 significance level of H0: μ= $85,000 versus Ha: μ > $85,000, where μ is the true mean income in the population of people who live near the restaurant. The power of the test to detect that μ= $85,000 is 0.64. Determine if the following change would increase or decrease the power of the test. Explain your answers. Use a random sample of 30 people instead of 50 people. ",Decrease; a smaller sample size gives less information about the true mean µ. ,Free Response,,,,
-TPS,9.2,62b,"You are thinking about opening a restaurant and are searching for a good location. From the research you have done, you know that the mean income of those living near the restaurant must be over $85,000 to support the type of upscale restaurant you wish to open. You decide to take a simple random sample of 50 people living near one potential site. Based on the mean income of this sample, you will perform a test at the α= 0.05 significance level of H0: μ= $85,000 versus Ha: μ > $85,000, where μ is the true mean income in the population of people who live near the restaurant. The power of the test to detect that μ= $85,000 is 0.64. Determine if the following change would increase or decrease the power of the test. Explain your answers. Try to detect that μ= $85,500 instead of μ= $86,000.",Decrease; it is harder to detect a smaller difference between the null and alternative parameter value. ,Free Response,,,,
-TPS,9.2,62c,"You are thinking about opening a restaurant and are searching for a good location. From the research you have done, you know that the mean income of those living near the restaurant must be over $85,000 to support the type of upscale restaurant you wish to open. You decide to take a simple random sample of 50 people living near one potential site. Based on the mean income of this sample, you will perform a test at the α= 0.05 significance level of H0: μ= $85,000 versus Ha: μ > $85,000, where μ is the true mean income in the population of people who live near the restaurant. The power of the test to detect that μ= $85,000 is 0.64. Determine if the following change would increase or decrease the power of the test. Explain your answers. Change the significance level to α 0.10",Increase; using a larger significance level makes it easier to reject H0 when Ha is true. ,Free Response,,,,
-TPS,9.2,63a,"A company that makes potato chips requires each shipment of potatoes to meet certain quality standards. If the company finds convincing evidence that more than 8% of the potatoes in the shipment have “blemishes,” the truck will be sent back to the supplier to get another load of potatoes. Otherwise, the entire truckload will be used to make potato chips. To make the decision, a supervisor will inspect a random sample of potatoes from the shipment. He will then perform a test of H0 p =: 0.08 versus Ha: p > 0.08, where p is the true proportion of potatoes with blemishes in a given truckload. The power of the test to detect that p = 0.11, based on a random sample of 500 potatoes and significance level α= 0.05 is 0.764.  Explain one disadvantage of using α= 0.10 α= 0.05 when performing the test.",The larger significance level will increase the probability of a Type I error. ,Free Response,,,,
-TPS,9.2,63b,"A company that makes potato chips requires each shipment of potatoes to meet certain quality standards. If the company finds convincing evidence that more than 8% of the potatoes in the shipment have “blemishes,” the truck will be sent back to the supplier to get another load of potatoes. Otherwise, the entire truckload will be used to make potato chips. To make the decision, a supervisor will inspect a random sample of potatoes from the shipment. He will then perform a test of H0 p =: 0.08 versus Ha: p > 0.08, where p is the true proportion of potatoes with blemishes in a given truckload. The power of the test to detect that p = 0.11, based on a random sample of 500 potatoes and significance level α= 0.05 is 0.764.  Explain one disadvantage of taking a random sample of 500 potatoes instead of 250 potatoes from the shipment.",The larger sample size would require more time and money. ,Free Response,,,,
-TPS,9.2,64a,"You are thinking about opening a restaurant and are searching for a good location. From the research you have done, you know that the mean income of those living near the restaurant must be over $85,000 to support the type of upscale restaurant you wish to open. You decide to take a simple random sample of 50 people living near one potential site. Based on the mean income of this sample, you will perform a test at the α= 0.05 significance level of H0: μ= $85,000 versus Ha: μ > $85,000, where μ is the true mean income in the population of people who live near the restaurant. The power of the test to detect that μ= $85,000 is 0.64.  Explain one disadvantage of using α= 0.10 instead of α= 0.05 when performing the test. ",The larger significance level will increase the probability of a Type I error. ,Free Response,,,,
-TPS,9.2,64b,"You are thinking about opening a restaurant and are searching for a good location. From the research you have done, you know that the mean income of those living near the restaurant must be over $85,000 to support the type of upscale restaurant you wish to open. You decide to take a simple random sample of 50 people living near one potential site. Based on the mean income of this sample, you will perform a test at the α= 0.05 significance level of H0: μ= $85,000 versus Ha: μ > $85,000, where μ is the true mean income in the population of people who live near the restaurant. The power of the test to detect that μ= $85,000 is 0.64.  Explain one disadvantage of taking a random sample of 50 people instead of 30 people.",The larger sample size would require more time and money. ,Free Response,,,,
-TPS,9.2,65a,"A local high school makes a change that should improve student satisfaction with the parking situation. Before the change, 37% of the school’s students approved of the parking that was provided. After the change, the principal surveys an SRS of students at the school. She would like to perform a test of H0 p =: 0.37 verses Ha: p >: 0.37 where p is the true proportion of students at school who are satisfied with the parking situation after the change. The power of the test to detect that p = 0.45 based on a random sample of 200 students and a significance level of α= 0.05 is 0.75. Interpret this value.",If the true proportion of students at the school who are satisfied with the parking situation after the change is p= 0.45 there is a 0.75 probability that the principal will find convincing evidence for Ha: p > 0.37. ,Free Response,,,,
-TPS,9.2,65b,"A local high school makes a change that should improve student satisfaction with the parking situation. Before the change, 37% of the school’s students approved of the parking that was provided. After the change, the principal surveys an SRS of students at the school. She would like to perform a test of H0 p =: 0.37 verses Ha: p >: 0.37 where p is the true proportion of students at school who are satisfied with the parking situation after the change. The power of the test to detect that p = 0.45 based on a random sample of 200 students and a significance level of α= 0.05 is 0.75. Find the probability of a Type I error and the probability of a Type II error for the test.",P(Type I error) 5 α = 0.05 and P(Type II error) = 1 - Power = 1 - 0.75 = 0.25. ,Free Response,,,,
-TPS,9.2,65c,"A local high school makes a change that should improve student satisfaction with the parking situation. Before the change, 37% of the school’s students approved of the parking that was provided. After the change, the principal surveys an SRS of students at the school. She would like to perform a test of H0 p =: 0.37 verses Ha: p >: 0.37 where p is the true proportion of students at school who are satisfied with the parking situation after the change. The power of the test to detect that p = 0.45 based on a random sample of 200 students and a significance level of α= 0.05 is 0.75. Describe two ways to increase the power of the test.",The power would increase by increasing the sample size or using a larger significance level.,Free Response,,,,
-TPS,9.2,66a,"A company that manufactures classroom chairs for high school students claims that the mean breaking strength of the chairs is 300 pounds. One of the chairs collapsed beneath a 220-pound student last week. You suspect that the manufacturer is exaggerating the breaking strength of the chairs, so you would like to perform a test of H0: μ = 300 verses Ha: μ < 300 where μ is the true mean breaking strength of this company’s classroom chairs. The power of the test to detect that μ= 294 based on a random sample of 30 chairs and a significance level of α= 0.05 is 0.71. Interpret this value.","If the true mean breaking strength of this company’s classroom chairs is µ = 294 pounds, there is a 0.71 probability that I will find convincing evidence for Ha: µ > 300. ",Free Response,,,,
-TPS,9.2,66b,"A company that manufactures classroom chairs for high school students claims that the mean breaking strength of the chairs is 300 pounds. One of the chairs collapsed beneath a 220-pound student last week. You suspect that the manufacturer is exaggerating the breaking strength of the chairs, so you would like to perform a test of H0: μ = 300 verses Ha: μ < 300 where μ is the true mean breaking strength of this company’s classroom chairs. The power of the test to detect that μ= 294 based on a random sample of 30 chairs and a significance level of α= 0.05 is 0.71. Find the probability of a Type I error and the probability of a Type II error for the test.",P(Type I error) = α = 0.05 and P(Type II error) = 1 - 0.71 = 0.29 ,Free Response,,,,
-TPS,9.2,66c,"A company that manufactures classroom chairs for high school students claims that the mean breaking strength of the chairs is 300 pounds. One of the chairs collapsed beneath a 220-pound student last week. You suspect that the manufacturer is exaggerating the breaking strength of the chairs, so you would like to perform a test of H0: μ = 300 verses Ha: μ < 300 where μ is the true mean breaking strength of this company’s classroom chairs. The power of the test to detect that μ= 294 based on a random sample of 30 chairs and a significance level of α= 0.05 is 0.71. Describe two ways to increase the power of the test.",The power would increase by increasing the sample size or using a larger significance level. ,Free Response,,,,
-TPS,9.2,67a,You read that a significance test at the α= 0.01 significance level has probability 0.14 of making a Type II error when a specific alternative is true. What is the power of the test against this alternative? ,(a) Power = 1 - P (Type II error) = 1 - 0.14 = 0.86 ,Free Response,,,,
-TPS,9.2,67b,You read that a significance test at the α= 0.01 significance level has probability 0.14 of making a Type II error when a specific alternative is true. What’s the probability of making a Type I error?,P (Type I error) = α = 0.01,Free Response,,,,
-TPS,9.2,68a,A scientist calculates that a test at the α= 0.05 significance level has probability 0.23 of making a Type II error when a specific alternative is true. What is the power of the test against this alternative? ,Power = 1 - P (Type II error) = 1 - 0.23 = 0.77 ,Free Response,,,,
-TPS,9.2,68b,A scientist calculates that a test at the α= 0.05 significance level has probability 0.23 of making a Type II error when a specific alternative is true. What’s the probability of making a Type I error? ,P (Type I error) = α = 0.05,Free Response,,,,
-TPS,9.2,70,"After once again losing a football game to the archrival, a college’s alumni association conducted a survey to see if alumni were in favor of firing the coach. An SRS of 100 alumni from the population of all living alumni was taken, and 64 of the alumni in the sample were in favor of firing the coach. Suppose you wish to see if a majority of all living alumni is in favor of firing the coach. The appropriate standardized test statistic is (a) z = (0.64 – 0.5) / sqrt((0.64(0.36))/100) (b) z = (0.5 – 0.64) / sqrt((0.64(0.36))/100) (c) z = (0.64 – 0.5) / sqrt((0.5(0.5))/100) (d) z = (0.64 – 0.5) / sqrt((0.64(0.36))/64) (e) z = (0.5 – 0.64) / sqrt((0.5(0.5))/100)",c,Multiple Choice,,,,
-TPS,9.2,71,"Which of choices (a) through (d) is not a condition for performing a significance test about a population proportion p? (a) The data should come from a random sample from the population of interest. (b) Both np0 and (1 – np0) should be at least 10. (c) If you are sampling without replacement from a finite population, then you should sample less than 10% of the population. (d) The population distribution should be approximately Normal, unless the sample size is large. (e) All of the above are conditions for performing a significance test about a population proportion.",d,Multiple Choice,,,,
-TPS,9.2,72,"The standardized test statistic for a test of H0: p = 0.4 versus Ha: p ≠ 0.4 is z = 2.43. This test is  (a) not significant at either α= 0.05 or α= 0.01.  (b) significant at α= 0.05, but not at α= 0.01. (c) significant at α= 0.01, but not at α= 0.05.  (d) significant at both α= 0.05 and α= 0.01. (e) inconclusive because we don’t know the value of p^. ",b,Multiple Choice,,,,
-TPS,9.2,73,"Which of the following 95% confidence intervals would lead us to reject H0: p =0.30 in favor of Ha: p ≠ 0.30 at the 5% significance level? (a) (0.19, 0.27) (b) (0.24, 0.30) (c) (0.27, 0.31) (d) (0.29, 0.38) (e) None of these",a,Multiple Choice,,,,
-TPS,9.2,74,A researcher plans to conduct a significance test at the α= 0.01 significance level. She designs her study to have a power of 0.90 at a particular alternative value of the parameter of interest. The probability that the researcher will commit a Type II error for the particular alternative value of the parameter she used is (a) 0.01. (b) 0.10. (c) 0.89. (d) 0.90. (e) 0.99.,b,Multiple Choice,,,,
-TPS,9.3,78,"A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used in an attempt to improve athletic performance. Researchers want to know if there is a difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids. State appropriate hypotheses for performing a significance test. Be sure to define the parameters of interest.","H0: p1 - p2 = 0, Ha: p1 - p2 ≠ 0, where p1 = the true proportion of high school freshmen in Illinois who have used anabolic steroids and p2 = the true proportion of high school seniors in Illinois who have used anabolic steroids.",Free Response,,,,
-TPS,9.3,80,"A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used in an attempt to improve athletic performance. Researchers want to know if there is a difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids. Check if the conditions for performing a significance test are met.","Random: Independent random samples. 10%: 1679 < 10% of all high school freshmen in Illinois and 1366 < 10% of all high school seniors in Illinois. Large Counts: p^C= (34 + 24) / (1679 + 1366) = 0.019; 31.9, 1647.1, 25.95, 1340.05 are ≥ 10.",Free Response,,,,
-TPS,9.3,81a,Fire is a serious threat to shrubs in dry climates. Some shrubs can resprout from their roots after their tops are destroyed. Researchers wondered if fire would help with resprouting. One study of resprouting took place in a dry area of Mexico. 21 The researchers randomly assigned shrubs to treatment and control groups. They clipped the tops of all the shrubs. They then applied a propane torch to the stumps of the treatment group to simulate a fire. All 12 of the shrubs in the treatment group resprouted. Only 8 of the 12 shrubs in the control group resprouted. State appropriate hypotheses for performing a significance test. Be sure to define the parameters of interest.,"H0: p1 - p2 = 0, Ha: p1 - p2 > 0, where p1 = true proportion of shrubs that would resprout after being clipped and burned and p2 = . . . after being clipped. ",Free Response,,,,
-TPS,9.3,81b,Fire is a serious threat to shrubs in dry climates. Some shrubs can resprout from their roots after their tops are destroyed. Researchers wondered if fire would help with resprouting. One study of resprouting took place in a dry area of Mexico. 21 The researchers randomly assigned shrubs to treatment and control groups. They clipped the tops of all the shrubs. They then applied a propane torch to the stumps of the treatment group to simulate a fire. All 12 of the shrubs in the treatment group resprouted. Only 8 of the 12 shrubs in the control group resprouted. Check if the conditions for performing a significance test are met.,Random: Random assignment. Large Counts: Not met because all are < 10,Free Response,,,,
-TPS,9.3,82a,"Lyme disease is spread in the northeastern United States by infected ticks. The ticks are infected mainly by feeding on mice, so more mice result in more infected ticks. The mouse population, in turn, rises and falls with the abundance of acorns, their favored food. Experimenters studied two similar forest areas in a year when the acorn crop failed. To see if mice are more likely to breed when there are more acorns, the researchers added hundreds of thousands of acorns to one area to imitate an abundant acorn crop, while leaving the other area untouched. The next spring, 54 of the 72 mice trapped in the first area were in breeding condition, versus 10 of the 17 mice trapped in the second area.  State appropriate hypotheses for performing a significance test. Be sure to define the parameters of interest.","H0: p1 - p2 = 0, Ha: p1 - p2 > 0, where p1 = true proportion of mice that are in breeding condition in the area of abundant acorn crop and p2 = . . . in the untouched area. ",Free Response,,,,
-TPS,9.3,82b,"Lyme disease is spread in the northeastern United States by infected ticks. The ticks are infected mainly by feeding on mice, so more mice result in more infected ticks. The mouse population, in turn, rises and falls with the abundance of acorns, their favored food. Experimenters studied two similar forest areas in a year when the acorn crop failed. To see if mice are more likely to breed when there are more acorns, the researchers added hundreds of thousands of acorns to one area to imitate an abundant acorn crop, while leaving the other area untouched. The next spring, 54 of the 72 mice trapped in the first area were in breeding condition, versus 10 of the 17 mice trapped in the second area. Check if the conditions for performing a significance test are met. ",Random: Unknown. Large Counts: Not met because 4.777 < 10.,Free Response,,,,
-TPS,9.3,84a,"A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used in an attempt to improve athletic performance. Researchers want to know if there is a difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.  Explain why the sample results give some evidence for the alternative hypothesis.",p^1 - p^2 = 0.0027 ≠ 0 ,Free Response,,,,
-TPS,9.3,84b,"A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used in an attempt to improve athletic performance. Researchers want to know if there is a difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.  Calculate the standardized test statistic and P -value. ","z= 0.54, P-value = 0.5904 ",Free Response,,,,
-TPS,9.3,84c,"A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used in an attempt to improve athletic performance. Researchers want to know if there is a difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.  A test was done and a P-value was found to be 0.5904. What conclusion would you make?","Because the P-value of 0.5904 > α= 0.05, we fail to reject H0. We do not have convincing evidence that the true proportion of high school freshmen in Illinois who have used anabolic steroids differs from the true proportion of high school seniors in Illinois who have. ",Free Response,,,,
-TPS,9.3,85a,Phoebe has a hunch that older students at her very large high school are more likely to bring a bag lunch than younger students because they have grown tired of cafeteria food. She takes a simple random sample of 80 sophomores and finds that 52 of them bring a bag lunch. A simple random sample of 104 seniors reveals that 78 of them bring a bag lunch. Do these data give convincing evidence to support Phoebe’s hunch at the α= 0.05 significance level?,"S: H0: p1 - p2 = 0, Ha: p1 - p2 < 0, where p1 = true proportion of sophomores who bring a bag lunch and p2 = . . . seniors . . . P: Two-sample z test for p1 - p2. Random: Independent random samples. 10%: 80 < 10% of all sophomores and 104 < 10% of all seniors. Large Counts: 56.56, 23.44, 73.528, 30.472 are all ≥ 10. D: z=-1.48, P-value = 0.0699. C: Because the P-value of 0.0699 > α= 0.05, we fail to reject H0. There is not convincing evidence that the true proportion of sophomores who bring a bag lunch is less than the true proportion of seniors who do. ",Free Response,,,,
-TPS,9.3,85b,Phoebe has a hunch that older students at her very large high school are more likely to bring a bag lunch than younger students because they have grown tired of cafeteria food. She takes a simple random sample of 80 sophomores and finds that 52 of them bring a bag lunch. A simple random sample of 104 seniors reveals that 78 of them bring a bag lunch. Interpret the P -value of 0.0699 in the context of this study.,"If there is no difference in the true proportion of sophomores and seniors who bring a bag lunch, there is a 0.0699 probability of getting a difference in the proportions as large as or larger than the one observed (0.65 - 0.75 =-0.10) by chance alone.",Free Response,,,,
-TPS,9.3,86a,"In a study of 3000 randomly selected teenagers in 1990, 450 showed some hearing loss. In a similar study of 1800 teenagers reported in 2010, 351 showed some hearing loss. Do these data give convincing evidence that the proportion of all teens with hearing loss has increased at the α= 0.01 significance level?","S: H0: p1 - p2 = 0, Ha: p1 - p2 < 0, where p1 = true proportion of teenagers in 1990 with some hearing loss and p2 = . . . in 2010 . . . P: Two-sample z test for p1 - p2. Random: Independent random samples. 10%: 3000 < 10% of all teenagers in 1990 and 1800 < 10% of all teenagers in 2010. Large Counts: 501, 2499, 300.6, 1499.4 ≥ 10. D: z=-4.05, P-value ~ 0. C: Because the P-value of approximately 0 < α= 0.05, we reject H0. There is convincing evidence that the true proportion of teenagers in 1990 with some hearing loss is less than the true proportion of teenagers in 2010 with some hearing loss. ",Free Response,,,,
-TPS,9.3,86b,"In a study of 3000 randomly selected teenagers in 1990, 450 showed some hearing loss. In a similar study of 1800 teenagers reported in 2010, 351 showed some hearing loss. Interpret the P -value of ~0 in the context of this study.","If there is no difference in the true proportion of teenagers in 1990 and 2010 with some hearing loss, there is about a 0 probability of getting a difference in the proportions",Free Response,,,,
-TPS,9.3,87a,"A recent study of peanut allergies—the LEAP trial—explored whether early exposure to peanuts helps or hurts subsequent development of an allergy to peanuts. Infants (4 to 11 months old) who had shown evidence of other kinds of allergies were randomly assigned to one of two groups. Group 1 consumed a baby-food form of peanut butter. Group 2 avoided peanut butter. At 5 years old, 10 of 307 children in the peanut-consumption group were allergic to peanuts, and 55 of 321 children in the peanut-avoidance group were allergic to peanuts. Does this study provide convincing evidence of a difference at the α= 0.05 significance level in the development of peanut allergies in infants like the ones in this study who consume or avoid peanut butter?","S: H0: p1 - p2 = 0, Ha: p1 - p2 ≠ 0, where p1 = true proportion of children like the ones in this study who are exposed to peanut butter as infants who are allergic to peanuts at age 5 and p2 = . . . not exposed . . . P: Two-sample z test for p1 - p2. Random: Random assignment. Large Counts: 31.928, 275.072, 33.384, 287.616 ≥ 10. D: z=-5.71, P-value ~0. C: Because the P-value of approximately 0 < α= 0.05, we reject H0. There is convincing evidence that the true proportion of children like the ones in this study who are exposed to peanut butter as infants who are allergic to peanuts at age 5 differs from the true proportion of children like the ones in this study who are not exposed to peanut butter as infants who are allergic to peanuts at age 5. ",Free Response,,,,
-TPS,9.3,87b,"A recent study of peanut allergies—the LEAP trial—explored whether early exposure to peanuts helps or hurts subsequent development of an allergy to peanuts. Infants (4 to 11 months old) who had shown evidence of other kinds of allergies were randomly assigned to one of two groups. Group 1 consumed a baby-food form of peanut butter. Group 2 avoided peanut butter. At 5 years old, 10 of 307 children in the peanut-consumption group were allergic to peanuts, and 55 of 321 children in the peanut-avoidance group were allergic to peanuts. Based on the conclusion of rejecting the H0, which mistake—a Type I error or a Type II error—could you have made? Explain your answer.",Type I error ,Free Response,,,,
-TPS,9.3,87c,"A recent study of peanut allergies—the LEAP trial—explored whether early exposure to peanuts helps or hurts subsequent development of an allergy to peanuts. Infants (4 to 11 months old) who had shown evidence of other kinds of allergies were randomly assigned to one of two groups. Group 1 consumed a baby-food form of peanut butter. Group 2 avoided peanut butter. At 5 years old, 10 of 307 children in the peanut-consumption group were allergic to peanuts, and 55 of 321 children in the peanut-avoidance group were allergic to peanuts. Should you generalize the result of a significance test to all infants? Why or why not?","No; when subjects are not randomly selected, we should not generalize the results of an experiment to some larger population. ",Free Response,,,,
-TPS,9.3,87d,"A recent study of peanut allergies—the LEAP trial—explored whether early exposure to peanuts helps or hurts subsequent development of an allergy to peanuts. Infants (4 to 11 months old) who had shown evidence of other kinds of allergies were randomly assigned to one of two groups. Group 1 consumed a baby-food form of peanut butter. Group 2 avoided peanut butter. At 5 years old, 10 of 307 children in the peanut-consumption group were allergic to peanuts, and 55 of 321 children in the peanut-avoidance group were allergic to peanuts. A 95% confidence interval for p1 – p2 is (0.185, 0.093). Explain how the confidence interval provides more information than a significance test.","The confidence interval does not include 0 as a plausible value for p1 - p2, which is consistent with the decision to reject H0: p1 - p2 = 0 in part (a). The confidence interval tells us that any value between 20.185 and 20.093 is plausible for p1 - p2 based on the sample data. ",Free Response,,,,
-TPS,9.3,88a,"Lowering bad cholesterol Which of two widely prescribed drugs—Lipitor or Pravachol—helps lower “bad cholesterol” more? In an experiment, called the PROVE-IT Study, researchers recruited about 4000 people with heart disease as subjects. These volunteers were randomly assigned to one of two treatment groups: Lipitor or Pravachol. At the end of the study, researchers compared the proportion of subjects in each group who died, had a heart attack, or suffered other serious consequences within two years. For the 2063 subjects using Pravachol, the proportion was 0.263. For the 2099 subjects using Lipitor, the proportion was 0.224. Does this study provide convincing evidence at the α= 0.05 significance level of a difference in the effectiveness of Lipitor and Pravachol for people like the ones in this study?","S: H0: p1-p2 =0, Ha: p1 - p2 ≠ 0, where p1 = true proportion of subjects like these who would die, have a heart attack, or suffer other serious consequences within 2 years if they took Pravachol and p2 = true proportion who would suffer serious consequences if they took Lipitor; α= 0.05. P: Two-sample z test for p1 - p2. Random: Volunteers were randomly assigned to the treatments. Large Counts: 510.057, 1588.943, 501.309, 1561.691 are ≥ 10. D: z=-2.95; P-value =0.0031. C: Because the P-value of 0.0031 < α= 0.05, we reject H0. There is convincing evidence the true proportion of subjects like these who would die, have a heart attack, or suffer other serious consequences within 2 years if they took Pravachol differs from the true proportion of subjects like these who would die, have a heart attack, or suffer other serious consequences within 2 years if they took Lipitor.",Free Response,,,,
-TPS,9.3,88b,"Lowering bad cholesterol Which of two widely prescribed drugs—Lipitor or Pravachol—helps lower “bad cholesterol” more? In an experiment, called the PROVE-IT Study, researchers recruited about 4000 people with heart disease as subjects. These volunteers were randomly assigned to one of two treatment groups: Lipitor or Pravachol. At the end of the study, researchers compared the proportion of subjects in each group who died, had a heart attack, or suffered other serious consequences within two years. For the 2063 subjects using Pravachol, the proportion was 0.263. For the 2099 subjects using Lipitor, the proportion was 0.224. Based on the conclusion of rejecting the H0, which mistake—a Type I error or a Type II error—could you have made? Explain your answer.","Because we rejected H0, it is possible we made a Type I error.",Free Response,,,,
-TPS,9.3,88c,"Lowering bad cholesterol Which of two widely prescribed drugs—Lipitor or Pravachol—helps lower “bad cholesterol” more? In an experiment, called the PROVE-IT Study, researchers recruited about 4000 people with heart disease as subjects. These volunteers were randomly assigned to one of two treatment groups: Lipitor or Pravachol. At the end of the study, researchers compared the proportion of subjects in each group who died, had a heart attack, or suffered other serious consequences within two years. For the 2063 subjects using Pravachol, the proportion was 0.263. For the 2099 subjects using Lipitor, the proportion was 0.224. Should you generalize the result of a significance test to all people with heart disease? Why or why not?",No; this experiment recruited volunteers as subjects. We should not generalize to some larger population of interest. ,Free Response,,,,
-TPS,9.3,88d,"Lowering bad cholesterol Which of two widely prescribed drugs—Lipitor or Pravachol—helps lower “bad cholesterol” more? In an experiment, called the PROVE-IT Study, researchers recruited about 4000 people with heart disease as subjects. These volunteers were randomly assigned to one of two treatment groups: Lipitor or Pravachol. At the end of the study, researchers compared the proportion of subjects in each group who died, had a heart attack, or suffered other serious consequences within two years. For the 2063 subjects using Pravachol, the proportion was 0.263. For the 2099 subjects using Lipitor, the proportion was 0.224. A 95% confidence interval for pPR – pL is (0.013, 0.065). Explain how the confidence interval provides more information than a significance test.","The confidence interval does not include 0 as a plausible value for p1 - p2, which is consistent with the decision to reject H0: p1 - p2 = 0 in part (a). The confidence interval tells us that any value between 0.013 and 0.065 is plausible for p1 - p2 based on the sample data.",Free Response,,,,
-TPS,9.3,89a,"A recent study of peanut allergies—the LEAP trial—explored whether early exposure to peanuts helps or hurts subsequent development of an allergy to peanuts. Infants (4 to 11 months old) who had shown evidence of other kinds of allergies were randomly assigned to one of two groups. Group 1 consumed a baby-food form of peanut butter. Group 2 avoided peanut butter. At 5 years old, 10 of 307 children in the peanut-consumption group were allergic to peanuts, and 55 of 321 children in the peanut-avoidance group were allergic to peanuts. Explain how the following change to the design of the experiment would affect the power of the test. Then give a drawback of making that change. Suppose that researchers had recruited twice as many infants for the LEAP trial.","Increasing the sample size will increase the power of the test. Increasing the sample size decreases the variability of both the null and alternative distributions, making it easier to reject the null hypothesis when it is false. A drawback is that an experiment that uses twice as many infants will be more expensive and will require a lot more work in following up with the parents of these infants when they are 5 years old. ",Free Response,,,,
-TPS,9.3,89b,"A recent study of peanut allergies—the LEAP trial—explored whether early exposure to peanuts helps or hurts subsequent development of an allergy to peanuts. Infants (4 to 11 months old) who had shown evidence of other kinds of allergies were randomly assigned to one of two groups. Group 1 consumed a baby-food form of peanut butter. Group 2 avoided peanut butter. At 5 years old, 10 of 307 children in the peanut-consumption group were allergic to peanuts, and 55 of 321 children in the peanut-avoidance group were allergic to peanuts. Explain how the following change to the design of the experiment would affect the power of the test. Then give a drawback of making that change. Suppose that researchers had used α= 0.10 instead of α= 0.05.","Using α = 0.10 instead of α = 0.05 will increase the power of the test. When α is larger, it is easier to reject the null hypothesis because the P-value doesn’t need to be as small. A drawback to increasing a is that doing so increases the probability of making a Type I error. ",Free Response,,,,
-TPS,9.3,89c,"A recent study of peanut allergies—the LEAP trial—explored whether early exposure to peanuts helps or hurts subsequent development of an allergy to peanuts. Infants (4 to 11 months old) who had shown evidence of other kinds of allergies were randomly assigned to one of two groups. Group 1 consumed a baby-food form of peanut butter. Group 2 avoided peanut butter. At 5 years old, 10 of 307 children in the peanut-consumption group were allergic to peanuts, and 55 of 321 children in the peanut-avoidance group were allergic to peanuts. Explain how the following change to the design of the experiment would affect the power of the test. Then give a drawback of making that change. Suppose that researchers had used 628 male subjects but no female subjects in the study.","Using male infants only would increase the power of the test by eliminating a source of variability, making it easier to reject the null hypothesis when it is false. A drawback is that this also limits the scope of inference to males only. ",Free Response,,,,
-TPS,9.3,90a,"Lowering bad cholesterol Which of two widely prescribed drugs—Lipitor or Pravachol—helps lower “bad cholesterol” more? In an experiment, called the PROVE-IT Study, researchers recruited about 4000 people with heart disease as subjects. These volunteers were randomly assigned to one of two treatment groups: Lipitor or Pravachol. At the end of the study, researchers compared the proportion of subjects in each group who died, had a heart attack, or suffered other serious consequences within two years. For the 2063 subjects using Pravachol, the proportion was 0.263. For the 2099 subjects using Lipitor, the proportion was 0.224. Explain how the following change to the design of the experiment would affect the power of the test. Then give a drawback of making that change. Suppose that researchers had recruited twice as many subjects for the PROVE-IT Study.","Increasing the sample size will increase the power of the test. Increasing the sample size decreases the variability of both the null and alternative distributions, making it easier to reject the null hypothesis when it is false. A drawback is that an experiment that uses twice as many subjects will be more expensive and will require a lot more work in following up with the subjects at the end of the 2-year long experiment. ",Free Response,,,,
-TPS,9.3,90b,"Lowering bad cholesterol Which of two widely prescribed drugs—Lipitor or Pravachol—helps lower “bad cholesterol” more? In an experiment, called the PROVE-IT Study, researchers recruited about 4000 people with heart disease as subjects. These volunteers were randomly assigned to one of two treatment groups: Lipitor or Pravachol. At the end of the study, researchers compared the proportion of subjects in each group who died, had a heart attack, or suffered other serious consequences within two years. For the 2063 subjects using Pravachol, the proportion was 0.263. For the 2099 subjects using Lipitor, the proportion was 0.224. Explain how the following change to the design of the experiment would affect the power of the test. Then give a drawback of making that change. Suppose that researchers had used α= 0.10 instead of α= 0.05.","Using α = 0.10 instead of α = 0.05 will increase the power of the test. When α is larger, it is easier to reject the null hypothesis because the P-value doesn’t need to be as small. A drawback to increasing a is that doing so increases the probability of making a Type I error.",Free Response,,,,
-TPS,9.3,90c,"Lowering bad cholesterol Which of two widely prescribed drugs—Lipitor or Pravachol—helps lower “bad cholesterol” more? In an experiment, called the PROVE-IT Study, researchers recruited about 4000 people with heart disease as subjects. These volunteers were randomly assigned to one of two treatment groups: Lipitor or Pravachol. At the end of the study, researchers compared the proportion of subjects in each group who died, had a heart attack, or suffered other serious consequences within two years. For the 2063 subjects using Pravachol, the proportion was 0.263. For the 2099 subjects using Lipitor, the proportion was 0.224. Explain how the following change to the design of the experiment would affect the power of the test. Then give a drawback of making that change. Suppose that researchers had used 4162 subjects under age 60 but no older subjects in the study.","Using only subjects under the age of 60 would increase the power of the test by eliminating a source of variability, making it easier to reject the null hypothesis when it is false. A drawback is that this also limits the scope of inference to those under the age of 60.",Free Response,,,,
-TPS,9.3,91a,"Some women would like to have children but cannot do so for medical reasons. One option for these women is a procedure called in vitro fertilization (IVF), which involves fertilizing an egg outside the woman’s body and implanting it in her uterus. Two hundred women who were about to undergo IVF served as subjects in an experiment. Each subject was randomly assigned to either a treatment group or a control group. Several people (called intercessors) prayed intentionally for the women in the treatment group, although they did not know the women, a process known as intercessory prayer. Prayers continued for 3 weeks following IVF. The intercessors did not pray for the women in the control group. Here are the results: 44 of the 88 women in the treatment group got pregnant, compared to 21 out of 81 in the control group. Is the pregnancy rate significantly higher for women who received intercessory prayer? To find out, researchers perform a test of H0: p1 = p2 verses Ha: p1> p2 where p1 and p2 are the actual pregnancy rates for women like those in the study who do and don’t receive intercessory prayer, respectively. Name the appropriate test and check that the conditions for carrying out this test are met.","Two-sample z test for p1 - p2. Random: Two groups in a randomized experiment. Large Counts: 33.88, 54.12, 31.185, 49.815 are ≥ 10. ",Free Response,,,,
-TPS,9.3,91b,"Some women would like to have children but cannot do so for medical reasons. One option for these women is a procedure called in vitro fertilization (IVF), which involves fertilizing an egg outside the woman’s body and implanting it in her uterus. Two hundred women who were about to undergo IVF served as subjects in an experiment. Each subject was randomly assigned to either a treatment group or a control group. Several people (called intercessors) prayed intentionally for the women in the treatment group, although they did not know the women, a process known as intercessory prayer. Prayers continued for 3 weeks following IVF. The intercessors did not pray for the women in the control group. Here are the results: 44 of the 88 women in the treatment group got pregnant, compared to 21 out of 81 in the control group. Is the pregnancy rate significantly higher for women who received intercessory prayer? To find out, researchers perform a test of H0: p1 = p2 verses Ha: p1> p2 where p1 and p2 are the actual pregnancy rates for women like those in the study who do and don’t receive intercessory prayer, respectively. The appropriate test yields a P-value of 0.0007. Interpret this P-value in context.","If there is no difference in the true pregnancy rates of women prayed for and those who are not, there is a 0.0007 probability of getting a difference in pregnancy rates as large as or larger than the one observed (0.500 - 0.259 = 0.241) by chance alone. ",Free Response,,,,
-TPS,9.3,91c,"Some women would like to have children but cannot do so for medical reasons. One option for these women is a procedure called in vitro fertilization (IVF), which involves fertilizing an egg outside the woman’s body and implanting it in her uterus. Two hundred women who were about to undergo IVF served as subjects in an experiment. Each subject was randomly assigned to either a treatment group or a control group. Several people (called intercessors) prayed intentionally for the women in the treatment group, although they did not know the women, a process known as intercessory prayer. Prayers continued for 3 weeks following IVF. The intercessors did not pray for the women in the control group. Here are the results: 44 of the 88 women in the treatment group got pregnant, compared to 21 out of 81 in the control group. Is the pregnancy rate significantly higher for women who received intercessory prayer? To find out, researchers perform a test of H0: p1 = p2 verses Ha: p1> p2 where p1 and p2 are the actual pregnancy rates for women like those in the study who do and don’t receive intercessory prayer, respectively. What conclusion should researchers draw at the α 0.05= significance level based on a P-level of 0.007?","Because the P-value of 0.0007 is less than α = 0.05, we reject H0. There is convincing evidence the pregnancy rates among women like these who are prayed for is higher than the rates for those not prayed for. ",Free Response,,,,
-TPS,9.3,91d,"Some women would like to have children but cannot do so for medical reasons. One option for these women is a procedure called in vitro fertilization (IVF), which involves fertilizing an egg outside the woman’s body and implanting it in her uterus. Two hundred women who were about to undergo IVF served as subjects in an experiment. Each subject was randomly assigned to either a treatment group or a control group. Several people (called intercessors) prayed intentionally for the women in the treatment group, although they did not know the women, a process known as intercessory prayer. Prayers continued for 3 weeks following IVF. The intercessors did not pray for the women in the control group. Here are the results: 44 of the 88 women in the treatment group got pregnant, compared to 21 out of 81 in the control group. Is the pregnancy rate significantly higher for women who received intercessory prayer? To find out, researchers perform a test of H0: p1 = p2 verses Ha: p1> p2 where p1 and p2 are the actual pregnancy rates for women like those in the study who do and don’t receive intercessory prayer, respectively. The women in the study did not know whether they were being prayed for. Explain why this is important.",Knowing might have affected their behavior in some way (even unconsciously) that would have affected if they became pregnant. Then we wouldn’t know if the prayer or other behaviors caused the higher pregnancy rate.,Free Response,,,,
-TPS,9.3,95,"A sample survey interviews SRSs of 500 female college students and 550 male college students. Researchers want to determine whether there is a difference in the proportion of male and female college students who worked for pay last summer. In all, 410 of the females and 484 of the males say they worked for pay last summer. Let pM and pF be the proportions of all college males and females who worked last summer. The hypotheses to be tested are (a) H0: pM – pF = 0 versus: Ha: pM – pF ≠ 0 (b) H0: pM – pF = 0 versus: Ha: pM – pF > 0  (c) H0: pM – pF = 0 versus: Ha: pM – pF < 0  (d) H0: pM – pF > 0 versus: Ha: pM – pF = 0  (e) H0: pM – pF ≠ 0 versus: Ha: pM – pF = 0",a,Multiple Choice,,,,
-TPS,9.3,96,"A sample survey interviews SRSs of 500 female college students and 550 male college students. Researchers want to determine whether there is a difference in the proportion of male and female college students who worked for pay last summer. In all, 410 of the females and 484 of the males say they worked for pay last summer. The researchers report that the results were statistically significant at the 1% level. Which of the following is the most appropriate conclusion? (a) Because the P-value is less than 1%, fail to reject H0. There is not convincing evidence that the proportion of male college students in the study who worked for pay last summer is different from the proportion of female college students in the study who worked for pay last summer. (b) Because the P-value is less than 1%, fail to reject H0. There is not convincing evidence that the proportion of all male college students who worked for pay last summer is different from the proportion of all female college students who worked for pay last summer. (c) Because the P-value is less than 1%, reject H0. There is convincing evidence that the proportion of all male college students who worked for pay last summer is the same as the proportion of all female college students who worked for pay last summer. (d) Because the P-value is less than 1%, reject H0. There is convincing evidence that the proportion of all male college students in the study who worked for pay last summer is different from the proportion of all female college students in the study who worked for pay last summer. (e) Because the P-value is less than 1%, reject H0. There is convincing evidence that the proportion of all male college students who worked for pay last summer is different from the proportion of all female college students who worked for pay last summer.",e,Multiple Choice,,,,
-TPS,9.3,97,"A sample survey interviews SRSs of 500 female college students and 550 male college students. Researchers want to determine whether there is a difference in the proportion of male and female college students who worked for pay last summer. In all, 410 of the females and 484 of the males say they worked for pay last summer. Let pM and pF be the proportions of all college males and females who worked last summer. The hypotheses to be tested are H0: pM – pF = 0 versus: Ha: pM – pF ≠ 0. Which of the following is the correct standard error for the test?  (a) sqrt((0.851(0.149))/1050)  (b) sqrt(((0.851(0.149))/550) + ((0.851(0.149))/500))  (c) sqrt(((0.880(0.120))/550) + ((0.820(0.180))/500))  (d) sqrt(((0.851(0.149))/1050) + ((0.851(0.149))/1050))  (e) sqrt(((0.880(0.120))/1050) + ((0.820(0.180))/1050))",b,Multiple Choice,,,,
-TPS,9.3,98,"In an experiment to learn whether substance M can help restore memory, the brains of 20 rats were treated to damage their memories. First, the rats were trained to run a maze. After a day, 10 rats (determined at random) were given substance M and 7 of them succeeded in the maze. Only 2 of the 10 control rats were successful. The two-sample z test for the difference in the true proportions (a) gives z = 2.25, P < 0.02.  (b) gives z = 2.60, P < 0.005.  (c) gives z = 2.25, P < 0.04 but not <0.02.  (d) should not be used because the Random condition is violated.  (e) should not be used because the Large Counts condition is violated.",e,Multiple Choice,,,,
-TPS,R9,1a,"Rob once read that one-quarter of all people have played/danced in the rain at some point in their lives. His friend Justin thinks that the proportion is higher than 0.25 for their high school. To settle their dispute, they ask a random sample of 80 students in their school and find out that 28 have played/danced in the rain. State the appropriate null and alternative hypotheses.","H0: p= 0.25; Ha: p> 0.25, where p = the true proportion of all students at this school who have played/ danced in the rain. ",Free Response,,,,
-TPS,R9,1b,"Rob once read that one-quarter of all people have played/danced in the rain at some point in their lives. His friend Justin thinks that the proportion is higher than 0.25 for their high school. To settle their dispute, they ask a random sample of 80 students in their school and find out that 28 have played/danced in the rain. Explain why the sample data give some evidence for Ha.",The sample data give some evidence for Ha because p^ = 28/80 = 0.35 > 0.25.,Free Response,,,,
-TPS,R9,1c,"Rob once read that one-quarter of all people have played/danced in the rain at some point in their lives. His friend Justin thinks that the proportion is higher than 0.25 for their high school. To settle their dispute, they ask a random sample of 80 students in their school and find out that 28 have played/danced in the rain. Identify the appropriate test to perform and show that the conditions for carrying out the test are met.",Appropriate test: One-sample z test for p. Random: We have a random sample of 80 students from this school. 10%: Assume the sample size (80) is less than 10% of all students at this school. Normal/Large Sample: np0 = 80(0.25) = 20 ≥ 10 and n(1 - p0) = 80(0.75) = 60 ≥ 10. ,Free Response,,,,
-TPS,R9,1d,"Rob once read that one-quarter of all people have played/danced in the rain at some point in their lives. His friend Justin thinks that the proportion is higher than 0.25 for their high school. To settle their dispute, they ask a random sample of 80 students in their school and find out that 28 have played/danced in the rain. Find the standardized test statistic and P-value, and make an appropriate conclusion.","z= 2.07; P-value = 0.0188 Conclusion: We will use α = 0.05. Because the P-value of 0.0118 > α = 0.05, we reject H0. We have convincing evidence that the true proportion of all students at this school who have played/ danced in the rain in their lifetime is greater than 0.25",Free Response,,,,
-TPS,R9,2a,"A drug company has developed a new vaccine for preventing the flu. The company claims that fewer than 5% of adults who use its vaccine will get the flu. To test the claim, researchers give the vaccine to a random sample of 1000 adults. State appropriate hypotheses for testing the company’s claim. Be sure to define your parameter.","H0: p= 0.05, Ha: p< 0.05, where p is the true proportion of adults who will get the flu after using the vaccine. We will use α = 0.05. ",Free Response,,,,
-TPS,R9,2b,"A drug company has developed a new vaccine for preventing the flu. The company claims that fewer than 5% of adults who use its vaccine will get the flu. To test the claim, researchers give the vaccine to a random sample of 1000 adults. Describe a Type I error and a Type II error in this setting, and give the consequences of each.","Type I: Finding convincing evidence that less than 5% of patients would get the flu, when in reality at least 5% would. Consequence: The company might get sued for false advertising. Type II: Failing to find convincing evidence that less than 5% would get the flu, when in reality less than 5% would. Consequence: Loss of potential income. ",Free Response,,,,
-TPS,R9,2c,"A drug company has developed a new vaccine for preventing the flu. The company claims that fewer than 5% of adults who use its vaccine will get the flu. To test the claim, researchers give the vaccine to a random sample of 1000 adults. Would you recommend a significance level of 0.01, 0.05, or 0.10 for this test? Justify your choice.","Because a Type I error is more serious in this case, and P(Type I error) = α, I would recommend a significance level of α = 0.01 to minimize the possibility of making this type of error. ",Free Response,,,,
-TPS,R9,2d,"A drug company has developed a new vaccine for preventing the flu. The company claims that fewer than 5% of adults who use its vaccine will get the flu. To test the claim, researchers give the vaccine to a random sample of 1000 adults. The power of the test to detect the fact that only 3% of adults who use this vaccine would develop flu using α= 0.05 is 0.9437. Interpret this value.","If the true proportion of adults who will get the flu after using the vaccine is p= 0.03, there is a 0.9437 probability that the researchers will find convincing evidence for Ha: p< 0.05. ",Free Response,,,,
-TPS,R9,2e,"A drug company has developed a new vaccine for preventing the flu. The company claims that fewer than 5% of adults who use its vaccine will get the flu. To test the claim, researchers give the vaccine to a random sample of 1000 adults. The power of the test to detect the fact that only 3% of adults who use this vaccine would develop flu using α= 0.05 is 0.9437. Explain two ways that you could increase the power of the test.",The power could be increased by increasing the sample size or increasing the significance level α. ,Free Response,,,,
-TPS,R9,3,"A drug company has developed a new vaccine for preventing the flu. The company claims that fewer than 5% of adults who use its vaccine will get the flu. To test the claim, researchers give the vaccine to a random sample of 1000 adults. Of the 1000 adults who were given the vaccine, 43 got the flu. Do these data provide convincing evidence to support the company’s claim?","S: H0: p= 0.05, Ha: p< 0.05, where p = the true proportion of adults who will get the flu after using the vaccine. We will use α = 0.05. P: One-sample z test for p. Random: We have a random sample of 1000 adults. 10%: The sample size (1000) is less than 10% of the population of adults. Large Counts: np0 = 1000(0.05) = 50 ≥ 10 and n(1 - p0) = 1000(0.95) = 950 ≥ 10. D: z=-1.02; P-value = 0.1539. C: Because the P-value of 0.1539 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true proportion of adults who will get the flu after using the vaccine is less than 0.05.",Free Response,,,,
-TPS,R9,4a,"An American roulette wheel has 18 red slots among its 38 slots. To test if a particular roulette wheel is fair, you spin the wheel 50 times and the ball lands in a red slot 31 times. The resulting P -value is 0.0384. Interpret the P -value. ","Assuming the roulette wheel is fair, there is a 0.0384 probability that we would get a sample proportion of reds (p^ = 31/50) at least this different from the expected proportion of reds (18/38) by chance alone. ",Free Response,,,,
-TPS,R9,4b,"An American roulette wheel has 18 red slots among its 38 slots. To test if a particular roulette wheel is fair, you spin the wheel 50 times and the ball lands in a red slot 31 times. The resulting P -value is 0.0384. What conclusion would you make at the α= 0.05 level?","Because the P-value of 0.0384 < α = 0.05, we reject H0. We have convincing evidence that the true proportion of reds is not equal to 18/38 = 0.474.",Free Response,,,,
-TPS,R9,4c,"An American roulette wheel has 18 red slots among its 38 slots. To test if a particular roulette wheel is fair, you spin the wheel 50 times and the ball lands in a red slot 31 times. The resulting P -value is 0.0384. The casino manager uses your data to produce a 99% confidence interval for p and gets (0.44, 0.80). He says that this interval provides convincing evidence that the wheel is fair. How do you respond?","Because 18/38 = 0.474 is one of the plausible values in the interval, this interval does not provide convincing evidence that the wheel is unfair. But it also does not prove that the wheel is fair; there are many plausible values in the interval that are not equal to 18/38. This conclusion is inconsistent with the conclusion in part (b) because the manager used a 99% CI, which is equivalent to a test using α = 0.01. If the manager had used a 95% CI, 18/38 would not be considered a plausible value.",Free Response,,,,
-TPS,R9,5a,"AZT was the first drug that seemed effective in delaying the onset of AIDS. Evidence of AZT’s effectiveness came from a large randomized comparative experiment. The subjects were 870 volunteers who were infected with HIV (the virus that causes AIDS), but who did not yet have AIDS. The study assigned 435 of the subjects at random to take 500 milligrams of AZT each day and another 435 sub-jects to take a placebo. At the end of the study, 38 of the placebo subjects and 17 of the AZT subjects had developed AIDS. If the results of the study are statistically significant, is it reasonable to conclude that AZT is the cause of the decrease in the proportion of people like these who will develop AIDS?","Yes, if the results are statistically significant, we have reason to conclude that AZT is the cause of the decrease in the proportion of people like these who will develop AIDS because the subjects were randomly assigned to the treatment and control groups in this randomized comparative experiment. ",Free Response,,,,
-TPS,R9,5b,"AZT was the first drug that seemed effective in delaying the onset of AIDS. Evidence of AZT’s effectiveness came from a large randomized comparative experiment. The subjects were 870 volunteers who were infected with HIV (the virus that causes AIDS), but who did not yet have AIDS. The study assigned 435 of the subjects at random to take 500 milligrams of AZT each day and another 435 sub-jects to take a placebo. At the end of the study, 38 of the placebo subjects and 17 of the AZT subjects had developed AIDS. Do the data provide convincing evidence at the α= 0.05 level that taking AZT lowers the proportion of infected people like the ones in this study who will develop AIDS in a given period of time?","S: H0: p1 - p2 = 0, Ha: p1 - p2 < 0, where p1 = true proportion of patients like these who take AZT and develop AIDS and p2 = true proportion of patients like these who take placebo and develop AIDS; α = 0.05. P: Two-sample z test for p1 - p2. Random: The subjects were assigned at random to take AZT or a placebo. Large Counts: 27.405, 407.595, 27.405, 407.595 are ≥ 10. D: z = -2.91; P-value = 0.0018. C: Because the P-value of 0.0018 < α = 0.05, we reject H0 . We have convincing evidence that taking AZT lowers the proportion of patients like these who develop AIDS compared to a placebo.",Free Response,,,,
-TPS,T9,1,An opinion poll asks a random sample of adults whether they favor banning ownership of handguns by private citizens. A commentator believes that more than half of all adults favor such a ban. The null and alternative hypotheses you would use to test this claim are  (a) H0: p^ = 0.5; Ha: p^ > 0.5.  (b) H0: p = 0.5; Ha: p > 0.5.  (c) H0: p = 0.5; Ha: p < 0.5.  (d) H0: p = 0.5; Ha: p ≠ 0.5.  (e) H0: p > 0.5; Ha: p = 0.5.,b,Multiple Choice,,,,
-TPS,T9,2,"The power takeoff driveline on tractors used in agriculture can be a serious hazard to operators of farm equipment. The driveline is covered by a shield in new tractors, but the shield is often missing on older tractors. Two types of shields are the bolt-on and the flip-up. It was believed that the bolt-on shield was perceived as a nuisance by the operators and deliberately removed, but the flip-up shield is easily lifted for inspection and maintenance and may be left in place. In a study by the U.S. National Safety Council, random samples of older tractors with both types of shields were taken to see what proportion of shields were removed. Of 183 tractors designed to have bolt-on shields, 35 had been removed. Of the 136 tractors with flip-up shields, 15 were removed. We wish to perform a test of H0: pB = pF versus Ha: pB > pF where pB and pF are the proportions of all tractors with the bolt-on and flip-up shields removed, respectively. Which of the following is not a condition for performing the significance test? (a) Both populations are Normally distributed. (b) The data come from two independent samples. (c) Both samples were chosen at random. (d) The expected counts of successes and failures are large enough to use Normal calculations. (e) Both populations are more than 10 times the corresponding sample sizes.",a,Multiple Choice,,,,
-TPS,T9,4,A significance test allows you to reject a null hypothesis H0 in favor of an alternative hypothesis Ha at the 5% significance level. What can you say about significance at the 1% level? (a) H0 can be rejected at the 1% significance level. (b) There is insufficient evidence to reject H0 at the 1% significance level. (c) There is sufficient evidence to accept H0 at the 1% significance level. (d) Ha can be rejected at the 1% significance level. (e) The answer can’t be determined from the information given.,e,Multiple Choice,,,,
-TPS,T9,5,A random sample of 100 likely voters in a small city produced 59 voters in favor of Candidate A. The value of the standardized test statistic for performing a test of H0: p = 0.5 versus Ha: p > 0.5 following?  (a) z = (0.59-0.5) / sqrt((0.59(0.41))/100) (b) z = (0.59-0.5) / sqrt((0.5(0.5))/100) (c) z = (0.5-0.59) / sqrt((0.59(0.41))/100) (d) z = (0.5-0.59) / sqrt((0.5(0.5))/100) (e) z = (0.59-0.5) / sqrt(100),b,Multiple Choice,,,,
-TPS,T9,6,"A researcher claims to have found a drug that causes people to grow taller. The coach of the basketball team at Brandon University has expressed interest but demands evidence. Over 1000 Brandon students volunteer to participate in an experiment to test this new drug. Fifty of the volunteers are randomly selected, their heights are measured, and they are given the drug. Two weeks later, their heights are measured again. The power of the test to detect an average increase in height of 1 inch could be increased by (a) using only the 12 members of the basketball team in the experiment. (b) using α 0.01= instead ofα= 0.05. (c) using α= 0.05 instead of α= 0.01. (d) giving the drug to 25 randomly selected students instead of 50. (e) using a two-sided test instead of a one-sided test.",c,Multiple Choice,,,,
-TPS,T9,7,"A 95% confidence interval for the proportion of viewers of a certain reality television show who are over 30 years old is (0.26, 0.35). Suppose the show’s producers want to test the hypothesis H0: p = 0.25 against Ha: p ≠ 0.25. Which of the following is an appropriate conclusion for them to draw at the α= 0.05 significance level? (a) Fail to reject H0; there is convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old equals 0.25. (b) Fail to reject H0; there is not convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old differs from 0.25. (c) Reject H0; there is not convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old differs from 0.25. (d) Reject H0; there is convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old is greater than 0.25. (e) Reject H0; there is convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old differs from 0.25.",e,Multiple Choice,,,,
-TPS,T9,8,"In a test of H0: p = 0.4 against Ha: p ≠ 0.4, a random sample of size 100 yields a standardized test statistic of z = 1.28. Which of the following is closest to the P-value for this test? (a) 0.90 (b) 0.40 (c) 0.05 (d) 0.20 (e) 0.10",d,Multiple Choice,,,,
-TPS,T9,9,Bags of a certain brand of tortilla chips claim to have a net weight of 14 ounces. A representative of a consumer advocacy group wishes to see if there is convincing evidence that the mean net weight is less than advertised and so intends to test the hypotheses H0: μ =14; Ha: μ < 14. A Type I error in this situation would mean concluding that the bags (a) are being underfilled when they really aren’t. (b) are being underfilled when they really are. (c) are not being underfilled when they really are. (d) are not being underfilled when they really aren’t. (e) are being overfilled when they are really underfilled.,a,Multiple Choice,,,,
-TPS,T9,10,"Conference organizers wondered whether posting a sign that says “Please take only one cookie” would reduce the proportion of conference attendees who take multiple cookies from the snack table during a break. To find out, the organizers randomly assigned 212 attendees to take their break in a room where the snack table had the sign posted, and 189 attendees to take their break in a room where the snack table did not have a sign posted. In the room without the sign posted, 24.3% of attendees took multiple cookies. In the room with the sign posted, 17.0% of attendees took multiple cookies. Is this decrease in proportions statistically significant at the α = 0.05 level? (a) No. The P-value is 0.034. (b) No. The P-value is 0.068. (c) Yes. The P-value is 0.034. (d) Yes. The P-value is 0.068. (e) Cannot be determined from the information given.",c,Multiple Choice,,,,
-TPS,T9,11a,"A software company is trying to decide whether to pro-duce an upgrade of one of its programs. Customers would have to pay $100 for the upgrade. For the upgrade to be profitable, the company must sell it to more than 20% of their customers. The company will survey a random sample of 60 customers about this issue. Which would be a more serious mistake in this setting—a Type I error or a Type II error? Justify your answer.","Type I: Finding convincing evidence that more than 20% of customers would pay for the upgrade, when in reality they would not. Type II: Finding convincing evidence that more than 20% of customers would pay for the upgrade, when in reality more than 20% would. For the company, a Type I error is worse because they would go ahead with the upgrade and lose money. ",Free Response,,,,
-TPS,T9,11b,"A software company is trying to decide whether to pro-duce an upgrade of one of its programs. Customers would have to pay $100 for the upgrade. For the upgrade to be profitable, the company must sell it to more than 20% of their customers. The company will survey a random sample of 60 customers about this issue. In the company’s survey, 16 customers are willing to pay $100 each for the upgrade. Do the sample data give convincing evidence that more than 20% of the company’s customers are willing to purchase the upgrade? Carry out an appropriate test at the significance level.","S: H0: p= 0.20, Ha: p≠ 0.20, where p = the true proportion of customers who would pay $100 for the upgrade using α = 0.05. P: One-sample z test for p. Random: We have a random sample of 60 customers. 10%: The sample size (60) is less than 10% of this company’s customers. Large Counts: np0 = 60(0.20) = 12 ≥ 10 and n(1 - p0) = 60(0.8) = 48 ≥ 10. D: z= 1.29; P-value = 0.0985. C: Because the P-value of 0.0985 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true proportion of customers who would pay $100 for the upgrade is greater than 0.20.",Free Response,,,,
-TPS,T9,12a,A random sample of 100 of last year’s model of a certain popular car found that 20 had a specific minor defect in the brakes. The automaker adjusted the production process to try to reduce the proportion of cars with the brake problem. A random sample of 350 of this year’s model found that 50 had the minor brake defect. Was the company’s adjustment successful? Carry out an appropriate test using α= 0.05 to support your answer.,"STATE: H0: p1 - p2 = 0, Ha: p1 - p2 > 0, where p1 = true proportion of cars that have the brake defect in last year’s model and p2 = true proportion of cars with the brake defect in this year’s model; α = 0.05. PLAN: Two-sample z test for p1 - p2. Random: Independent random samples. 10%: n1 = 100 < 10% of last year’s model and n2 = 350 < 10% of this year’s model. Large Counts: 15.6, 84.4, 54.6, and 295.4 are ≥ 10. DO: z= 1.39; P-value = 0.0822. CONCLUDE: Because the P-value of 0.0822 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true proportion of brake defects is smaller in this year’s model compared to last year’s model. ",Free Response,,,,
-TPS,T9,12b,A random sample of 100 of last year’s model of a certain popular car found that 20 had a specific minor defect in the brakes. The automaker adjusted the production process to try to reduce the proportion of cars with the brake problem. A random sample of 350 of this year’s model found that 50 had the minor brake defect. Suppose that the proportion of cars with the defect was reduced by 0.10 from last year to this year. The power of the test to detect this decrease is 0.72. Interpret this value. ,"If the true proportion of cars with the defect has decreased by 0.10 from last year to this year, there is a 0.72 probability that the automaker will find convincing evidence for Ha: p1 - p2 > 0. ",Free Response,,,,
-TPS,T9,12c,"A random sample of 100 of last year’s model of a certain popular car found that 20 had a specific minor defect in the brakes. The automaker adjusted the production process to try to reduce the proportion of cars with the brake problem. A random sample of 350 of this year’s model found that 50 had the minor brake defect. Other than increasing the sample sizes, identify one way of increasing the power of the test.","Besides increasing the sample sizes, the power of the test can be increased by using a larger value for a, the significance level of the test.",Free Response,,,,
-TPS,10.1,1a,What critical value *t should be used for a confidence interval for the population mean for the following situation? A 95% confidence interval based on n =10 randomly selected observations,"df = 9, t* = 2.262; Tech: invT(area: 0.025, df: 9) 522.262, so t*= 2.262.",Free Response,,,,
-TPS,10.1,1b,What critical value *t should be used for a confidence interval for the population mean for the following situation? A 99% confidence interval from an SRS of 20 observations,"df = 19, t* = 2.861; Tech: invT(area: 0.005, df: 19) 522.861, so t* = 2.861.",Free Response,,,,
-TPS,10.1,1c,What critical value *t should be used for a confidence interval for the population mean for the following situation? A 90% confidence interval based on a random sample of 77 individuals,"df = 60, t* = 1.671; Tech: invT(area: 0.05, df: 76) 521.665, so t* = 1.665.",Free Response,,,,
-TPS,10.1,2a,What critical value *t should be used for a confidence interval for the population mean for the following situation? A 90% confidence interval based on n =12 randomly selected observations,"df = 11, t* = 1.796; Tech: invT(area: 0.05, df: 11) 521.796, so t* = 1.796.",Free Response,,,,
-TPS,10.1,2b,What critical value *t should be used for a confidence interval for the population mean for the following situation? A 95% confidence interval from an SRS of 30 observations,"df = 29, t* = 2.045; Tech: invT(area: 0.025, df: 29) 522.045, so t*= 2.045.",Free Response,,,,
-TPS,10.1,2c,What critical value *t should be used for a confidence interval for the population mean for the following situation? A 99% confidence interval based on a random sample of size 58,"df = 50, t* = 2.678’ Tech: invT(area: 0.005, df: 57) 522.665, so t* = 2.665.",Free Response,,,,
-TPS,10.1,5a,Determine if the conditions are met for constructing a confidence interval for the population mean in the following setting: How much time do students at your school spend on the Internet? You collect data from the 32 members of your AP® Statistics class and calculate the mean amount of time that these students spent on the Internet yesterday.,Random: No; the AP® Statistics students are not a random sample of all students. 10%: 32 < 10% of all students in the school. (Condition met if the number of students at the school is at least 320). Normal/Large Sample: Yes; n= 32 ≥ 30.,Free Response,,,,
-TPS,10.1,6a,Determine if the conditions are met for constructing a confidence interval for the population mean in the following setting: We want to estimate the average age at which U.S. presidents have died. So we obtain a list of all U.S. presidents who have died and their ages at death.,Random: No; a list of all U.S. presidents is not a random sample. 10%: No; the n presidents who have died is not less than 10% of all presidents that have died. Normal/Large Sample: Yes; n≥ 30.,Free Response,,,,
-TPS,10.1,7,Blood pressure A medical study finds that x= 114.9 and sx = 9.3 for the seated systolic blood pressure of 27 randomly selected adults. What is the standard error of the mean? Interpret this value in context.,"SEx = sx / sqrt(n) = 9.3 / sqrt(27) = 1.7898. In many random samples of size 27, the sample mean blood pressure will typically vary by about 1.7898 from the population mean blood pressure.",Free Response,,,,
-TPS,10.1,8,A study of commuting times reports the travel times to work of a random sample of 20 employed adults in New York State. The mean is x = 31.25 minutes and the standard deviation is sx = 21.88 minutes. What is the standard error of the mean? Interpret this value in context.,"SEx = sx / sqrt(n) = 21.88 / sqrt(20) = 4.8925. In many random samples of size 20, the sample mean commute time will typically vary by about 4.8925 from the population mean commute time.",Free Response,,,,
-TPS,10.1,9a,"Breast-feeding mothers secrete calcium into their milk. Some of the calcium may come from their bones, so mothers may lose bone mineral. Researchers measured the percent change in bone mineral content (BMC) of the spines of 47 randomly selected mothers during three months of breast-feeding. The mean change in BMC was −3.587% and the standard deviation was 2.506%. Construct and interpret a 99% confidence interval to estimate the mean percent change in BMC in the population of breast-feeding mothers.","S: µ = the true mean percent change in BMC for breast-feeding mothers. P: One-sample t interval. Random: The mothers were randomly selected. 10%: 47 is less than 10% of all breast-feeding mothers. Normal/Large Sample: n= 47 ≥ 30. D: df = 40 (24.575, 22.599); Tech: (24.569, 22.605) with df = 40. C: We are 99% confident that the interval from 24.569 to 22.605 captures µ = the true mean percent change in BMC for breast-feeding mothers. ",Free Response,,,,
-TPS,10.1,9b,"Breast-feeding mothers secrete calcium into their milk. Some of the calcium may come from their bones, so mothers may lose bone mineral. Researchers measured the percent change in bone mineral content (BMC) of the spines of 47 randomly selected mothers during three months of breast-feeding. The mean change in BMC was −3.587% and the standard deviation was 2.506%. A 99% confidence interval was found to be (-4.569, -2.599). Do these data give convincing evidence that nursing mothers lose bone mineral, on average? Explain your answer.","Because all the plausible values in the interval are negative (indicating bone loss), the data give convincing evidence. ",Free Response,,,,
-TPS,10.1,10a,The Trial Urban District Assessment (TUDA) is a government-sponsored study of student achievement in large urban school districts. TUDA gives a reading test scored from 0 to 500. A score of 243 is a “basic” reading level and a score of 281 is “proficient.” Scores for a random sample of 1470 eighth-graders in Atlanta had a mean of 240 with standard deviation of 42.17.  Construct and interpret a 99% confidence interval for the mean reading test score of all Atlanta eighth-graders.,"S: µ = the true mean reading test score for all Atlanta eighth-graders. P: One-sample t interval. Random: Students were randomly selected. 10%: 1470 is less than 10% of eighth-graders in Atlanta. Normal/Large Sample: n= 1470 ≥ 30. D: df = 1469 (237.16, 242.84); Tech: (237.16, 242.84) with df = 1469. C: We are 99% confident that the interval from 237.16 to 242.84 captures µ = the true mean reading test score for all Atlanta eighth-graders. ",Free Response,,,,
-TPS,10.1,10b,"The Trial Urban District Assessment (TUDA) is a government-sponsored study of student achievement in large urban school districts. TUDA gives a reading test scored from 0 to 500. A score of 243 is a “basic” reading level and a score of 281 is “proficient.” Scores for a random sample of 1470 eighth-graders in Atlanta had a mean of 240 with standard deviation of 42.17. A 99% confidence interval was found to be (237.16, 242.84).  Is there convincing evidence that the mean reading test score for all Atlanta eighth-graders is less than the basic level? Explain your answer.","Because all the plausible values in the interval are less than 243, there is convincing evidence.",Free Response,,,,
-TPS,10.1,11a,Ann and Tori wanted to estimate the average weight of an Oreo cookie to determine if it was less than advertised (34 grams for 3 cookies). They selected a random sample of 36 cookies and found the weight of each cookie (in grams). The mean weight was x¯ =11.3921 grams with a standard deviation of sx = 0.0817 grams. Construct and interpret a 90% confidence interval for the true mean weight of an Oreo cookie.,"S: µ = the true mean weight of an Oreo cookie. P: One-sample t interval. Random: The cookies were randomly selected. 10%: 36 is less than 10% of all Oreo cookies. Normal/Large Sample: n= 36 ≥ 30. D: df = 30 (11.369, 11.4152); Tech: (11.369, 11.415) with df = 30. C: We are 90% confident that the interval from 11.369 to 11.4152 captures µ = the true mean weight for all Oreo cookies. ",Free Response,,,,
-TPS,10.1,11b,"Ann and Tori wanted to estimate the average weight of an Oreo cookie to determine if it was less than advertised (34 grams for 3 cookies). They selected a random sample of 36 cookies and found the weight of each cookie (in grams). The mean weight was x¯ =11.3921 grams with a standard deviation of sx = 0.0817 grams. A 90% confidence interval was found to be (11.369, 11.4152). Interpret the confidence level. ","If we were to select many random samples of size 36 from the population of all Oreo cookies and construct a 90% confidence interval using each sample, about 90% of the intervals would capture the true mean weight of an Oreo cookie.",Free Response,,,,
-TPS,10.1,12a,"Fruit flies are used frequently in genetic research because of their quick reproductive cycle. The length of the thorax (in millimeters) was measured for each fly in a random sample of 49 male fruit flies. The mean length was x¯ = 0.8004 mm, with a standard deviation of sx = 0.0782 mm. Construct and interpret a 90% confidence interval for the true mean thorax length of a male fruit fly."," S: µ = the true mean thorax length of a male fruit fly. P: One-sample t interval. Random: Flies were randomly selected. 10%: 49 is less than 10% of all male fruit flies. Normal/ Large Sample: n= 49 ≥ 30. D: df = 48 (0.7816, 0.8192); Tech: (0.7817, 0.8191) with df = 48. C: We are 90% confident that the interval from 0.7817 to 0.8191 captures µ = the true mean thorax length for all male fruit flies. ",Free Response,,,,
-TPS,10.1,12b,"Fruit flies are used frequently in genetic research because of their quick reproductive cycle. The length of the thorax (in millimeters) was measured for each fly in a random sample of 49 male fruit flies. The mean length was x¯ = 0.8004 mm, with a standard deviation of sx = 0.0782 mm. A 90% confidence interval was found to be (0.7816, 0.8192). Interpret the confidence level. ","If we were to select many random samples of size 49 from the population of all male fruit flies and construct a 90% confidence interval using each sample, about 90% of the intervals would capture the true mean thorax length. ",Free Response,,,,
-TPS,10.1,13,"Melissa and Madeline love pepperoni pizza, but sometimes they are disappointed with the small number of pepperonis on their pizza. To investigate, they went to their favorite pizza restaurant at 10 random times during the week and ordered a large pepperoni pizza. 8 Here are the number of pepperonis on each pizza: 47 36 25 37 46 36 49 32 32 34 Construct and interpret a 95% confidence interval for the true mean number of pepperonis on a large pizza at this restaurant.","S: m 5 the true mean number of pepperonis on a large pizza at this restaurant. P: One-sample t interval. Random: Pizzas were randomly selected. 10%: 10 is less than 10% of all pepperoni pizzas made at this restaurant. Normal/ Large Sample: The dotplot doesn’t show any outliers or strong skewness. D: x¯= 37.4, sx= 7.662, and n= 10; df = 9 (31.919, 42.881); Tech: (31.919, 42.881) with df = 9. C: We are 95% confident that the interval from 31.919 to 42.881 captures µ = the true mean.",Free Response,,,,
-TPS,10.1,14,"Carly and Maysem plan to be preschool teachers after they graduate from college. To prepare for snack time, they want to know the mean number of goldfish crackers in a bag of original-flavored goldfish. To estimate this value, they randomly selected 12 bags of original-flavored goldfish and counted the number of crackers in each bag. 9 Here are their data: 317 330 325 323 332 337 324 342 330 349 335 333 Construct and interpret a 95% confidence interval for the true mean number of crackers in a bag of original-flavored goldfish.","S: µ = the true mean number of crackers in a bag of original goldfish. P: One-sample t interval. Random: Bags of goldfish were randomly selected. 10%: 12 is less than 10% of all bags of original-flavored goldfish. Normal/Large Sample: The dotplot doesn’t show any outliers or strong skewness. D: x¯= 331.417, sx= 8.775, and n= 12; df = 11 (325.842, 336.992); Tech: (325.84, 336.99) with df = 11. C: We are 95% confident that the interval from 325.84 to 336.99 captures µ = the true mean. ",Free Response,,,,
-TPS,10.1,15a,"Melissa and Madeline love pepperoni pizza, but sometimes they are disappointed with the small number of pepperonis on their pizza. To investigate, they went to their favorite pizza restaurant at 10 random times during the week and ordered a large pepperoni pizza. 8 Here are the number of pepperonis on each pizza: 47 36 25 37 46 36 49 32 32 34. A 95% confidence interval was found to be (31.919, 42.881). Explain why it was necessary to inspect a graph of the sample data when checking the Normal/Large Sample condition.","It was necessary because the sample size was not large (<30). When the sample size is less than 30, we must assume the population is Normally distributed. ",Free Response,,,,
-TPS,10.1,15b,"Melissa and Madeline love pepperoni pizza, but sometimes they are disappointed with the small number of pepperonis on their pizza. To investigate, they went to their favorite pizza restaurant at 10 random times during the week and ordered a large pepperoni pizza. 8 Here are the number of pepperonis on each pizza: 47 36 25 37 46 36 49 32 32 34. A 95% confidence interval was found to be (31.919, 42.881). According to the manager of the restaurant, there should be an average of 40 pepperonis on a large pizza. Based on the interval, is there convincing evidence that the average number of pepperonis is less than 40? Explain your answer.","The value 40 is a plausible value found within the confidence interval, so we do not have convincing evidence that the average number of pepperonis is less than 40.",Free Response,,,,
-TPS,10.1,15c,"Melissa and Madeline love pepperoni pizza, but sometimes they are disappointed with the small number of pepperonis on their pizza. To investigate, they went to their favorite pizza restaurant at 10 random times during the week and ordered a large pepperoni pizza. 8 Here are the number of pepperonis on each pizza: 47 36 25 37 46 36 49 32 32 34. A 95% confidence interval was found to be (31.919, 42.881). Explain two ways that Melissa and Madeline could reduce the margin of error of their estimate. Why might they object to these changes?",Melissa and Madeline could increase the sample size or decrease the confidence level. They may not want to increase the sample size because it would be expensive and they also may not want to eat that much pizza. They may not want to decrease the confidence level because it is desirable to have a high degree of confidence that the population parameter has been captured by the interval. ,Free Response,,,,
-TPS,10.1,16a,"Carly and Maysem plan to be preschool teachers after they graduate from college. To prepare for snack time, they want to know the mean number of goldfish crackers in a bag of original-flavored goldfish. To estimate this value, they randomly selected 12 bags of original-flavored goldfish and counted the number of crackers in each bag. 9 Here are their data: 317 330 325 323 332 337 324 342 330 349 335 333. A 95% confidence interval was found to be (325.842, 336.992). Explain why it was necessary to inspect a graph of the sample data when checking the Normal/Large Sample condition.","It was necessary because the sample size was not large (<30). When the sample size is less than 30, we must assume the population is Normally distributed. ",Free Response,,,,
-TPS,10.1,16b,"Carly and Maysem plan to be preschool teachers after they graduate from college. To prepare for snack time, they want to know the mean number of goldfish crackers in a bag of original-flavored goldfish. To estimate this value, they randomly selected 12 bags of original-flavored goldfish and counted the number of crackers in each bag. 9 Here are their data: 317 330 325 323 332 337 324 342 330 349 335 333. A 95% confidence interval was found to be (325.842, 336.992). According to the packaging, there are supposed to be 330 goldfish in each bag of crackers. Based on the interval, is there convincing evidence that the average number of goldfish is less than 330? Explain your answer.","The value 330 is a plausible value found within the confidence interval, so we do not have convincing evidence that the average number of goldfish is less than 330.",Free Response,,,,
-TPS,10.1,16c,"Carly and Maysem plan to be preschool teachers after they graduate from college. To prepare for snack time, they want to know the mean number of goldfish crackers in a bag of original-flavored goldfish. To estimate this value, they randomly selected 12 bags of original-flavored goldfish and counted the number of crackers in each bag. 9 Here are their data: 317 330 325 323 332 337 324 342 330 349 335 333. A 95% confidence interval was found to be (325.842, 336.992). Explain two ways that Carly and Maysem could reduce the margin of error of their estimate. Why might they object to these changes?",Carly and Maysem could increase the sample size or decrease the confidence level. They may not want to increase the sample size because that would become expensive. They may not want to decrease the confidence level because it is desirable to have a high degree of confidence that the population parameter has been captured by the interval.,Free Response,,,,
-TPS,10.1,17,The body mass index (BMI) of all American young women is believed to follow a Normal distribution with a standard deviation of about 7.5. How large a sample would be needed to estimate the mean BMI μ in this population to within ±1 with 99% confidence?,Solving n≥ (2.576(7.5) / 1)^2 = 373.26. Select an SRS of 374 women. ,Free Response,,,,
-TPS,10.1,18,The SAT again High school students who take the SAT Math exam a second time generally score higher than on their first try. Past data suggest that the score increase has a standard deviation of about 50 points. How large a sample of high school students would be needed to estimate the mean change in SAT score to within 2 points with 95% confidence?,Solving n≥ (1.96(50) / 2)^2 = 2401. Select a random sample of 2401 students.,Free Response,,,,
-TPS,10.1,19a,Writers in some fields summarize data by giving x and its standard error rather than x¯ and sx. Biologists studying willow plants in Yellowstone National Park reported their results in a table with columns labeled x¯ ± SE. The table entry for the heights of willow plants (in centimeters) in one region of the park was 61.55 ± 19.03. The researchers measured a total of 23 plants. Find the sample standard deviation sx for these measurements.,"SEx= 19.03 = sx/sqrt(n) = sx/sqrt(23), so sx= 19.03sqrt(23) = 91.26. ",Free Response,,,,
-TPS,10.1,19b,Writers in some fields summarize data by giving x and its standard error rather than x¯ and sx. Biologists studying willow plants in Yellowstone National Park reported their results in a table with columns labeled x¯ ± SE. The table entry for the heights of willow plants (in centimeters) in one region of the park was 61.55 ± 19.03. The researchers measured a total of 23 plants. A hasty reader believes that the interval given in the table is a 95% confidence interval for the mean height of willow plants in this region of the park. Find the actual confidence level for the given interval.,"Because the researchers are using a critical value of t* = 1. With df = 23 - 1 = 22, the area between t=-1 and t= 1 is approximately tcdf(lower: 21, upper: 1, df: 22) = 0.67. So, the confidence level is 67%.",Free Response,,,,
-TPS,10.1,20a,"When two lights that are close together blink alternately, we “see” one light moving back and forth if the time between blinks is short. What is the longest interval of time between blinks that preserves the illusion of motion? Ask subjects to turn a knob that slows the blinking until they “see” two lights rather than one light moving. A report gives the results in the form “mean plus or minus the standard error of the mean.”  Data for 12 subjects are summarized as 251 ± 45 (in milliseconds). Find the sample standard deviation sx for these measurements.","SEx= 15 = sx/sqrt(n) = sx/sqrt(12), so sx= 45sqrt(12) = 155.88. ",Free Response,,,,
-TPS,10.1,20b,"When two lights that are close together blink alternately, we “see” one light moving back and forth if the time between blinks is short. What is the longest interval of time between blinks that preserves the illusion of motion? Ask subjects to turn a knob that slows the blinking until they “see” two lights rather than one light moving. A report gives the results in the form “mean plus or minus the standard error of the mean.”  Data for 12 subjects are summarized as 251 ± 45 (in milliseconds). A hasty reader believes that the interval given in the report is a 95% confidence interval for the population mean. Find the actual confidence level for the given interval.","Because the researchers are using a critical value of t* = 1. With df = 12 - 1 = 11, the area between t=-1 and t= 1 is approximately tcdf(lower: 21, upper: 1, df: 11) = 0.66. So, the confidence level is 66%.",Free Response,,,,
-TPS,10.1,21,One reason for using a t distribution instead of the standard Normal distribution to find critical values when calculating a level C confidence interval for a population mean is that (a) z can be used only for large samples. (b) z requires that you know the population standard deviation σ. (c) z requires that you can regard your data as an SRS from the population. (d) z requires that the sample size is less than 10% of the population size. (e) a z critical value will lead to a wider interval than a t critical value.,b,Multiple Choice,,,,
-TPS,10.1,22,You have an SRS of 23 observations from a large population. The distribution of sample values is roughly symmetric with no outliers. What critical value would you use to obtain a 98% confidence interval for the mean of the population? (a) 2.177 (b) 2.183 (c) 2.326 (d) 2.500 (e) 2.508,e,Multiple Choice,,,,
-TPS,10.1,23,A quality control inspector will measure the salt content (in milligrams) in a random sample of bags of potato chips from an hour of production. Which of the following would result in the smallest margin of error in estimating the mean salt content μ? (a) 90% confidence; n = 25 (b) 90% confidence; n =50 (c) 95% confidence; n = 25 (d) 95% confidence; n =50 (e) n =100 at any confidence level ,b,Multiple Choice,,,,
-TPS,10.1,24,Scientists collect data on the blood cholesterol levels (milligrams per deciliter of blood) of a random sample of 24 laboratory rats. A 95% confidence interval for the mean blood cholesterol level μ is 80.2 to 89.8. Which of the following would cause the most worry about the validity of this interval? (a) There is a clear outlier in the data. (b) A stemplot of the data shows a mild right skew. (c) You do not know the population standard deviation σ. (d) The population distribution is not exactly Normal. (e) None of these are a problem when using a t interval.,a,Multiple Choice,,,,
-TPS,10.2,27,"How many pairs of shoes do teenagers have? To find out, a group of AP® Statistics students conducted a survey. They selected a random sample of 20 female students and a separate random sample of 20 male students from their school. Then they recorded the number of pairs of shoes that each student reported having. Here are their data: Males 14 7 6 5 12 38 8 7 10 10 10 11 4 5 22 7 5 10 35 7 Females 50 26 26 31 57 19 24 22 23 38 13 50 13 34 23 30 49 13 15 51. Let μ1 = the true mean number of pairs of shoes that male students at the school have and μ2 = the true mean number of pairs of shoes that female students at the school have. Check if the conditions for calculating a confidence interval for μ1- μ2 are met.",Random: Met because these are two independent random samples. 10%: Met because 20 < 10% of all males at the school and 20 < 10% of all females at the school. Normal/Large Sample: Not met; there are fewer than 30 observations in each group and a dotplot for males shows several outliers.,Free Response,,,,
-TPS,10.2,28,"For their final project, a group of AP® Statistics students wanted to compare the texting habits of males and females. They asked a random sample of students from their school to record the number of text messages sent and received over a 2-day period. Here are their data: Males 127 44 28 83 0 6 78 6 5 213 73 20 214 28 11 Females 112 203 102 54 379 305 179 24 127 65 41 27 298 6 130 0 Let μ1 = the true mean number of texts sent by male students at the school and μ2 = the true mean number of texts sent by female students at the school. Check if the conditions for calculating a confidence interval for μ1-μ2 are met.","Random: Met; even though the data came from a single random sample, it is reasonable to consider the two samples independent because knowing the response of a male shouldn’t predict the response of a female. 10%: Met because 15 < 10% of all males at the school and 16 < 10% of all females at the school. Normal/Large Sample: Not met; there are fewer than 30 observations in each group and a dotplot for males shows several outliers",Free Response,,,,
-TPS,10.2,33a,"College financial aid offices expect students to use summer earnings to help pay for college. But how large are these earnings? One large university studied this question by asking a random sample of 1296 students who had summer jobs how much they earned. The financial aid office separated the responses into two groups based on gender, so these can be viewed as independent samples. Here are the data in summary form:  Males: n=675, x¯=$1884.52, sx=1360.39 Females n=621, x¯=$1368.37, sx= $1037.46. How can you tell from the summary statistics that the distribution of earnings in each group is strongly skewed to the right? The use of two-sample t procedures is still justified. Why?","Skewed to the right because the earnings cannot be negative, yet the standard deviation is almost as large as the distance between the mean and 0. The use of the two-sample t procedures is justified because the sample sizes are both very large (675 ≥ 30 and 621 ≥ 30). ",Free Response,,,,
-TPS,10.2,33b,"College financial aid offices expect students to use summer earnings to help pay for college. But how large are these earnings? One large university studied this question by asking a random sample of 1296 students who had summer jobs how much they earned. The financial aid office separated the responses into two groups based on gender, so these can be viewed as independent samples. Here are the data in summary form:  Males: n=675, x¯=$1884.52, sx=1360.39 Females n=621, x¯=$1368.37, sx= $1037.46. Construct and interpret a 90% confidence interval for the difference between the true mean summer earnings of male and female students at this university.","S: µ1 = true mean summer earnings of male students and µ2 = true mean for female students. P: Two-sample t interval for µ1 - µ2. Random: It is reasonable to consider the two samples independent because knowing the response of a male shouldn’t predict the response of a female. 10%: n1 = 675 < 10% of male students and n2 = 621 < 10% of female students. Normal/Large Sample: n1 = 675 ≥ 30 and n2 = 621 ≥ 30. D: Using df = 1249.21, (413.62, 634.64); using df = 620, (413.52, 634.72). C: We are 90% confident the interval from $413.62 to $634.64 captures µ1 - µ2 = true difference in mean summer earnings of male and female students at this university. ",Free Response,,,,
-TPS,10.2,33c,"College financial aid offices expect students to use summer earnings to help pay for college. But how large are these earnings? One large university studied this question by asking a random sample of 1296 students who had summer jobs how much they earned. The financial aid office separated the responses into two groups based on gender, so these can be viewed as independent samples. Here are the data in summary form:  Males: n=675, x¯=$1884.52, sx=1360.39 Females n=621, x¯=$1368.37, sx= $1037.46. Interpret the 90% confidence level in the context of this study.","If we took many random samples of 675 males and 621 females from this university and each time constructed a 90% confidence interval in this same way, about 90% of the resulting intervals would capture the true difference in mean earnings for males and females.",Free Response,,,,
-TPS,10.2,34a,"In a study of heart surgery, one issue was the effect of drugs called beta blockers on the pulse rate of patients during surgery. The available subjects were randomly assigned into two groups. One group received a beta blocker; the other group received a placebo. The pulse rate of each patient at a critical point during the operation was recorded. Here are the data in summary form: Beta blocker: n=30, x¯=65.2, sx=7.8; Placebo: n=30, x¯=70.3, sx=8.3. The distribution of pulse rate in each group is not Normal. The use of two-sample t procedures is still justified. Why?",Because the sample sizes are both large (30 ≥ 30 and 30 ≥ 30).,Free Response,,,,
-TPS,10.2,34b,"In a study of heart surgery, one issue was the effect of drugs called beta blockers on the pulse rate of patients during surgery. The available subjects were randomly assigned into two groups. One group received a beta blocker; the other group received a placebo. The pulse rate of each patient at a critical point during the operation was recorded. Here are the data in summary form: Beta blocker: n=30, x¯=65.2, sx=7.8; Placebo: n=30, x¯=70.3, sx=8.3. Construct and interpret a 99% confidence interval for the difference in mean pulse rates for patients like these who receive a beta blocker or a placebo.","S: µ1 = true mean pulse rate of patients like those in the experiment who take beta-blockers and µ2 = mean pulse rate for those who do not. P: Two-sample t interval for µ1 - µ2. Random: Two groups in a randomized experiment. Normal/Large Sample: n1 = 30 ≥ 30 and n2 = 30 ≥ 30. D: Using df = 57.78, (210.64, 0.44); using df = 29, (210.83, 0.63). C: We are 99% confident the interval from 210.64 to 0.44 captures µ1 - µ2 = true difference in the mean pulse rate of patients receiving a beta blocker during surgery and those who take a placebo. ",Free Response,,,,
-TPS,10.2,34c,"In a study of heart surgery, one issue was the effect of drugs called beta blockers on the pulse rate of patients during surgery. The available subjects were randomly assigned into two groups. One group received a beta blocker; the other group received a placebo. The pulse rate of each patient at a critical point during the operation was recorded. Here are the data in summary form: Beta blocker: n=30, x¯=65.2, sx=7.8; Placebo: n=30, x¯=70.3, sx=8.3. Interpret the 99% confidence level in the context of this study.","If we took 60 patients and repeated many random assignments of them to the treatments of a beta blocker and a placebo and each time constructed a 99% confidence interval in this same way, about 99% of the resulting intervals would capture the true difference in mean pulse rates for patients given a beta blocker or placebo. ",Free Response,,,,
-TPS,10.2,35a,"Catherine and Ana wanted to know if student athletes (students on at least one varsity team) have faster reaction times than non-athletes. They took separate random samples of 33 athletes and 30 non-athletes from their school and tested their reaction time using an online reaction test, which measured the time (in seconds) between when a green light went on and the subject pressed a key on the computer keyboard. A 95% confidence interval for the difference (Non-athlete − Athlete) in the mean reaction time was 0.018 ±0.034 seconds. Does the interval provide convincing evidence of a difference in the true mean reaction time of athletes and non-athletes? Explain your answer.","No, because our interval includes a difference of 0 (no difference) as a plausible value. ",Free Response,,,,
-TPS,10.2,35b,"Catherine and Ana wanted to know if student athletes (students on at least one varsity team) have faster reaction times than non-athletes. They took separate random samples of 33 athletes and 30 non-athletes from their school and tested their reaction time using an online reaction test, which measured the time (in seconds) between when a green light went on and the subject pressed a key on the computer keyboard. A 95% confidence interval for the difference (Non-athlete − Athlete) in the mean reaction time was 0.018 ±0.034 seconds. Does the interval provide convincing evidence that the true mean reaction time of athletes and non-athletes is the same? Explain your answer. ","No; instead, we don’t have convincing evidence the mean reaction times differ. Zero is a plausible value for the difference in means, but many other plausible values besides 0 are in the confidence interval. ",Free Response,,,,
-TPS,10.2,35c,"Catherine and Ana wanted to know if student athletes (students on at least one varsity team) have faster reaction times than non-athletes. They took separate random samples of 33 athletes and 30 non-athletes from their school and tested their reaction time using an online reaction test, which measured the time (in seconds) between when a green light went on and the subject pressed a key on the computer keyboard. A 95% confidence interval for the difference (Non-athlete − Athlete) in the mean reaction time was 0.018 ±0.034 seconds. Identify two ways Catherine and Ana could reduce the width of their interval. Describe any drawbacks to these actions.",They could increase the sample sizes or decrease the confidence level. Drawbacks: Increasing the sample sizes would require more work; decreasing the confidence level would give them less certainty that they captured the difference in the true mean reaction time of athletes and non-athletes.,Free Response,,,,
-TPS,10.2,36a,A researcher wants to see if birds that build larger nests lay larger eggs. She selects two random samples of nests: one of small nests and the other of large nests. Then she weighs one egg (chosen at random if there is more than one egg) from each nest. A 95% confidence interval for the difference (Large Small) between the mean mass (in grams) of eggs in small and large nests is 1.6±2.0. Does the interval provide convincing evidence of a difference in the true mean egg mass of birds with small nests and birds with large nests? Explain your answer.,"No, because our interval includes a difference of 0 (no difference) as a plausible value. ",Free Response,,,,
-TPS,10.2,36b,A researcher wants to see if birds that build larger nests lay larger eggs. She selects two random samples of nests: one of small nests and the other of large nests. Then she weighs one egg (chosen at random if there is more than one egg) from each nest. A 95% confidence interval for the difference (Large Small) between the mean mass (in grams) of eggs in small and large nests is 1.6±2.0. Does the interval provide convincing evidence that the true mean egg mass of birds with small nests and birds with large nests is the same? Explain your answer.,"No; instead, we don't have convincing evidence the mean reaction times differ. Zero is a plausible value for the difference in means, but many other plausible values besides 0 are in the confidence interval.",Free Response,,,,
-TPS,10.2,36c,A researcher wants to see if birds that build larger nests lay larger eggs. She selects two random samples of nests: one of small nests and the other of large nests. Then she weighs one egg (chosen at random if there is more than one egg) from each nest. A 95% confidence interval for the difference (Large Small) between the mean mass (in grams) of eggs in small and large nests is 1.6±2.0. Identify two ways the researcher could reduce the width of her interval. Describe any drawbacks to these actions.,The researcher could increase the sample sizes or decrease the confidence level. Drawbacks: Increasing the sample sizes could make it difficult to find additional nests with eggs; decreasing the confidence level decreases the certainty of capturing the difference in the true mean egg mass of birds with small nests and birds with large nests. ,Free Response,,,,
-TPS,10.2,43a,Do piano lessons improve the spatial-temporal reasoning of preschool children? A study designed to investigate this question measured the spatial-temporal reasoning of a random sample of 34 preschool children before and after 6 months of piano lessons. The difference (After - Before) in the reasoning scores for each student has mean 3.618 and standard deviation 3.055. Construct and interpret a 90% confidence interval for the true mean difference.,"S: µdiff 5 true mean difference (After − Before) in reasoning scores for all preschool students who take 6 months of piano lessons. P: One-sample t interval for µdiff. Random: Random sample of 34 preschool children. 10%: 34 < 10% of all preschool children. Normal/Large Sample: ndiff = 34 ≥ 30. D: x¯diff = 3.618, sdiff = 3.055, and ndiff = 34; df = 33; (2.731, 4.505). C: We are 90% confident the interval from 2.731 to 4.505 captures the true mean difference (After − Before) in reasoning scores for all preschool students who take 6 months of piano lessons. ",Free Response,,,,
-TPS,10.2,43b,"Do piano lessons improve the spatial-temporal reasoning of preschool children? A study designed to investigate this question measured the spatial-temporal reasoning of a random sample of 34 preschool children before and after 6 months of piano lessons. The difference (After - Before) in the reasoning scores for each student has mean 3.618 and standard deviation 3.055. An 90% interval was calculated to be (2.731, 4.505). Can you conclude that taking 6 months of piano lessons would cause an increase in preschool students’ average reasoning scores? Why or why not?",No; a randomized experiment is needed to show causation.,Free Response,,,,
-TPS,10.2,44a,"A bank wonders if omitting the annual credit card fee for customers who charge at least $2400 in a year will increase the amount charged on its credit cards. The bank makes this offer to an SRS of 200 of its credit card customers. It then compares how much these customers charge this year with the amount that they charged last year. The mean increase in the sample is $332, and the standard deviation is $108. Construct and interpret a 99% confidence interval for the true mean increase.","S: µdiff 5 true mean increase in the amount this bank’s credit card customers would spend with no annual fee. P: One-sample t interval for µdiff. Random: SRS of 200 credit card customers. 10%: 200 < 10% of the population of all credit card customers. Normal/Large Sample: ndiff = 200 ≥ 30. D: x¯diff = 332, sdiff = 108, and ndiff = 200; df = 199; (312.14, 351.86). C: We are 99% confident that the interval from $312.14 to $351.86 captures the true mean increase in the amount this bank’s credit card customers would spend with no annual fee. ",Free Response,,,,
-TPS,10.2,44b,"A bank wonders if omitting the annual credit card fee for customers who charge at least $2400 in a year will increase the amount charged on its credit cards. The bank makes this offer to an SRS of 200 of its credit card customers. It then compares how much these customers charge this year with the amount that they charged last year. The mean increase in the sample is $332, and the standard deviation is $108. A 99% confidence interval was calculated to be (312.14, 351.86). Can you conclude that dropping the annual fee would cause an increase in the average amount spent by this bank’s credit card customers? Why or why not?",No; there is no control group to compare results.,Free Response,,,,
-TPS,10.2,45a,"After hearing that students can improve short-term memory by chewing the same flavor of gum while studying for and taking a test, Leila and Valerie designed an experiment to investigate Using 30 volunteers, they randomly assigned 15 to chew gum while they studied a list of 40 words for 90 seconds. Immediately after the 90-second study period—and while chewing the same gum—the subjects wrote down as many words as they could remember. The remaining 15 followed the same procedure without chewing gum. Two weeks later, each of the 30 subjects did the oppo-site treatment. The number of words correctly remembered for each test was recorded for each subject. Explain why these are paired data.",They result from recording two values for each individual. ,Free Response,,,,
-TPS,10.2,45b,"After hearing that students can improve short-term memory by chewing the same flavor of gum while studying for and taking a test, Leila and Valerie designed an experiment to investigate Using 30 volunteers, they randomly assigned 15 to chew gum while they studied a list of 40 words for 90 seconds. Immediately after the 90-second study period—and while chewing the same gum—the subjects wrote down as many words as they could remember. The remaining 15 followed the same procedure without chewing gum. Two weeks later, each of the 30 subjects did the oppo-site treatment. The number of words correctly remembered for each test was recorded for each subject. Verify that the conditions for constructing a one-sample t interval for a mean difference are satisfied.",PLAN: One-sample t interval for µdiff. Random: The students were randomly assigned a treatment order. Normal/ Large Sample: ndiff = 30 ≥ 30. ,Free Response,,,,
-TPS,10.2,45c,"After hearing that students can improve short-term memory by chewing the same flavor of gum while studying for and taking a test, Leila and Valerie designed an experiment to investigate Using 30 volunteers, they randomly assigned 15 to chew gum while they studied a list of 40 words for 90 seconds. Immediately after the 90-second study period—and while chewing the same gum—the subjects wrote down as many words as they could remember. The remaining 15 followed the same procedure without chewing gum. Two weeks later, each of the 30 subjects did the oppo-site treatment. The number of words correctly remembered for each test was recorded for each subject. The 95% confidence interval for the true mean difference (Gum – No gum) in number of words remembered is –0.67 to 1.54. Interpret the confidence level.","If they repeated this experiment many times with these 30 volunteers and constructed a 95% CI using the results of each experiment, about 95% of the intervals would capture the true mean difference (Gum – No gum) in the number of words remembered for students like these. ",Free Response,,,,
-TPS,10.2,45d,"After hearing that students can improve short-term memory by chewing the same flavor of gum while studying for and taking a test, Leila and Valerie designed an experiment to investigate Using 30 volunteers, they randomly assigned 15 to chew gum while they studied a list of 40 words for 90 seconds. Immediately after the 90-second study period—and while chewing the same gum—the subjects wrote down as many words as they could remember. The remaining 15 followed the same procedure without chewing gum. Two weeks later, each of the 30 subjects did the oppo-site treatment. The number of words correctly remembered for each test was recorded for each subject. The 95% confidence interval for the true mean difference (Gum – No gum) in number of words remembered is –0.67 to 1.54. Is there convincing evidence that chewing gum helps subjects like these with short-term memory? Explain your answer.","Because 0 is included in the CI, there is not convincing evidence that chewing gum helps subjects like these with short-term memory.",Free Response,,,,
-TPS,10.2,46a,"Do people get stressed out when other people watch them work? To find out, Sean and Shelby recruited 30 volunteers to take part in an experiment.25 Fifteen of the subjects were randomly assigned to complete a word search puzzle while Sean and Shelby stood close by and visibly took notes. The remaining 15 were assigned to complete a word search puzzle while Sean and Shelby stood at a distance. After each subject completed the word search, they completed a second word search under the opposite treatment. The amount of time required to complete each puzzle was recorded for each subject. Explain why these are paired data.",They result from recording two values for each individual.,Free Response,,,,
-TPS,10.2,46b,"Do people get stressed out when other people watch them work? To find out, Sean and Shelby recruited 30 volunteers to take part in an experiment.25 Fifteen of the subjects were randomly assigned to complete a word search puzzle while Sean and Shelby stood close by and visibly took notes. The remaining 15 were assigned to complete a word search puzzle while Sean and Shelby stood at a distance. After each subject completed the word search, they completed a second word search under the opposite treatment. The amount of time required to complete each puzzle was recorded for each subject. Verify that the conditions for constructing a one-sample t interval for a mean difference are satisfied.",PLAN: One-sample t interval for µdiff. Random: The students were randomly assigned a treatment order. Normal/Large Sample: ndiff = 30 ≥ 30. ,Free Response,,,,
-TPS,10.2,46c,"Do people get stressed out when other people watch them work? To find out, Sean and Shelby recruited 30 volunteers to take part in an experiment.25 Fifteen of the subjects were randomly assigned to complete a word search puzzle while Sean and Shelby stood close by and visibly took notes. The remaining 15 were assigned to complete a word search puzzle while Sean and Shelby stood at a distance. After each subject completed the word search, they completed a second word search under the opposite treatment. The amount of time required to complete each puzzle was recorded for each subject. The 95% confidence interval for the true mean difference (Close by – At a distance) in amount of time needed to complete the puzzle is –12.7 seconds to 119.4 seconds. Interpret the confidence level.","If they conducted this experiment many times with these 30 volunteers and constructed a 95% CI using the results of each experiment, about 95% of the intervals would capture the true mean difference (Close by - At a distance) in the amount of time needed to complete the puzzle for subjects like these. ",Free Response,,,,
-TPS,10.2,46d,"Do people get stressed out when other people watch them work? To find out, Sean and Shelby recruited 30 volunteers to take part in an experiment.25 Fifteen of the subjects were randomly assigned to complete a word search puzzle while Sean and Shelby stood close by and visibly took notes. The remaining 15 were assigned to complete a word search puzzle while Sean and Shelby stood at a distance. After each subject completed the word search, they completed a second word search under the opposite treatment. The amount of time required to complete each puzzle was recorded for each subject. The 95% confidence interval for the true mean difference (Close by – At a distance) in amount of time needed to complete the puzzle is –12.7 seconds to 119.4 seconds. Is there convincing evidence that standing close by causes subjects like these to take longer to complete a word search? Explain your answer.","Because 0 is included in the CI, there is not convincing evidence that standing close by causes subjects like these to take longer to complete a word search.",Free Response,,,,
-TPS,10.2,49,"Researchers suspect that Variety A tomato plants have a different average yield than Variety B tomato plants. To find out, researchers randomly select 10 Variety A and 10 Variety B tomato plants. Then the researchers divide in half each of 10 small plots of land in different locations. For each plot, a coin toss deter-mines which half of the plot gets a Variety A plant; a Variety B plant goes in the other half. After harvest, they compare the yield in pounds for the plants at each location. The 10 differences (Variety A - Variety B) in yield are recorded. A graph of the differences looks roughly symmetric and unimodal with no outliers. The mean difference is x¯(A-B) = 0.34 and the standard deviation of the differences is s(A-B) = 0.83. Let µ(A-B) = the true mean difference (Variety A -Variety B) in yield for tomato plants of these two varieties. Which of the following is the best reason to use a one-sample t interval for a mean difference rather than a two-sample t interval for a difference in means to analyze these data? (a) The number of plots is the same for Variety A and Variety B plants. (b) The response variable, yield of tomatoes, is quantitative. (c) This is an experiment with randomly assigned treatments. (d) Each plot is given both varieties of tomato plant. (e) The sample size is less than 30 for both treatments. ",d,Multiple Choice,,,,
-TPS,10.2,50,"Researchers suspect that Variety A tomato plants have a different average yield than Variety B tomato plants. To find out, researchers randomly select 10 Variety A and 10 Variety B tomato plants. Then the researchers divide in half each of 10 small plots of land in different locations. For each plot, a coin toss deter-mines which half of the plot gets a Variety A plant; a Variety B plant goes in the other half. After harvest, they compare the yield in pounds for the plants at each location. The 10 differences (Variety A - Variety B) in yield are recorded. A graph of the differences looks roughly symmetric and unimodal with no outliers. The mean difference is x¯(A-B) = 0.34 and the standard deviation of the differences is s(A-B) = 0.83. Let µ(A-B) = the true mean difference (Variety A -Variety B) in yield for tomato plants of these two varieties.  A 95% confidence interval for μA − µB is given by  (a) 0.34 ± 1.96(0.83) (b) 0.34 ± 1.96(0.83/sqrt(10)) (c) 0.34 ± 1.812(0.83/sqrt(10)) (d) 0.34 ± 2.262(0.83) (e) 0.34 ± 2.262(0.83/sqrt(10))",e,Multiple Choice,,,,
-TPS,10.2,51,"Jordan wondered if the bean burritos at Restaurant A tend to be heavier than the bean burritos at Restaurant B. To investigate, she visited each restaurant at 10 randomly selected times, ordered a bean burrito, and weighed the burrito. The 95% confidence interval for the difference (A – B) in the mean weight of burrito is 0.06 ounce ± 0.20 ounce. Based on the confidence interval, which conclusion is most appropriate? (a) Because 0 is included in the interval, there is convincing evidence that the mean weight is the same at both restaurants. (b) Because 0 is included in the interval, there isn’t convincing evidence that the mean weight is different at the two restaurants. (c) Because 0.06 is included in the interval, there is convincing evidence that the mean weight is greater at Restaurant A than Restaurant B. (d) Because 0.06 is included in the interval, there isn’t convincing evidence that the mean weight is greater at Restaurant A than Restaurant B. (e) Because there are more positive values in the interval than negative values, there is convincing evidence that the mean weight is greater at Restaurant A than Restaurant B.",b,Multiple Choice,,,,
-TPS,10.2,52,A random sample of 30 words from Jane Austen’s Pride and Prejudice had a mean length of 4.08 letters with a standard deviation of 2.40. A random sample of 30 words from Henry James’s What Maisie Knew had a mean length of 3.85 letters with a standard deviation of 2.26. Which of the following is a correct expression for a 95% confidence interval for the difference in mean word length for these two novels? (a) (4.08 – 3.85) ± 2.576((2.40/sqrt(30)) + (2.26/sqrt(30)) (b) (4.08 – 3.85) ± 2.045((2.40/sqrt(30)) + (2.26/sqrt(30)) (c) (4.08 – 3.85) ± 2.045sqrt((2.40^2/29) + (2.26^2/29) (d) (4.08 – 3.85) ± 2.045sqrt((2.40^2/30) + (2.26^2/30) (e) (4.08 – 3.85) ± 2.576sqrt((2.40^2/30) + (2.26^2/30),d,Multiple Choice,,,,
-TPS,R10,1a,Find the appropriate critical value for constructing a confidence interval in the following setting: Estimating a population mean μ at a 95% confidence level based on an SRS of size 12.,"df = 11, t* = 2.201 ",Free Response,,,,
-TPS,R10,1b,Find the appropriate critical value for constructing a confidence interval in the following setting: Estimating a true mean difference μdiff at a 99% confidence level based on paired data set with 75 differences.,"df = 74, using Table B and df = 60, t* = 2.660. Using technology and df = 74, t* = 2.644. ",Free Response,,,,
-TPS,R10,2a,"A company that produces AA batteries tests the lifetime of a random sample of 30 batteries using a special device designed to imitate real-world use. Based on the testing, the company makes the following statement: “Our AA batteries last an average of 430 to 470 minutes, and our confidence in that interval is 95%.” Determine the point estimate, margin of error, standard error, and sample standard deviation.","Point estimate = (430 1 470)/2 = 450 minutes; margin of error = 470 – 450 = 20. SE = sx / sqrt(30) = 9.780 because 20 = 2.045 sx / sqrt(30). Finally, sx = 53.57 because sx / sqrt(30) = 9.780. ",Free Response,,,,
-TPS,R10,2b,"A company that produces AA batteries tests the lifetime of a random sample of 30 batteries using a special device designed to imitate real-world use. Based on the testing, the company makes the following statement: “Our AA batteries last an average of 430 to 470 minutes, and our confidence in that interval is 95%.” Explain the phrase “our confidence in that interval is 95%.”","If we were to select many samples of 30 batteries from this population and compute a 95% confidence interval for the mean lifetime from each sample, about 95% of these intervals will capture the true mean lifetime of the batteries. ",Free Response,,,,
-TPS,R10,2c,"A company that produces AA batteries tests the lifetime of a random sample of 30 batteries using a special device designed to imitate real-world use. Based on the testing, the company makes the following statement: “Our AA batteries last an average of 430 to 470 minutes, and our confidence in that interval is 95%.” Explain two ways the company could reduce the margin of error.",The company could reduce the margin of error by increasing the sample size or decreasing the confidence level.,Free Response,,,,
-TPS,R10,3a,A random sample of 16 of the more than 200 auto engine crankshafts produced in one day was selected. Here are measurements (in millimeters) of a critical component on these crankshafts: 224.120 224.001 224.017 223.982 223.989 223.961 223.960 224.089 223.987 223.976 223.902 223.980 224.098 224.057 223.913 223.999 Construct and interpret a 95% confidence interval for the mean length of this component on all the crankshafts produced on that day.,"S: m 5 the true mean measurement of the critical dimension for the engine crankshafts produced in one day. P: One-sample t interval. Random: Random sample. 10%: 16 < 10% of all crankshafts produced in one day. Normal/ Large Sample: The histogram shows no strong skewness or outliers. D: x = 224.002, sx= 0.0618, and n= 16. Thus, df = 15 and t* = 2.131. (223.969, 224.035). C: We are 95% confident that the interval from 223.969 to 224.035 mm captures m 5 the true mean measurement of the critical dimension for engine crankshafts produced on this day. ",Free Response,,,,
-TPS,R10,3b,"A random sample of 16 of the more than 200 auto engine crankshafts produced in one day was selected. Here are measurements (in millimeters) of a critical component on these crankshafts: 224.120 224.001 224.017 223.982 223.989 223.961 223.960 224.089 223.987 223.976 223.902 223.980 224.098 224.057 223.913 223.999 The mean length is supposed to be μ = 224 mm but can drift away from this target during production. Does the interval (223.969, 224.035) provide convincing evidence that the mean has drifted from 224 mm? Explain your answer.","Because 224 is in this interval, it is a plausible value for the true mean. We don’t have convincing evidence that the process mean has drifted.",Free Response,,,,
-TPS,R10,4,"The National Assessment of Educational Progress (NAEP) Young Adult Literacy Assessment Survey interviewed separate random samples of 840 men and 1077 women aged 21 to 25 years. 29 The mean and standard deviation of scores on the NAEP’s test of quantitative skills were x¯1 = 272.401 and s1 = 59.21 for the men in the sample. For the women, the results were x¯2 = 274.732 and s2 = 57.52. Construct and interpret a 90% confidence interval for the difference in mean score for male and female young adults.","S: µ1 = true mean NAEP quantitative skills test score for young men and µ2 = true mean NAEP quantitative skills test score for young women. P: Two-sample t interval for µ1 - µ2. Random: It is reasonable to consider the two samples independent because knowing the response of a male shouldn’t predict the response of a female. 10%: n1 = 840 < 10% of all young men and n2 = 10.77 < 10% of all young women. Normal/Large Sample: n1 = 840 ≥ 30 and n2 = 1077 ≥ 30. D: x¯1 = 272.40, s1 = 59.2, n1 = 840, x¯2 = 274.73, s2 = 57.5, n2 =1077. Using df = 1777.52, (26.76, 2.10); using df = 839, (26.76, 2.10). C: We are 90% confident that the interval from 26.76 to 2.10 captures m1 2 m2 5 true difference in the mean NAEP quantitative skills test score for young men and the mean test score for young women. ",Free Response,,,,
-TPS,T10,1,"Anne claims that a store-brand fertilizer works better than homemade compost as a soil enhancement when growing tomatoes. To test her theory, she plants two tomato plants in each of five planters. One plant in each planter is grown in soil with store-brand fertilizer and the other plant is grown in soil with homemade compost, with the choice of soil determined at random. In three months, she will harvest and weigh the tomatoes from each plant. Which of the following is the correct confidence interval Anne should use to analyze these data? (a) Two-sample z interval for μ1 – μ2 (b) Paired z interval for µdiff (c) Two-sample t interval for μ μ−12 (d) Paired t interval for µdiff (e) The correct interval cannot be determined without the data.",d,Multiple Choice,,,,
-TPS,T10,2,"The weights (in pounds) of three adult males are 160, 215, and 195. What is the standard error of the mean for these data? (a) 190 (b) 27.84 (c) 22.73 (d) 16.07 (e) 13.13",d,Multiple Choice,,,,
-TPS,T10,3,You want to compute a 90% confidence interval for the mean difference in height for mothers and their adult daughters using a random sample of 30 mothers who have an adult daughter. What critical value should you use for this interval? (a) 1.645 (b) 1.671 (c) 1.697 (d) 1.699 (e) 1.761,d,Multiple Choice,,,,
-TPS,T10,4,We want to construct a one-sample t interval for a population mean using data from a population with unknown shape. In which of the following circumstances would it be inappropriate to construct the interval based on an SRS of size 14 from the population? (a) A stemplot of the data is roughly bell-shaped. (b) A histogram of the data shows slight skewness. (c) A boxplot shows that the values above the median are much more variable than the values below the median. (d) The sample standard deviation is large. (e) The sample standard deviation is small.,c,Multiple Choice,,,,
-TPS,T10,5,"A 90% confidence interval for the mean μ of a population is computed from a random sample and is found to be 90±30. Which of the following could be the 95% confidence interval based on the same data? (a) 90±21 (b) 90±30 (c) 90±39 (d) 90±70 (e) Without knowing the sample size, any of the above answers could be the 95% confidence interval.",c,Multiple Choice,,,,
-TPS,T10,6,"Do high school seniors with part-time jobs spend less time doing homework per week, on average, than seniors without part-time jobs? For a random sample of 45 seniors with part-time jobs, the mean amount of homework time is 4.2 hours with a standard deviation of 3.8 hours. For a random sample of 45 seniors without part time jobs, the mean amount of homework time is 5.8 hours with a standard deviation of 4.9 hours. Assuming the conditions are met, which of the following is the correct standard error for a 95% confidence interval for a difference in the population means? (a) sqrt(4.9^2 / 45 – 3.8^2 / 45) (b) sqrt(4.9^2 / 45 + 3.8^2 / 45) (c) (4.9-3.8) / sqrt(45) (d) sqrt(5.8^2 / 45 – 4.2^2 / 45) (e) sqrt(5.8^2 / 45 + 4.2^2 / 45)",b,Multiple Choice,,,,
-TPS,T10,7,"Few people enjoy melted ice cream. Being from the sunny state of Arizona, Megan and Jenna decided to test if generic vanilla ice cream melts faster than Breyers vanilla ice cream. At 10 different times during the day and night, the girls put a single scoop of each type of ice cream in the same location out-side and timed how long it took for each scoop to melt completely. When constructing a paired t interval for a mean difference using these data, which of the following distributions should Megan and Jenna check for Normality? i. The distribution of melt time for the generic ice cream ii. The distribution of melt time for the Breyers ice cream iii. The distribution of difference in melt time (a) I only (b) II only (c) III only (d) I and II only (e) I, II, and III",c,Multiple Choice,,,,
-TPS,T10,8,"A Census Bureau report on the income of Americans says that, with 90% confidence, the median income of all U.S. households in a recent year was $57,005 with a margin of error of $742. Which of the following is the most appropriate conclusion? (a) 90% of all households had incomes in the interval $57,005 ± $742. (b) We can be sure that the median income for all households in the country lies in the interval $57,005 ± $742. (c) 90% of the households in the sample interviewed by the Census Bureau had incomes in the interval $57,005 ± $742. (d) The Census Bureau got the result $57,005 ± $742 using a method that will capture the true median income 90% of the time when used repeatedly. (e) 90% of all possible samples of this same size would result in a sample median that falls within $742 of $57,005.",d,Multiple Choice,,,,
-TPS,T10,9,A quiz question gives random samples of n =10 observations from each of two Normally distributed populations. Tom uses a table of t distribution critical values and 9 degrees of freedom to calculate a 95% confidence interval for the difference in the two population means. Janelle uses her calculator’s two-sample t interval with 16.87 degrees of freedom to compute the 95% confidence interval. Assume that both students calculate the intervals correctly. Which of the following is true? (a) Tom’s confidence interval is wider. (b) Janelle’s confidence interval is wider. (c) Both confidence intervals are the same width. (d) There is insufficient information to determine which confidence interval is wider. (e) Janelle made a mistake; degrees of freedom has to be a whole number.,a,Multiple Choice,,,,
-TPS,T10,10,"The makers of a specialty brand of bottled water claim that their “mini” bottles contain 8 ounces of water. To investigate this claim, a consumer advocate randomly selected a sample of 10 bottles and carefully measured the amount of water in each bottle. The mean volume was 7.98 ounces and the 95% confidence interval for true mean volume is 7.93 to 8.03 ounces. Based on the sample, which of the following conclusions best addresses the makers’ claim? (a) Because 7.98 is in the interval, there is convincing evidence that their claim is correct. (b) Because 7.98 is in the interval, there is not convincing evidence that their claim is incorrect. (c) Because 8 is in the interval, there is convincing evidence that their claim is correct. (d) Because 8 is in the interval, there is not convincing evidence that their claim is incorrect. (e) Because 0 is not the interval, there is convincing evidence that their claim is incorrect.",d,Multiple Choice,,,,
-TPS,T10,11a,"Many people have asked the question, but few have been patient enough to collect the data. How many licks does it take to get to the center of a Tootsie Pop? After some intense research, a researcher revealed a 95% confidence interval for the mean number of licks to be 317.64 licks to 394.56 licks. Interpret the confidence level.","If we were to select many random samples of the same size from the population of all Tootsie Pops and construct a 95% confidence interval using each sample, about 95% of the intervals would capture the true mean number of licks required to get to the center. ",Free Response,,,,
-TPS,T10,11b,"Many people have asked the question, but few have been patient enough to collect the data. How many licks does it take to get to the center of a Tootsie Pop? After some intense research, a researcher revealed a 95% confidence interval for the mean number of licks to be 317.64 licks to 394.56 licks. Calculate the point estimate and margin of error used to construct the interval.",The point estimate is (317.64 + 394.56) / 2 = 356.1. The margin of error is 394.56 - 356.1 = 38.46. ,Free Response,,,,
-TPS,T10,11c,"Many people have asked the question, but few have been patient enough to collect the data. How many licks does it take to get to the center of a Tootsie Pop? After some intense research, a researcher revealed a 95% confidence interval for the mean number of licks to be 317.64 licks to 394.56 licks. Name two things the researcher could do to decrease the margin of error. Discuss a drawback of each.",The researcher could reduce the margin of error by decreasing the confidence level. The drawback is that we can’t be as confident that our interval will capture the true mean. The researcher could also reduce the margin of error by increasing the sample size. The drawback is that larger samples cost more time and money to obtain.,Free Response,,,,
-TPS,T10,12,A milk processor monitors the number of bacteria per milliliter in raw milk received at the factory. A random sample of 10 one-milliliter specimens of milk supplied by one producer gives the following data: 5370 4890 5100 4500 5260 5150 4900 4760 4700 4870 Construct and interpret a 90% confidence interval for the population mean μ.,"S: µ = the true mean number of bacteria per milliliter in raw milk received at the factory. P: One-sample t interval. Random: The data come from a random sample. 10%: n= 10 is less than 10% of all 1-milliliter specimens that arrive at the factory. Normal/Large Sample: The dotplot shows that there is no strong skewness or outliers. D: x= 4950.0, sx= 268.5, and n= 10; df = 9 and t* = 1.833. (4794.37, 5105.63). C: We are 90% confident that the interval from 4794.37 to 5105.63 bacteria per ml captures µ = the true mean number of bacteria in the milk received at this factory.",Free Response,,,,
-TPS,11.1,1,The Survey of Study Habits and Attitudes (SSHA) is a psychological test with scores that range from 0 to 200. The mean score for U.S. college students is 115. A teacher suspects that older students have better attitudes toward school. She gives the SSHA to an SRS of 45 students from the more than 1000 students at her college who are at least 30 years of age. The teacher wants to perform a test at the α= 0.05 significance level of  H0: μ = 115; Ha: μ > 115 where μ = the mean SSHA score in the population of students at her college who are at least 30 years old. Check if the conditions for performing the test are met.,Random: Random sample of students. 10%: 45 < 10% of 1000. Normal/Large Sample: n= 45 ≥ 30. ,Free Response,,,,
-TPS,11.1,2,A machine is supposed to fill bags with an average of 19.2 ounces of candy. The manager of the candy factory wants to be sure that the machine does not consistently underfill or overfill the bags. So the manager plans to conduct a significance test at the α = 0.10 significance level of  H0: μ = 19.2; Ha: μ ≠ 19.2 where μ = the true mean amount of candy (in ounces) that the machine put in all bags filled that day. The manager takes a random sample of 75 bags of candy produced that day and weighs each bag. Check if the conditions for performing the test are met.,Random: Random sample of bags. 10%: Assume 75 < 10% of population. Normal/Large Sample: n= 75 ≥ 30. ,Free Response,,,,
-TPS,11.1,3a,A tablet computer manufacturer claims that its batteries last an average of 11.5 hours when playing videos. The quality-control department randomly selects 20 tablets from each day’s production and tests the fully charged batteries by playing a video repeatedly until the battery dies. The quality-control department will discard the batteries from that day’s production run if they find convincing evidence that the mean battery life is less than 11.5 hours. Here are the summary statistics of the data from one day: n= 20; Mean=11.07; SD= 1.097; Min= 10; Q1= 10.3; Med= 10.6; Q3= 11.85; Max= 13.9. State appropriate hypotheses for the quality-control department to test. Be sure to define your parameter.,"H0: µ = 11.5, Ha: µ < 11.5, where µ = the true mean battery life when playing videos for all tablets. ",Free Response,,,,
-TPS,11.1,3b,A tablet computer manufacturer claims that its batteries last an average of 11.5 hours when playing videos. The quality-control department randomly selects 20 tablets from each day’s production and tests the fully charged batteries by playing a video repeatedly until the battery dies. The quality-control department will discard the batteries from that day’s production run if they find convincing evidence that the mean battery life is less than 11.5 hours. Here are the summary statistics of the data from one day: n= 20; Mean=11.07; SD= 1.097; Min= 10; Q1= 10.3; Med= 10.6; Q3= 11.85; Max= 13.9. Check if the conditions for performing the test are met.,Random: Random sample. 10%: 20 < 10% of population. Normal/Large Sample: Not met because the sample size is less than 30 and the dotplot of the distribution of battery life is strongly skewed to the right. ,Free Response,,,,
-TPS,11.1,4a,"A retailer entered into an exclusive agreement with a supplier who guaranteed to provide all products at competitive prices. To be sure the supplier honored the terms of the agreement, the retailer had an audit performed on a random sample of 25 invoices. The percent of purchases on each invoice for which an alternative supplier offered a lower price than the original supplier was recorded. 4 For example, a data value of 38 means that the price would be lower with a different supplier for 38% of the items on the invoice. Some numerical summaries of the data are shown here. The retailer would like to determine if there is convincing evidence that the mean percent of purchases for which an alternative supplier offered lower prices is greater than 50% in the population of this company’s invoices. n= 25; Mean 77.76; SD 32.6768; Min 0; Q1 68; Med 100; Q3 100; Max 100. State appropriate hypotheses for the retailer’s test. Be sure to define your parameter.","H0: µ = 50, Ha: µ > 50, where µ = the true mean percent of purchases for which an alternative supplier offered lower prices. ",Free Response,,,,
-TPS,11.1,4b,"A retailer entered into an exclusive agreement with a supplier who guaranteed to provide all products at competitive prices. To be sure the supplier honored the terms of the agreement, the retailer had an audit performed on a random sample of 25 invoices. The percent of purchases on each invoice for which an alternative supplier offered a lower price than the original supplier was recorded. 4 For example, a data value of 38 means that the price would be lower with a different supplier for 38% of the items on the invoice. Some numerical summaries of the data are shown here. The retailer would like to determine if there is convincing evidence that the mean percent of purchases for which an alternative supplier offered lower prices is greater than 50% in the population of this company’s invoices. n= 25; Mean 77.76; SD 32.6768; Min 0; Q1 68; Med 100; Q3 100; Max 100. Check if the conditions for performing the test are met.",Random: Random sample. 10%: 25 < 10% of population. Normal/Large Sample: Not met because the sample size is less than 30 and the histogram of the data is strongly skewed to the left.,Free Response,,,,
-TPS,11.1,5a,Suppose you want to perform a test of H0: μ = 64 versus Ha: μ ≠ 64 at the α = 0.05 significance level. A random sample of size n = 25 from the population of interest yields x¯ = 62.8 and sx=5.36. Assume that the conditions for carrying out the test are met. Explain why the sample result gives some evidence for the alternative hypothesis.,x¯= 62.8 ≠ 64 ,Free Response,,,,
-TPS,11.1,5b,Suppose you want to perform a test of H0: μ = 64 versus Ha: μ ≠ 64 at the α = 0.05 significance level. A random sample of size n = 25 from the population of interest yields x¯ = 62.8 and sx=5.36. Assume that the conditions for carrying out the test are met. Calculate the standardized test statistic and P -value.,t= (62.8 – 64) / (5.36 / sqrt(25) = -1.12; P-value: df = 24 Table B: Between 0.20 and 0.30; Tech: 0.2738 ,Free Response,,,,
-TPS,11.1,6a,Suppose you want to perform a test of H0: μ =5 versus Ha: α < 5 at the α = 0.05 significance level. A random sample of size n= 20 population of interest yields x¯=4.7 and sx=.47. Assume that the conditions for carrying out the test are met. Explain why the sample result gives some evidence for the alternative hypothesis.,x¯= 4.7 < 5 ,Free Response,,,,
-TPS,11.1,6b,Suppose you want to perform a test of H0: μ =5 versus Ha: α < 5 at the α = 0.05 significance level. A random sample of size n= 20 population of interest yields x¯=4.7 and sx=.47. Assume that the conditions for carrying out the test are met. Calculate the standardized test statistic and P -value.,t= (4.7 – 5) / (0.74 / sqrt(20) = -1.81; P-value: df = 19 Table B: Between 0.025 and 0.05; Tech: 0.0431,Free Response,,,,
-TPS,11.1,7a,The Survey of Study Habits and Attitudes (SSHA) is a psychological test with scores that range from 0 to 200. The mean score for U.S. college students is 115. A teacher suspects that older students have better attitudes toward school. She gives the SSHA to an SRS of 45 students from the more than 1000 students at her college who are at least 30 years of age. The teacher wants to perform a test at the α= 0.05 significance level of H0: μ = 115; Ha: μ > 115 where μ = the mean SSHA score in the population of students at her college who are at least 30 years old. The sample mean SSHA score was 125.7 and the sample standard deviation was 29.8. Calculate the standardized test statistic.,t= (125.7 – 115) / (29.8 / sqrt(45) = 2.41,Free Response,,,,
-TPS,11.1,7b,The Survey of Study Habits and Attitudes (SSHA) is a psychological test with scores that range from 0 to 200. The mean score for U.S. college students is 115. A teacher suspects that older students have better attitudes toward school. She gives the SSHA to an SRS of 45 students from the more than 1000 students at her college who are at least 30 years of age. The teacher wants to perform a test at the α= 0.05 significance level of H0: μ = 115; Ha: μ > 115 where μ = the mean SSHA score in the population of students at her college who are at least 30 years old. The sample mean SSHA score was 125.7 and the sample standard deviation was 29.8. Find and interpret the P -value. ,"P-value: df = 44 Table B: Between 0.01 and 0.02; Tech: 0.0101. Assuming that the true mean SSHA score for older students is 115, there is a 0.0101 probability of getting a sample mean of at least 125.7 by chance alone. ",Free Response,,,,
-TPS,11.1,7c,The Survey of Study Habits and Attitudes (SSHA) is a psychological test with scores that range from 0 to 200. The mean score for U.S. college students is 115. A teacher suspects that older students have better attitudes toward school. She gives the SSHA to an SRS of 45 students from the more than 1000 students at her college who are at least 30 years of age. The teacher wants to perform a test at the α= 0.05 significance level of H0: μ = 115; Ha: μ > 115 where μ = the mean SSHA score in the population of students at her college who are at least 30 years old. The sample mean SSHA score was 125.7 and the sample standard deviation was 29.8. What conclusion would you make?,"Because the P-value of 0.0101 < α = 0.05, we reject H0. We have convincing evidence that the true mean SSHA score in the population of students at her college who are at least 30 years old is greater than 115.",Free Response,,,,
-TPS,11.1,8a,A machine is supposed to fill bags with an average of 19.2 ounces of candy. The manager of the candy factory wants to be sure that the machine does not consistently underfill or overfill the bags. So the manager plans to conduct a significance test at the α = 0.10 significance level of H0: μ = 19.2; Ha: μ ≠ 19.2 where μ = the true mean amount of candy (in ounces) that the machine put in all bags filled that day. The manager takes a random sample of 75 bags of candy produced that day and weighs each bag.  The sample mean weight for the bags of candy was 19.28 ounces and the sample standard deviation was 0.81 ounce. Calculate the standardized test statistic.,t= (19.28 - 19.2) / (0.81 / sqrt(75) = 0.86,Free Response,,,,
-TPS,11.1,8b,A machine is supposed to fill bags with an average of 19.2 ounces of candy. The manager of the candy factory wants to be sure that the machine does not consistently underfill or overfill the bags. So the manager plans to conduct a significance test at the α = 0.10 significance level of H0: μ = 19.2; Ha: μ ≠ 19.2 where μ = the true mean amount of candy (in ounces) that the machine put in all bags filled that day. The manager takes a random sample of 75 bags of candy produced that day and weighs each bag.  The sample mean weight for the bags of candy was 19.28 ounces and the sample standard deviation was 0.81 ounce. Find and interpret the P -value. ,"P-value: df = 74 Table B: Using df = 60, between 2(0.15) 5 0.30 and 2(0.20) = 0.40; Tech: 0.3926. Assuming that the true mean weight of candy bags is 19.2 ounces, there is a 0.3926 probability of getting a sample mean as different from 19.2 as 19.28 by chance alone.",Free Response,,,,
-TPS,11.1,8c,A machine is supposed to fill bags with an average of 19.2 ounces of candy. The manager of the candy factory wants to be sure that the machine does not consistently underfill or overfill the bags. So the manager plans to conduct a significance test at the α = 0.10 significance level of H0: μ = 19.2; Ha: μ ≠ 19.2 where μ = the true mean amount of candy (in ounces) that the machine put in all bags filled that day. The manager takes a random sample of 75 bags of candy produced that day and weighs each bag.  The sample mean weight for the bags of candy was 19.28 ounces and the sample standard deviation was 0.81 ounce. What conclusion would you make? ,"Because the P-value of 0.3926 < α = 0.10, we fail to reject H0. We do not have convincing evidence that the true mean amount of candy (in ounces) that the machine put in all bags filled that day is different from 19.2.",Free Response,,,,
-TPS,11.1,9a,"Every road has one at some point—construction zones that have much lower speed limits. To see if drivers obey these lower speed limits, a police officer uses a radar gun to measure the speed (in miles per hour, or mph) of a random sample of 10 drivers in a 25 mph construction zone. Here are the data: 27 33 32 21 30 30 29 25 27 34. Is there convincing evidence at the α = 0.01 significance level that the average speed of drivers in this construction zone is greater than the posted speed limit?","S: H0: µ = 25, Ha: µ > 25, where µ = the true mean speed of all drivers in a construction zone using α = 0.01. P: One-sample t test for µ. Random: Random sample. 10%: 10 < 10% of population. Normal/Large Sample: There is no strong skewness or outliers in the sample. D: x= 28.8, sx= 3.94, t= 3.05, df = 9, P-value = 0.0069. C: Because the P-value of 0.0069 < α = 0.01, we reject H0. We have convincing evidence that the true mean speed of all drivers in the construction zone is greater than 25 mph. ",Free Response,,,,
-TPS,11.1,9b,"Every road has one at some point—construction zones that have much lower speed limits. To see if drivers obey these lower speed limits, a police officer uses a radar gun to measure the speed (in miles per hour, or mph) of a random sample of 10 drivers in a 25 mph construction zone. Here are the data: 27 33 32 21 30 30 29 25 27 34.A test results in a conclusion to reject Ho. Which kind of mistake—a Type I error or a Type II error—could you have made? Explain what this mistake would mean in context.",Type I error: Finding convincing evidence that the true mean speed is greater than 25 mph when it really isn’t. ,Free Response,,,,
-TPS,11.1,10a,A study was carried out with a random sample of 10 patients who suffer from insomnia to investigate the effectiveness of a drug designed to increase sleep time. The following data show the number of additional hours of sleep per night gained by each subject after taking the drug. 5 A negative value indicates that the subject got less sleep after taking the drug. 1.9 0.8 1.1 0.1 −0.1 4.4 5.5 1.6 4.6 3.4. Is there convincing evidence at the α = 0.01 significance level that the average sleep increase is greater than 0 for insomnia patients when taking this drug?,"S: H0: α = 0, Ha: µ > 0, where µ = the true mean number of additional hours of sleep per night gained by using the drug for all people who would take it using α = 0.01. P: One-sample t test for µ. Random: Random sample. 10%: 10 < 10% of population. Normal/ Large Sample: The dotplot doesn’t show any outliers or strong skewness. D: x= 2.33, sx= 2.002, t= 3.68, df = 9, P-value = 0.00254. C: Because the P-value of 0.00254 < α = 0.05, we reject H0. There is convincing evidence that the drug is effective at increasing the average sleep time for patients who suffer from insomnia. ",Free Response,,,,
-TPS,11.1,10b,A study was carried out with a random sample of 10 patients who suffer from insomnia to investigate the effectiveness of a drug designed to increase sleep time. The following data show the number of additional hours of sleep per night gained by each subject after taking the drug. 5 A negative value indicates that the subject got less sleep after taking the drug. 1.9 0.8 1.1 0.1 −0.1 4.4 5.5 1.6 4.6 3.4. A test results in a conclusion to reject H0. Which kind of mistake—a Type I error or a Type II error—could you have made? Explain what this mistake would mean in context.,Type I error: Finding convincing evidence that the true mean sleep increase is greater than 0 when it really isn’t. ,Free Response,,,,
-TPS,11.1,11,"A school librarian purchases a novel for her library. The publisher claims that the book is written at a fifth-grade reading level, but the librarian suspects that the reading level is lower than that. The librarian selects a random sample of 40 pages and uses a standard readability test to assess the reading level of each page. The mean reading level of these pages is 4.8 with a standard deviation of 0.8. Do these data give convincing evidence at the α = 0.05 significance level that the average reading level of this novel is less than 5?","S: H0: µ = 5, Ha: µ < 5, where µ = the true mean reading level of all pages in this novel using α = 0.05. P: One-sample t test for µ. Random: Random sample. 10%: Assume 40 < 10% of population. Normal/ Large Sample: n= 40 ≥ 30. D: x= 4.8, sx= 0.8, t= -1.58, df = 39, and P-value = 0.0610. C: Because the P-value of 0.0610 > α = 0.05, we fail to reject H0. There is not convincing evidence that the true mean reading level for this novel is less than 5.",Free Response,,,,
-TPS,11.1,12,"One company’s bottles of grapefruit juice are filled by a machine that is set to dispense an average of 180 milliliters (ml) of liquid. The company has been getting negative feedback from customers about underfilled bottles. To investigate, a quality-control inspector takes a random sample of 40 bottles and measures the volume of liquid in each bottle. The mean amount of liquid in the bottles is 179.6 ml and the standard deviation is 1.3 ml. Do these data provide convincing evidence at the α = 0.05 significance level that the machine is underfilling the bottles, on average?","S: H0: µ = 180, Ha: µ < 180, where µ = the true mean amount (ml) of grapefruit juice dispensed using α = 0.05. P: One-sample t test for µ. Random: Random sample. 10%: Assume 40 < 10% of population. Normal/Large Sample: n= 40 ≥ 30. D: x= 179.6, sx= 1.3, t= -1.95, df = 39, P-value = 0.0292. C: Because the P-value of 0.0292 < α = 0.05, we reject H0. There is convincing evidence that the true mean amount of grapefruit juice dispensed is less than 180 ml.",Free Response,,,,
-TPS,11.1,13,A drug manufacturer forms tablets by compressing a granular material that contains the active ingredient and various fillers. The hardness of a sample from each batch of tablets produced is measured to control the compression process. The target value for the hardness is μ =11.5. The hardness data for a random sample of 20 tablets from one large batch are 11.627 11.374 11.383 11.477 11.613 11.592 11.715 11.570 11.493 11.458 11.485 11.623 11.602 11.552 11.509 11.472 11.360 11.463 11.429 11.531 Is there convincing evidence at the 5% level that the mean hardness of the tablets in this batch differs from the target value?,"S: H0: µ = 11.5, Ha: µ ≠ 11.5, where µ = the true mean hardness of the tablets using α = 0.05. P: One-sample t test for µ. Random: Random sample. 10%: 20 < 10% of population. Normal/Large Sample: There is no strong skewness or outliers in the sample. D: x= 11.5164, sx= 0.095, t= 0.77, df = 19, and P-value = 0.4494. C: Because the P-value of 0.4494 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true mean hardness of these tablets is different from 11.5.",Free Response,,,,
-TPS,11.1,14,"Student researchers Haley, Jeff, and Nathan saw an article on the Internet claiming that the average vertical jump for teens was 15 inches. They wondered if the average vertical jump of students at their school differed from 15 inches, so they obtained a list of student names and selected a random sample of 20 students. After contacting these students several times, they finally convinced them to allow their vertical jumps to be measured. Here are the data (in inches): 11.0 11.5 12.5 26.5 15.0 12.5 22.0 15.0 13.5 12.0 23.0 19.0 15.5 21.0 12.5 23.0 20.0 8.5 25.5 20.5 Do these data provide convincing evidence at the α = 0.10 level that the average vertical jump of students at this school differs from 15 inches?","S: H0: µ = 15, Ha: µ ≠ 15, where µ = the true mean vertical jump of all students at this school using a 5 0.10. P: One-sample t test for m. Random: Random sample. 10%: 20 < 10% of population. Normal/Large Sample: There is no strong skewness or outliers in the sample, so it is reasonable to use a t procedure. D: x¯ = 17, sx= 5.368, t= 1.67, df = 19, and P-value = 0.1121. C: Because the P-value of 0.1121 > α = 0.10, we fail to reject H0. We do not have convincing evidence that the true mean vertical jump distance for all students at this school is different from 15 inches.",Free Response,,,,
-TPS,11.1,15a,"A drug manufacturer forms tablets by compressing a granular material that contains the active ingredient and various fillers. The hardness of a sample from each batch of tablets produced is measured to control the compression process. The target value for the hardness is μ =11.5. The hardness data for a random sample of 20 tablets from one large batch are 11.627 11.374 11.383 11.477 11.613 11.592 11.715 11.570 11.493 11.458 11.485 11.623 11.602 11.552 11.509 11.472 11.360 11.463 11.429 11.531  A 95% confidence interval for the true mean hardness μ is (11.472, 11.561). Explain how this interval gives more information than a significance test.","The test in Exercise 13 only allowed us to fail to reject H0: µ = 11.5. The confidence interval tells us that any value of m between 11.472 and 11.561 is plausible based on the sample data. This is consistent with, but gives more information than, the test in Exercise 13. ",Free Response,,,,
-TPS,11.1,15b,A drug manufacturer forms tablets by compressing a granular material that contains the active ingredient and various fillers. The hardness of a sample from each batch of tablets produced is measured to control the compression process. The target value for the hardness is μ =11.5. The hardness data for a random sample of 20 tablets from one large batch are 11.627 11.374 11.383 11.477 11.613 11.592 11.715 11.570 11.493 11.458 11.485 11.623 11.602 11.552 11.509 11.472 11.360 11.463 11.429 11.531  The power of the test to detect that μ =11.55 is 0.61. Interpret this value.,"If the true mean hardness of the tablets is = 11.55, there is a 0.61 probability that the drug manufacturer will find convincing evidence for Ha: µ ≠ 11.5. ",Free Response,,,,
-TPS,11.1,15c,A drug manufacturer forms tablets by compressing a granular material that contains the active ingredient and various fillers. The hardness of a sample from each batch of tablets produced is measured to control the compression process. The target value for the hardness is μ =11.5. The hardness data for a random sample of 20 tablets from one large batch are 11.627 11.374 11.383 11.477 11.613 11.592 11.715 11.570 11.493 11.458 11.485 11.623 11.602 11.552 11.509 11.472 11.360 11.463 11.429 11.531  Find the probability of a Type II error if μ =11.55.,1 - 0.61 = 0.39 ,Free Response,,,,
-TPS,11.1,15d,A drug manufacturer forms tablets by compressing a granular material that contains the active ingredient and various fillers. The hardness of a sample from each batch of tablets produced is measured to control the compression process. The target value for the hardness is μ =11.5. The hardness data for a random sample of 20 tablets from one large batch are 11.627 11.374 11.383 11.477 11.613 11.592 11.715 11.570 11.493 11.458 11.485 11.623 11.602 11.552 11.509 11.472 11.360 11.463 11.429 11.531  The probability of a Type II error was found to be 0.39. Describe two ways to decrease this probability.,Increase the sample size or use a larger significance level. ,Free Response,,,,
-TPS,11.1,16a,"Student researchers Haley, Jeff, and Nathan saw an article on the Internet claiming that the average vertical jump for teens was 15 inches. They wondered if the average vertical jump of students at their school differed from 15 inches, so they obtained a list of student names and selected a random sample of 20 students. After contacting these students several times, they finally convinced them to allow their vertical jumps to be measured. Here are the data (in inches): 11.0 11.5 12.5 26.5 15.0 12.5 22.0 15.0 13.5 12.0 23.0 19.0 15.5 21.0 12.5 23.0 20.0 8.5 25.5 20.5  A 90% confidence interval for the true mean vertical jump μ is (14.924, 19.076). Explain how this interval gives more information than a significance test.","The test in Exercise 14 only allowed us to fail to reject H0: µ = 15. The confidence interval tells us that any value of m between 14.472 and 19.076 is plausible based on the sample data. This is consistent with, but gives more information than, the test in Exercise 14. ",Free Response,,,,
-TPS,11.1,16b,"Student researchers Haley, Jeff, and Nathan saw an article on the Internet claiming that the average vertical jump for teens was 15 inches. They wondered if the average vertical jump of students at their school differed from 15 inches, so they obtained a list of student names and selected a random sample of 20 students. After contacting these students several times, they finally convinced them to allow their vertical jumps to be measured. Here are the data (in inches): 11.0 11.5 12.5 26.5 15.0 12.5 22.0 15.0 13.5 12.0 23.0 19.0 15.5 21.0 12.5 23.0 20.0 8.5 25.5 20.5  The power of the test to detect that μ =17 inches is 0.49. Interpret this value.","If the true mean vertical jump of the students at this school is µ = 17 inches, there is a 0.49 probability that the student researchers will find convincing evidence for Ha: µ ≠ 15. ",Free Response,,,,
-TPS,11.1,16c,"Student researchers Haley, Jeff, and Nathan saw an article on the Internet claiming that the average vertical jump for teens was 15 inches. They wondered if the average vertical jump of students at their school differed from 15 inches, so they obtained a list of student names and selected a random sample of 20 students. After contacting these students several times, they finally convinced them to allow their vertical jumps to be measured. Here are the data (in inches): 11.0 11.5 12.5 26.5 15.0 12.5 22.0 15.0 13.5 12.0 23.0 19.0 15.5 21.0 12.5 23.0 20.0 8.5 25.5 20.5  Find the probability of a Type II error if μ =17. ",1 - 0.49 = 0.51 ,Free Response,,,,
-TPS,11.1,16d,"Student researchers Haley, Jeff, and Nathan saw an article on the Internet claiming that the average vertical jump for teens was 15 inches. They wondered if the average vertical jump of students at their school differed from 15 inches, so they obtained a list of student names and selected a random sample of 20 students. After contacting these students several times, they finally convinced them to allow their vertical jumps to be measured. Here are the data (in inches): 11.0 11.5 12.5 26.5 15.0 12.5 22.0 15.0 13.5 12.0 23.0 19.0 15.5 21.0 12.5 23.0 20.0 8.5 25.5 20.5  The probability of a Type II error was found to be 0.51. Describe two ways to decrease this probability.",Increase the sample size or use a larger significance level. ,Free Response,,,,
-TPS,11.1,17a,"How long does it take for a chunk of information to travel from one server to another and back on the Internet? According to the site internettrafficreport.com, the average response time is 200 milliseconds (about one-fifth of a second). Researchers wonder if this claim is true, so they collect data on response times (in milliseconds) for a random sample of 14 servers in Europe. A graph of the data reveals no strong skewness or outliers. State an appropriate pair of hypotheses for a significance test in this setting. Be sure to define the parameter of interest.","H0: µ = 200, Ha: α ≠ 200, where µ = the true mean response time of European servers (in milliseconds). ",Free Response,,,,
-TPS,11.1,17b,"How long does it take for a chunk of information to travel from one server to another and back on the Internet? According to the site internettrafficreport.com, the average response time is 200 milliseconds (about one-fifth of a second). Researchers wonder if this claim is true, so they collect data on response times (in milliseconds) for a random sample of 14 servers in Europe. A graph of the data reveals no strong skewness or outliers. Check conditions for performing a significance test.",Random: Random sample. 10%: 14 < 10% of population. Normal/Large Sample: A graph of the data reveals no strong skewness or outliers. ,Free Response,,,,
-TPS,11.1,17c,"How long does it take for a chunk of information to travel from one server to another and back on the Internet? According to the site internettrafficreport.com, the average response time is 200 milliseconds (about one-fifth of a second). Researchers wonder if this claim is true, so they collect data on response times (in milliseconds) for a random sample of 14 servers in Europe. A graph of the data reveals no strong skewness or outliers. The 95% confidence interval for the mean response time is 158.22 to 189.64 milliseconds. Based on this interval, what conclusion would you make for a test of the hypotheses of a significance test at the 5% significance level?","Because the 95% confidence interval does not contain 200, we reject H0 at the α = 0.05 significance level. We have convincing evidence that the mean response time of European servers is different from 200 milliseconds. ",Free Response,,,,
-TPS,11.1,17d,"How long does it take for a chunk of information to travel from one server to another and back on the Internet? According to the site internettrafficreport.com, the average response time is 200 milliseconds (about one-fifth of a second). Researchers wonder if this claim is true, so they collect data on response times (in milliseconds) for a random sample of 14 servers in Europe. A graph of the data reveals no strong skewness or outliers. Do we have convincing evidence that the mean response time of servers in the United States is different from 200 milliseconds? Justify your answer.",No! We cannot draw any conclusions about the United States because we only collected information from a random sample of servers in Europe. ,Free Response,,,,
-TPS,11.1,18a,"A blogger claims that U.S. adults drink an average of 40 ounces (that’s five 8-ounce glasses) of water per day. Researchers wonder if this claim is true, so they ask a random sample of 24 U.S. adults about their daily water intake. A graph of the data shows a roughly symmetric shape with no outliers. State an appropriate pair of hypotheses for a significance test in this setting. Be sure to define the parameter of interest.","H0: µ = 5, Ha: µ ≠ 5, where µ = the true mean number of 8-ounce glasses of water that a U.S. adult drinks per day, or H0: µ = 40 vs. Ha: µ ≠ 40, where µ = the true mean daily water intake (in ounces) for all U.S. adults. ",Free Response,,,,
-TPS,11.1,18b,"A blogger claims that U.S. adults drink an average of 40 ounces (that’s five 8-ounce glasses) of water per day. Researchers wonder if this claim is true, so they ask a random sample of 24 U.S. adults about their daily water intake. A graph of the data shows a roughly symmetric shape with no outliers. Check conditions for performing a significance test. ",Random: Random sample. 10%: 24 < 10% of population. Normal/Large Sample: A graph shows a roughly symmetric shape with no outliers. ,Free Response,,,,
-TPS,11.1,18c,"A blogger claims that U.S. adults drink an average of 40 ounces (that’s five 8-ounce glasses) of water per day. Researchers wonder if this claim is true, so they ask a random sample of 24 U.S. adults about their daily water intake. A graph of the data shows a roughly symmetric shape with no outliers. The 90% confidence interval for the mean daily water intake is 30.35 to 36.92 ounces. Based on this interval, what conclusion would you make for a test of the hypotheses at the 10% significance level?","Because the 90% confidence interval does not contain 40 ounces, we reject H0 at the α = 0.10 significance level. We have convincing evidence that the true mean daily water intake for U.S. adults is different from 40 ounces. ",Free Response,,,,
-TPS,11.1,18d,"A blogger claims that U.S. adults drink an average of 40 ounces (that’s five 8-ounce glasses) of water per day. Researchers wonder if this claim is true, so they ask a random sample of 24 U.S. adults about their daily water intake. A graph of the data shows a roughly symmetric shape with no outliers. Do we have convincing evidence that the amount of water U.S. children drink per day differs from 40 ounces? Justify your answer.",No! We cannot draw any conclusions about the mean amount of water U.S. children drink per day because we only collected information from a random sample of U.S. adults. ,Free Response,,,,
-TPS,11.1,19a,The P -value for a two-sided test of the null hypothesis H0: μ = 10 is 0.06. Does the 95% confidence interval for μ include 10? Why or why not?,"Yes; because the P-value of 0.06 > α = 0.05, we fail to reject H0: µ = 10 at the 5% level of significance. The 95% confidence interval will include 10. ",Free Response,,,,
-TPS,11.1,19b,The P -value for a two-sided test of the null hypothesis H0: μ = 10 is 0.06. Does the 90% confidence interval for μ include 10? Why or why not?,"No; because the P-value of 0.06 < α = 0.10, we reject H0: µ = 10 at the 10% level of significance. The 90% confidence interval would not include 10 as a plausible value. ",Free Response,,,,
-TPS,11.1,20a,The P -value for a two-sided test of the null hypothesis H0: μ =15is 0.03. Does the 99% confidence interval for μ include 15? Why or why not?,"Yes; because the P-value of 0.03 > α = 0.01, we fail to reject H0: µ = 15 at the 1% level of significance. The 99% confidence interval will include 15. ",Free Response,,,,
-TPS,11.1,20b,The P -value for a two-sided test of the null hypothesis H0: μ =15is 0.03. Does the 95% confidence interval for μ include 15? Why or why not?,"No; because the P-value of 0.03 < α = 0.05, we reject H0: µ = 15 at the 5% level of significance. The 95% confidence interval would not include 15 as a plausible value.",Free Response,,,,
-TPS,11.1,21a,A researcher looking for evidence of extrasensory perception (ESP) tests 500 subjects. Four of these subjects do significantly better (P<0.01) than random guessing. Is it proper to conclude that these four people have ESP? Explain your answer.," No; in a sample of size n= 500, we expect to see about (500)(0.01) = 5 people who do better than random guessing, with a significance level of 0.01. These four might have ESP, or they may simply be among the “lucky” ones we expect to see just by chance. ",Free Response,,,,
-TPS,11.1,21b,A researcher looking for evidence of extrasensory perception (ESP) tests 500 subjects. Four of these subjects do significantly better (P<0.01) than random guessing. What should the researcher now do to test whether any of these four subjects has ESP?,The researcher should repeat the procedure on these four people to see if they again perform well. ,Free Response,,,,
-TPS,11.1,22,"A medical experiment investigated whether taking the herb echinacea could help prevent colds. The study measured 50 different response variables usually associated with colds, such as low-grade fever, congestion, frequency of coughing, and so on. At the end of the study, those taking echinacea displayed significantly better responses at the α = 0.05 level than those taking a placebo for 3 of the 50 response variables studied. Should we be convinced that echinacea helps prevent colds? Why or why not?","At the α = 0.05 level, we expect to have a Type I error (getting a significant result when the null hypothesis is true) 5% of the time. If 50 tests are performed at this significance level and all 50 null hypotheses were true, we would expect, on average, (50)(0.05) = 2.5 Type I errors. The three significant results in this situation may be such errors.",Free Response,,,,
-TPS,11.1,23,"A national chain of SAT-preparation schools wants to know if using a smart-phone app in addition to its regular program will help increase student scores more than using just the regular program. On average, the students in the regular program increase their scores by 128 points during the 3-month class. To investigate using the smartphone app, the prep schools have 5000 students use the app along with the regular program and measure their improvement. Then the schools will test the following hypotheses: H0: μ =128 versus Ha: μ > 128, where µ is the true mean improvement in the SAT score for students who attend these prep schools. After 3 months, the average improvement was x¯ =130 with a standard deviation of sx = 65. The standardized test statistic is t = 2.18 with a P-value of 0.0148. Explain why this result is statistically significant, but not practically important.","Although the hypothesis test shows that the results are statistically significant (we have convincing evidence that µ > 128), this is not practically significant. A test with such a large sample size will often produce a significant result for a very small departure from the null value. There is little practical significance to an increase in average SAT score of only 2 points. ",Free Response,,,,
-TPS,11.1,24,"Music and mazes A researcher wishes to determine if people are able to complete a certain pencil and paper maze more quickly while listening to classical music. Suppose previous research has established that the mean time needed for people to complete a certain maze (without music) is 40 seconds. The researcher decides to test the hypotheses H0: µ = 40 versus Ha: µ < 40, where μ = the average time in seconds to complete the maze while listening to classical music. To do so, the researcher has 10,000 people complete the maze with classical music playing. The mean time for these people is x¯ = 39.92 seconds, and the P-value of his significance test is 0.0002. Explain why this result is statistically significant, but not practically important.","Although the hypothesis test shows that the results are statistically significant (we have convincing evidence that µ < 40), this is not practically significant. A test with such a large sample size will often produce a significant result for a very small departure from the null value. There is little practical significance to a change in average finish time of 0.08 second.",Free Response,,,,
-TPS,11.1,25,"A marketing consultant observes 50 consecutive shoppers at a supermarket, recording how much each shopper spends in the store. Explain why it would not be wise to use these data to carry out a significance test about the mean amount spent by all shoppers at this supermarket.","It would not be wise to use these data to carry out a significance test about the mean amount spent by all shoppers at the supermarket because this sample wasn’t randomly selected—it was a convenience sample. Depending on the time of day or the day of the week, certain types of shoppers may be underrepresented. ",Free Response,,,,
-TPS,11.1,26,"Joe is writing a report on the backgrounds of American presidents. He looks up the ages of all the presidents when they entered office. Because Joe took a statistics course, he uses these numbers to perform a significance test about the mean age of all U.S. presidents. Explain why this makes no sense.",It makes no sense to perform a significance test about the mean age of all U.S. presidents because Joe looked up the ages of every president when he entered office. This is not information taken from a sample. We have information about all presidents—the whole population of interest—so Joe can calculate the exact value of µ.,Free Response,,,,
-TPS,11.1,27,The reason we use t procedures instead of z procedures when carrying out a test about a population mean is that (a) z requires that the sample size be large. (b) z requires that you know the population standard deviation σ. (c) z requires that the data come from a random sample. (d) z requires that the population distribution be Normal. (e) z can only be used for proportions. ,b,Multiple Choice,,,,
-TPS,11.1,28,You are testing H0: μ =75 against Ha: µ < 75 based on an SRS of 20 observations from a Normal population. The t statistic is =−2.25. The P-value (a) falls between 0.01 and 0.02. (b) falls between 0.02 and 0.04. (c) falls between 0.04 and 0.05. (d) falls between 0.05 and 0.25. (e) is greater than 0.25. ,a,Multiple Choice,,,,
-TPS,11.1,29,You are testing H0: μ = 10 against Ha: µ ≠ 10 based on an SRS of 15 observations from a Normal population. What values of the t statistic are statistically significant at the α = 0.005 level? (a) t >3.326 (b) t >3.286 (c) t >2.977 (d) t <−3.326 or t >3.326 (e) t <−3.286 or t >3.286,d,Multiple Choice,,,,
-TPS,11.1,30,"After checking that conditions are met, you perform a significance test of H0: μ = 1 versus Ha: μ ≠ 1. You obtain a P-value of 0.022. Which of the following must be true? (a) A 95% confidence interval for μ will include the value 1. (b) A 95% confidence interval for μ will include the value 0.022. (c) A 99% confidence interval for μ will include the value 1. (d) A 99% confidence interval for μ will include the value 0.022. (e) None of these is necessarily true. ",c,Multiple Choice,,,,
-TPS,11.1,31,The most important condition for making an inference about a population mean from a significance test is that (a) the data come from a random sample. (b) the population distribution is exactly Normal. (c) the data contain no outliers. (d) the sample size is less than 10% of the population size. (e) the sample size is at least 30.,a,Multiple Choice,,,,
-TPS,11.1,32,Vigorous exercise helps people live several years longer (on average). Whether mild activities like slow walking extend life is not clear. Suppose that the added life expectancy from regular slow walking is just 2 months. A significance test is more likely to find a significant increase in mean life expectancy with regular slow walking if (a) it is based on a very large random sample and a 5% significance level is used. (b) it is based on a very large random sample and a 1% significance level is used. (c) it is based on a very small random sample and a 5% significance level is used. (d) it is based on a very small random sample and a 1% significance level is used. (e) the size of the sample doesn’t have any effect on the significance of the test.,a,Multiple Choice,,,,
-TPS,11.2,41a,"To see if fish oil can help reduce blood pressure, males with high blood pressure were recruited and randomly assigned to different treatments. Seven of the men were assigned to a 4-week diet that included fish oil. Seven other men were assigned to a 4-week diet that included a mixture of oils that approximated the types of fat in a typical diet. Each man’s blood pressure was measured at the beginning of the study. At the end of the 4 weeks, each volunteer’s blood pressure was measured again and the reduction in diastolic blood pressure was recorded. These differences are shown in the table. Note that a negative value means that the subject’s blood pressure increased. Fish oil: 8 12 10 14 2 0 0 Regular oil: −6 0 1 2 -3 -4 2. Do these data provide convincing evidence that fish oil helps reduce blood pressure more, on average, than regular oil for men like these?","S: H0: µ1 - µ2 = 0, Ha: µ1 - µ2 > 0, where µ1 = true mean reduction in blood pressure for subjects like these who take fish oil and µ2 = true mean reduction . . . regular oil. P: Two-sample t test for µ1 - µ2. Random: Randomized assignment. Normal/Large Sample: The dotplots show no strong skewness and no outliers. D: x¯1 = 6.571, s1 = 5.855, n1 = 7, x¯2 = =1.143, s2 = 3.185, n2 = 7, and t= 3.06. Using df = 9.264, P-value = 0.0065; using df = 6, P-value = 0.0111. C: Because the P-value of 0.0065 < α = 0.05, we reject H0. We have convincing evidence fish oil helps reduce blood pressure more, on average, than regular oil for subjects like these. ",Free Response,,,,
-TPS,11.2,41b,"To see if fish oil can help reduce blood pressure, males with high blood pressure were recruited and randomly assigned to different treatments. Seven of the men were assigned to a 4-week diet that included fish oil. Seven other men were assigned to a 4-week diet that included a mixture of oils that approximated the types of fat in a typical diet. Each man’s blood pressure was measured at the beginning of the study. At the end of the 4 weeks, each volunteer’s blood pressure was measured again and the reduction in diastolic blood pressure was recorded. These differences are shown in the table. Note that a negative value means that the subject’s blood pressure increased. Fish oil: 8 12 10 14 2 0 0 Regular oil: −6 0 1 2 -3 -4 2. A significance test was done with a P-value of 0.0065. Interpret the P -value from part (a) in the context of this study.","Assuming the true mean reduction in blood pressure is the same regardless of whether the individual takes fish oil or regular oil, there is a 0.0065 probability that we would observe a difference in sample means of 7.714 or greater by chance alone.",Free Response,,,,
-TPS,11.2,42a,"Do birds learn to time their breeding? Blue titmice eat caterpillars. The birds would like lots of caterpillars around when they have young to feed, but they must breed much earlier. Do the birds learn from one year’s experience when to time their breeding next year? Researchers randomly assigned 7 pairs of birds to have the natural caterpillar supply supplemented while feeding their young and another 6 pairs to serve as a control group relying on natural food supply. The next year, they measured how many days after the caterpillar peak the birds produced their nestlings. 13 The investigators expected the control group to adjust their breeding date the following year, whereas the well-fed supplemented group had no reason to change. Here are the data (days after caterpillar peak): Control: 4.6 2.3 7.7 6.0 4.6 −1.2 Supplemented: 15.5 11.3 5.4 16.5 11.3 11.4 7.7. Do the data provide convincing evidence that birds like these that have to rely on the natural food supply produce their nestlings closer to the caterpillar peak, on average, than birds like these that have the caterpillar supply supplemented?","S: H0: µ1 - µ2 = 0, Ha: µ1 - µ2 < 0, where µ1 = true mean time to breeding for birds relying on natural food supply and µ2 = true mean . . . food supplements. P: Two-sample t test for µ1 - µ2. Random: Random assignment. Normal/Large Sample: The dotplots show no strong skewness or outliers. D: x¯1 = 4.0, s1 = 3.11, n1 = 6, x¯2 = 11.3, s2 = 3.93, n2 = 7, and t=-3.74. Using df = 10.955, P-value = 0.0016; using df = 5, P-value = 0.0067. C: Because the P-value of 0.0016 < α = 0.05, we reject H0. We have convincing evidence the true mean time to breeding is less for birds relying on natural food supply than for birds with food supplements ",Free Response,,,,
-TPS,11.2,42b,"Do birds learn to time their breeding? Blue titmice eat caterpillars. The birds would like lots of caterpillars around when they have young to feed, but they must breed much earlier. Do the birds learn from one year’s experience when to time their breeding next year? Researchers randomly assigned 7 pairs of birds to have the natural caterpillar supply supplemented while feeding their young and another 6 pairs to serve as a control group relying on natural food supply. The next year, they measured how many days after the caterpillar peak the birds produced their nestlings. 13 The investigators expected the control group to adjust their breeding date the following year, whereas the well-fed supplemented group had no reason to change. Here are the data (days after caterpillar peak): Control: 4.6 2.3 7.7 6.0 4.6 −1.2 Supplemented: 15.5 11.3 5.4 16.5 11.3 11.4 7.7. A significance test was done and P-value was found to be 0.0016. Interpret the P-value from part (a) in the context of this study.","Assuming the true mean time to breeding is the same for birds relying on natural food supply and birds with food supplements, there is a 0.0016 probability that we would observe a difference in sample means of 27.3 or smaller by chance alone. ",Free Response,,,,
-TPS,11.2,43,"Researchers equipped random samples of 56 male and 56 female students from a large university with a small device that secretly records sound for a random 30 seconds during each 12.5-minute period over 2 days. Then they counted the number of words spoken by each subject during each recording period and, from this, estimated how many words per day each subject speaks. The female estimates had a mean of 16,177 words per day with a standard deviation of 7520 words per day. For the male estimates, the mean was 16,569 and the standard deviation was 9108. Do these data provide convincing evidence at the α = 0.05 significance level of a difference in the average number of words spoken in a day by all male and all female students at this university?","S: H0: µ1 - µ2 = 0, Ha: µ1 - µ2 ≠ 0, where µ1 = true mean number of words spoken per day by female students and µ2 = true mean number . . . by male students. P: Two-sample t test for µ1 - µ2. Random: Independent random samples. 10%: n1 = 56 < 10% of females at a large university and n2 = 56 < 10% of males at a large university. Normal/Large Sample: n1 = 56 ≥ 30 and n2 = 56 $≥ 30. D: t= -0.25; using df = 106.195, P-value = 0.8043. Using df = 55, P-value = 0.8035. C: Because the P-value of 0.8043 > α = 0.05, we fail to reject H0. We do not have convincing evidence the true mean number of words spoken per day by female students differs from the true mean number of words spoken per day by male students at this university. ",Free Response,,,,
-TPS,11.2,44,"In many parts of the northern United States, two color variants of the Eastern Gray Squirrel—gray and black—are found in the same habitats. A scientist studying squirrels in a large forest wonders if there is a difference in the sizes of the two color variants. He collects random samples of 40 squirrels of each color from a large forest and weighs them. The 40 black squirrels have a mean weight of 20.3 ounces and a standard deviation of 2.1 ounces. The 40 gray squirrels have a mean weight of 19.2 ounces and a standard deviation of 1.9 ounces. Do these data provide convincing evidence at the α = 0.01 significance level of a difference in the mean weights of all gray and black Eastern Gray Squirrels in this forest?","S: H0: µ1 - µ2 = 0, Ha: µ1 - µ2 ≠ 0, where µ1 =true mean weight of black squirrels and µ2 = that of gray squirrels. P: Two-sample t test for µ1 - µ2. Random: Independent random samples. 10%: n1 = 40 < 10% of black squirrels and n2 = 40 < 10% of gray squirrels. Normal/ Large Sample: n1 = 40 ≥ 30 and n2 = 40 ≥ 30. D: t= 2.46; using df = 77.232, P-value = 0.0163. Using df = 39, P-value = 0.0184. C: Because the P-value of 0.0163 > α = 0.01, we fail to reject H0. We do not have convincing evidence the true mean weight of black squirrels differs from the true mean weight of gray squirrels",Free Response,,,,
-TPS,11.2,45a,"Researchers equipped random samples of 56 male and 56 female students from a large university with a small device that secretly records sound for a random 30 seconds during each 12.5-minute period over 2 days. Then they counted the number of words spoken by each subject during each recording period and, from this, estimated how many words per day each subject speaks. The female estimates had a mean of 16,177 words per day with a standard deviation of 7520 words per day. For the male estimates, the mean was 16,569 and the standard deviation was 9108. Construct and interpret a 95% confidence interval for the difference between the true means.","D: x¯1 = 16,177, s1 = 7520, n1 = 56, x¯2 = 16,569, s2 = 9108, n2 = 56. Using df = 106.2, (-3521, 2737); using df = 55, (-3555, 2771). C: We are 95% confident that the interval from 23521 to 2737 words captures µ1 - µ2 = true difference in mean number of words spoken per day by female students versus male students at this university. ",Free Response,,,,
-TPS,11.2,45b,"Researchers equipped random samples of 56 male and 56 female students from a large university with a small device that secretly records sound for a random 30 seconds during each 12.5-minute period over 2 days. Then they counted the number of words spoken by each subject during each recording period and, from this, estimated how many words per day each subject speaks. The female estimates had a mean of 16,177 words per day with a standard deviation of 7520 words per day. For the male estimates, the mean was 16,569 and the standard deviation was 9108. Explain how the confidence interval provides more information than the test.","The two-sided test only allows us to reject (or fail to reject) a difference of 0, whereas a confidence interval provides a set of plausible values for the true difference in means.",Free Response,,,,
-TPS,11.2,46a,"In many parts of the northern United States, two color variants of the Eastern Gray Squirrel—gray and black—are found in the same habitats. A scientist studying squirrels in a large forest wonders if there is a difference in the sizes of the two color variants. He collects random samples of 40 squirrels of each color from a large forest and weighs them. The 40 black squirrels have a mean weight of 20.3 ounces and a standard deviation of 2.1 ounces. The 40 gray squirrels have a mean weight of 19.2 ounces and a standard deviation of 1.9 ounces. Construct and interpret a 99% confidence interval for the difference between the true means. ","D: x¯1 = 20.3, s1 = 2.1, n1 = 40, x¯2 = 19.2, s2 = 1.9, n2 = 40. Using df = 77.23, (-0.0826, 2.2826); using df = 30, (-0.131, 2.331). C: We are 99% confident that the interval from 20.0826 to 2.2826 ounces captures µ1 - µ2 5 true difference in mean weight of all black and gray squirrels. ",Free Response,,,,
-TPS,11.2,46b,"In many parts of the northern United States, two color variants of the Eastern Gray Squirrel—gray and black—are found in the same habitats. A scientist studying squirrels in a large forest wonders if there is a difference in the sizes of the two color variants. He collects random samples of 40 squirrels of each color from a large forest and weighs them. The 40 black squirrels have a mean weight of 20.3 ounces and a standard deviation of 2.1 ounces. The 40 gray squirrels have a mean weight of 19.2 ounces and a standard deviation of 1.9 ounces. Explain how the confidence interval provides more information than the test.","The two-sided test only allows us to reject (or fail to reject) a difference of 0, whereas a confidence interval provides a set of plausible values for the true difference in mean weight of all black and gray squirrels.",Free Response,,,,
-TPS,11.2,49a,"In a pilot study, a company’s new cholesterol-reducing drug outperforms the currently available drug. If the data provide convincing evidence that the mean cholesterol reduction with the new drug is more than 10 milligrams per deciliter of blood (mg/dl) greater than with the current drug, the company will begin the expensive process of mass-producing the new drug. For the 14 subjects who were assigned at random to the current drug, the mean cholesterol reduction was 54.1 mg/dl with a standard deviation of 11.93 mg/dl. For the 15 sub-jects who were randomly assigned to the new drug, the mean cholesterol reduction was 68.7 mg/dl with a standard deviation of 13.3 mg/dl. Graphs of the data reveal no outliers or strong skewness. Researchers want to perform a test of H0: μnew - µcur = 10 versus Ha: μnew – μcur > 10. Calculate the standardized test statistic and P-value. ","t = 0.98, P-value = 0.1675 using df = 26.96. ",Free Response,,,,
-TPS,11.2,49b,"In a pilot study, a company’s new cholesterol-reducing drug outperforms the currently available drug. If the data provide convincing evidence that the mean cholesterol reduction with the new drug is more than 10 milligrams per deciliter of blood (mg/dl) greater than with the current drug, the company will begin the expensive process of mass-producing the new drug. For the 14 subjects who were assigned at random to the current drug, the mean cholesterol reduction was 54.1 mg/dl with a standard deviation of 11.93 mg/dl. For the 15 sub-jects who were randomly assigned to the new drug, the mean cholesterol reduction was 68.7 mg/dl with a standard deviation of 13.3 mg/dl. Graphs of the data reveal no outliers or strong skewness. Researchers want to perform a test of H0: μnew - µcur = 10 versus Ha: μnew – μcur > 10. Interpret the P-value. What conclusion would you make? ","Assuming that the true mean cholesterol reduction for subjects who take the new drug is 10 mg/dl more than subjects who take the old drug, there is a 0.1675 probability that we would observe a difference in sample means that is at least 4.6 mg/dl or greater beyond a difference of 10 mg/dl by chance alone. Because the P-value of 0.1675 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true mean cholesterol reduction is more than 10 mg/dl greater for the new drug than for the current drug.",Free Response,,,,
-TPS,11.2,49c,"In a pilot study, a company’s new cholesterol-reducing drug outperforms the currently available drug. If the data provide convincing evidence that the mean cholesterol reduction with the new drug is more than 10 milligrams per deciliter of blood (mg/dl) greater than with the current drug, the company will begin the expensive process of mass-producing the new drug. For the 14 subjects who were assigned at random to the current drug, the mean cholesterol reduction was 54.1 mg/dl with a standard deviation of 11.93 mg/dl. For the 15 sub-jects who were randomly assigned to the new drug, the mean cholesterol reduction was 68.7 mg/dl with a standard deviation of 13.3 mg/dl. Graphs of the data reveal no outliers or strong skewness. Researchers want to perform a test of H0: μnew - µcur = 10 versus Ha: μnew – μcur > 10. Describe two ways to increase the power of the test.","To increase the power of this test, we could increase the sample sizes or the significance level. ",Free Response,,,,
-TPS,11.2,50a,"A company that makes hotel toilets claims that its new pressure-assisted toilet reduces the average amount of water used by more than 0.5 gallon per flush when compared to its current model. To test this claim, the company randomly selects 30 toilets of each type and measures the amount of water that is used when each toilet is flushed once. For the current-model toilets, the mean amount of water used is 1.64 gal with a standard deviation of 0.29 gal. For the new toilets, the mean amount of water used is 1.09 gal with a standard deviation of 0.18 gal. Researchers want to perform a test of H0: μcur – μnew = 0.5 versus Ha: μcur – μnew > 0.5. Calculate the standardized test statistic and P-value. ","t = 0.80, P-value = 0.2131 using df = 48.46",Free Response,,,,
-TPS,11.2,50b,"A company that makes hotel toilets claims that its new pressure-assisted toilet reduces the average amount of water used by more than 0.5 gallon per flush when compared to its current model. To test this claim, the company randomly selects 30 toilets of each type and measures the amount of water that is used when each toilet is flushed once. For the current-model toilets, the mean amount of water used is 1.64 gal with a standard deviation of 0.29 gal. For the new toilets, the mean amount of water used is 1.09 gal with a standard deviation of 0.18 gal. Researchers want to perform a test of H0: μcur – μnew = 0.5 versus Ha: μcur – μnew > 0.5. Interpret the P-value. What conclusion would you make? ","Assuming that the true mean amount of water used by the new toilets is 0.5 gallon less than the current toilets, there is a 0.2131 probability that we would observe a difference in sample means that is at least 0.05 gallon or greater beyond a difference of 0.5 gallon by chance alone. Because the P-value of 0.2131 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the new model toilet reduces the amount of water used by greater than 0.50 gallon per flush than the current-model toilet, on average. ",Free Response,,,,
-TPS,11.2,50c,"A company that makes hotel toilets claims that its new pressure-assisted toilet reduces the average amount of water used by more than 0.5 gallon per flush when compared to its current model. To test this claim, the company randomly selects 30 toilets of each type and measures the amount of water that is used when each toilet is flushed once. For the current-model toilets, the mean amount of water used is 1.64 gal with a standard deviation of 0.29 gal. For the new toilets, the mean amount of water used is 1.09 gal with a standard deviation of 0.18 gal. Researchers want to perform a test of H0: μcur – μnew = 0.5 versus Ha: μcur – μnew > 0.5. Describe two ways to increase the power of the test.","To increase the power of this test, we could increase the sample sizes or the significance level.",Free Response,,,,
-TPS,11.2,55a,"Does listening to music while studying help or hinder students’ learning? Two statis-tics students designed an experiment to find out. They selected a random sample of 30 students from their medium-sized high school to participate. Each subject was given 10 minutes to memorize two different lists of 20 words, once while listening to music and once in silence. The order of the two word lists was determined at random; so was the order of the treatments. The difference (Silence - Music) in the number of words recalled was recorded for each subject. The mean difference was 1.57 and the standard deviation of the differences was 2.70. If the result of this study is statistically significant, can you conclude that the difference in the ability to memorize words was caused by whether students were performing the task in silence or with music playing? Why or why not?","Yes! The data arose from a randomized comparative experiment, so if the result of this study is statistically significant, we can conclude that the difference in the ability to memorize words was caused by whether students were performing the task in silence or with music playing. ",Free Response,,,,
-TPS,11.2,55b,"Does listening to music while studying help or hinder students’ learning? Two statis-tics students designed an experiment to find out. They selected a random sample of 30 students from their medium-sized high school to participate. Each subject was given 10 minutes to memorize two different lists of 20 words, once while listening to music and once in silence. The order of the two word lists was determined at random; so was the order of the treatments. The difference (Silence - Music) in the number of words recalled was recorded for each subject. The mean difference was 1.57 and the standard deviation of the differences was 2.70. Do the data provide convincing evidence at the α = 0.01 significance level that the number of words recalled in silence or when listening to music differs, on average, for students at this school?","S: H0: µdiff = 0, Ha: µdiff ≠ 0, where µdiff = true mean difference (Music – Silence) in the average number of words recalled by students at this school; α = 0.01. P: Paired t test for µdiff. Random: The treatments were assigned in a random order. Normal/Large Sample: ndiff = 30 ≠ 30. D: x¯diff = 1.57, sdiff = 2.70, and ndiff = 30; t= 3.18; df = 29; P-value = 0.0034. C: Because the P-value of 0.0034 < α = 0.01, we reject H0. We have convincing evidence that the true mean difference (Music – Silence) in the average number of words recalled by students at this school is different from 0. ",Free Response,,,,
-TPS,11.2,55c,"Does listening to music while studying help or hinder students’ learning? Two statis-tics students designed an experiment to find out. They selected a random sample of 30 students from their medium-sized high school to participate. Each subject was given 10 minutes to memorize two different lists of 20 words, once while listening to music and once in silence. The order of the two word lists was determined at random; so was the order of the treatments. The difference (Silence - Music) in the number of words recalled was recorded for each subject. The mean difference was 1.57 and the standard deviation of the differences was 2.70. A test was run a conclusion was made to reject theH0. Which type of error—a Type I error or a Type II error—could you have made? Explain your answer.","Because we rejected H0, it is possible that we made a Type I error.",Free Response,,,,
-TPS,11.2,56a,"Do people behave differently on Friday the 13th? Researchers collected data on the number of shoppers at a random sample of 45 grocery stores on Friday the 6th and Friday the 13th in the same month. Then they calculated the difference (subtracting in the order 6th minus 13th) in the number of shoppers at each store on these 2 days. The mean difference is −46.5 and the standard deviation of the differences is 178.0. If the result of this study is statistically significant, can you conclude that the difference in shopping behavior is due to the effect of Friday the 13th on people’s behavior? Why or why not?"," No; this is an observational study, not a controlled experiment. So even if the result of this study is statistically significant, we cannot conclude that the difference in shopping behavior is due to the effect of Friday the 13th. ",Free Response,,,,
-TPS,11.2,56b,"Do people behave differently on Friday the 13th? Researchers collected data on the number of shoppers at a random sample of 45 grocery stores on Friday the 6th and Friday the 13th in the same month. Then they calculated the difference (subtracting in the order 6th minus 13th) in the number of shoppers at each store on these 2 days. The mean difference is −46.5 and the standard deviation of the differences is 178.0.Do these data provide convincing evidence at the α= 0.05 10 level that the number of shoppers at grocery stores on these 2 days differs, on average?","S: H0: µdiff = 0, Ha: µdiff ≠ 0, where µdiff = true mean difference (6th – 13th) in the mean number of shoppers at grocery stores; α = 0.10. P: Paired t test for µdiff. Random: Paired data; random sample of 45 grocery stores. 10%: 45 < 10% of all grocery stores. Normal/Large Sample: ndiff = 45 ≥ 30. D: x¯diff = -46.5, sdiff = 178.0, and ndiff = 45; t= -1.75; df = 44; P-value = 0.0867. C: Because the P-value of 0.0867 > α = 0.10, we reject H0. We have convincing evidence that the true mean difference (6th 2 13th) in the number of shoppers at grocery stores on these two days differs from 0. ",Free Response,,,,
-TPS,11.2,56c,Do people behave differently on Friday the 13th? Researchers collected data on the number of shoppers at a random sample of 45 grocery stores on Friday the 6th and Friday the 13th in the same month. Then they calculated the difference (subtracting in the order 6th minus 13th) in the number of shoppers at each store on these 2 days. The mean difference is −46.5 and the standard deviation of the differences is 178.0. A test was done and a conclusion was made to reject the H0. Which type of error—a Type I error or a Type II error—could you have made? Explain your answer.,"Because we rejected Ha, it is possible that we made a Type I error.",Free Response,,,,
-TPS,11.2,57a,"Does listening to music while studying help or hinder students’ learning? Two statis-tics students designed an experiment to find out. They selected a random sample of 30 students from their medium-sized high school to participate. Each subject was given 10 minutes to memorize two different lists of 20 words, once while listening to music and once in silence. The order of the two word lists was determined at random; so was the order of the treatments. The difference (Silence - Music) in the number of words recalled was recorded for each subject. The mean difference was 1.57 and the standard deviation of the differences was 2.70. Construct and interpret a 99% confidence interval for the true mean difference.","S: µdiff = true mean difference (Music - Silence) in the average number of words recalled by students at this school. P: One-sample t interval for µdiff. Random: The treatments were assigned in a random order. Normal/Large Sample: ndiff = 30 ≥ 30. D: x¯diff = 1.57, sdiff = 2.70, ndiff = 30; df = 29, (0.211, 2.929). C: We are 99% confident the interval from 0.211 to 2.929 captures the true mean difference (Music - Silence) in the number of words recalled by students at this school. ",Free Response,,,,
-TPS,11.2,57b,"Does listening to music while studying help or hinder students’ learning? Two statis-tics students designed an experiment to find out. They selected a random sample of 30 students from their medium-sized high school to participate. Each subject was given 10 minutes to memorize two different lists of 20 words, once while listening to music and once in silence. The order of the two word lists was determined at random; so was the order of the treatments. The difference (Silence - Music) in the number of words recalled was recorded for each subject. The mean difference was 1.57 and the standard deviation of the differences was 2.70. Explain how the confidence interval provides more information than the test.","The two-sided test only allows us to reject a difference of 0, where the confidence interval provided a set of plausible values (0.211, 2.929) for the true mean difference (Music - Silence) in the number of words recalled by students at this school. ",Free Response,,,,
-TPS,11.2,58a,Do people behave differently on Friday the 13th? Researchers collected data on the number of shoppers at a random sample of 45 grocery stores on Friday the 6th and Friday the 13th in the same month. Then they calculated the difference (subtracting in the order 6th minus 13th) in the number of shoppers at each store on these 2 days. The mean difference is −46.5 and the standard deviation of the differences is 178.0. Construct and interpret a 90% confidence interval for the true mean difference. ,"S: µdiff = true mean difference (6th - 13th) in the number of shoppers at grocery stores. P: One-sample t interval for µdiff. Random: Paired data; random sample of 45 grocery stores. 10%: 45 < 10% of all grocery stores. Normal/Large Sample: ndiff = 45 ≥ 30. D: x¯diff = -46.5, sdiff = 178.0, ndiff = 45; using df = 40, (-90.284, -0.916). Using df = 44, (-91.08, -1.92). C: We are 90% confident that the interval from -91.08 to 21.92 captures the true mean difference (6th - 13th) in the number of shoppers at grocery stores.",Free Response,,,,
-TPS,11.2,58b,Do people behave differently on Friday the 13th? Researchers collected data on the number of shoppers at a random sample of 45 grocery stores on Friday the 6th and Friday the 13th in the same month. Then they calculated the difference (subtracting in the order 6th minus 13th) in the number of shoppers at each store on these 2 days. The mean difference is −46.5 and the standard deviation of the differences is 178.0. Explain how the confidence interval provides more information than the test.,"The two-sided test only allowed us to reject a difference of zero, where the confidence interval provided a set of plausible values (291.08, 21.92) for true mean difference (6th - 13th) in the number of shoppers at grocery stores. ",Free Response,,,,
-TPS,11.2,59a,"In the following setting, decide whether you should use two-sample t procedures to perform inference about a difference in means or paired t procedures to perform inference about a mean difference. Explain your choice:  To test the wear characteristics of two tire brands, A and B, each of 50 cars of the same make and model is randomly assigned Brand A tires or Brand B tires.",Two-sample t test; the data are being produced using two distinct groups of cars in a randomized experiment. ,Free Response,,,,
-TPS,11.2,59b,"In the following setting, decide whether you should use two-sample t procedures to perform inference about a difference in means or paired t procedures to perform inference about a mean difference. Explain your choice:  To test the effect of background music on productivity, factory workers are observed. For one month, each subject works without music. For another month, the subject works while listening to music on an MP3 player. The month in which each subject listens to music is determined by a coin toss.",Paired t test; this is a matched pairs experimental design in which both treatments are applied to each subject in a random order. ,Free Response,,,,
-TPS,11.2,59c,"In the following setting, decide whether you should use two-sample t procedures to perform inference about a difference in means or paired t procedures to perform inference about a mean difference. Explain your choice:  How do young adults look back on adolescent romance? Investigators interviewed a random sample of 40 couples in their mid-twenties. The female and male partners were interviewed separately. Each was asked about his or her current relationship and also about a romantic relationship that lasted at least 2 months when they were aged 15 or 16. One response variable was a measure on a numerical scale of how much the attractiveness of the adolescent partner mattered. You want to find out how much men and women differ on this measure.",Paired t test; the data were collected from the male and female partners in 40 couples.,Free Response,,,,
-TPS,11.2,60a,"In the following setting, decide whether you should use two-sample t procedures to perform inference about a difference in means or paired t procedures to perform inference about a mean difference. Explain your choice: To compare the average weight gain of pigs fed two different diets, nine pairs of pigs were used. The pigs in each pair were littermates. A coin toss was used to decide which pig in each pair got Diet A and which got Diet B.",Paired t test; this is a matched pairs design using pairs of pigs that were littermates. One pig in each pair received one treatment and the other pig in the pair received the other treatment. ,Free Response,,,,
-TPS,11.2,60b,"In the following setting, decide whether you should use two-sample t procedures to perform inference about a difference in means or paired t procedures to perform inference about a mean difference. Explain your choice: Separate random samples of male and female college professors are taken. We wish to compare the average salaries of male and female teachers.",Two-sample t test; the data come from independent random samples of male and female college professors. ,Free Response,,,,
-TPS,11.2,60c,"In the following setting, decide whether you should use two-sample t procedures to perform inference about a difference in means or paired t procedures to perform inference about a mean difference. Explain your choice: To test the effects of a new fertilizer, 100 plots are treated with the new fertilizer, and 100 plots are treated with another fertilizer. A computer’s random number generator is used to determine which plots get which fertilizer.",Two-sample t test; the data are being produced using two distinct groups of plots in a randomized experiment.,Free Response,,,,
-TPS,11.2,65,"A study of road rage asked random samples of 596 men and 523 women about their behavior while driving. Based on their answers, each person was assigned a road rage score on a scale of 0 to 20. The participants were chosen by random digit dialing of phone numbers. The researchers performed a test of the following hypotheses: H0: µM = µF versus Ha: µM ≠µF. Which of the following describes a Type II error in the context of this study? (a) Finding convincing evidence that the true means are different for males and females, when in reality the true means are the same (b) Finding convincing evidence that the true means are different for males and females, when in reality the true means are different (c) Not finding convincing evidence that the true means are different for males and females, when in reality the true means are the same (d) Not finding convincing evidence that the true means are different for males and females, when in reality the true means are different (e) Not finding convincing evidence that the true means are different for males and females, when in reality there is convincing evidence that the true means are different",d,Multiple Choice,,,,
-TPS,11.2,66,"A study of road rage asked random samples of 596 men and 523 women about their behavior while driving. Based on their answers, each person was assigned a road rage score on a scale of 0 to 20. The participants were chosen by random digit dialing of phone numbers. The researchers performed a test of the following hypotheses: H0: µM = µF versus Ha: µM ≠µF. The P-value for the stated hypotheses is 0.002. Interpret this value in the context of this study. (a) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability of getting a difference in sample means equal to the one observed in this study. (b) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability of getting a difference in sample means at least as large in either direction as the one observed in this study. (c) Assuming that the true mean road rage score is different for males and females, there is a 0.002 probability of getting a difference in sample means at least as large in either direction as the one observed in this study. (d) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability that the null hypothesis is true. (e) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability that the alternative hypothesis is true.",b,Multiple Choice,,,,
-TPS,11.2,67,"A study of road rage asked random samples of 596 men and 523 women about their behavior while driving. Based on their answers, each person was assigned a road rage score on a scale of 0 to 20. The participants were chosen by random digit dialing of phone numbers. The researchers performed a test of the following hypotheses: H0: µM = µF versus Ha: µM ≠µF. Based on the P-value in Exercise 66, which of the following must be true? (a) A 90% confidence interval for µM - µF will contain 0. (b) A 95% confidence interval for µM - µF will contain 0. (c) A 99% confidence interval for µM - µF will contain 0. (d) A 99.9% confidence interval for µM - µF will contain 0. (e) It is impossible to determine whether any of these statements is true based only on the P-value.",d,Multiple Choice,,,,
-TPS,R11,1a,The average height of 18-year-old American women is 64.2 inches. You wonder whether the mean height of this year’s female graduates from a large local high school differs from the national average. You measure an SRS of 48 female graduates and find that x¯ = 63.5 inches and sx = 3.7 inches. State the appropriate null and alternative hypotheses for performing a significance test.,"H0: µ = 64.2, Ha: µ ≠ 64.2, where µ = the true mean height of all female graduates from the large local high school this year. ",Free Response,,,,
-TPS,R11,1b,The average height of 18-year-old American women is 64.2 inches. You wonder whether the mean height of this year’s female graduates from a large local high school differs from the national average. You measure an SRS of 48 female graduates and find that x¯ = 63.5 inches and sx = 3.7 inches. Explain why the sample data give some evidence for Ha. ,"The observed sample mean is x= 63.5 inches, which does not equal 64.2 inches. ",Free Response,,,,
-TPS,R11,1c,The average height of 18-year-old American women is 64.2 inches. You wonder whether the mean height of this year’s female graduates from a large local high school differs from the national average. You measure an SRS of 48 female graduates and find that x¯ = 63.5 inches and sx = 3.7 inches. Identify the appropriate test and show that the conditions for carrying out the test are met.,One-sample t test for a population mean. Random: Random sample of 48 female graduates. 10%: n = 48 < 10% of all female graduates from this large high school. Normal/ Large Sample: n = 48 ≥ 30 ,Free Response,,,,
-TPS,R11,1d,The average height of 18-year-old American women is 64.2 inches. You wonder whether the mean height of this year’s female graduates from a large local high school differs from the national average. You measure an SRS of 48 female graduates and find that x¯ = 63.5 inches and sx = 3.7 inches. Find the standardized test statistic and P-value. ,"t = -1.31, df = 47, P-value = 0.1966 ",Free Response,,,,
-TPS,R11,1e,The average height of 18-year-old American women is 64.2 inches. You wonder whether the mean height of this year’s female graduates from a large local high school differs from the national average. You measure an SRS of 48 female graduates and find that x¯ = 63.5 inches and sx = 3.7 inches. Interpret the P-value. What conclusion would you make?,"P-value interpretation: Assuming that the true mean height of all female graduates is 64.2 inches, there is a 0.1966 probability of getting a sample mean as different as or more different than 63.5 inches (in either direction) by chance alone. Conclusion: Because the P-value of 0.1966 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true mean height of all female graduates from this high school is different from the national average of 64.2 inches.",Free Response,,,,
-TPS,R11,2,"Does the use of fancy type fonts slow down the reading of text on a computer screen? Adults can read four paragraphs of a certain text in the common Times New Roman font in an average of 22 seconds. Researchers asked a random sample of 24 adults to read this text in the ornate font named Gigi. Here are their times (in seconds). 23.2 21.2 28.9 27.7 29.1 27.3 16.1 22.6 25.6 34.2 23.9 26.8 20.5 34.3 21.4 32.6 26.2 34.1 31.5 24.6 23.0 28.6 24.4 28.1 Do these data provide convincing evidence that it takes adults longer than 22 seconds, on average, to read these four paragraphs in Gigi font? ","S: H0: µ = 22, Ha: µ > 22, where µ = the true mean amount of time (in seconds) it takes adults to read four paragraphs of text in the ornate font Gigi. We will use α = 0.05. P: One-sample t test for µ. Random: We have a random sample of 24 adults. 10%: The sample size (24) is less than 10% of the population of adults. Normal/ Large Sample: Because the sample size is small, we need to graph the sample data. The histogram shows that the distribution is roughly symmetric with no outliers, so using a t procedure is appropriate. D: x¯= 26.496, sx= 4.728, t= 4.66, df = 23, and P-value = 0.000054. C: Because the P-value of 0.000054 < α = 0.05, we reject H0 . There is convincing evidence that the true mean amount of time it takes adults to read four paragraphs of text in the ornate font Gigi is greater than 22 seconds. ",Free Response,,,,
-TPS,R11,3a,"In the children’s game Don’t Break the Ice, small plastic ice cubes are squeezed into a square frame. Each child takes turns tapping out a cube of “ice” with a plastic hammer, hoping that the remaining cubes don’t collapse. For the game to work correctly, the cubes must be big enough so that they hold each other in place in the plastic frame, but not so big that they are too difficult to tap out. The machine that produces the plastic cubes is designed to make cubes that are 25.4 millimeters (mm) wide, but the width varies a little. To ensure that the machine is working well, a supervisor inspects a random sample of 50 cubes every hour and measures their width. If the sample provides convincing evidence at the α= 0.05 significance level that the true mean width μ of the cubes produced that hour differs from 25.4 mm, the supervisor will discard all of the cubes. Describe a Type II error in this setting.","Type II error: Failing to find convincing evidence that the true mean width of the cubes differs from 25.4 mm, when the true mean width of the cubes differs from 25.4 mm.",Free Response,,,,
-TPS,R11,3b,"In the children’s game Don’t Break the Ice, small plastic ice cubes are squeezed into a square frame. Each child takes turns tapping out a cube of “ice” with a plastic hammer, hoping that the remaining cubes don’t collapse. For the game to work correctly, the cubes must be big enough so that they hold each other in place in the plastic frame, but not so big that they are too difficult to tap out. The machine that produces the plastic cubes is designed to make cubes that are 25.4 millimeters (mm) wide, but the width varies a little. To ensure that the machine is working well, a supervisor inspects a random sample of 50 cubes every hour and measures their width. If the sample provides convincing evidence at the α= 0.05 significance level that the true mean width μ of the cubes produced that hour differs from 25.4 mm, the supervisor will discard all of the cubes. Identify one benefit and one drawback of changing the significance level to α= 0.10.",This change will reduce the probability of making a Type II error. A drawback to this change is an increased chance of rejecting batches of cubes that are of the appropriate width. ,Free Response,,,,
-TPS,R11,3c," In the children’s game Don’t Break the Ice, small plastic ice cubes are squeezed into a square frame. Each child takes turns tapping out a cube of “ice” with a plastic hammer, hoping that the remaining cubes don’t collapse. For the game to work correctly, the cubes must be big enough so that they hold each other in place in the plastic frame, but not so big that they are too difficult to tap out. The machine that produces the plastic cubes is designed to make cubes that are 25.4 millimeters (mm) wide, but the width varies a little. To ensure that the machine is working well, a supervisor inspects a random sample of 50 cubes every hour and measures their width. If the sample provides convincing evidence at the α= 0.05 significance level that the true mean width μ of the cubes produced that hour differs from 25.4 mm, the supervisor will discard all of the cubes. The data from a sample taken during 1 hour result in a 95% confidence interval of (25.39, 25.441). What conclusion should the supervisor make?","Because m 5 25.4 mm is one of the plausible values in this 95% confidence interval, the supervisor should not reject this batch as being significantly different from the intended mean width of 25.4 mm.",Free Response,,,,
-TPS,T11,1,Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. The mean time is 18 seconds for one particular maze. A researcher thinks that a loud noise will cause the mice to complete the maze faster. She measures how long each of 10 mice takes with a noise stimulus. What are the null and alternative hypotheses for the appropriate significance test? (a) H0: μ = 18 vs. Ha: μ ≠ 18 (b) H0: μ = 18 vs. Ha: μ< 18 (c) H0: μ = 18 vs. Ha: μ> 18 (d) H0: μ< 18 vs. Ha: μ = 18 (e) H0: μ≠ 18 vs. Ha: μ = 18,b,Multiple Choice,,,,
-TPS,T11,2,"You are thinking of conducting a one-sample t test about a population mean μ using a 0.05 significance level. Which of the following statements is correct? (a) You should not carry out the test if the sample does not have a Normal distribution. (b) You can safely carry out the test if there are no outliers, regardless of the sample size. (c) You can carry out the test if a graph of the data shows no strong skewness, regardless of the sample size. (d) You can carry out the test only if the population standard deviation is known. (e) You can safely carry out the test if your sample size is at least 30.",e,Multiple Choice,,,,
-TPS,T11,3,"A 95% confidence interval for μ based on n= 15 observations from a Normal population is (–0.73, 1.92). If we use this confidence interval to test the hypothesis H0: μ= 0 against Ha: μ ≠ 0, which of the following is the most appropriate conclusion? (a) Reject H0 at the α= 0.05 level of significance. (b) Fail to reject H0 at the α= 0.05 level of significance. (c) Reject H0 at the α= 0.10 level of significance. (d) Fail to reject H0 at the α= 0.10 level of significance.(e) We cannot perform the required test since we do not know the value of the standardized test statistic.",b,Multiple Choice,,,,
-TPS,T11,4,Which of the following has the smallest probability?  (a)P(t>2) if t has 5 degrees of freedom  (b) P(t>2) if t has 2 degrees of freedom.  (c) P(z>2) if z is a standard Normal random variable.  (d) P(t<2) if t has 5 degrees of freedom. (e) P(z<2) if z is a standard Normal random variable.,c,Multiple Choice,,,,
-TPS,T11,5,"A significance test was performed to test H0: μ = 2 versus the alternative Ha: μ ≠ 2. A sample of size 28 produced a standardized test statistic of t= 2.051. Assuming all conditions for inference were met, which of the following intervals contains the P-value for this test? (a) 0.01 < P < 0.02  (b) 0.02 < P <0.025  (c) 0.025 < P < 0.05  (d) 0.05 < P < 0.10  (e) P > 0.10",d,Multiple Choice,,,,
-TPS,T11,6,"A study of road rage asked separate random samples of 596 men and 523 women about their behavior while driving. Based on their answers, each respondent was assigned a road rage score on a scale of 0 to 20. Are the conditions for performing a two-sample t test satisfied? (a) Maybe; we have independent random samples, but we should look at the data to check Normality. (b) No; road rage scores on a scale from 0 to 20 can’t be Normal. (c) No; we don’t know the population standard deviations. (d) Yes; the large sample sizes guarantee that the corresponding population distributions will be Normal. (e) Yes; we have two independent random samples and large sample sizes.",e,Multiple Choice,,,,
-TPS,T11,7,A researcher wished to compare the average amount of time spent in extracurricular activities by high school students in a suburban school district with the average time spent by students in a large city school district. The researcher obtained an SRS of 60 high school students in a large suburban school district and found the mean time spent in extracurricular activities per week to be 6 hours with a standard deviation of 3 hours. The researcher also obtained an independent SRS of 40 high school students in a large city school district and found the mean time spent in extracurricular activities per week to be 5 hours with a standard deviation of 2 hours. Suppose that the researcher decides to carry out a significance test of H0: µsuburban = µcity versus a two-sided alternative. Which is the correct standardized test statistic?  (a) z = ((6-5)-0) / sqrt((3/60) + (2/40)) (b) z = ((6-5)-0) / sqrt((3^2/60) + (2^2/40)) (c) t = ((6-5)-0) / (3 / sqrt(60) + 2 / sqrt(40)) (d) t = ((6-5)-0) / sqrt((3/60) + (2/40)) (e) t = ((6-5)-0) / sqrt((3^2/60) + (2^2/40)),e,Multiple Choice,,,,
-TPS,T11,8,A researcher wished to compare the average amount of time spent in extracurricular activities by high school students in a suburban school district with the average time spent by students in a large city school district. The researcher obtained an SRS of 60 high school students in a large suburban school district and found the mean time spent in extracurricular activities per week to be 6 hours with a standard deviation of 3 hours. The researcher also obtained an independent SRS of 40 high school students in a large city school district and found the mean time spent in extracurricular activities per week to be 5 hours with a standard deviation of 2 hours. Suppose that the researcher decides to carry out a significance test of H0: µsuburban = µcity versus a two-sided alternative. The P-value for the test is 0.048. A correct conclusion is to (a) fail to reject H0 because 0.048<α= 0.05. There is convincing evidence of a difference in the average time spent on extracurricular activities by students in the suburban and city school districts. (b) fail to reject H0 because 0.048<α= 0.05. There is not convincing evidence of a difference in the average time spent on extracurricular activities by students in the suburban and city school districts. (c) fail to reject H0 because 0.048<α= 0.05. There is convincing evidence that the average time spent on extracurricular activities by students in the suburban and city school districts is the same. (d) reject H0 because 0.048<α= 0.05. There is not convincing evidence of a difference in the average time spent on extracurricular activities by students in the suburban and city school districts. (e) reject H0 because 0.048<α= 0.05. There is convincing evidence of a difference in the average time spent on extracurricular activities by students in the suburban and city school districts.,e,Multiple Choice,,,,
-TPS,T11,9,"Are TV commercials louder than their surrounding programs? To find out, researchers collected data on 50 randomly selected commercials in a given week. With the television’s volume at a fixed setting, they measured the maximum loudness of each commercial and the maximum loudness in the first 30 seconds of regular programming that followed. Assuming conditions for inference are met, the most appropriate method for answering the question of interest is (a) a two-sample t test for a difference in means. (b) a two-sample t interval for a difference in means. (c) a paired t test for a mean difference. (d) a paired t interval for a mean difference. (e) a two-sample z test for a difference in proportions.",c,Multiple Choice,,,,
-TPS,T11,10,"Researchers want to evaluate the effect of a natural product on reducing blood pressure. They plan to carry out a randomized experiment to compare the mean reduction in blood pressure of a treatment (natural product) group and a placebo group. Then they will use the data to perform a test of H0: µT - µP = 0 versus Ha: µT - µP > 0, where μT = the true mean reduction in blood pressure when taking the natural product and μP = the true mean reduction in blood pressure when taking a placebo for subjects like the ones in the experiment. The researchers would like to detect whether the natural product reduces blood pressure by at least 7 points more, on average, than the placebo. If groups of size 50 are used in the experiment, a two-sample t test using α= 0.01 will have a power of 80% to detect a 7-point difference in mean blood pressure reduction. If the researchers want to be able to detect a 5-point difference instead, then the power of the test(a) would be less than 80%. (b) would be greater than 80%. (c) would still be 80%. (d) could be either less than or greater than 80%. (e) would vary depending on the values of μT and μP.",a,Multiple Choice,,,,
-TPS,T11,11a,"A government report says that the average amount of money spent per U.S. household per week on food is about $158. A random sample of 50 households in a small city is selected, and their weekly spending on food is recorded. The sample data have a mean of $165 and a standard deviation of $32. Is there convincing evidence that the mean weekly spending on food in this city differs from the national figure of $158? State appropriate hypotheses for performing a significance test in this setting. Be sure to define the parameter of interest.","We want to test H0: µ = $158, Ha: µ ≠ $158, where µ = the true mean amount spent on food by households in this city. We will perform the test at the α = 0.05 significance level. ",Free Response,,,,
-TPS,T11,11b,"A government report says that the average amount of money spent per U.S. household per week on food is about $158. A random sample of 50 households in a small city is selected, and their weekly spending on food is recorded. The sample data have a mean of $165 and a standard deviation of $32. Is there convincing evidence that the mean weekly spending on food in this city differs from the national figure of $158? The distribution of household spending in this small city is heavily skewed to the right. Explain why the Normal/Large Sample condition is met in this case.",The Normal/Large sample condition is met because n = 50 ≥ 30. ,Free Response,,,,
-TPS,T11,11c,"A government report says that the average amount of money spent per U.S. household per week on food is about $158. A random sample of 50 households in a small city is selected, and their weekly spending on food is recorded. The sample data have a mean of $165 and a standard deviation of $32. Is there convincing evidence that the mean weekly spending on food in this city differs from the national figure of $158? The P-value of the test is 0.128. Interpret this value. What conclusion would you make?","Assuming that the true mean amount of money spent on food per household in this city is $158, there is a 0.1283 probability of getting a sample mean as different as or more different than $165 (in either direction) by chance alone. Conclusion: Because the P-value of 0.128 > α = 0.05, we fail to reject H0. We do not have convincing evidence that the true mean amount spent on food per household in this city is different from the national average of $158.",Free Response,,,,
-LFD,9.1,1,"For estimating a population characteristic, why is an unbiased statistic with a small standard error preferred over an unbiased statistic with a larger standard error?",An unbiased statistic with a smaller standard error is preferred because it is likely to result in an estimate that is closer to the actual value of the population characteristic than an unbiased statistic that has a larger standard error.,Free Response,,,,
-LFD,9.1,4,A researcher wants to estimate the proportion of students enrolled at a university who eat fast food more than three times in a typical week. Would the standard error of the sample proportion p^ be smaller for random samples of size n = 50 or random samples of size n = 200?,n = 200,Free Response,,,,
-LFD,9.1,5a,Use the formula for the standard error of p^ to explain why the standard error is greater when the value of the population proportion p is near 0.5 than when it is near 1. 459.,The formula for the standard error of p^ is σp^ = sqrt((p(1 - p))/n). The quantity p(1-p) reaches a maximum value when p = 0.5.,Free Response,,,,
-LFD,9.1,5b,Use the formula for the standard error of p^ to explain why the standard error of p^ is the same when the value of the population proportion is p = 0.2 as it is when p =0.8.,"The standard error of p^ is the same when p = 0.2 as when p = 0.8 because when p = 0.2, (1 - p) = (1 – 0.2) = 0.8. Similarly, when p = 0.8, (1 - p) = (1 – 0.8) = 0.2. So, the quantity p(1-p)  is the same in both cases.",Free Response,,,,
-LFD,9.1,6,"A random sample will be selected from the population of all adult residents of a particular city. The sample proportion p^ will be used to estimate p, the proportion of all adult residents who are employed full time. For which of the following situations will the estimate tend to be closest to the actual value of p? i. n = 500, p = 0.6 ii. n = 450, p = 0.7 iii. n = 400, p = 0.8",n = 400 and p = 0.8,Free Response,,,,
-LFD,9.1,13,"If two statistics are available for estimating a population characteristic, under what circumstances might you choose a biased statistic over an unbiased statistic?","A biased statistic might be chosen over an unbiased statistic if the bias is not too large, and the standard error of the biased statistic is much smaller than the standard error of the unbiased statistic. In this case, the observed value of the biased statistic might be closer to the actual value than the value of an unbiased statistic.",Free Response,,,,
-LFD,9.1,15,A researcher wants to estimate the proportion of city residents who favor spending city funds to promote tourism. Would the standard error of the sample proportion be smaller for random samples of size n = 100 or random samples of size n = 200?,n = 200,Free Response,,,,
-LFD,9.2,17a,"A large online retailer is interested in learning about the proportion of customers making a purchase during a particular month who were satisfied with the online ordering process. A random sample of 600 of these customers included 492 who indicated they were satisfied. For the following statement, indicate if the statement is correct or incorrect. If the statement is incorrect, explain what makes it incorrect. Statement: It is unlikely that the estimate p^=0.82 differs from the value of the actual population proportion by more than 0.0157.","Incorrect, because the value 0.0157 is the standard error of p^, and therefore approximately 32% of all possible values of p^ would differ from the value of the actual population proportion by more than 0.0157 (using properties of the normal distribution).",Free Response,,,,
-LFD,9.2,17b,"A large online retailer is interested in learning about the proportion of customers making a purchase during a particular month who were satisfied with the online ordering process. A random sample of 600 of these customers included 492 who indicated they were satisfied. For the following statement, indicate if the statement is correct or incorrect. If the statement is incorrect, explain what makes it incorrect. Statement: It is unlikely that the estimate p^=0.82 differs from the value of the actual population proportion by more than 0.0307.",Correct,Free Response,,,,
-LFD,9.2,17c,"A large online retailer is interested in learning about the proportion of customers making a purchase during a particular month who were satisfied with the online ordering process. A random sample of 600 of these customers included 492 who indicated they were satisfied. For the following statement, indicate if the statement is correct or incorrect. If the statement is incorrect, explain what makes it incorrect. Statement: The estimate p^=0.82 will never differ from the value of the actual population proportion by more than 0.0307.","Incorrect, because the phrase “will never differ from the value of the actual population proportion” is wrong. The value 0.0307 is the margin of error and indicates that in about 95% of all possible random samples, the estimation error will be less than the margin of error. In about 5% of the random samples, the estimation error will be greater than the margin of error.",Free Response,,,,
-LFD,9.2,18a,Consider taking a random sample from a population with p = 0.40.  What is the standard error of p^ for random samples of size 100? ,0.049,Free Response,,,,
-LFD,9.2,18b,Consider taking a random sample from a population with p = 0.40.  Would the standard error of p^ be larger for samples of size 100 or samples of size 200? ,n = 100,Free Response,,,,
-LFD,9.2,18c,"Consider taking a random sample from a population with p = 0.40.  If the sample size were doubled from 100 to 200, by what factor would the standard error of p^ decrease?",1 / sqrt(2) = 0.707,Free Response,,,,
-LFD,9.2,19a,"The report “2005 Electronic Monitoring & Surveillance Survey: Many Companies Monitoring, Recording, Videotaping—and Firing—Employees” (American Management Association, 2005) summarized a survey of 526 U.S. businesses. The report stated that 137 of the 526 businesses had fired workers for misuse of the Internet. Assume that this sample is representative of businesses in the United States. Estimate the proportion of all businesses in the U.S. that have fired workers for misuse of the Internet. What statistic did you use? ",p^ = 0.260,Free Response,,,,
-LFD,9.2,19b,"The report “2005 Electronic Monitoring & Surveillance Survey: Many Companies Monitoring, Recording, Videotaping—and Firing—Employees” (American Management Association, 2005) summarized a survey of 526 U.S. businesses. The report stated that 137 of the 526 businesses had fired workers for misuse of the Internet. Assume that this sample is representative of businesses in the United States. Use the sample data to estimate the standard error of p^. ",σp^ = 0.019,Free Response,,,,
-LFD,9.2,19c,"The report “2005 Electronic Monitoring & Surveillance Survey: Many Companies Monitoring, Recording, Videotaping—and Firing—Employees” (American Management Association, 2005) summarized a survey of 526 U.S. businesses. The report stated that 137 of the 526 businesses had fired workers for misuse of the Internet. Assume that this sample is representative of businesses in the United States. Compute and interpret the margin of error.",margin of error = 0.037. The estimate of the proportion of all businesses that have fired workers for misuse of the Internet is unlikely to differ from the actual population proportion by more than 0.037.,Free Response,,,,
-LFD,9.2,20a,"The use of the formula for margin of error requires a large sample. For the following combination of n and p^, indicate whether the sample size is large enough for use of this formula to be appropriate: n = 50 and p^ = 0.30.",yes,Free Response,,,,
-LFD,9.2,20b,"The use of the formula for margin of error requires a large sample. For the following combination of n and p^, indicate whether the sample size is large enough for use of this formula to be appropriate: n = 50 and p^ = 0.05.",no,Free Response,,,,
-LFD,9.2,20c,"The use of the formula for margin of error requires a large sample. For the following combination of n and p^, indicate whether the sample size is large enough for use of this formula to be appropriate: n= 15 and p^ = 0.45.",no,Free Response,,,,
-LFD,9.2,20d,"The use of the formula for margin of error requires a large sample. For the following combination of n and p^, indicate whether the sample size is large enough for use of this formula to be appropriate: n = 100 and p^ = 0.01.",no,Free Response,,,,
-LFD,9.2,21a,"Most American college students make use of the Internet for both academic and social purposes. What proportion of students use it for more than 3 hours a day? The authors of the paper “U.S. College Students’ Internet Use: Race, Gender and Digital Divides” (Journal of Computer-Mediated Communication [2009]: 244–264) describe a survey of 7,421 students at 40 colleges and universities. The sample was selected to reflect general demographics of U.S. college students. Of the students surveyed, 2,998 reported Internet use of more than 3 hours per day. Use the given information to estimate the proportion of college students who use the Internet more than 3 hours per day. ",p^ = 0.404,Free Response,,,,
-LFD,9.2,21b,"Most American college students make use of the Internet for both academic and social purposes. What proportion of students use it for more than 3 hours a day? The authors of the paper “U.S. College Students’ Internet Use: Race, Gender and Digital Divides” (Journal of Computer-Mediated Communication [2009]: 244–264) describe a survey of 7,421 students at 40 colleges and universities. The sample was selected to reflect general demographics of U.S. college students. Of the students surveyed, 2,998 reported Internet use of more than 3 hours per day. Verify that the conditions needed in order for the margin of error formula to be appropriate are met. ","The sample was selected in such a way that makes it representative of the population of U.S. college students. Additionally, there are 2,998 successes and 4,423 failures in the sample, which are both at least 10. ",Free Response,,,,
-LFD,9.2,21c,"Most American college students make use of the Internet for both academic and social purposes. What proportion of students use it for more than 3 hours a day? The authors of the paper “U.S. College Students’ Internet Use: Race, Gender and Digital Divides” (Journal of Computer-Mediated Communication [2009]: 244–264) describe a survey of 7,421 students at 40 colleges and universities. The sample was selected to reflect general demographics of U.S. college students. Of the students surveyed, 2,998 reported Internet use of more than 3 hours per day. Compute the margin of error.",margin of error = 0.011 ,Free Response,,,,
-LFD,9.2,21d,"Most American college students make use of the Internet for both academic and social purposes. What proportion of students use it for more than 3 hours a day? The authors of the paper “U.S. College Students’ Internet Use: Race, Gender and Digital Divides” (Journal of Computer-Mediated Communication [2009]: 244–264) describe a survey of 7,421 students at 40 colleges and universities. The sample was selected to reflect general demographics of U.S. college students. Of the students surveyed, 2,998 reported Internet use of more than 3 hours per day. Interpret the margin of error in the context of this problem.",It is unlikely that the estimated proportion of U.S. college students who use the Internet more than 3 hours per day (p^ = 0.404) will differ from the actual population proportion by more than 0.011 (or 1.1%).,Free Response,,,,
-LFD,9.2,22a,"Consumption of fast food is a topic of interest to researchers in the field of nutrition. The article “Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children” (Pediatrics [2004]: 112–118) reported that 1,720 of those in a random sample of 6,212 American children indicated that they ate fast food on a typical day. Use the given information to estimate the proportion of American children who eat fast food on a typical day. ",p^ = 0.277,Free Response,,,,
-LFD,9.2,22b,"Consumption of fast food is a topic of interest to researchers in the field of nutrition. The article “Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children” (Pediatrics [2004]: 112–118) reported that 1,720 of those in a random sample of 6,212 American children indicated that they ate fast food on a typical day. Verify that the conditions needed in order for the margin of error formula to be appropriate are met. ","The sample is a random sample from the population of American children. Additionally, there are 1,720 successes and 4,492 failures in the sample, which are both greater than 10. ",Free Response,,,,
-LFD,9.2,22c,"Consumption of fast food is a topic of interest to researchers in the field of nutrition. The article “Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children” (Pediatrics [2004]: 112–118) reported that 1,720 of those in a random sample of 6,212 American children indicated that they ate fast food on a typical day. Compute the margin of error. ",margin of error = 0.011 ,Free Response,,,,
-LFD,9.2,22d,"Consumption of fast food is a topic of interest to researchers in the field of nutrition. The article “Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children” (Pediatrics [2004]: 112–118) reported that 1,720 of those in a random sample of 6,212 American children indicated that they ate fast food on a typical day. Interpret the margin of error in the context of this problem.",It is unlikely that the estimated proportion of American children who indicated that they eat fast food on a typical day (p^ = 0.277) will differ from the actual population proportion by more than 0.011.,Free Response,,,,
-LFD,9.2,29,"The article “Viewers Speak Out Against Reality TV” (Associated Press, September 12, 2005) included the following statement: “Few people believe there’s much reality in reality TV: a total of 82% said the shows are either ‘totally made up’ or ‘mostly distorted.’” This statement was based on a survey of 1,002 randomly selected adults. Compute and interpret the margin of error for the reported percentage.",margin of error = 0.024. It is unlikely that the estimated proportion of adults who believe that the shows are mostly or totally made up (p^ = 0.82) will differ from the actual population proportion by more than 0.024.,Free Response,,,,
-LFD,9.2,31,"The Gallup Organization conducts and annual survey on crime. It was reported that 25% of all households experienced some sort of crime during the past year. This estimate was based on a sample of 1,002 selected adults. The report states, “One can say with 95% confidence that the margin of sampling error is ±3 percentage points.” Explain how this statement is justified.","In this case, margin of error = 0.027, which rounds to 3%.",Free Response,,,,
-LFD,9.2,33,"An article in the Chicago Tribue (August 29, 1999) reported that in a poll of residents of the Chicago suburbs, 43% felt that their financial situation had improved during the past year. The following state is from the article: “The findings of this Tribune poll are based on interviews with 930 randomly selected suburban residents. The sample included suburban Cook County plus DuPage, Kane, Lake, McHenry, and Will Counties. In a sample of this size, one can say with 95% certainty that results will differ by no more than 3% from results obtained if all residents had been included in the poll.” Give a statistical argument to justify the claim that the estimate of 43% is within 3% of the actual percentage of all residents who feel that their financial situation has improved.","In this case, margin of error = 0.032, which rounds to 3%. The margin of error tells you that it is unlikely that the estimate will differ from the actual population proportion by more than 0.03.",Free Response,,,,
-LFD,9.3,34a,"Suppose that a city planning commission wants to know the proportion of city residents who support installing streetlights in the downtown area. Two different people independently selected random samples of city residents and used their sample data to construct the following confidence intervals for the population proportion: Interval 1: (0.28, 0.34) Interval 2: (0.31, 0.33). Explain how it is possible that the two confidence intervals are not centered in the same place. ","The confidence intervals are centered at p^. In this case, the intervals are not centered in the same place because two different samples were taken, each yielding a different value of p^. ",Free Response,,,,
-LFD,9.3,34b,"Suppose that a city planning commission wants to know the proportion of city residents who support installing streetlights in the downtown area. Two different people independently selected random samples of city residents and used their sample data to construct the following confidence intervals for the population proportion: Interval 1: (0.28, 0.34) Interval 2: (0.31, 0.33). Which of the two intervals conveys more precise information about the value of the population proportion? ",Interval 2 conveys more precise information about the value of the population proportion because Interval 2 is narrower than Interval 1. ,Free Response,,,,
-LFD,9.3,34c,"Suppose that a city planning commission wants to know the proportion of city residents who support installing streetlights in the downtown area. Two different people independently selected random samples of city residents and used their sample data to construct the following confidence intervals for the population proportion: Interval 1: (0.28, 0.34) Interval 2: (0.31, 0.33). If both confidence intervals have a 95% confidence level, which confidence interval was based on the smaller sample size? How can you tell? ","A smaller sample size produces a larger margin of error. In this case, Interval 1 (being wider than Interval 2) was based on the smaller sample size. ",Free Response,,,,
-LFD,9.3,34d,"Suppose that a city planning commission wants to know the proportion of city residents who support installing streetlights in the downtown area. Two different people independently selected random samples of city residents and used their sample data to construct the following confidence intervals for the population proportion: Interval 1: (0.28, 0.34) Interval 2: (0.31, 0.33). If both confidence intervals were based on the same sample size, which interval has the higher confidence level? How can you tell?","Interval 1 would have the higher confidence level, because the z critical value for higher confidence is larger, resulting in a wider confidence interval.",Free Response,,,,
-LFD,9.3,35a,Explain which one would result in a wider large-sample confidence interval for p:  90% confidence level or 95% confidence level ,95%,Free Response,,,,
-LFD,9.3,35b,Explain which one would result in a wider large-sample confidence interval for p: n = 100 or n = 400,n = 100,Free Response,,,,
-LFD,9.3,36,"Based on data from a survey of 1,200 randomly selected Facebook users (USA Today, March 24, 2010), a 95% confidence interval for the proportion of all Facebook users who say it is OK for someone to “friend” his or her boss is (0.41, 0.47). What is the meaning of the confidence level of 95% that is associated with this interval?",The method used to construct this interval estimate is successful in capturing the actual value of the population proportion about 95% of the time.,Free Response,,,,
-LFD,9.3,37a,"Appropriate use of the interval p^ ± (z critical values) sqrt((p^(1-p^))/n) requires a large sample. For the following combination of n and p^, indicate whether the sample size is large enough for this interval to be appropriate: n = 50 and p^ = 0.30",yes,Free Response,,,,
-LFD,9.3,37b,"Appropriate use of the interval p^ ± (z critical values) sqrt((p^(1-p^))/n) requires a large sample. For the following combination of n and p^, indicate whether the sample size is large enough for this interval to be appropriate: n = 50 and p^ = 0.05",no,Free Response,,,,
-LFD,9.3,37c,"Appropriate use of the interval p^ ± (z critical values) sqrt((p^(1-p^))/n) requires a large sample. For the following combination of n and p^, indicate whether the sample size is large enough for this interval to be appropriate: n = 15 and p^ = 0.45",no,Free Response,,,,
-LFD,9.3,37d,"Appropriate use of the interval p^ ± (z critical values) sqrt((p^(1-p^))/n) requires a large sample. For the following combination of n and p^, indicate whether the sample size is large enough for this interval to be appropriate: n = 100 and p^ = 0.01",no,Free Response,,,,
-LFD,9.3,38a,The formula used to compute a large-sample confidence interval for p is p^ ± (z critical values) sqrt((p^(1-p^))/n). What is the appropriate z critical value for a 90% confidence level?,1.1645,Free Response,,,,
-LFD,9.3,38b,The formula used to compute a large-sample confidence interval for p is p^ ± (z critical values) sqrt((p^(1-p^))/n). What is the appropriate z critical value for a 99% confidence level? ,2.58,Free Response,,,,
-LFD,9.3,38c,The formula used to compute a large-sample confidence interval for p is p^ ± (z critical values) sqrt((p^(1-p^))/n). What is the appropriate z critical value for a 80% confidence level? ,1.28,Free Response,,,,
-LFD,9.3,39,"The article “Career Expert Provides DOs and DON'Ts for Job Seekers on Social Networking” (CareerBuilder.com, August 19, 2009) included data from a survey of 2,667 hiring managers and human resource professionals. The article noted that more employers are now using social networks to screen job applicants. Of the 2,667 people who participated in the survey, 1,200 indicated that they use social networking sites such as Facebook, MySpace, and LinkedIn to research job applicants. Assume that the sample is representative of hiring managers and human resource professionals. Answer the four key questions (QSTN) to confirm that the suggested method in this situation is a confidence interval for a population proportion.",Question type (Q): Estimation; Study type (S): Sample data; Type of data (T): One categorical variable; Number of samples or treatments (N): One sample.,Free Response,,,,
-LFD,9.3,40,"The article “Career Expert Provides DOs and DON'Ts for Job Seekers on Social Networking” (CareerBuilder.com, August 19, 2009) included data from a survey of 2,667 hiring managers and human resource professionals. The article noted that more employers are now using social networks to screen job applicants. Of the 2,667 people who participated in the survey, 1,200 indicated that they use social networking sites such as Facebook, MySpace, and LinkedIn to research job applicants. Assume that the sample is representative of hiring managers and human resource professionals. Use the five-step process for estimation problems (EMC) to construct and interpret a 95% confidence interval for the proportion of hiring managers and human resource professionals who use social networking sites to research job applicants. Identify each of the five steps in your solution.","Estimate (E): The proportion of hiring managers and human resources professionals who use social networking sites to research job applicants, p, will be estimated. Method (M): Because the answers to the four key questions are estimation, sample data, one categorical variable, and one sample, consider a 95% confidence interval for the proportion of hiring managers and human resource professionals who use social networking sites to research job applicants. Check (C): The sample is representative of hiring managers and human resource professionals. In addition, the sample includes 1,200 successes and 1,467 failures, which are both greater than 10. The two required conditions are satisfied. Calculations (C): (0.4311, 0.4689) Communicate Results (C): Interpret confidence interval: You can be 95% confident that the actual proportion of hiring managers and human resources professionals is somewhere between 0.4311 and 0.4689. Interpret confidence level: The method used to construct this interval estimate is successful in capturing the actual value of the population proportion about 95% of the time.",Free Response,,,,
-LFD,9.3,41a,"The article “Students Increasingly Turn to Credit Cards” (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college freshmen and 48% of college seniors carry a credit card balance from month to month. Suppose that the reported percentages were based on random samples of 1,000 college freshmen and 1,000 college seniors. Construct and interpret a 90% confidence interval for the proportion of college freshmen who carry a credit card balance from month to month. ","(0.3449, 0.3951). You can be 90% confident that the actual proportion of college freshmen who carry a credit card balance is somewhere between 0.3449 and 0.3951. ",Free Response,,,,
-LFD,9.3,41b,"The article “Students Increasingly Turn to Credit Cards” (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college freshmen and 48% of college seniors carry a credit card balance from month to month. Suppose that the reported percentages were based on random samples of 1,000 college freshmen and 1,000 college seniors. Construct and interpret a 90% confidence interval for the proportion of college seniors who carry a credit card balance from month to month. ","(0.454, 0.506). You can be 90% confident that the actual proportion of college seniors who carry a credit card balance is somewhere between 0.454 and 0.506. ",Free Response,,,,
-LFD,9.3,41c,"The article “Students Increasingly Turn to Credit Cards” (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college freshmen and 48% of college seniors carry a credit card balance from month to month. Suppose that the reported percentages were based on random samples of 1,000 college freshmen and 1,000 college seniors. Explain why the 90% confidence intervals for freshman and seniors are not the same width.","The two confidence intervals do not have the same width because the standard errors, and hence the margins of error, are based on two different values for p^.",Free Response,,,,
-LFD,9.3,42a,"In a survey of 1,000 randomly selected adults in the United States, participants were asked what their most favorite and least favorite subjects were when they were in school (Associated Press, August 17, 2005). In what might seem like a contradiction, math was chosen more often than any other subject in both categories. Math was chosen by 230 of the 1,000 as their most favorite subject and chosen by 370 of the 1,000 as their least favorite subject. Construct and interpret a 95% confidence interval for the proportion of U.S. adults for whom math was their most favorite subject.","(0.2039, 0.2561). You can be 95% confident that the actual proportion of U.S. adults for whom math was their most favorite subject is somewhere between 0.2039 and 0.2561. ",Free Response,,,,
-LFD,9.3,42b,"In a survey of 1,000 randomly selected adults in the United States, participants were asked what their most favorite and least favorite subjects were when they were in school (Associated Press, August 17, 2005). In what might seem like a contradiction, math was chosen more often than any other subject in both categories. Math was chosen by 230 of the 1,000 as their most favorite subject and chosen by 370 of the 1,000 as their least favorite subject. Construct and interpret a 95% confidence interval for the proportion of U.S. adults for whom math was their least favorite subject.","(0.3401, 0.3999). You can be 95% confident that the actual proportion of U.S. adults for whom math was their least favorite subject is somewhere between 0.3401 and 0.3999.",Free Response,,,,
-LFD,9.3,43a,"It probably wouldn’t surprise you to know that Valentine’s Day means big business for florists, jewelry stores, and restaurants. But did you know that it is also a big day for pet stores? In January 2010, the National Retail Federation conducted a survey of consumers in a representative sample of adult Americans (“This Valentine's Day, Couples Cut Back on Gifts to Each Other, According to NRF Survey,” www.nrf.com). One of the questions in the survey asked if the respondent planned to spend money on a Valentine’s Day gift for his or her pet. The proportion who responded that they did plan to purchase a gift for their pet was 0.173. Suppose that the sample size for this survey was n = 200. Construct and interpret a 95% confidence interval for the proportion of all adult Americans who planned to purchase a Valentine’s Day gift for their pet. ","(0.1206, 0.2254). You can be 95% confident that the actual proportion of all adult Americans who planned to purchase a Valentine’s Day gift for their pet is somewhere between 0.1206 and 0.2254. ",Free Response,,,,
-LFD,9.3,43b,"It probably wouldn’t surprise you to know that Valentine’s Day means big business for florists, jewelry stores, and restaurants. But did you know that it is also a big day for pet stores? In January 2010, the National Retail Federation conducted a survey of consumers in a representative sample of adult Americans (“This Valentine's Day, Couples Cut Back on Gifts to Each Other, According to NRF Survey,” www.nrf.com). One of the questions in the survey asked if the respondent planned to spend money on a Valentine’s Day gift for his or her pet. A confidence interval was calculated using n = 200. The actual sample size for the survey was much larger than 200. Would a 95% confidence interval computed using the actual sample size have been narrower or wider than the confidence interval computed?","The 95% confidence interval computed for the actual sample size would have been narrower than the 95% confidence interval computed with n = 200 because the standard error, and hence the margin of error, would have been smaller.",Free Response,,,,
-LFD,9.3,55a,"One thousand randomly selected adult Americans participated in a survey conducted by the Associated Press (June, 2006). When asked “Do you think it is sometimes justified to lie, or do you think lying is never justified?” 52% responded that lying was never justified. When asked about lying to avoid hurting someone’s feelings, 650 responded that this was often or sometimes OK. Construct and interpret a 90% confidence interval for the proportion of adult Americans who would say that lying is never justified.","(0.494, 0.546). You can be 90% confident that the actual proportion of adult Americans who would say that lying is never justified is somewhere between 0.494 and 0.546. ",Free Response,,,,
-LFD,9.3,55b,"One thousand randomly selected adult Americans participated in a survey conducted by the Associated Press (June, 2006). When asked “Do you think it is sometimes justified to lie, or do you think lying is never justified?” 52% responded that lying was never justified. When asked about lying to avoid hurting someone’s feelings, 650 responded that this was often or sometimes OK. Construct and interpret a 90% confidence interval for the proportion of adult Americans who think that it is often or sometimes OK to lie to avoid hurting someone’s feelings. ","(0.6252, 0.6748). You can be 90% confident that the actual proportion of adult Americans who would say that it is often or sometimes OK to lie to avoid hurting someone’s feelings is somewhere between 0.6252 and 0.6748. ",Free Response,,,,
-LFD,9.3,55c,"One thousand randomly selected adult Americans participated in a survey conducted by the Associated Press (June, 2006). When asked “Do you think it is sometimes justified to lie, or do you think lying is never justified?” 52% responded that lying was never justified. When asked about lying to avoid hurting someone’s feelings, 650 responded that this was often or sometimes OK. Using the 90% confidence intervals for the proportion of adult Americans who think lying is never justified and who think it is often or sometimes OK to lie, comment on the apparent inconsistency in the responses.","The confidence interval for the proportion of adult Americans who would say that lying is never justified indicates that it is plausible that at least 50% of adult Americans would say that lying is never justified, and the confidence interval for proportion of adult Americans who would say that it is often or sometimes OK to lie indicates that it is also plausible that well over 50% of adult Americans would say that it is often or sometimes OK to lie to avoid hurting someone’s feelings. These are contradictory responses.",Free Response,,,,
-LFD,9.3,57,"In a study of 1,710 schoolchildren in Australia (Herald Sun, October 27, 1994), 1,060 children indicated that they normally watch TV before school in the morning. (Interestingly, only 35% of the parents said their children watched TV before school.) Construct and interpret a 95% confidence interval for the proportion of all Australian children who say they watch TV before school. In order for the method used to construct the interval to be valid, what assumption about the sample must be reasonable?","(0.597, 0.643). You can be 95% confident that the actual proportion of all Australian children who would say that they watch TV before school is between 0.597 and 0.643. For the method used to construct the interval to be valid, the sample must have either been randomly selected from the population of interest or the sample must have been selected in such a way that it should result in a sample that is representative of the population.",Free Response,,,,
-LFD,9.3,59,"An Associated Press article on potential violent behavior reported the results of a survey of 750 workers who were employed full time (San Luis Obispo Tribune, September 7, 1999). Of those surveyed, 125 indicated that they were so angered by a coworker during the past year that they felt like hitting the coworker (but didn’t). Assuming that it is reasonable to regard this sample as representative of the population of full-time workers, use this information to construct and interpret a 90% confidence interval estimate of p, the proportion of all full-time workers so angered in the last year that they wanted to hit a coworker.","(0.1446, 0.1894). You can be 90% confident that the actual proportion of all full-time workers so angered in the past year that they wanted to hit a co-worker is between 0.1446 and 0.1894.",Free Response,,,,
-LFD,9.4,61,"A discussion of digital ethics appears in the article “Academic Cheating, Aided by Cell Phones or Web, Shown to be Common” (Los Angeles Times, June 17, 2009). One question posed in the article is: What proportion of college students have used cell phones to cheat on an exam? Suppose you have been asked to estimate this proportion for students enrolled at a large college. How many students should you include in your sample if you want to estimate this proportion with a margin of error of 0.02?","Assuming a 95% confidence level, and using a conservative estimate of p = 0.5, n = 2,401. ",Free Response,,,,
-LFD,9.4,62,"The article “Consumers Show Increased Liking for Diesel Autos” (USA Today, January 29, 2003) reported that 27% of U.S. consumers would opt for a diesel car if it ran as cleanly and performed as well as a car with a gas engine. Suppose that you suspect that the proportion might be different in your area. You decide to conduct a survey to estimate this proportion for the adult residents of your city. What is the required sample size if you want to estimate this proportion with a margin of error of 0.05? Compute the required sample size first using 0.27 as a preliminary estimate of p and then using the conservative value of 0.5. How do the two sample sizes compare? What sample size would you recommend for this study?","Assuming a 95% confidence level, and using the preliminary estimate of p = 0.27, n = 303. Using the conservative estimate of p = 0.5, n = 385. ",Free Response,,,,
-LFD,9.4,63,"A manufacturer of small appliances purchases plastic handles for coffeepots from an outside vendor. If a handle is cracked, it is considered defective and can’t be used. A large shipment of plastic handles is received. How many handles from the shipment should be inspected in order to estimate p, the proportion of defective handles in the shipment, with a margin of error of 0.1?","Assuming a 95% confidence level, and using a conservative estimate of p = 0.5, n = 97.",Free Response,,,,
-LFD,9.4,67,"Data from a representative sample were used to estimate that 32% of all computer users in 2011 had tried to get on a Wi-Fi network that was not their own in order to save money (USA Today, May 16, 2011). You decide to conduct a survey to estimate this proportion for the current year. What is the required sample size if you want to estimate this proportion with a margin of error of 0.05? Compute the required sample size first using 0.32 as a preliminary estimate of p and then using the conservative value of 0.5. How do the two sample sizes compare? What sample size would you recommend for this study?",,Free Response,,,,
-LFD,9.4,69a,"The article “Hospitals Dispute Medtronic Data on Wires” (The Wall Street Journal, February 4, 2010) describes several studies of the failure rate of defibrillators used in the treatment of heart problems. In one study conducted by the Mayo Clinic, it was reported that failures within the first 2 years were experienced by 18 of 89 patients under 50 years old and 13 of 362 patients age 50 and older. Assume that these two samples are representative of patients who receive this type of defibrillator in the two age groups. Construct and interpret a 95% confidence interval for the proportion of patients under 50 years old who experience a failure within the first 2 years. ",,Free Response,,,,
-LFD,9.4,69b,"The article “Hospitals Dispute Medtronic Data on Wires” (The Wall Street Journal, February 4, 2010) describes several studies of the failure rate of defibrillators used in the treatment of heart problems. In one study conducted by the Mayo Clinic, it was reported that failures within the first 2 years were experienced by 18 of 89 patients under 50 years old and 13 of 362 patients age 50 and older. Assume that these two samples are representative of patients who receive this type of defibrillator in the two age groups. Construct and interpret a 99% confidence interval for the proportion of patients age 50 and older who experience a failure within the first 2 years. ",,Free Response,,,,
-LFD,9.4,69c,"The article “Hospitals Dispute Medtronic Data on Wires” (The Wall Street Journal, February 4, 2010) describes several studies of the failure rate of defibrillators used in the treatment of heart problems. In one study conducted by the Mayo Clinic, it was reported that failures within the first 2 years were experienced by 18 of 89 patients under 50 years old and 13 of 362 patients age 50 and older. Assume that these two samples are representative of patients who receive this type of defibrillator in the two age groups. Suppose that the researchers wanted to estimate the proportion of patients under 50 years old who experience this type of failure with a margin of error of 0.03. How large a sample should be used? Use the given study results to obtain a preliminary estimate of the population proportion.",,Free Response,,,,
-LFD,9R,73,Will p^ from a random sample of size 400 tend to be closer to the actual value of the population proportion when p = 0.4 or when p = 0.7? Provide an explanation for your choice.,p^ from a random sample of size 400 tends to be closer to the actual value when p = 0.7. ,Free Response,,,,
-LFD,9R,75a,"In response to budget cuts, county officials are interested in learning about the proportion of county residents who favor closure of a county park rather than closure of a county library. In a random sample of 500 county residents, 198 favored closure of a county park. For the statement below, indicate if the statement is correct or incorrect. If the statement is incorrect, explain what makes it incorrect. Statement: It is unlikely that the estimate p^ = 0.396 differs from the value of the actual population proportion by more than 0.0429.",correct,Free Response,,,,
-LFD,9R,75b,"In response to budget cuts, county officials are interested in learning about the proportion of county residents who favor closure of a county park rather than closure of a county library. In a random sample of 500 county residents, 198 favored closure of a county park. For the statement below, indicate if the statement is correct or incorrect. If the statement is incorrect, explain what makes it incorrect. Statement: The estimate p^ = 0.396 will never differ from the value of the actual population proportion by more than 0.0429.",Incorrect. About 95% of all possible sample proportions computed from random samples of size 500 will be within 0.0429 of the actual population proportion (this is the margin of error for a 95% confidence level). About 5% of the sample proportions would differ from the actual population proportion by more than 0.0429. ,Free Response,,,,
-LFD,9R,75c,"In response to budget cuts, county officials are interested in learning about the proportion of county residents who favor closure of a county park rather than closure of a county library. In a random sample of 500 county residents, 198 favored closure of a county park. For the statement below, indicate if the statement is correct or incorrect. If the statement is incorrect, explain what makes it incorrect. Statement: It is unlikely that the estimate p^ = 0.396  differs from the value of the actual population proportion by more than 0.0219.",Incorrect. About 68% of all possible sample proportions computed from random samples of size 500 will be within 0.0219 of the actual population proportion (this is the standard deviation of the sampling distribution). About 32% of the sample proportions would differ from the actual population proportion by more than 0.0219.,Free Response,,,,
-LFD,9R,77a,"In a survey on supernatural experiences, 722 of 4,013 adult Americans reported that they had seen a ghost (""What Supernatural Experiences We've Had,"" USA Today, February 8, 2010). Assume that this sample is representative of the population of adult Americans. Use the given information to estimate the proportion of adult Americans who would say they have seen a ghost. ",p^ = 0.18 ,Free Response,,,,
-LFD,9R,77b,"In a survey on supernatural experiences, 722 of 4,013 adult Americans reported that they had seen a ghost (""What Supernatural Experiences We've Had,"" USA Today, February 8, 2010). Assume that this sample is representative of the population of adult Americans. Verify that the conditions for use of the margin of error formula to be appropriate are met. ","The two conditions are: (1) Large sample size: The number of successes and failures in the sample are at least 10. In this case, there are 722 successes, and 3,291 failures, which are both much greater than 10. (2) Random selection of the sample, or the sample is representative of the population: We are told that the sample is representative of the population. ",Free Response,,,,
-LFD,9R,77c,"In a survey on supernatural experiences, 722 of 4,013 adult Americans reported that they had seen a ghost (""What Supernatural Experiences We've Had,"" USA Today, February 8, 2010). Assume that this sample is representative of the population of adult Americans. Compute the margin of error. ",margin of error = 0.0119 ,Free Response,,,,
-LFD,9R,77d,"In a survey on supernatural experiences, 722 of 4,013 adult Americans reported that they had seen a ghost (""What Supernatural Experiences We've Had,"" USA Today, February 8, 2010). Assume that this sample is representative of the population of adult Americans. Interpret the margin of error in context. ",It is unlikely that the estimate p^ = 0.18 differs from the value of the actual proportion of adults Americans who say they have seen a ghost by more than 0.0119. ,Free Response,,,,
-LFD,9R,77e,"In a survey on supernatural experiences, 722 of 4,013 adult Americans reported that they had seen a ghost (""What Supernatural Experiences We've Had,"" USA Today, February 8, 2010). Assume that this sample is representative of the population of adult Americans. Construct and interpret a 90% confidence interval for the proportion of all adult Americans who have seen a ghost. ","(0.17, 0.19). You can be 90% confident that the actual proportion of all adult Americans who have seen a ghost is somewhere between 0.17 and 0.19.",Free Response,,,,
-LFD,9R,77f,"In a survey on supernatural experiences, 722 of 4,013 adult Americans reported that they had seen a ghost (""What Supernatural Experiences We've Had,"" USA Today, February 8, 2010). Assume that this sample is representative of the population of adult Americans. Would a 99% confidence interval be narrower or wider than a 90% confidence interval? Justify your answer.",A 99% confidence interval would be wider because the z critical value for 99% confidence (z = 2.58) is larger than the z critical value for 90% confidence (z = 1.645).,Free Response,,,,
-LFD,9R,79a,Describe how the confidence level affects the width of the large-sample confidence interval for p.,"As the confidence level increases, the width of the confidence interval also increases. ",Free Response,,,,
-LFD,9R,79b,Describe how the sample size affects the width of the large-sample confidence interval for p.,"As the sample size increases, the width of the confidence interval decreases. ",Free Response,,,,
-LFD,9R,79c,Describe how the value of p^ affects the width of the large-sample confidence interval for p.,"When p^ = 0.5, the margin of error (for a fixed sample size) is maximum and decreases symmetrically as p^ decreases toward 0 or increases toward 1. A larger margin of error results in a wider confidence interval.",Free Response,,,,
-LFD,9R,81,"The study ""Digital Footprints"" (Pew Internet & American Life Project, www.pewinternet.org, 2007) reported that 47% of Internet users have searched for information about themselves online. The 47% figure was based on a random sample of Internet users. Suppose that the sample size was n = 300 (the actual sample size was much larger). Answer the four key questions (QSTN) to confirm that the suggested method in this situation is a large-sample confidence interval for a population proportion.",Question type (Q): Estimation; Study type (S): Sample data: Type of data (T): One categorical variable; Number of samples or treatments (N): One sample.,Free Response,,,,
-LFD,9R,83a,"The article ""Kids Digital Day: Almost 8 Hours"" (USA Today, January 20, 2010) summarized a national survey of 2,002 Americans ages 8 to 18. The sample was selected to be representative of Americans in this age group. Of those surveyed, 1,321 reported owning a cell phone. Use this information to construct and interpret a 90% confidence interval for the proportion of all Americans ages 8 to 18 who own a cell phone. ","(0.6426, 0.6774). You can be 90% confident that the actual proportion of all Americans ages 8 to 18 who own a cell phone is somewhere between 0.6426 and 0.6774. ",Free Response,,,,
-LFD,9R,83b,"The article ""Kids Digital Day: Almost 8 Hours"" (USA Today, January 20, 2010) summarized a national survey of 2,002 Americans ages 8 to 18. The sample was selected to be representative of Americans in this age group. Of those surveyed, 1,522 reported owning an MP3 music player. Use this information to construct and interpret a 90% confidence interval for the proportion of all Americans ages 8 to 18 who own an MP3 music player. ","(0.7443, 0.7757). You can be 90% confident that the actual proportion of all Americans ages 8 to 18 who own an MP3 music player is somewhere between 0.7443 and 0.7757.",Free Response,,,,
-LFD,9R,83c,"The article ""Kids Digital Day: Almost 8 Hours"" (USA Today, January 20, 2010) summarized a national survey of 2,002 Americans ages 8 to 18. The sample was selected to be representative of Americans in this age group. Explain why the 90% confidence interval for the proportion of all Americans ages 8 to 18 who own a cell phone is narrower than the 90% confidence interval for the proportion of all Americans ages 8 to 18 who own an MP3 music player even though the confidence levels and the sample sizes used to compute the two intervals were the same.","The interval of all Americans ages 8 to 18 who own a cell phone is narrower because p^ of 0.76 is farther from 0.5 than p^ of 0.66. For a fixed sample size and confidence level, the confidence interval is widest when p^ 0.5 and decreases symmetrically as p^ moves closer to 0 or 1.",Free Response,,,,
-LFD,10.1,1,Explain why the statement p^ = 0.40 is not a legitimate hypothesis.,p^ is a sample statistic. Hypotheses are about population characteristics.,Free Response,,,,
-LFD,10.1,2,"CareerBuilder.com conducted a survey to learn about the proportion of employers who had ever sent an employee home because they were dressed inappropriately (June 17, 2008, www.careerbuilder.com). Suppose you are interested in determining if the resulting data provide strong evidence in support of the claim that more than one-third of employers have sent an employee home to change clothes. To answer this question, what null and alternative hypotheses should you test?",H0: p = 1/3 and Ha: p > 1/3,Free Response,,,,
-LFD,10.1,3,"The report “How Teens Use Media” (Nielsen, June 2009) says that 67% of U.S. teens have a Facebook page that they update at least once a week. Suppose you plan to select a random sample of 400 students at the local high school. You will ask each student in the sample if he or she has a Facebook page and if it is updated at least once per week. You plan to use the resulting data to decide if there is evidence that the proportion of students at the high school who have a Facebook page that they update at least once a week differs from the national figure given in the Nielsen report. What hypotheses should you test?",H0: p = 0.67 and Ha: p ≠ 0.67,Free Response,,,,
-LFD,10.1,4a,"The article “Poll Finds Most Oppose Return to Draft, Wouldn’t Encourage Children to Enlist” (Associated Press, December 18, 2005) reports that in a random sample of 1,000 American adults, 430 answered yes to the following question: “If the military draft were reinstated, would you favor drafting women as well as men?” The data were used to test H0: p=0.5 versus Ha: p<0.5, and the null hypothesis was rejected. Based on the result of the hypothesis test, what can you conclude about the proportion of American adults who favor drafting women if a military draft were reinstated? ",There is convincing evidence that the proportion of American adults who favor drafting women is less than 0.5.,Free Response,,,,
-LFD,10.1,4b,"The article “Poll Finds Most Oppose Return to Draft, Wouldn’t Encourage Children to Enlist” (Associated Press, December 18, 2005) reports that in a random sample of 1,000 American adults, 430 answered yes to the following question: “If the military draft were reinstated, would you favor drafting women as well as men?” The data were used to test H0: p=0.5 versus Ha: p<0.5, and the null hypothesis was rejected. Is it reasonable to say that the data provide strong support for the alternative hypothesis? ",Yes,Free Response,,,,
-LFD,10.1,4c,"The article “Poll Finds Most Oppose Return to Draft, Wouldn’t Encourage Children to Enlist” (Associated Press, December 18, 2005) reports that in a random sample of 1,000 American adults, 430 answered yes to the following question: “If the military draft were reinstated, would you favor drafting women as well as men?” The data were used to test H0: p=0.5 versus Ha: p<0.5, and the null hypothesis was rejected. Is it reasonable to say that the data provide strong evidence against the null hypothesis?",Yes,Free Response,,,,
-LFD,10.1,5a,"Do state laws that allow private citizens to carry concealed weapons result in a reduced crime rate? The author of a study carried out by the Brookings Institution is reported as saying, “The strongest thing I could say is that I don’t see any strong evidence that they are reducing crime” (San Luis Obispo Tribune, January 23, 2003). Is this conclusion consistent with testing H0: concealed weapons laws reduce crime versus Ha: concealed weapons laws do not reduce crime or with testing H0: concealed weapons laws do not reduce crime versus Ha: concealed weapons laws reduce crime Explain.",The conclusion is consistent with testing H0; concealed weapons laws do not reduce crime versus Ha; concealed weapons laws reduce crime. ,Free Response,,,,
-LFD,10.1,5b,"Do state laws that allow private citizens to carry concealed weapons result in a reduced crime rate? The author of a study carried out by the Brookings Institution is reported as saying, “The strongest thing I could say is that I don’t see any strong evidence that they are reducing crime” (San Luis Obispo Tribune, January 23, 2003). Does the stated conclusion indicate that the null hypothesis was rejected or not rejected? Explain.",The null hypothesis was not rejected because no evidence was found that the laws were reducing crime.,Free Response,,,,
-LFD,10.1,6,"In a hypothesis test, what does it mean to say that the null hypothesis was rejected?","The sample data provide convincing evidence against the null hypothesis. If the null hypothesis were true, the sample data would be very unlikely.",Free Response,,,,
-LFD,10.1,13a,Is p = 0.65 a legitimate hypotheses?,legitimate ,Free Response,,,,
-LFD,10.1,13b,Is p^ = 0.90 a legitimate hypotheses?,not legitimate ,Free Response,,,,
-LFD,10.1,13c,Is p^ = 0.10 a legitimate hypotheses?,not legitimate ,Free Response,,,,
-LFD,10.1,13d,Is p = 0.5 a legitimate hypotheses?,legitimate ,Free Response,,,,
-LFD,10.1,13e,Is p > 4.30 a legitimate hypotheses?,not legitimate,Free Response,,,,
-LFD,10.1,15,"A college has decided to introduce the use of plus and minus with letter grades, as long as there is convincing evidence that more than 60% of the faculty favor the change. A random sample of faculty will be selected, and the resulting data will be used to test the relevant hypotheses. If p represents the proportion of all faculty who favor a change to plus–minus grading, which of the following pairs of hypotheses should be tested? H0: p = 0.6 verses Ha: p < 0.6 or H0: p = 0.6 verses Ha: p > 0.6. Explain your choice.","H0: p = 0.6 versus Ha: p > 0.6. In order to make the change, the university requires evidence that more than 60% of the faculty are in favor of the change.",Free Response,,,,
-LFD,10.2,17,One type of error in a hypothesis test is rejecting the null hypothesis when it is true. What is the other type of error that might occur when a hypothesis test is carried out?,Not rejecting the null hypothesis when it is not true,Free Response,,,,
-LFD,10.2,18,"Suppose that for a particular hypothesis test, the consequences of a Type I error are very serious. Would you want to carry out the test using a small significance level α (such as 0.01) or a larger significance level (such as 0.10)? Explain the reason for your choice.",A small significance level because α is the probability of a Type I error.,Free Response,,,,
-LFD,10.2,19a,"Occasionally, warning flares of the type contained in most automobile emergency kits fail to ignite. A consumer group wants to investigate a claim that the proportion of defective flares made by a particular manufacturer is much higher than the advertised value of 0.1. A large number of flares will be tested, and the results will be used to decide between H0: H0: p = 0.01 and Ha: p > 0.01, where p represents the actual proportion of defective flares made by this manufacturer. If H0 is rejected, charges of false advertising will be filed against the manufacturer. Explain why the alternative hypothesis was chosen to be Ha: p > 0.01.","Before filing charges of false advertising against the company, the consumer advocacy group would require convincing evidence that more than 10% of the flares are defective. ",Free Response,,,,
-LFD,10.2,19b,"Occasionally, warning flares of the type contained in most automobile emergency kits fail to ignite. A consumer group wants to investigate a claim that the proportion of defective flares made by a particular manufacturer is much higher than the advertised value of 0.1. A large number of flares will be tested, and the results will be used to decide between H0: H0: p = 0.01 and Ha: p > 0.01, where p represents the actual proportion of defective flares made by this manufacturer. If H0 is rejected, charges of false advertising will be filed against the manufacturer. Give a consequence of a Type I and Type II error.",A Type I error is thinking that more than 10% of the flares are defective when in fact 10% (or fewer) of the flares are defective. This would result in the expensive and time-consuming process of filing charges of false advertising against the company when the company advertising is not false. A Type II error is not thinking that more than 10% of the flares are defective when in fact more than 10% of the flares are defective. This would result in the consumer advocacy group not filing charges when the company advertising was false.,Free Response,,,,
-LFD,10.2,20a,"A television manufacturer states that at least 90% of its TV sets will not need service during the first 3 years of operation. A consumer group wants to investigate this statement. A random sample of n = 100 purchasers is selected and each person is asked if the set purchased needed repair during the first 3 years. Let p denote the proportion of all sets made by this manufacturer that will not need service in the first 3 years. The consumer group does not want to claim false advertising unless there is strong evidence that p < 0.09. The appropriate hypotheses are then H0: p = 0.09 versus Ha: p < 0.09. In the context of this problem, describe Type I and Type II errors, and discuss the possible consequences of each. ",A Type I error would be thinking that less than 90% of the TV sets need no repair when in fact (at least) 90% need no repair. The consumer agency might take action against the manufacturer when the manufacturer is not at fault. A Type II error would be not thinking that less than 90% of the TV sets need no repair when in fact less than 90% need no repair. The consumer agency would not take action against the manufacturer when the manufacturer is making untrue claims about the reliability of the TV sets. ,Free Response,,,,
-LFD,10.2,20b,A television manufacturer states that at least 90% of its TV sets will not need service during the first 3 years of operation. A consumer group wants to investigate this statement. A random sample of n = 100 purchasers is selected and each person is asked if the set purchased needed repair during the first 3 years. Let p denote the proportion of all sets made by this manufacturer that will not need service in the first 3 years. The consumer group does not want to claim false advertising unless there is strong evidence that p < 0.09. The appropriate hypotheses are then H0: p = 0.09 versus Ha: p < 0.09. Would you recommend a test procedure that uses α = 0.01 or one that uses α = 0.10? Explain.,Taking action against the manufacturer when the manufacturer is not at fault could involve large and unnecessary legal costs to the consumer agency. α = 0.01 should be recommended.,Free Response,,,,
-LFD,10.2,27,Describe the two types of errors that might be made when a hypothesis test is carried out.,"A Type I error is rejecting a true null hypothesis, and a Type II error is not rejecting a false null hypothesis",Free Response,,,,
-LFD,10.2,31,"The article “Fewer Parolees Land Back Behind Bars” (Associated Press, April 11, 2006) includes the following statement: “Just over 38% of all felons who were released from prison in 2003 landed back behind bars by the end of the following year, the lowest rate since 1979.” Explain why it would not be necessary to carry out a hypothesis test to determine if the proportion of felons released in 2003 who landed back behind bars by the end of the following year was less than 0.40.","The “38%” value given in the article is the proportion of all felons; in other words, it is a population proportion. Therefore, you know that the population proportion is less than 0.4, and there is no need for a hypothesis test.",Free Response,,,,
-LFD,10.3,32a,"The article “Breast-Feeding Rates Up Early” (USA Today, Sept. 14, 2010) summarizes a survey of mothers whose babies were born in 2009. The Center for Disease Control sets goals for the proportion of mothers who will still be breast-feeding their babies at various ages. The goal for 12 months after birth is 0.25 or more. Suppose that the survey used a random sample of 1,200 mothers and that you want to use the survey data to decide if there is evidence that the goal is not being met. Let p denote the proportion of all mothers of babies born in 2009 who were still breast-feeding at 12 months. Describe the shape, center, and spread of the sampling distribution of p^ for random samples of size 1,200 if the null hypothesis H0: p = 0.25 is true. ",The sampling distribution of p^ will be approximately normal with a mean of 0.25 and a standard deviation of 0.0125. ,Free Response,,,,
-LFD,10.3,32b,"The article “Breast-Feeding Rates Up Early” (USA Today, Sept. 14, 2010) summarizes a survey of mothers whose babies were born in 2009. The Center for Disease Control sets goals for the proportion of mothers who will still be breast-feeding their babies at various ages. The goal for 12 months after birth is 0.25 or more. Suppose that the survey used a random sample of 1,200 mothers and that you want to use the survey data to decide if there is evidence that the goal is not being met. Let p denote the proportion of all mothers of babies born in 2009 who were still breast-feeding at 12 months. Would you be surprised to observe a sample proportion of p^ = 0.24 for a sample of size 1,200 if the null hypothesis H0: p = 0.25 were true? Explain why or why not. ","A sample proportion of p^ = 0.24 would not be surprising if p = 0.25, because it is less than one standard deviation below 0.25. This is not unusual for a normal distribution. ",Free Response,,,,
-LFD,10.3,32c,"The article “Breast-Feeding Rates Up Early” (USA Today, Sept. 14, 2010) summarizes a survey of mothers whose babies were born in 2009. The Center for Disease Control sets goals for the proportion of mothers who will still be breast-feeding their babies at various ages. The goal for 12 months after birth is 0.25 or more. Suppose that the survey used a random sample of 1,200 mothers and that you want to use the survey data to decide if there is evidence that the goal is not being met. Let p denote the proportion of all mothers of babies born in 2009 who were still breast-feeding at 12 months. Would you be surprised to observe a sample proportion as small as p^ = 0.20 for a sample of size 1,200 if the null hypothesis H0: p = 0.25  were true? Explain why or why not. ","A sample proportion of p^ = 0.20 would be surprising if p = 0.25, because it is 4 standard deviations below 0.25. This would be unusual for a normal distribution. ",Free Response,,,,
-LFD,10.3,32d,"The article “Breast-Feeding Rates Up Early” (USA Today, Sept. 14, 2010) summarizes a survey of mothers whose babies were born in 2009. The Center for Disease Control sets goals for the proportion of mothers who will still be breast-feeding their babies at various ages. The goal for 12 months after birth is 0.25 or more. Suppose that the survey used a random sample of 1,200 mothers and that you want to use the survey data to decide if there is evidence that the goal is not being met. Let p denote the proportion of all mothers of babies born in 2009 who were still breast-feeding at 12 months. The actual sample proportion observed in the study was p^ = 0.22. Based on this sample proportion, is there convincing evidence that the goal is not being met, or is the observed sample proportion consistent with what you would expect to see when the null hypothesis is true? Support your answer with a probability calculation.","If p = 0.25, P(p^≤ 0.22) = P(z ≤ -2.4) = 0.0082. Because this probability is small, there is convincing evidence that the goal is not being met.",Free Response,,,,
-LFD,10.3,33a,"The article “The Benefits of Facebook Friends: Social Capital and College Students’ Use of Online Social Network Sites” (Journal of Computer-Mediated Communication [2007]: 1143–1168) describes a study of n = 286 undergraduate students at Michigan State University. Suppose that it is reasonable to regard this sample as a random sample of undergraduates at Michigan State. You want to use the survey data to decide if there is evidence that more than 75% of the students at this university have a Facebook page that includes a photo of themselves. Let p denote the proportion of all Michigan State undergraduates who have such a page. Describe the shape, center, and spread of the sampling distribution of p^ for random samples of size 286 if the null hypothesis H0: p = 0.75 is true. ",The sampling distribution of p^ will be approximately normal with a mean of 0.75 and a standard deviation of 0.0256. ,Free Response,,,,
-LFD,10.3,33b,The article “The Benefits of Facebook Friends: Social Capital and College Students’ Use of Online Social Network Sites” (Journal of Computer-Mediated Communication [2007]: 1143–1168) describes a study of n = 286 undergraduate students at Michigan State University. Suppose that it is reasonable to regard this sample as a random sample of undergraduates at Michigan State. You want to use the survey data to decide if there is evidence that more than 75% of the students at this university have a Facebook page that includes a photo of themselves. Let p denote the proportion of all Michigan State undergraduates who have such a page. Would you be surprised to observe a sample proportion of p^ = 0.83 for a sample of size 286 if the null hypothesis H0: p = 0.75  were true? Explain why or why not. ,"A sample proportion of p^ = 0.83 would be surprising if p = 0.75, because it is more than 3 standard deviations above 0.75. This would be unusual for a normal distribution. ",Free Response,,,,
-LFD,10.3,33c,The article “The Benefits of Facebook Friends: Social Capital and College Students’ Use of Online Social Network Sites” (Journal of Computer-Mediated Communication [2007]: 1143–1168) describes a study of n = 286 undergraduate students at Michigan State University. Suppose that it is reasonable to regard this sample as a random sample of undergraduates at Michigan State. You want to use the survey data to decide if there is evidence that more than 75% of the students at this university have a Facebook page that includes a photo of themselves. Let p denote the proportion of all Michigan State undergraduates who have such a page. Would you be surprised to observe a sample proportion of p^ = 0.79 for a sample of size 286 if the null hypothesis H0: p = 0.75  were true? Explain why or why not. ,"A sample proportion of p^ = 0.79 would not be surprising if p = 0.75, because it is about 1.5 standard deviations above 0.75. This is not unusual for a normal distribution. ",Free Response,,,,
-LFD,10.3,33d,"The article “The Benefits of Facebook Friends: Social Capital and College Students’ Use of Online Social Network Sites” (Journal of Computer-Mediated Communication [2007]: 1143–1168) describes a study of n = 286 undergraduate students at Michigan State University. Suppose that it is reasonable to regard this sample as a random sample of undergraduates at Michigan State. You want to use the survey data to decide if there is evidence that more than 75% of the students at this university have a Facebook page that includes a photo of themselves. Let p denote the proportion of all Michigan State undergraduates who have such a page. The actual sample proportion observed in the study was p^ = 0.80. Based on this sample proportion, is there convincing evidence that the null hypothesis H0: p = 0.75  is not true, or is p^ consistent with what you would expect to see when the null hypothesis is true? Support your answer with a probability calculation.","If p = 0.75, P(p^ ≥ 0.80) = P(z ≥ 1.95) = 0.0256. Because this probability is small, there is convincing evidence that the null hypothesis is not true.",Free Response,,,,
-LFD,10.3,37a,"Every year on Groundhog Day (February 2), the famous groundhog Punxsutawney Phil tries to predict whether there will be 6 more weeks of winter. The article “Groundhog Has Been Off Target” (USA Today, Feb. 1, 2011) states that “based on weather data, there is no predictive skill for the groundhog.” Suppose that you plan to take a random sample of 20 years and use weather data to determine the proportion of these years the groundhog’s prediction was correct. Describe the shape, center, and spread of the sampling distribution of p^ for samples of size 20 if the groundhog has only a 50–50 chance of making a correct prediction.",The sampling distribution of p^ will be approximately normal with a mean of 0.5 and a standard deviation of 0.1118. ,Free Response,,,,
-LFD,10.3,37b,"Every year on Groundhog Day (February 2), the famous groundhog Punxsutawney Phil tries to predict whether there will be 6 more weeks of winter. The article “Groundhog Has Been Off Target” (USA Today, Feb. 1, 2011) states that “based on weather data, there is no predictive skill for the groundhog.” Suppose that you plan to take a random sample of 20 years and use weather data to determine the proportion of these years the groundhog’s prediction was correct. What sample proportion values would convince you that the groundhog’s predictions have a better than 50–50 chance of being correct?",One reasonable answer would be sample proportions greater than 0.7236 (these are sample proportions that are more than two standard deviations above 0.5).,Free Response,,,,
-LFD,10.4,38a,Use the definition of the P-value to explain why H0 would be rejected if P-value = 0.0003.,"A P-value of 0.0003 means that it is very unlikely (probability = 0.0003), assuming that H0 is true, that you would get a sample result at least as inconsistent with H0 as the one obtained in the study. H0 would be rejected. ",Free Response,,,,
-LFD,10.4,38b,Use the definition of the P-value to explain why H0 would not be rejected if P-value = 0.350.,"A P-value of 0.350 means that it is not particularly unlikely (probability = 0.350), assuming that H0 is true, that you would get a sample result at least as inconsistent with H0 as the one obtained in the study. There is no reason to reject H0.",Free Response,,,,
-LFD,10.4,39a,"The article “Poll Finds Most Oppose Return to Draft, Wouldn’t Encourage Children to Enlist” (Associated Press, December 18, 2005) reports that in a random sample of 1,000 American adults, 700 indicated that they oppose the reinstatement of a military draft. Suppose you want to use this information to decide if there is convincing evidence that the proportion of American adults who oppose reinstatement of the draft is greater than two-thirds. What hypotheses should be tested in order to answer this question? ",H0: p = 2/3 and Ha: p > 2/3,Free Response,,,,
-LFD,10.4,39b,"The article “Poll Finds Most Oppose Return to Draft, Wouldn’t Encourage Children to Enlist” (Associated Press, December 18, 2005) reports that in a random sample of 1,000 American adults, 700 indicated that they oppose the reinstatement of a military draft. Suppose you want to use this information to decide if there is convincing evidence that the proportion of American adults who oppose reinstatement of the draft is greater than two-thirds. The P-value for this test is 0.013. What conclusion would you reach if α = 0.05?",The null hypothesis would be rejected because the P-value (0.013) is less than α=0.05.,Free Response,,,,
-LFD,10.4,45,Explain why a P-value of 0.0002 would be interpreted as strong evidence against the null hypothesis.,"A P-value of 0.0002 means that it is very unlikely (probability = 0.0002), assuming that H0 is true, that you would get a sample result at least as inconsistent with H0 as the one obtained in the study. This is strong evidence against the null hypothesis.",Free Response,,,,
-LFD,10.5,47,Answer the following four key questions (Q: Estimating or hypothesis testing? S: Sample data or experiment data? T: One variable or two? Categorical or numerical? N: How many samples or treatments?) on the following scenario and indicate whether the method that you would consider would be a large-sample hypothesis test for a population proportion: The paper “College Students’ Social Networking Experiences on Facebook” (Journal of Applied Developmental Psychology [2009]: 227–238) summarized a study in which 92 students at a private university were asked how much time they spent on Facebook on a typical weekday. The researchers were interested in estimating the average time spent on Facebook by students at this university.,"Estimation, sample data, one numerical variable, one sample. A hypothesis test for a population proportion would not be appropriate.",Free Response,,,,
-LFD,10.5,48,"Answer the following four key questions (Q: Estimating or hypothesis testing? S: Sample data or experiment data? T: One variable or two? Categorical or numerical? N: How many samples or treatments?) on the following scenario and indicate whether the method that you would consider would be a large-sample hypothesis test for a population proportion: The article “iPhone Can Be Addicting, Says New Survey” (msnbc.com, March 8, 2010) described a survey administered to 200 college students who owned an iPhone. One of the questions on the survey asked students if they slept with their iPhone in bed with them. You would like to use the data from this survey to determine if there is convincing evidence that a majority of college students with iPhones sleep with their phones.","Hypothesis testing, sample data, one categorical variable, one sample. A hypothesis test for a population proportion would be appropriate.",Free Response,,,,
-LFD,10.5,49,"Answer the following four key questions (Q: Estimating or hypothesis testing? S: Sample data or experiment data? T: One variable or two? Categorical or numerical? N: How many samples or treatments?) on the following scenario and indicate whether the method that you would consider would be a large-sample hypothesis test for a population proportion: USA Today (Feb. 17, 2011) reported that 10% of 1,008 American adults surveyed about their use of e-mail said that they had ended a relationship by e-mail. You would like to use this information to estimate the proportion of all adult Americans who have used e-mail to end a relationship.","Estimation, sample data, one categorical variable, one sample. A hypothesis test for a population proportion would not be appropriate.",Free Response,,,,
-LFD,10.5,50a,"Assuming a random sample from a large population, is the following null hypotheses and sample sizes is the large-sample z test appropriate? H0: p = 0.2, n = 25",Large-sample z test is not appropriate. ,Free Response,,,,
-LFD,10.5,50b,"Assuming a random sample from a large population, is the following null hypotheses and sample sizes is the large-sample z test appropriate? H0: p = 0.6, n = 200",Large-sample z test is appropriate. ,Free Response,,,,
-LFD,10.5,50c,"Assuming a random sample from a large population, is the following null hypotheses and sample sizes is the large-sample z test appropriate? H0: p = 0.9, n = 100",Large-sample z test is appropriate. ,Free Response,,,,
-LFD,10.5,50d,"Assuming a random sample from a large population, is the following null hypotheses and sample sizes is the large-sample z test appropriate? H0: p = 0.05, n = 75",Large-sample z test is not appropriate.,Free Response,,,,
-LFD,10.5,51a,"Let p denote the proportion of students at a large university who plan to purchase a campus meal plan in the next academic year. For a large-sample z test of H0: p = 0.20 versus Ha: p =<0.20 , find the P-value associated with a z test statistic of –0.55.",0.2912,Free Response,,,,
-LFD,10.5,51b,"Let p denote the proportion of students at a large university who plan to purchase a campus meal plan in the next academic year. For a large-sample z test of H0: p = 0.20 versus Ha: p =<0.20 , find the P-value associated with a z test statistic of –0.92.",0.1788,Free Response,,,,
-LFD,10.5,51c,"Let p denote the proportion of students at a large university who plan to purchase a campus meal plan in the next academic year. For a large-sample z test of H0: p = 0.20 versus Ha: p =<0.20 , find the P-value associated with a z test statistic of –1.99.",0.0233,Free Response,,,,
-LFD,10.5,51d,"Let p denote the proportion of students at a large university who plan to purchase a campus meal plan in the next academic year. For a large-sample z test of H0: p = 0.20 versus Ha: p =<0.20 , find the P-value associated with a z test statistic of –2.24.",0.0125,Free Response,,,,
-LFD,10.5,51e,"Let p denote the proportion of students at a large university who plan to purchase a campus meal plan in the next academic year. For a large-sample z test of H0: p = 0.20 versus Ha: p =<0.20 , find the P-value associated with a z test statistic of 1.40.",0.9192,Free Response,,,,
-LFD,10.5,52a,"The paper “Debt Literacy, Financial Experiences and Over-Indebtedness” (Social Science Research Network, Working paper W14808, 2008) included data from a survey of 1,000 Americans. One question on the survey was: “You owe $3,000 on your credit card. You pay a minimum payment of $30 each month. At an Annual Percentage Rate of 12% (or 1% per month), how many years would it take to eliminate your credit card debt if you made no additional charges?” Answer options for this question were: (a) less than 5 years; (b) between 5 and 10 years; (c) between 10 and 15 years; (d) never—you will continue to be in debt; (e) don’t know; and (f) prefer not to answer. Only 354 of the 1,000 respondents chose the correct answer of never. Assume that the sample is representative of adult Americans. Is there convincing evidence that the proportion of adult Americans who can answer this question correctly is less than 0.40 (40%)? Use the five-step process for hypothesis testing (HMCCC) described in this section and α = 0.05 to test the appropriate hypotheses.","H0: p = 0.4, Ha: p < 0.4, z = -2.969, P-value = 0.001, reject H0. There is convincing evidence that the proportion of all adult Americans who would answer the question correctly is less than 0.4. ",Free Response,,,,
-LFD,10.5,52b,"The paper “Debt Literacy, Financial Experiences and Over-Indebtedness” (Social Science Research Network, Working paper W14808, 2008) included data from a survey of 1,000 Americans. One question on the survey was: “You owe $3,000 on your credit card. You pay a minimum payment of $30 each month. At an Annual Percentage Rate of 12% (or 1% per month), how many years would it take to eliminate your credit card debt if you made no additional charges?” Answer options for this question were: (a) less than 5 years; (b) between 5 and 10 years; (c) between 10 and 15 years; (d) never—you will continue to be in debt; (e) don’t know; and (f) prefer not to answer. The paper also reported that 37.8% of those in the sample chose one of the wrong answers (a, b, or c) as their response to this question. Is it reasonable to conclude that more than one-third of adult Americans would select a wrong answer to this question? Use α = 0.05.","H0: p = 1/3, Ha: p < 1/3, z = 2.996, P-value = 0.001, reject H0. There is convincing evidence that more than one-third of adult Americans would select a wrong answer.",Free Response,,,,
-LFD,10.5,53,"The paper “Teens and Distracted Driving” (Pew Internet & American Life Project, 2009) reported that in a representative sample of 283 American teens ages 16 to 17, there were 74 who indicated that they had sent a text message while driving. For purposes of this exercise, assume that this sample is a random sample of 16- to 17-year-old Americans. Do these data provide convincing evidence that more than a quarter of Americans ages 16 to 17 have sent a text message while driving? Test the appropriate hypotheses using a significance level of 0.01.","H0: p = 0.25, Ha: p > 0.25, z = 0.45, P-value = 0.328, reject H0, fail to reject H0. There is no convincing evidence that more than 25% of Americans ages 16 to 17 have sent a text message while driving.",Free Response,,,,
-LFD,10.5,54,"The article “Theaters Losing Out to Living Rooms” (San Luis Obispo Tribune, June 17, 2005) states that movie attendance declined in 2005. The Associated Press found that 730 of 1,000 randomly selected adult Americans prefer to watch movies at home rather than at a movie theater. Is there convincing evidence that a majority of adult Americans prefer to watch movies at home? Test the relevant hypotheses using a 0.05 significance level.","H0: p = 0.5, Ha: p > 0.5, z = 14.546, P-value ≈ 0, reject H0. There is convincing evidence that a majority of adult Americans prefer to watch movies at home.",Free Response,,,,
-LFD,10.5,55a,"The article referenced in the previous exercise also reported that 470 of 1,000 randomly selected adult Americans thought that the quality of movies being produced is getting worse. Is there convincing evidence that less than half of adult Americans believe that movie quality is getting worse? Use a significance level of 0.05.","z = -1.897, P-value = 0.0289, reject H0 ",Free Response,,,,
-LFD,10.5,55b,"The article referenced in the previous exercise also reported that 470 of 1,000 randomly selected adult Americans thought that the quality of movies being produced is getting worse. Suppose that the sample size had been 100 instead of 1,000, and that 47 thought that movie quality was getting worse (so that the sample proportion is still 0.47). Based on this sample of 100, is there convincing evidence that less than half of adult Americans believe that movie quality is getting worse? Use a significance level of 0.05.","z = -0.6, P-value = 0.274, fail to reject H0 ",Free Response,,,,
-LFD,10.5,56,"In a survey of 1,005 adult Americans, 46% indicated that they were somewhat interested or very interested in having Web access in their cars (USA Today, May 1, 2009). Suppose that the marketing manager of a car manufacturer claims that the 46% is based only on a sample and that 46% is close to half, so there is no reason to believe that the proportion of all adult Americans who want car Web access is less than 0.50. Is the marketing manager correct in his claim? Provide statistical evidence to support your answer. For purposes of this exercise, assume that the sample can be considered representative of adult Americans."," H0: p = 0.5, Ha: = p < 0.5, z = -2.536, P-value = 0.006, reject H0. There is convincing evidence that the proportion of all adult Americans who want car Web access is less than 0.5. The marketing manager is not correct in his claim.",Free Response,,,,
-LFD,10.5,67a,"In a survey of 1,000 women ages 22 to 35 who work full-time, 540 indicated that they would be willing to give up some personal time in order to make more money (USA Today, March 4, 2010). The sample was selected to be representative of women in the targeted age group. Do the sample data provide convincing evidence that a majority of women ages 22 to 35 who work full-time would be willing to give up some personal time for more money? Test the relevant hypotheses using α = 0.01. ","z = 2.530, P-value = 0.0057, reject H0",Free Response,,,,
-LFD,10.5,67b,"In a survey of 1,000 women ages 22 to 35 who work full-time, 540 indicated that they would be willing to give up some personal time in order to make more money (USA Today, March 4, 2010). The sample was selected to be representative of women in the targeted age group. Would it be reasonable to generalize the conclusion of a significance test to all working women? Explain why or why not.",No. The survey only included women ages 22 to 35.,Free Response,,,,
-LFD,10.5,69,"In a representative sample of 2,013 American adults, 1,590 indicated that lack of respect and courtesy in American society is a serious problem (Associated Press, April 3, 2002). Is there convincing evidence that more than three-quarters of American adults believe that lack of respect and courtesy is a serious problem? Test the relevant hypotheses using a significance level of 0.05.","H0: p = 0.75, Ha: p > 0.75, z = 4.13, P-value = 0.0057, reject H0. There is convincing evidence that more than three-quarters of American adults believe that lack of respect and courtesy is a serious problem.",Free Response,,,,
-LFD,10.5,71,"A number of initiatives on the topic of legalized gambling have appeared on state ballots. A political candidate has decided to support legalization of casino gambling if he is convinced that more than two-thirds of American adults approve of casino gambling. Suppose that 1,035 of the people in a random sample of 1,523 American adults said they approved of casino gambling. Is there convincing evidence that more than two-thirds approve?","z = 1.069, P-value = 0.143, fail to reject H0",Free Response,,,,
-LFD,10.5,73,"The survey described in the previous exercise also asked the individuals in the sample what they thought was their best chance to obtain more than $500,000 in their lifetimes. Twenty-eight percent responded “win a lottery or sweepstakes.” Does this provide convincing evidence that more than one-fourth of U.S. adults see a lottery or sweepstakes win as their best chance of accumulating $500,000? Carry out a test using a significance level of 0.01.","H0: p = 0.25, Ha: p > 0.25, z = 2.202, P-value = 0.014, reject H0. There is not convincing evidence that more than one-fourth of American adults see a lottery or sweepstakes win as their best chance of accumulating $500,000.",Free Response,,,,
-LFD,10.5,75,"The article “Americans Seek Spiritual Guidance on Web” (San Luis Obispo Tribune, October 12, 2002) reported that 68% of the U.S. population belongs to a religious community. In a survey on Internet use, 84% of “religion surfers” (defined as those who seek spiritual help online or who have used the Web to search for prayer and devotional resources) belong to a religious community. Suppose that this result was based on a representative sample of 512 religion surfers. Is there convincing evidence that the proportion of religion surfers who belong to a religious community is different from 0.68, the proportion for the U.S. population? Use α = 0.05.","H0: p = 0.68, Ha: p ≠ 0.25, z = 7.761, P-value ≈ 0, reject H0. There is not convincing evidence that more than one-fourth of American adults see a lottery or sweepstakes win as their best chance of accumulating $500,000. There is convincing evidence that the proportion of religion surfers who belong to a religious community is different from 0.68.",Free Response,,,,
-LFD,10.6,76,"Suppose you plan to test H0: p = 0.40 versus Ha: p > 0.40 . Assuming the same significance level, would the power of the test be greater if n = 50 or if n = 100? Explain your choice.",n = 100; the power is greater for a larger sample size.,Free Response,,,,
-LFD,10.6,77,"Suppose you plan to test H0: p = 0.25 versus Ha: p ≠ 0.25 . Assuming the same sample size and significance level, would the power of the test be smaller if the actual value of the population proportion were 0.15 or 0.30? Explain your choice.",p = 0.30; the power is smaller when the actual value of p is closer to the hypothesized value.,Free Response,,,,
-LFD,10.6,78,"Suppose you plan to test H0: p = 0.82 versus Ha: p > 0.82. Assuming the same sample size, would the power of the test be greater for a significance level of α = 0.01 or α = 0.05? Explain your choice.",α = 0.05. The power is greater when the significance level is larger.,Free Response,,,,
-LFD,10.6,83a,"The city council in a large city has become concerned about the trend toward exclusion of renters with children in apartments within the city. The housing coordinator has decided to select a random sample of 125 apartments and determine for each whether children are permitted. Let p be the proportion of all apartments that prohibit children. If the city council is convinced that p is greater than 0.75, it will consider appropriate legislation. If 102 of the 125 sampled apartments exclude renters with children, would a test with α=0.05 lead you to the conclusion that more than 75% of all apartments exclude children? ","Testing H0: p = 0.75 against Ha: p > 0.75, z = 1.704, P-value = 0.044. H0 is rejected, there is convincing evidence that more than 75% of apartments exclude children. ",Free Response,,,,
-LFD,10.6,83b,"The city council in a large city has become concerned about the trend toward exclusion of renters with children in apartments within the city. The housing coordinator has decided to select a random sample of 125 apartments and determine for each whether children are permitted. Let p be the proportion of all apartments that prohibit children. If the city council is convinced that p is greater than 0.75, it will consider appropriate legislation. What is the power of the test when p=0.8 and α=0.05?",0.351,Free Response,,,,
-LFD,10R,85,"The report “How Teens Use Media” (Nielsen, June 2009) says that 37% of U.S. teens access the Internet from a mobile phone. Suppose you plan to select a random sample of students at the local high school and will ask each student in the sample if he or she accesses the Internet from a mobile phone. You want to determine if there is evidence that the proportion of students at the high school who access the Internet using a mobile phone differs from the national figure of 0.37 given in the Nielsen report. What hypotheses should you test?","H0: p = 0.37, Ha: p ≠ 0.37",Free Response,,,,
-LFD,10R,87,Explain why failing to reject the null hypothesis in a hypothesis test does not mean there is convincing evidence that the null hypothesis is true.,Failing to reject the null hypothesis means that you are not convinced that the null hypothesis is false. This is not the same as being convinced that it is true.,Free Response,,,,
-LFD,10R,89a,"The National Cancer Institute conducted a 2-year study to determine whether cancer death rates for areas near nuclear power plants are higher than for areas without nuclear facilities (San Luis Obispo Telegram-Tribune, September 17, 1990). A spokesperson for the Cancer Institute said, “From the data at hand, there was no convincing evidence of any increased risk of death from any of the cancers surveyed due to living near nuclear facilities. However, no study can prove the absence of an effect.” Suppose p denotes the true proportion of the population in areas near nuclear power plants who die of cancer during a given year. The researchers at the Cancer Institute might have considered two rival hypotheses of the form H0: p is equal to the corresponding value for areas without nuclear facilities ha: p is greater than the corresponding value for areas without nuclear facilities Did the researchers reject H0 or fail to reject H0? ",The researchers failed to reject H0 ,Free Response,,,,
-LFD,10R,89b,"The National Cancer Institute conducted a 2-year study to determine whether cancer death rates for areas near nuclear power plants are higher than for areas without nuclear facilities (San Luis Obispo Telegram-Tribune, September 17, 1990). A spokesperson for the Cancer Institute said, “From the data at hand, there was no convincing evidence of any increased risk of death from any of the cancers surveyed due to living near nuclear facilities. However, no study can prove the absence of an effect.” If the Cancer Institute researchers are incorrect in their conclusion that there is no evidence of increased risk of death from cancer associated with living near a nuclear power plant, are they making a Type I or a Type II error? Explain. ",Type II error ,Free Response,,,,
-LFD,10R,89c,"The National Cancer Institute conducted a 2-year study to determine whether cancer death rates for areas near nuclear power plants are higher than for areas without nuclear facilities (San Luis Obispo Telegram-Tribune, September 17, 1990). A spokesperson for the Cancer Institute said, “From the data at hand, there was no convincing evidence of any increased risk of death from any of the cancers surveyed due to living near nuclear facilities. However, no study can prove the absence of an effect.” Comment on the spokesperson’s last statement that no study can prove the absence of an effect. Do you agree with this statement?",Yes,Free Response,,,,
-LFD,10R,91a,"The article “Cops Get Screened for Digital Dirt” (USA Today, Nov. 12, 2010) summarizes a report on law enforcement agency use of social media to screen applicants for employment. The report was based on a survey of 728 law enforcement agencies. One question on the survey asked if the agency routinely reviewed applicants’ social media activity during background checks. For purposes of this exercise, suppose that the 728 agencies were selected at random and that you want to use the survey data to decide if there is convincing evidence that more than 25% of law enforcement agencies review applicants’ social media activity as part of routine background checks. Describe the shape, center, and spread of the sampling distribution of p^ for samples of size 728 if the null hypothesis H0: p = 0.25 is true. ",The sampling distribution of p^ will be approximately normal with a mean of 0.25 and a standard deviation of 0.0160. ,Free Response,,,,
-LFD,10R,91b,"The article “Cops Get Screened for Digital Dirt” (USA Today, Nov. 12, 2010) summarizes a report on law enforcement agency use of social media to screen applicants for employment. The report was based on a survey of 728 law enforcement agencies. One question on the survey asked if the agency routinely reviewed applicants’ social media activity during background checks. For purposes of this exercise, suppose that the 728 agencies were selected at random and that you want to use the survey data to decide if there is convincing evidence that more than 25% of law enforcement agencies review applicants’ social media activity as part of routine background checks. Would you be surprised to observe a sample proportion of p^ = 0.27 for a sample of size 728 if the null hypothesis H0: p = 0.25 were true? Explain why or why not. ","A sample proportion of p^ = 0.27 would not be surprising if p = 0.25, because it is only 1.25 standard deviations above 0.25. This is not unusual for a normal distribution. ",Free Response,,,,
-LFD,10R,91c,"The article “Cops Get Screened for Digital Dirt” (USA Today, Nov. 12, 2010) summarizes a report on law enforcement agency use of social media to screen applicants for employment. The report was based on a survey of 728 law enforcement agencies. One question on the survey asked if the agency routinely reviewed applicants’ social media activity during background checks. For purposes of this exercise, suppose that the 728 agencies were selected at random and that you want to use the survey data to decide if there is convincing evidence that more than 25% of law enforcement agencies review applicants’ social media activity as part of routine background checks. Would you be surprised to observe a sample proportion of p^ = 0.31 for a sample of size 728 if the null hypothesis H0: p = 0.25 were true? Explain why or why not. ","A sample proportion of p^ = 0.31 would be surprising if p = 0.25, because it is 3.75 standard deviations above 0.25. This would be unusual for a normal distribution. ",Free Response,,,,
-LFD,10R,91d,"The article “Cops Get Screened for Digital Dirt” (USA Today, Nov. 12, 2010) summarizes a report on law enforcement agency use of social media to screen applicants for employment. The report was based on a survey of 728 law enforcement agencies. One question on the survey asked if the agency routinely reviewed applicants’ social media activity during background checks. For purposes of this exercise, suppose that the 728 agencies were selected at random and that you want to use the survey data to decide if there is convincing evidence that more than 25% of law enforcement agencies review applicants’ social media activity as part of routine background checks. The actual sample proportion observed in the study was p^ = 0.33. Based on this sample proportion, is there convincing evidence that more than 25% of law enforcement agencies review social media activity as part of background checks, or is this sample proportion consistent with what you would expect to see when the null hypothesis is true?","If p = 0.25, P(p^ ≥ 0.33) = P(z ≥ 5) ≈ 0. Because this probability is approximately 0, there is convincing evidence that more than 25% of law enforcement agencies review social media activity.",Free Response,,,,
-LFD,10R,93,"The report “Credit Card Statistics, Industry Facts, Debt Statistics” (creditcards.com) summarizes a survey of college undergraduates. Each person in a sample of students was asked about his or her credit card balance, and the average balance was calculated to be $3,173. You are interested in determining if there is evidence that the mean credit card debt for undergraduates is greater than the known average from the previous year.","Hypothesis testing, sample data, one numerical variable, one sample. A hypothesis test for a population proportion would not be appropriate.",Free Response,,,,
-LFD,10R,95a,Suppose that the sample of 899 college students described in the previous exercise can be regarded as representative of college students in the United States. What hypotheses would you test to answer the question posed in the previous exercise? ,"H0: p = 0.25, Ha: p > 0.25",Free Response,,,,
-LFD,10R,95b,Suppose that the sample of 899 college students described in the previous exercise can be regarded as representative of college students in the United States. Is the sample size large enough for a large-sample test for a population proportion to be appropriate? ,Yes ,Free Response,,,,
-LFD,10R,95c,Suppose that the sample of 899 college students described in the previous exercise can be regarded as representative of college students in the United States. What is the value of the test statistic and the associated P-value for this test?,"z = 3.33, P-value ≈ 0",Free Response,,,,
-LFD,10R,95d,"Suppose that the sample of 899 college students described in the previous exercise can be regarded as representative of college students in the United States. If a significance level of 0.05 were chosen, would you reject the null hypothesis or fail to reject the null hypothesis?",reject H0,Free Response,,,,
-LFD,10R,97,"Past experience is that when individuals are approached with a request to fill out and return a particular questionnaire in a provided stamped and addressed envelope, the response rate is 40%. An investigator believes that if the person distributing the questionnaire were stigmatized in some obvious way, potential respondents would feel sorry for the distributor and thus tend to respond at a rate higher than 40%. To test this theory, a distributor wore an eye patch. Of the 200 questionnaires distributed by this individual, 109 were returned. Does this provide evidence that the response rate in this situation is greater than the previous rate of 40%? State and test the appropriate hypotheses using a significance level of 0.05.","H0: p = 0.40, Ha: p > 0.40, z = 4.186, P-value ≈ 0, reject H0. There is convincing evidence that the response rate is greater than 40%.",Free Response,,,,
-LFD,11.1,1a,"USA Today (February 16, 2012) reported that the percentage of U.S. residents living in poverty was 12.5% for men and 15.1% for women. These percentages were estimates based on data from large representative samples of men and women. Suppose that the sample sizes were 1,200 for men and 1,000 for women. You would like to use the survey data to estimate the difference in the proportion living in poverty for men and women. Answer the four key questions (QSTN) for this problem. What method would you consider based on the answers to these questions? ","Estimation, sample data, one categorical variable, two samples. A large-sample confidence interval for a difference in proportions should be considered.",Free Response,,,,
-LFD,11.1,1b,"USA Today (February 16, 2012) reported that the percentage of U.S. residents living in poverty was 12.5% for men and 15.1% for women. These percentages were estimates based on data from large representative samples of men and women. Suppose that the sample sizes were 1,200 for men and 1,000 for women. You would like to use the survey data to estimate the difference in the proportion living in poverty for men and women. Use the five-step process for estimation problems (EMCCC) to compute and interpret a 90% large-sample confidence interval for the difference in the proportion living in poverty for men and women.","(-0.050, -0.002). You can be 90% confident that the actual difference between the proportion of male U.S. residents living in poverty and this proportion for females is between -0.050 and -0.002. Because both endpoints of this interval are negative, you would estimate that the proportion of men living in poverty is smaller than the proportion of women living in poverty by somewhere between 0.002 and 0.050.",Free Response,,,,
-LFD,11.1,2a,"The article ""Portable MP3 Player Ownership Reaches New High"" (Ipsos Insight, June 29, 2006) reported that in 2006, 20% of those in a random sample of 1,112 Americans ages 12 and older indicated that they owned an MP3 player. In a similar survey conducted in 2005, only 15% reported owning an MP3 player. Suppose that the 2005 figure was also based on a random sample of size 1,112. Are the sample sizes large enough to use the large-sample confidence interval for a difference in population proportions?","Yes. p^1 = 0.20 and p^2 = 0.15. Then n1*p^1 = 222.4, n1(1 - p^1) = 889.6, n2*p^2 =166.8, n2(1 - p^2) = 945.2 are all greater than 10.",Free Response,,,,
-LFD,11.1,2b,"The article ""Portable MP3 Player Ownership Reaches New High"" (Ipsos Insight, June 29, 2006) reported that in 2006, 20% of those in a random sample of 1,112 Americans ages 12 and older indicated that they owned an MP3 player. In a similar survey conducted in 2005, only 15% reported owning an MP3 player. Suppose that the 2005 figure was also based on a random sample of size 1,112. Estimate the difference in the proportion of Americans ages 12 and older who owned an MP3 player in 2006 and the corresponding proportion for 2005 using a 95% confidence interval. ","(0.018, 0.082) ",Free Response,,,,
-LFD,11.1,2c,"The article ""Portable MP3 Player Ownership Reaches New High"" (Ipsos Insight, June 29, 2006) reported that in 2006, 20% of those in a random sample of 1,112 Americans ages 12 and older indicated that they owned an MP3 player. In a similar survey conducted in 2005, only 15% reported owning an MP3 player. Suppose that the 2005 figure was also based on a random sample of size 1,112. Is zero included in the 95% confidence interval? What does this suggest about the change in this proportion from 2005 to 2006? ",Zero is not included in the confidence interval. This means that you can be confident that the proportion of Americans ages 12 and older who owned an MP3 player was greater in 2006 than in 2005. ,Free Response,,,,
-LFD,11.1,2d,"The article ""Portable MP3 Player Ownership Reaches New High"" (Ipsos Insight, June 29, 2006) reported that in 2006, 20% of those in a random sample of 1,112 Americans ages 12 and older indicated that they owned an MP3 player. In a similar survey conducted in 2005, only 15% reported owning an MP3 player. Suppose that the 2005 figure was also based on a random sample of size 1,112. Interpret the 95% confidence interval in the context of this problem.",You can be 95% confident that the proportion of Americans ages 12 and older who owned an MP3 player was greater in 2006 than in 2005 by somewhere between 0.018 and 0.082.,Free Response,,,,
-LFD,11.1,3a,"""Mountain Biking May Reduce Fertility in Men. Study Says"" was the headline of an article appearing in the San Luis Obispo Tribune (December 3, 2002). This conclusion was based on an Austrian study that compared sperm counts of avid mountain bikers (those who ride at least 12 hours per week) and nonbikers. Ninety percent of the avid mountain bikers studied had low sperm count compared to 26% of the nonbikers. Suppose that these percentages were based on independent samples of 100 avid mountain bikers and 100 nonbikers and that these samples are representative of avid mountain bikers and nonbikers. Using a confidence level of 95%, estimate the difference between the proportion of avid mountain bikers with low sperm count and the proportion for nonbikers. ","(0.536, 0.744). You can be 95% confident that the proportion of avid mountain bikers who have low sperm count is higher than the proportion for nonbikers by somewhere between 0.536 and 0.744. ",Free Response,,,,
-LFD,11.1,3b,"""Mountain Biking May Reduce Fertility in Men. Study Says"" was the headline of an article appearing in the San Luis Obispo Tribune (December 3, 2002). This conclusion was based on an Austrian study that compared sperm counts of avid mountain bikers (those who ride at least 12 hours per week) and nonbikers. Ninety percent of the avid mountain bikers studied had low sperm count compared to 26% of the nonbikers. Suppose that these percentages were based on independent samples of 100 avid mountain bikers and 100 nonbikers and that these samples are representative of avid mountain bikers and nonbikers. Is it reasonable to conclude that mountain biking 12 hours per week or more causes low sperm count based on a95% confidence interval? Explain.","No, because this was an observational study, and it is not a good idea to draw cause-and-effect conclusions from an observational study.",Free Response,,,,
-LFD,11.1,7,"To learn about the proportion of college students who are registered to vote, surveys were conducted in 2010 and in 2011. Suppose that in 2010, 78% of the people in a representative sample of 2,300 college students were registered to vote. In 2011, 70% of the people in a representative sample of 2,294 college students were registered to vote. Use the five-step process for estimation problems (EMC³) to construct and interpret a 99% large-sample confidence interval for the difference in the proportion of college students who were registered to vote in 2010 and this proportion in 2011.","(0.047, 0.113). You can be 99% confident that the proportion of high school students who were registered to vote in 2010 was greater than this proportion in 2011 by somewhere between 0.047 and 0.114.",Free Response,,,,
-LFD,11.1,9a,The article referenced in the previous exercise also reported that 5.7% of high school graduates were unemployed in 2008 and 9.7% of high school graduates were unemployed in 2009. Suppose that the reported percentages were based on independently selected representative samples of 400 high school graduates in each of these 2 years. Construct and interpret a 99% large-sample confidence interval for the difference in the proportions of high school graduates who were unemployed in these 2 years. ,"(-0.089, 0.009). Because 0 is included in this interval, there may be no difference in the proportion of high school graduates who were unemployed in 2008 and the proportion who were unemployed in 2009. ",Free Response,,,,
-LFD,11.2,12,"A hotel chain is interested in evaluating reservation processes. Guests can reserve a room by using either a telephone system or an online system that is accessed through the hotel’s web site. Independent random samples of 80 guests who reserved a room by phone and 60 guests who reserved a room online were selected. Of those who reserved by phone, 57 reported that they were satisfied with the reservation process. Of those who reserved online, 50 reported that they were satisfied. Based on these data, is it reasonable to conclude that the proportion who are satisfied is greater for those who reserve a room online? Test the appropriate hypotheses using a significance level of 0.05.","H0: pr - po = 0, Ha: pr – po < 0, z = -1.667, P-value = 0.048, reject H0. There is convincing evidence that the proportion who are satisfied is higher for those who reserve a room online.",Free Response,,,,
-LFD,11.2,13,"“Smartest People Often Dumbest About Sunburns” is the headline of an article that appeared in the San Luis Obispo Tribune (July 19, 2006). The article states that “those with a college degree reported a higher incidence of sunburn than those without a high school degree—43% versus 25%.” Suppose that these percentages were based on random samples of size 200 from each of the two groups of interest (college graduates and those without a high school degree). Is there convincing evidence that the proportion experiencing a sunburn is greater for college graduates than it is for those without a high school degree? Test the appropriate hypotheses using a significance level of 0.01.","H0: pc - ph = 0, Ha: pc – ph < 0, z = 3.800, P-value ≈ 0, reject H0. There is convincing evidence that the proportion experiencing sunburn is higher for college graduates than for those without a high school degree.",Free Response,,,,
-LFD,11.2,17,"The report ""Young People Living on the Edge"" (Greenberg Quinlan Rosner Research, 2008) summarizes a survey of people in two independent random samples. One sample consisted of 600 young adults (ages 19 to 35), and the other sample consisted of 300 parents of young adults ages 19 to 35. The young adults were presented with a variety of situations (such as getting married or buying a house) and were asked if they thought that their parents were likely to provide financial support in that situation. The parents of young adults were presented with the same situations and asked if they would be likely to provide financial support in that situation. When asked about getting married, 41% of the young adults said they thought parents would provide financial support and 43% of the parents said they would provide support. Carry out a hypothesis test to determine if there is convincing evidence that the proportion of young adults who think their parents would provide financial support and the proportion of parents who say they would provide support are different.","H0: p1 – p2 = 0, Ha: p1 – p2 ≠ 0, z = -0.574, P-value = 0.566. Fail to reject H0. There is not convincing evidence of a difference between the proportion of young adults who think that their parents would provide financial support for marriage and the proportion of parents who say they would provide financial support for marriage.",Free Response,,,,
-LFD,11.2,19,"In December 2001, the Department of Veterans' Affairs announced that it would begin paying benefits to soldiers suffering from Lou Gehrig's disease who had served in the Gulf War (The New York Times, December 11, 2001). This decision was based on an analysis in which the Lou Gehrig's disease incidence rate (the proportion developing the disease) for the approximately 700,000 soldiers sent to the Persian Gulf between August 1990 and July 1991 was compared to the incidence rate for the approximately 1.8 million other soldiers who were not in the Gulf during this time period. Based on these data, explain why it is not appropriate to perform a hypothesis test in this situation and yet it is still reasonable to conclude that the incidence rate is higher for Gulf War veterans than for those who did not serve in the Gulf War.","Since the values given are population characteristics, an inference procedure is not applicable. It is known that the rate of Lou Gehrig’s disease among soldiers sent to the war is higher than for those not sent to the war.",Free Response,,,,
-LFD,11R,21,"The news release referenced in the previous exercise also included data from independent samples of teenage drivers and parents of teenage drivers. In response to a question asking if they approved of laws banning the use of cell phones and texting while driving, 74% of the teens surveyed and 95% of the parents surveyed said they approved. The sample sizes were not given in the news release, but suppose that 600 teens and 400 parents of teens were surveyed and that these samples are representative of the two populations. Do the data provide convincing evidence that the proportion of teens who approve of banning cell phone and texting while driving is less than the proportion of parents of teens who approve? Test the relevant hypotheses using a significance level of 0.05.","H0: p1 – p2 = 0, Ha: p1 – p2 < 0, z = -8.543, P-value ≈ 0, reject H0. There is convincing evidence that the proportion of teens who approve of the proposed laws is less than the proportion of parents of teens who approve.",Free Response,,,,
-LFD,11R,23a,"The report “Audience Insights: Communicating to Teens (Aged 12–17)” (www.cdc.gov, 2009) described teens’ attitudes about traditional media, such as TV, movies, and newspapers. In a representative sample of American teenage girls, 41% said newspapers were boring. In a representative sample of American teenage boys, 44% said newspapers were boring. Sample sizes were not given in the report. Suppose that the percentages reported were based on samples of 58 girls and 41 boys. Is there convincing evidence that the proportion who think that newspapers are boring is different for teenage girls and boys? Carry out a hypothesis test using α = 0.05. ","H0: pG – pB = 0, Ha: pG – pB < 0, z = -0.298, P-value = 0.766, reject H0, fail to reject H0. There is not convincing evidence that the proportion who think that newspapers are boring is different for teenage girls and boys.",Free Response,,,,
-LFD,11R,23b,"The report “Audience Insights: Communicating to Teens (Aged 12–17)” (www.cdc.gov, 2009) described teens’ attitudes about traditional media, such as TV, movies, and newspapers. In a representative sample of American teenage girls, 41% said newspapers were boring. In a representative sample of American teenage boys, 44% said newspapers were boring. Sample sizes were not given in the report. Suppose that the percentages reported were based on samples of 2,000 girls and 2,500 boys. Is there convincing evidence that the proportion who think that newspapers are boring is different for teenage girls and boys? Carry out a hypothesis test using α = 0.05.","H0: pG – pB = 0, Ha: pG – pB ≠ 0, z = -2.022, P-value = 0.043, reject H0. There is convincing evidence that the proportion who think that newspapers are boring is different for teenage girls and boys.",Free Response,,,,
-LFD,12.1,1a,A random sample is selected from a population with mean µ = 100 and standard deviation σ = 10. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 9.,"100, 3.333 ",Free Response,,,,
-LFD,12.1,1b,A random sample is selected from a population with mean µ = 100 and standard deviation σ = 10. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 15.,"100, 2.582 ",Free Response,,,,
-LFD,12.1,1c,A random sample is selected from a population with mean µ = 100 and standard deviation σ = 10. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 36.,"100, 1.667 ",Free Response,,,,
-LFD,12.1,1d,A random sample is selected from a population with mean µ = 100 and standard deviation σ = 10. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 50.,"100, 1.414 ",Free Response,,,,
-LFD,12.1,1e,A random sample is selected from a population with mean µ = 100 and standard deviation σ = 10. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 100.,"100, 1.000 ",Free Response,,,,
-LFD,12.1,1f,A random sample is selected from a population with mean µ = 100 and standard deviation σ = 10. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 400.,"100, 0.500",Free Response,,,,
-LFD,12.1,3a,"The article “How Business Students Spend Their Time—Do They Really Know?” (Research in Higher Education Journal [May 2009]: 1–10) describes a study of 212 business students at a large public university. Each student kept a log of time spent on various activities during a 1-week period. For these 212 students, the mean time spent studying was 9.66 hours and the sample standard deviation was 6.62 hours. Suppose that this sample was a random sample from the population of all business majors at this university and that you are interested in learning about the value of µ, the mean time spent studying for this population. How do you know the sampling distribution of x-bar is centered at the actual (but unknown) value of the population mean.",The mean of the sampling distribution of x-bar is equal to the population mean µ. ,Free Response,,,,
-LFD,12.1,3b,"The article “How Business Students Spend Their Time—Do They Really Know?” (Research in Higher Education Journal [May 2009]: 1–10) describes a study of 212 business students at a large public university. Each student kept a log of time spent on various activities during a 1-week period. For these 212 students, the mean time spent studying was 9.66 hours and the sample standard deviation was 6.62 hours. Suppose that this sample was a random sample from the population of all business majors at this university and that you are interested in learning about the value of µ, the mean time spent studying for this population. How do you know an estimate of the standard deviation of x-bar, which describes how the x-bar values spread out around the population mean µ, is 0.455.",σx = σ / sqrt(n) = 6.62 / sqrt(212),Free Response,,,,
-LFD,12.1,3c,"The article “How Business Students Spend Their Time—Do They Really Know?” (Research in Higher Education Journal [May 2009]: 1–10) describes a study of 212 business students at a large public university. Each student kept a log of time spent on various activities during a 1-week period. For these 212 students, the mean time spent studying was 9.66 hours and the sample standard deviation was 6.62 hours. Suppose that this sample was a random sample from the population of all business majors at this university and that you are interested in learning about the value of µ, the mean time spent studying for this population. How do you know the sampling distribution of x-bar is approximately normal.",The sampling distribution of x-bar is approximately normal because the sample size (n = 212) is greater than 30.,Free Response,,,,
-LFD,12.1,4,Explain the difference between µ and µx.,"The quantity µ is the population mean, while µx is the mean of the x distribution. It is the mean value of x-bar for all possible random samples of size n.",Free Response,,,,
-LFD,12.1,5a,"The time that people have to wait for an elevator in an office building has a uniform distribution over the interval from 0 to 1 minute. For this distribution, µ = 0.5 and σ = 0.289. Let x-bar be the average waiting time for a random sample of 16 waiting times. What are the mean and standard deviation of the sampling distribution of x-bar? ","µx = µ = 0.5, σx = σ / sqrt(n) = 0.289 / sqrt(16) = 0.07225",Free Response,,,,
-LFD,12.1,5b,"The time that people have to wait for an elevator in an office building has a uniform distribution over the interval from 0 to 1 minute. For this distribution, µ = 0.5 and σ = 0.289. Let x-bar be the average waiting time for a random sample of 50 waiting times. What are the mean and standard deviation of the sampling distribution of x-bar?","When  n = 50, µx = µ = 0.5, and σx = σ / sqrt(n) = 0.289 / sqrt(50) = 0.07225. Since n ≥ 30, the distribution of x-bar is approximately normal",Free Response,,,,
-LFD,12.1,11a,A random sample is selected from a population with mean µ = 200 and standard deviation σ = 15. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 12.,"200, 4.330 ",Free Response,,,,
-LFD,12.1,11b,A random sample is selected from a population with mean µ = 200 and standard deviation σ = 15. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 20.,"200, 3.354 ",Free Response,,,,
-LFD,12.1,11c,A random sample is selected from a population with mean µ = 200 and standard deviation σ = 15. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 25.,"200, 3.000 ",Free Response,,,,
-LFD,12.1,11d,A random sample is selected from a population with mean µ = 200 and standard deviation σ = 15. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 40.,"200, 2.371 ",Free Response,,,,
-LFD,12.1,11e,A random sample is selected from a population with mean µ = 200 and standard deviation σ = 15. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 90.,"200, 1.581 ",Free Response,,,,
-LFD,12.1,11f,A random sample is selected from a population with mean µ = 200 and standard deviation σ = 15. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 300.,"200, 0.866",Free Response,,,,
-LFD,12.1,13,Explain the difference between x-bar and µx.,"x-bar is the mean of a single sample, while µ is the mean of the x-bar distribution. It is the mean value of x-bar for all possible random samples of size n.",Free Response,,,,
-LFD,12.1,15a,"Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is 65 mm and that the population standard deviation is 5 mm. If the distribution of interpupillary distance is normal and a random sample of n = 25 adult males is to be selected, what is the probability that the sample mean distance x-bar for these 25 will be between 64 and 67 mm? At least 68 mm? ","0.8185, 0.0013 ",Free Response,,,,
-LFD,12.1,15b,"Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is 65 mm and that the population standard deviation is 5 mm. Suppose that a random sample of 100 adult males is to be selected. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample mean distance will be between 64 and 67 mm? At least 68 mm?","0.9772, 0.0000",Free Response,,,,
-LFD,12.1,17,"A manufacturing process is designed to produce bolts with a diameter of 0.5 inches. Once each day, a random sample of 36 bolts is selected and the bolt diameters are recorded. If the resulting sample mean is less than 0.49 inches or greater than 0.51 inches, the process is shut down for adjustment. The standard deviation of bolt diameters is 0.02 inches. What is the probability that the manufacturing line will be shut down unnecessarily? (Hint: Find the probability of observing an x-bar in the shutdown range when the actual process mean is 0.5 inches.)",P(0.49 < x-bar < 0.51 = 0.9974: the probability that the manufacturing line will be shut down unnecessarily is 1 – 0.9974 = 0.0026.,Free Response,,,,
-LFD,12.2,19a,What percentage of the time will a variable that has a t distribution with 10 degrees of freedom fall between −1.81 and 1.81 ,90%,Free Response,,,,
-LFD,12.2,19b,What percentage of the time will a variable that has a t distribution with 24 degrees of freedom fall between −2.06 and 2.06 ,95%,Free Response,,,,
-LFD,12.2,19c,What percentage of the time will a variable that has a t distribution with 24 degrees of freedom fall outside the interval from −2.80 to 2.80 ,1%,Free Response,,,,
-LFD,12.2,19d,What percentage of the time will a variable that has a t distribution with 10 degrees of freedom fall to the left of −1.81,5%,Free Response,,,,
-LFD,12.2,20a,"What is the appropriate t critical value a 95% confidence, with a sample size of n = 17? ",2%,Free Response,,,,
-LFD,12.2,20b,"What is the appropriate t critical value a 99% confidence, with a sample size of n = 24? ",3%,Free Response,,,,
-LFD,12.2,20c,"What is the appropriate t critical value a 90% confidence, with a sample size of n = 13?",2%,Free Response,,,,
-LFD,12.2,21a,"The two intervals (114.4, 115.6) and (114.1, 115.9) are confidence intervals for µ = mean resonance frequency (in hertz) for all tennis rackets of a certain type. The two intervals were computed using the same sample data. What is the value of the sample mean resonance frequency?",115%,Free Response,,,,
-LFD,12.2,21b,"The two intervals (114.4, 115.6) and (114.1, 115.9) are confidence intervals for µ = mean resonance frequency (in hertz) for all tennis rackets of a certain type. The two intervals were computed using the same sample data. The confidence level for one of these intervals is 90%, and for the other it is 99%. Which is which, and how can you tell?","The 99% confidence interval is wider than the 90% confidence interval. The 90% interval is (114.4, 115.6), and the 99% interval is (114.1, 115.9).",Free Response,,,,
-LFD,12.2,22a,"In June, 2009, Harris Interactive conducted its Great Schools Survey. In this survey, the sample consisted of 1,086 adults who were parents of school-aged children. The sample was selected to be representative of the population of parents of school-aged children. One question on the survey asked respondents how much time per month (in hours) they spent volunteering at their children’s school during the previous school year. The following summary statistics for time volunteered per month were given: n = 1,086, x-bar = 5.6, median = 1. What does the fact that the mean is so much larger than the median tell you about the distribution of time spent volunteering at school per month?",That the mean is much greater than the median suggests that the distribution of times spent volunteering in the sample was positively skewed. ,Free Response,,,,
-LFD,12.2,22b,"In June, 2009, Harris Interactive conducted its Great Schools Survey. In this survey, the sample consisted of 1,086 adults who were parents of school-aged children. The sample was selected to be representative of the population of parents of school-aged children. One question on the survey asked respondents how much time per month (in hours) they spent volunteering at their children’s school during the previous school year. The following summary statistics for time volunteered per month were given: n = 1,086, x-bar = 5.6, median = 1. Explain why it is not reasonable to assume that the population distribution of time spent volunteering is approximately normal.","With the sample mean much greater than the sample median, and with the sample regarded as representative of the population, it seems very likely that the population is strongly positively skewed and, therefore, not normally distributed. ",Free Response,,,,
-LFD,12.2,22c,"In June, 2009, Harris Interactive conducted its Great Schools Survey. In this survey, the sample consisted of 1,086 adults who were parents of school-aged children. The sample was selected to be representative of the population of parents of school-aged children. One question on the survey asked respondents how much time per month (in hours) they spent volunteering at their children’s school during the previous school year. The following summary statistics for time volunteered per month were given: n = 1,086, x-bar = 5.6, median = 1. Explain why it is appropriate to use the one-sample t confidence interval to estimate the mean time spent volunteering for the population of parents of school-aged children even though the population distribution is not approximately normal.","Since n = 1086 ≥ 30, the sample size is large enough for the one-sample t confidence interval to be appropriate, even though the population distribution is not approximately normal. ",Free Response,,,,
-LFD,12.2,22d,"In June, 2009, Harris Interactive conducted its Great Schools Survey. In this survey, the sample consisted of 1,086 adults who were parents of school-aged children. The sample was selected to be representative of the population of parents of school-aged children. One question on the survey asked respondents how much time per month (in hours) they spent volunteering at their children’s school during the previous school year. The following summary statistics for time volunteered per month were given: n = 1,086, x-bar = 5.6, median = 1. Suppose that the sample standard deviation was s = 5.2 . Use the five-step process for estimation problems (EMC3) to compute and interpret a 98% confidence interval for µ, the mean time spent volunteering for the population of parents of school-aged children.","(5.232, 5.968). You can be 98% confident that the mean time spent volunteering for the population of parents of school-age children is between 5.232 and 5.968 hours.",Free Response,,,,
-LFD,12.2,23,"In a study of academic procrastination, the authors of the paper “Correlates and Consequences of Behavioral Procrastination” (Procrastination, Current Issues and New Directions [2000]) reported that for a sample of 411 undergraduate students at a mid-size public university, the mean time spent studying for the final exam in an introductory psychology course was 7.74 hours and the standard deviation of study times was 3.40 hours. Assume that this sample is representative of students taking introductory psychology at this university. Construct a 95% confidence interval estimate of µ, the mean time spent studying for the introductory psychology final exam.","(7.411, 8.069). You can be 95% confident that the mean time spent studying for the final exam is between 7.411 and 8.069 hours.",Free Response,,,,
-LFD,12.2,24,"The paper referenced in the previous exercise also gave the following sample statistics for the percentage of study time that occurred in the 24 hours prior to the final exam: n = 411, x-bar = 43.18, s = 21.46 Construct and interpret a 90% confidence interval for the mean percentage of study time that occurs in the 24 hours prior to the final exam.","(41.439, 44.921). You can be 90% confident that the mean percentage of study time that occurs in the 24 hours prior to the final exam is between 41.439% and 44.921%.",Free Response,,,,
-LFD,12.2,25,"The Bureau of Alcohol, Tobacco, and Firearms (BATF) has been concerned about lead levels in California wines. In a previous testing of wine specimens, lead levels ranging from 50 to 700 parts per billion were recorded. How many wine specimens should be tested if the BATF wishes to estimate the mean lead level for California wines with a margin of error of 10 parts per billion?","Using (sample range)/4 = 162.5 as an estimate of the population standard deviation, a sample size of 1,015 is needed",Free Response,,,,
-LFD,12.2,26a,"The article “Most Canadians Plan to Buy Treats, Many Will Buy Pumpkins, Decorations and/or Costumes” (Ipsos-Reid, October 24, 2005) summarized a survey of 1,000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween. The resulting sample mean and standard deviation were $46.65 and $83.70, respectively. Explain how it could be possible for the standard deviation of the anticipated Halloween expense to be larger than the mean anticipated expense.",This could happen if the sample distribution was very positively skewed. ,Free Response,,,,
-LFD,12.2,26b,"The article “Most Canadians Plan to Buy Treats, Many Will Buy Pumpkins, Decorations and/or Costumes” (Ipsos-Reid, October 24, 2005) summarized a survey of 1,000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween. The resulting sample mean and standard deviation were $46.65 and $83.70, respectively. Is it reasonable to think that the distribution of anticipated Halloween expense is approximately normal? Explain why or why not.","No. Since the sample distribution is very skewed to the right, it is very unlikely that the variable anticipated Halloween expense is approximately normally distributed. ",Free Response,,,,
-LFD,12.2,26c,"The article “Most Canadians Plan to Buy Treats, Many Will Buy Pumpkins, Decorations and/or Costumes” (Ipsos-Reid, October 24, 2005) summarized a survey of 1,000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween. The resulting sample mean and standard deviation were $46.65 and $83.70, respectively. Is it appropriate to use the one-sample t confidence interval to estimate the mean anticipated Halloween expense for Canadian residents? Explain why or why not.","Yes, since the sample is large. ",Free Response,,,,
-LFD,12.2,26d,"The article “Most Canadians Plan to Buy Treats, Many Will Buy Pumpkins, Decorations and/or Costumes” (Ipsos-Reid, October 24, 2005) summarized a survey of 1,000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween. The resulting sample mean and standard deviation were $46.65 and $83.70, respectively. If appropriate, construct and interpret a 99% confidence interval for the mean anticipated Halloween expense for Canadian residents.","(39.821, 53.479). You can be 99% confident that the mean anticipated Halloween expense for the population of Canadian residents is between 39.821 and 53.479 dollars.",Free Response,,,,
-LFD,12.2,35a,What is the appropriate t critical value for a 95% confidence and a sample size of n = 15. ,2.15,Free Response,,,,
-LFD,12.2,35b,What is the appropriate t critical value for a 99% confidence and a sample size of n = 20.  ,2.86,Free Response,,,,
-LFD,12.2,35c,What is the appropriate t critical value for a 90% confidence and a sample size of n = 26.,1.71,Free Response,,,,
-LFD,12.2,37,"Five students visiting the student health center for a free dental examination during National Dental Hygiene Month were asked how many months had passed since their last visit to a dentist. Their responses were: 6 17 11 22 29 Assuming that these five students can be considered as representative of all students participating in the free checkup program, construct a 95% confidence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program.","(5.776, 28.224). You can be 95% confident that the mean number of months elapsed since the last visit for the population of students participating in the program is between 5.776 and 28.224.",Free Response,,,,
-LFD,12.2,39,"Use the information given in the previous exercise to construct a 95% confidence interval for the mean number of partners on a mating flight for queen honeybees. For purposes of this exercise, assume these 30 queen honeybees are representative of the population of queen honeybees.","(3.301, 5.899). You can be 95% confident that the mean number of partners on a mating flight is between 3.301 and 5.899.",Free Response,,,,
-LFD,12.3,41a,"Give as much information as you can about the P-value of a t test with the following information: Upper-tailed test, df = 8, t = 2.0",0.04,Free Response,,,,
-LFD,12.3,41b,"Give as much information as you can about the P-value of a t test with the following information: Lower-tailed test, df = 10, t = -2.4",0.019,Free Response,,,,
-LFD,12.3,41c,"Give as much information as you can about the P-value of a t test with the following information: Lower-tailed test, n = 22, t = -4.2",0,Free Response,,,,
-LFD,12.3,41d,"Give as much information as you can about the P-value of a t test with the following information: Two-tailed test, df = 15, t = -1.6",0.13,Free Response,,,,
-LFD,12.3,42a,"The paper “Playing Active Video Games Increases Energy Expenditure in Children” (Pediatrics [2009]: 534–539) describes a study of the possible cardiovascular benefits of active video games. Mean heart rate for healthy boys ages 10 to 13 after walking on a treadmill at 2.6 km/hour for 6 minutes is known to be 98 beats per minute (bpm). For each of 14 boys, heart rate was measured after 15 minutes of playing Wii Bowling. The resulting sample mean and standard deviation were 101 bpm and 15 bpm, respectively. Assume that the sample of boys is representative of boys ages 10 to 13 and that the distribution of heart rates after 15 minutes of Wii Bowling is approximately normal. Does the sample provide convincing evidence that the mean heart rate after 15 minutes of Wii Bowling is different from the known mean heart rate after 6 minutes walking on the treadmill? Carry out a hypothesis test using α = 0.05.","H0: µ = 98, Ha: µ > 98, t = 0.748, P-value = 0.468, fail to reject H0. There is not convincing evidence that the mean heart rate after 15 minutes of Wii Bowling is different from the mean after 6 minutes of walking on a treadmill.",Free Response,,,,
-LFD,12.3,43a,"The paper “Playing Active Video Games Increases Energy Expenditure in Children” (Pediatrics [2009]: 534–539) describes a study of the possible cardiovascular benefits of active video games. Mean heart rate for healthy boys ages 10 to 13 after walking on a treadmill at 2.6 km/hour for 6 minutes is known to be 98 beats per minute (bpm). For each of 14 boys, heart rate was measured after 15 minutes of playing Wii Bowling. The resulting sample mean and standard deviation were 101 bpm and 15 bpm, respectively. Assume that the sample of boys is representative of boys ages 10 to 13 and that the distribution of heart rates after 15 minutes of Wii Bowling is approximately normal. The paper also states that the known resting mean heart rate for boys in this age group is 66 bpm. Is there convincing evidence that the mean heart rate after Wii Bowling for 15 minutes is higher than the known mean resting heart rate for boys of this age? Use α = 0.01. ","H0: µ = 66, Ha: µ > 66, t = 8.731, P-value ≈ 0, reject H0. There is convincing evidence that the mean heart rate after 15 minutes of Wii Bowling is higher than the mean resting heart rate.",Free Response,,,,
-LFD,12.3,44,"The Economist collects data each year on the price of a Big Mac in various countries around the world. A sample of McDonald’s restaurants in Europe in May 2009 resulted in the following Big Mac prices (after conversion to U.S. dollars):  3.80  5.89  4.92  3.88  2.65  5.57  6.39  3.24  The mean price of a Big Mac in the U.S. in May 2009 was $3.57. For purposes of this exercise, you can assume it is reasonable to regard the sample as representative of European McDonald’s restaurants. Does the sample provide convincing evidence that the mean May 2009 price of a Big Mac in Europe is greater than the reported U.S. price? Test the relevant hypotheses using α=0.05.","H0: µ = 3.57, Ha: µ > 3.57, t = 2.042, P-value = 0.040,  reject H0. There is convincing evidence that the mean price of a Big Mac in Europe is higher than $3.57.",Free Response,,,,
-LFD,12.3,45a,"In a study of computer use, 1,000 randomly selected Canadian Internet users were asked how much time they spend online in a typical week (Ipsos Reid, August 9, 2005). The sample mean was 12.7 hours. The sample standard deviation was not reported, but suppose that it was 5 hours. Carry out a hypothesis test with a significance level of 0.05 to decide if there is convincing evidence that the mean time spent online by Canadians in a typical week is greater than 12.5 hours.","H0: µ = 12.5, Ha: µ > 12.5, t = 1.26, P-value = 0.103, fail to reject H0. There is not convincing evidence that the mean time spent online is greater than 12.5 hours. ",Free Response,,,,
-LFD,12.3,45b,"In a study of computer use, 1,000 randomly selected Canadian Internet users were asked how much time they spend online in a typical week (Ipsos Reid, August 9, 2005). The sample mean was 12.7 hours. Now suppose that the sample standard deviation was 2 hours. Carry out a hypothesis test with a significance level of 0.05 to decide if there is convincing evidence that the mean time spent online by Canadians in a typical week is greater than 12.5 hours.","H0: µ = 12.5, Ha: µ > 12.5, t = 3.16, P-value = 0.0008, reject H0. There is convincing evidence that the mean time spent online is greater than 12.5 hours. ",Free Response,,,,
-LFD,12.3,46,"The paper titled ""Music for Pain Relief"" (The Cochrane Database of Systematic Reviews, April 19, 2006) concluded, based on a review of 51 studies of the effect of music on pain intensity, that ""Listening to music reduces pain intensity levels ... However, the magnitude of these positive effects is small, and the clinical relevance of music for pain relief in clinical practice is unclear."" Are the authors of this paper claiming that the pain reduction attributable to listening to music is not statistically significant, not practically significant, or neither statistically nor practically significant? Explain.","By saying that listening to music reduces pain levels, the authors are telling you that the study resulted in convincing evidence that pain levels are reduced when music is being listened to. (In other words, the results of the study were statistically significant.) By saying, however, that the magnitude of the positive effects was small, the authors are telling you that the effect was not practically significant.",Free Response,,,,
-LFD,12.3,53,"The eating habits of 12 vampire bats were examined in the article ""Foraging Behavior of the Indian False Vampire Bat"" (Biotropica [1991]: 63–67). These bats consume insects and frogs. For these 12 bats, the mean time to consume a frog was x-bar = 21.9 minutes. Suppose that the standard deviation was s = 7.7  minutes. Is there convincing evidence that the mean supper time of a vampire bat whose meal consists of a frog is greater than 20 minutes? What assumptions must be reasonable for the one-sample t test to be appropriate?","H0: µ = 20, Ha: µ > 20, t = 0.85, P-value = 0.205, fail to reject H0. There is not convincing evidence that the mean suppertime is greater than 2 minutes.",Free Response,,,,
-LFD,12.3,55,"An article titled ""Teen Boys Forget Whatever It Was"" appeared in the Australian newspaper The Mercury (April 21, 1997). It described a study of academic performance and attention span and reported that the mean time to distraction for teenage boys working on an independent task was 4 minutes. Although the sample size was not given in the article, suppose that this mean was based on a random sample of 50 teenage boys and that the sample standard deviation was 1.4 minutes. Is there convincing evidence that the average attention span for teenage boys is less than 5 minutes? Test the relevant hypotheses using α = 0.01.","H0: µ = 5, Ha: µ < 5, t = -5.051, P-value = 0.000, reject H0. There is convincing evidence that the mean attention span for teenage boys is less than 5 minutes.",Free Response,,,,
-LFD,12.3,57,"An automobile manufacturer decides to carry out a fuel efficiency test to determine if it can advertise that one of its models achieves 30 mpg (miles per gallon). Six people each drive a car from Phoenix to Los Angeles. The resulting fuel efficiencies (in miles per gallon) are: 27.2 29.3 31.2 28.4 30.3 29.6 Assuming that fuel efficiency is normally distributed, do these data provide evidence against the claim that actual mean fuel efficiency for this model is (at least) 30 mpg?","H0: µ = 30, Ha: µ < 30, t = -1.160, P-value = 0.149, fail to reject H0. There is not convincing evidence that the mean fuel efficiency under these circumstances is less than 30 miles per gallon.",Free Response,,,,
-LFD,12.3,59,"A hot tub manufacturer advertises that a water temperature of 100°F can be achieved in 15 minutes or less. A random sample of 25 tubs is selected, and the time necessary to achieve a 100°F temperature is determined for each tub. The sample mean time and sample standard deviation are 17.5 minutes and 2.2 minutes, respectively. Does this information cast doubt on the company's claim? Carry out a test of hypotheses using significance level 0.05.","H0: µ = 15, Ha: µ > 15, t = 5.682, P-value = 0.000, reject H0. There is convincing evidence that the mean time to 100°F is greater than 15 minutes.",Free Response,,,,
-LFD,12R,61a,"Let x denote the time (in minutes) that it takes a fifth-grade student to read a certain passage. Suppose that the mean value and standard deviation of the x distribution are µ = 2  minutes and σ = 0.8 minutes, respectively. If x-bar is the sample mean time for a random sample of n = 9 students, where is the x-bar distribution centered, and what is the standard deviation of the x-bar distribution?","µx = 2, σx = 0.267",Free Response,,,,
-LFD,12R,61b,"Let x denote the time (in minutes) that it takes a fifth-grade student to read a certain passage. Suppose that the mean value and standard deviation of the x distribution are µ = 2  minutes and σ = 0.8 minutes, respectively. If x-bar is the sample mean time for a random sample of size of n = 20, where is the x-bar distribution centered, and what is the standard deviation of the x-bar distribution? How about for a sample of size n = 100. How do the centers and spreads of the three x-bar distributions (include n = 9) compare to one another? Which sample size would be most likely to result in an x-bar value close to µ, and why?","In each case µi = 2. When n = 20, σx = 0.179, and when n = 100, σx = 0.08 . All three centers are the same, and the larger the sample size, the smaller the standard deviation of x-bar. Since the distribution of x-bar when n = 100 is the one with the smallest standard deviation of the three, this sample size is most likely to result in a value of x-bar close to µ.",Free Response,,,,
-LFD,12R,63a,"Suppose that a random sample of 50 bottles of a particular brand of cough medicine is selected, and the amount of cough medicine in each bottle is determined. Let µ denote the mean amount of cough medicine for the population of all bottles of this brand. Suppose that this sample of 50 results in a 95% confidence interval for µ of (7.8, 9.4). Would a 90% confidence interval have been narrower or wider than the given interval? Explain your answer. ",Narrower,Free Response,,,,
-LFD,12R,63b,"Suppose that a random sample of 50 bottles of a particular brand of cough medicine is selected, and the amount of cough medicine in each bottle is determined. Let µ denote the mean amount of cough medicine for the population of all bottles of this brand. Suppose that this sample of 50 results in a 95% confidence interval for µ of (7.8, 9.4). Consider the following statement: There is a 95% chance that µ is between 7.8 and 9.4. Is this statement correct? Why or why not? ","The statement is not correct. The population mean, µ, is a constant, and it is not appropriate to talk about the probability that it falls within a certain interval. ",Free Response,,,,
-LFD,12R,63c,"Suppose that a random sample of 50 bottles of a particular brand of cough medicine is selected, and the amount of cough medicine in each bottle is determined. Let µ denote the mean amount of cough medicine for the population of all bottles of this brand. Suppose that this sample of 50 results in a 95% confidence interval for µ of (7.8, 9.4). Consider the following statement: If the process of selecting a random sample of size 50 and then computing the corresponding 95% confidence interval is repeated 100 times, exactly 95 of the resulting intervals will include µ. Is this statement correct? Why or why not?","The statement is not correct. You can say that in the long run about 95 out of every 100 samples will result in confidence intervals that will contain µ, but we cannot say that in 100 such samples, exactly 95 will result in confidence intervals that contain µ.",Free Response,,,,
-LFD,12R,65,"How much money do people spend on graduation gifts? In 2007, the National Retail Federation (www.nrf.com) surveyed 2,815 consumers who reported that they bought one or more graduation gifts that year. The sample was selected to be representative of adult Americans who purchased graduation gifts in 2007. For this sample, the mean amount spent per gift was $55.05. Suppose that the sample standard deviation was $20. Construct and interpret a 98% confidence interval for the mean amount of money spent per graduation gift in 2007.","(54.172, 55.928). You can be 98% confident that the mean amount of money spent per graduation gift in 2007 was between $54.172 and $55.928.",Free Response,,,,
-LFD,12R,67,"Suppose that the researchers who carried out the study described in the previous exercise wanted to estimate the mean reaction time with a margin of error of 5 msec. Using the sample standard deviation as a preliminary estimate of the population standard deviation, compute the required sample size.",753,Free Response,,,,
-LFD,12R,69,"Medical research has shown that repeated wrist extension beyond 20 degrees increases the risk of wrist and hand injuries. Each of 24 students at Cornell University used a proposed new computer mouse design, and while using the mouse, each student's wrist extension was recorded. Data consistent with summary values given in the paper ""Comparative Study of Two Computer Mouse Designs"" (Cornell Human Factors Laboratory Technical Report RP7992) are given. Use these data to test the hypothesis that the mean wrist extension for people using this new mouse design is greater than 20 degrees. Are any assumptions required in order for it to be appropriate to generalize the results of your test to the population of all Cornell students? To the population of all university students?","H0: µ = 20, Ha: µ > 20, t = 14.836, P-value = 0.000, reject H0. There is convincing evidence that the mean wrist extension for all people using the new mouse design is greater than 20 degrees. To generalize the result to the population of Cornell students, you need to assume that the 24 students used in the study are representative of all students at the university. To generalize the result to the population of all university students, you need to assume that the 24 students used in the study are representative of all university students.",Free Response,,,,
-LFD,13.1,1a,"In the following study, the two populations of interest are the students at a particular university who live on campus and the students who live off campus. Which of these studies have samples that are independently selected? Study 1: To determine if there is evidence that the mean amount of money spent on food each month differs for the two populations, a random sample of 45 students who live on campus and a random sample of 50 students who live off campus are selected.",Independently selected,Free Response,,,,
-LFD,13.1,1b,"In the following study, the two populations of interest are the students at a particular university who live on campus and the students who live off campus. Which of these studies have samples that are independently selected? Study 2: To determine if the mean number of hours spent studying differs for the two populations, a random sample students who live on campus is selected. Each student in this sample is asked how many hours he or she spends working each week. For each of these students who live on campus, a student who lives off campus and who works the same number of hours per week is identified and included in the sample of students who live off campus.",Not independently selected,Free Response,,,,
-LFD,13.1,1c,"In the following study, the two populations of interest are the students at a particular university who live on campus and the students who live off campus. Which of these studies have samples that are independently selected? Study 3: To determine if the mean number of hours worked per week differs for the two populations, a random sample of students who live on campus and who have a brother or sister who also attends the university but who lives off campus is selected. The sibling who lives on campus is included in the on campus sample, and the sibling who lives off campus is included in the off-campus sample.",Not independently selected,Free Response,,,,
-LFD,13.1,1d,"In the following study, the two populations of interest are the students at a particular university who live on campus and the students who live off campus. Which of these studies have samples that are independently selected? Study 4: To determine if the mean amount spent on textbooks differs for the two populations, a random sample of students who live on campus is selected. A separate random sample of the same size is selected from the population of students who live off campus.",Independently selected,Free Response,,,,
-LFD,13.1,2a,"For the following hypothesis testing scenario, indicate whether or not the appropriate hypothesis test would be about a difference in population means. If not, explain why not. Scenario 1: The international polling organization Ipsos reported data from a survey of 2,000 randomly selected Canadians who carry debit cards (Canadian Account Habits Survey, July 24, 2006). Participants in this survey were asked what they considered the minimum purchase amount for which it would be acceptable to use a debit card. You would like to determine if there is convincing evidence that the mean minimum purchase amount for which Canadians consider the use of a debit card to be acceptable is less than $10.",The appropriate test is not about a difference in population means. There is only one sample. ,Free Response,,,,
-LFD,13.1,2b,"For the following hypothesis testing scenario, indicate whether or not the appropriate hypothesis test would be about a difference in population means. If not, explain why not. Scenario 2: Each person in a random sample of 247 male working adults and a random sample of 253 female working adults living in Calgary, Canada, was asked how long, in minutes, his or her typical daily commute was (""Calgary Herald Traffic Study,"" Ipsos, September 17, 2005). You would like to determine if there is convincing evidence that the mean commute times differ for male workers and female workers.",The appropriate test is about a difference in population means. ,Free Response,,,,
-LFD,13.1,2c,"For the following hypothesis testing scenario, indicate whether or not the appropriate hypothesis test would be about a difference in population means. If not, explain why not. Scenario 3: A hotel chain is interested in evaluating reservation processes. Guests can reserve a room using either a telephone system or an online system. Independent random samples of 80 guests who reserved a room by phone and 60 guests who reserved a room online were selected. Of those who reserved by phone, 57 reported that they were satisfied with the reservation process. Of those who reserved online, 50 reported that they were satisfied. You would like to determine if it reasonable to conclude that the proportion who are satisfied is higher for those who reserve a room online.","The appropriate test is not about a difference in population means. There are two samples, but the variable is categorical rather than numerical",Free Response,,,,
-LFD,13.1,3,"Do male college students spend more time using a computer than female college students? This was one of the questions investigated by the authors of the paper ""An Ecological Momentary Assessment of the Physical Activity and Sedentary Behaviour Patterns of University Students"" (Health Education Journal [2010]: 116–125). Each student in a random sample of 46 male students at a university in England and each student in a random sample of 38 female students from the same university kept a diary of how he or she spent time over a three-week period. For the sample of males, the mean time spent using a computer per day was 45.8 minutes and the standard deviation was 63.3 minutes. For the sample of females, the mean time spent using a computer was 39.4 minutes and the standard deviation was 57.3 minutes. Is there convincing evidence that the mean time male students at this university spend using a computer is greater than the mean time for female students? Test the appropriate hypotheses using α = 0.05.","H0: µM - µF = 0, Ha: µM - µF > 0, t = 0.49, P-value = 0.314, fail to reject H0. There is not convincing evidence that the mean time is greater for males than for females.",Free Response,,,,
-LFD,13.1,4,"A study compared students who use Facebook and students who do not use Facebook (""Facebook and Academic Performance,"" Computers in Human Behavior [2010]: 1237–1245). In addition to asking the students in the samples about GPA, each student was also asked how many hours he or she spent studying each day. The two samples (141 students who were Facebook users and 68 students who were not Facebook users) were independently selected from students at a large, public Midwestern university. Although the samples were not selected at random, they were selected to be representative of the two populations. For the sample of Facebook users, the mean number of hours studied per day was 1.47 hours and the standard deviation was 0.83 hours. For the sample of students who do not use Facebook, the mean was 2.76 hours and the standard deviation was 0.99 hours. Do these sample data provide convincing evidence that the mean time spent studying for Facebook users is less than the mean time spent studying for students who do not use Facebook? Use a significance level of 0.01.","H0: µNF - µSF = 0, Ha: µNF - µSF < 0, t = -9.29, P-value ≈ 0.000, reject H0. There is convincing evidence that the mean time spent studying is less for Facebook users than for those who do not use Facebook.",Free Response,,,,
-LFD,13.1,5,"The paper ""Ladies First?"" A Field Study of Discrimination in Coffee Shops"" (Applied Economics [2008]: 1-19) describes a study in which researchers observed wait times in coffee shops in Boston. Both wait time and gender of the customer were observed. The mean wait time for a sample of 145 male customers was 85.2 seconds. The mean wait time for a sample of 141 female customers was 113.7 seconds. The sample standard deviations (estimated from graphs in the paper) were 50 seconds for the sample of males and 75 seconds for the sample of females. Suppose that these two samples are representative of the populations of wait times for female coffee shop customers and for male coffee shop customers. Is there convincing evidence that the mean wait time differs for males and females? Test the relevant hypotheses using a significance level of 0.05.","H0: µM - µF = 0, Ha: µM - µF ≠ 0, t = -3.77, P-value ≈ 0.000, reject H0. There is convincing evidence that the mean wait time differs for males and females.",Free Response,,,,
-LFD,13.1,6,"Some people believe that talking on a cell phone while driving slows reaction time, increasing the risk of accidents. The study described in the paper ""A Comparison of the Cell Phone Driver and the Drunk Driver"" (Human Factors [2006]: 381–391) investigated the braking reaction time of people driving in a driving simulator. Drivers followed a pace car in the simulator, and when the pace car's brake lights came on, the drivers were supposed to step on the brake. The time between the pace car brake lights coming on and the driver stepping on the brake was measured. Two samples of 40 drivers participated in the study. The 40 people in one sample used a cell phone while driving. The 40 people in the second sample drank a mixture of orange juice and alcohol in an amount calculated to achieve a blood alcohol level of 0.08% (a value considered legally drunk in most states). For the cell phone sample, the mean braking reaction time was 779 milliseconds and the standard deviation was 209 milliseconds. For the alcohol sample, the mean breaking reaction time was 849 milliseconds and the standard deviation was 228. Is there convincing evidence that the mean braking reaction time is different for the population of drivers talking on a cell phone and the population of drivers who have a blood alcohol level of 0.08%? For purposes of this exercise, you can assume that the two samples are representative of the two populations of interest.","H0: µC - µA = 0, Ha: µC - µA ≠ 0, t = -1.43, P-value = 0.156 , fail to reject H0. There is not convincing evidence that the mean reaction time differs for those talking on a cell phone and those who have a blood alcohol level of 0.08%.",Free Response,,,,
-LFD,13.1,15,"Research has shown that, for baseball players, good hip range of motion results in improved performance and decreased body stress. The article “Functional Hip Characteristics of Baseball Pitchers and Position Players” (The American Journal of Sports Medicine, 2010: 383–388) reported on a study of independent samples of 40 professional pitchers and 40 professional position players. For the pitchers, the sample mean hip range of motion was 75.6 degrees and the sample standard deviation was 5.9 degrees, whereas the sample mean and sample standard deviation for position players were 79.6 degrees and 7.6 degrees, respectively. Assuming that the two samples are representative of professional baseball pitchers and position players, test hypotheses appropriate for determining if mean range of motion for pitchers is less than the mean for position players.","H0: µpf - µpp = 0, Ha: µpf - µpp < 0, t = -2.63, P-value = 0.005, for α = 0.05 or 0.01 reject H0. There is convincing evidence that the mean range of motion is less for pitchers than for position players.",Free Response,,,,
-LFD,13.1,19a,"The article “Plugged In, but Tuned Out” (USA Today, January 20, 2010) summarizes data from two surveys of kids ages 8 to 18. One survey was conducted in 1999 and the other was conducted in 2009. Data on number of hours per day spent using electronic media, consistent with summary quantities in the article, are given below (the actual sample sizes for the two surveys were much larger). For purposes of this exercise, you can assume that the two samples are representative of kids ages 8 to 18 in each of the 2 years when the surveys were conducted. 2009: 5 9 5 8 7 6 7 9 7 9 6 9 10 9 8  1999: 4 5 7 7 5 7 5 6 5 6 7 8 5 6 6 Because the given sample sizes are small, what assumption must be made about the distributions of electronic media use times for the two-sample t test to be appropriate? Use the given data to construct graphical displays that would be useful in determining whether this assumption is reasonable. Do you think it is reasonable to use these data to carry out a two-sample t test? ","Since boxplots are roughly symmetrical and since there are no outliers in either sample, the assumption of normality is plausible, and it is reasonable to carry out a two-sample t test. ",Free Response,,,,
-LFD,13.1,19b,"The article “Plugged In, but Tuned Out” (USA Today, January 20, 2010) summarizes data from two surveys of kids ages 8 to 18. One survey was conducted in 1999 and the other was conducted in 2009. Data on number of hours per day spent using electronic media, consistent with summary quantities in the article, are given below (the actual sample sizes for the two surveys were much larger). For purposes of this exercise, you can assume that the two samples are representative of kids ages 8 to 18 in each of the 2 years when the surveys were conducted. 2009: 5 9 5 8 7 6 7 9 7 9 6 9 10 9 8  1999: 4 5 7 7 5 7 5 6 5 6 7 8 5 6 6  Do the given data provide convincing evidence that the mean number of hours per day spent using electronic media was greater in 2009 than in 1999? Test the relevant hypotheses using a significance level of 0.01.","H0: µ2009 - µ1999 = 0, Ha: µ2009 - µ1999 > 0, t = 3.32, P-value = 0.001, reject H0. There is convincing evidence that the mean time spent using electronic media was greater in 2009 than in 1999.",Free Response,,,,
-LFD,13.1,21a,"In a study of malpractice claims where a settlement had been reached, two random samples were selected: a random sample of 515 closed malpractice claims that were found not to involve medical errors and a random sample of 889 claims that were found to involve errors (New England Journal of Medicine [2006]: 2024–2033). The following statement appeared in the paper: “When claims not involving errors were compensated, payments were significantly lower on average than were payments for claims involving errors ($313,205 vs. $521,560, P = 0.004).” What hypotheses did the researchers test to reach the stated conclusion? ","µ1 = mean payment for claims not involving errors; µ2 = mean payment for claims involving errors; H0: µ1 - µ2 = 0, Ha: µ1 - µ2 < 0.",Free Response,,,,
-LFD,13.1,21b,"In a study of malpractice claims where a settlement had been reached, two random samples were selected: a random sample of 515 closed malpractice claims that were found not to involve medical errors and a random sample of 889 claims that were found to involve errors (New England Journal of Medicine [2006]: 2024–2033). The following statement appeared in the paper: “When claims not involving errors were compensated, payments were significantly lower on average than were payments for claims involving errors ($313,205 vs. $521,560, P = 0.004).” Which of the following could have been the value of the test statistic for the hypothesis test? Explain your reasoning. i. t = 5.00 iii. t = 2.33 ii. t = 2.65 iv. t = 1.47","Since the samples are large, a t distribution with a large number of degrees of freedom is used, which can be approximated with the standard normal distribution. P(z > 2.65 = 0.004 , which is the P-value given. None of the other possible values of t gives the correct P-value.",Free Response,,,,
-LFD,13.3,47,"The paper “Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children in a National Household Survey” (Pediatrics [2004]: 112–118) investigated the effect of fast-food consumption on other dietary variables. For a representative sample of 663 teens who reported that they did not eat fast food during a typical day, the mean daily calorie intake was 2,258 and the sample standard deviation was 1,519. For a representative sample of 413 teens who reported that they did eat fast food on a typical day, the mean calorie intake was 2,637 and the standard deviation was 1,138. Use the given information and a 95% confidence interval to estimate the difference in mean daily calorie intake for teens who do eat fast food on a typical day and those who do not.","(−538.606, −219.394). You can be 95% confident that the mean daily calorie intake for teens who do not eat fast food on a typical day minus the mean daily calorie intake for teens who do eat fast food on a typical day is between −538.606 and −219.394.",Free Response,,,,
-LFD,13.3,48,"In the study described in the paper “Exposure to Diesel Exhaust Induces Changes in EEG in Human Volunteers” (Particle and Fibre Toxicology [2007]), 10 healthy men were exposed to diesel exhaust for 1 hour. A measure of brain activity (called median power frequency, or MPF) was recorded at two different locations in the brain both before and after the diesel exhaust exposure. The resulting data are given in the accompanying table. For purposes of this exercise, assume that the sample of 10 men is representative of healthy adult males. Construct and interpret a 90% confidence interval estimate for the difference in mean MPF at brain location 2 before and after exposure to diesel exhaust.","(−3.069, −0.791). You can be 90% confident that the mean difference in MPF at brain location 1 is between −3.069 and −0.791.",Free Response,,,,
-LFD,13.3,49,"In the study described in the paper “Exposure to Diesel Exhaust Induces Changes in EEG in Human Volunteers” (Particle and Fibre Toxicology [2007]), 10 healthy men were exposed to diesel exhaust for 1 hour. A measure of brain activity (called median power frequency, or MPF) was recorded at two different locations in the brain both before and after the diesel exhaust exposure. The resulting data are given in the accompanying table. For purposes of this exercise, assume that the sample of 10 men is representative of healthy adult males. Construct and interpret a 90% confidence interval estimate for the difference in mean MPF at brain location 2 before and after exposure to diesel exhaust.","(−2.228, −0.852). You can be 90% confident that the mean difference in MPF at brain location 1 is between −2.228 and −0.852.",Free Response,,,,
-LFD,13.3,50a,"Do girls think they don’t need to take as many science classes as boys? The article “Intentions of Young Students to Enroll in Science Courses in the Future: An Examination of Gender Differences” (Science Education [1999]: 55–76) describes a survey of randomly selected children in grades 4, 5, and 6. The 224 girls participating in the survey each indicated the number of science courses they intended to take in the future, and they also indicated the number of science courses they thought boys their age should take in the future. For each girl, the authors calculated the difference between the number of science classes she intends to take and the number she thinks boys should take. Explain why these data are paired. ","There is a single sample of girls, with two data values for each girl. ",Free Response,,,,
-LFD,13.3,50b,"Do girls think they don’t need to take as many science classes as boys? The article “Intentions of Young Students to Enroll in Science Courses in the Future: An Examination of Gender Differences” (Science Education [1999]: 55–76) describes a survey of randomly selected children in grades 4, 5, and 6. The 224 girls participating in the survey each indicated the number of science courses they intended to take in the future, and they also indicated the number of science courses they thought boys their age should take in the future. For each girl, the authors calculated the difference between the number of science classes she intends to take and the number she thinks boys should take. The mean of the differences was -0.83 (indicating girls intended, on average, to take fewer science classes than they thought boys should take), and the standard deviation was 1.51. Construct and interpret a 95% confidence interval for the mean difference.","(−1.029, −0.631). You can be 95% confident that the mean difference between the number of science classes a girl intends to take and the number she thinks boys should take is between −1.029 and −0.631.",Free Response,,,,
-LFD,13R,61a,"An individual can take either a scenic route to work or a nonscenic route. She decides that use of the nonscenic route can be justified only if it reduces the mean travel time by more than 10 minutes.  If µ1 refers to the mean travel time for scenic route and µ2 to the mean travel time for nonscenic route, what hypotheses should be tested?",H0: µ1 - µ2 = 10 versus Ha: µ1 - µ2 > 10,Free Response,,,,
-LFD,13R,61b,"An individual can take either a scenic route to work or a nonscenic route. She decides that use of the nonscenic route can be justified only if it reduces the mean travel time by more than 10 minutes. If µ1 refers to the mean travel time for nonscenic route and µ2 to the mean travel time for scenic route, what hypotheses should be tested?",H0: µ1 - µ2 = -10 versus Ha: µ1 - µ2 < -10,Free Response,,,,
-LFD,13R,63a,"Indicate whether or not the appropriate hypothesis test would be for a difference in population means for the following scenario. If not, explain why not. Scenario 1: The authors of the paper “Adolescents and MP3 Players: Too Many Risks, Too Few Precautions” (Pediatrics [2009]: e953–e958) studied independent random samples of 764 Dutch boys and 748 Dutch girls ages 12 to 19. Of the boys, 397 reported that they almost always listen to music at a high volume setting. Of the girls, 331 reported listening to music at a high volume setting. You would like to determine if there is convincing evidence that the proportion of Dutch boys who listen to music at high volume is greater than this proportion for Dutch girls.",no,Free Response,,,,
-LFD,13R,63b,"Indicate whether or not the appropriate hypothesis test would be for a difference in population means for the following scenario. If not, explain why not. Scenario 2: The report “Highest Paying Jobs for 2009–10 Bachelor’s Degree Graduates” (National Association of Colleges and Employers, February 2010) states that the mean yearly salary offer for students graduating with accounting degrees in 2010 is $48,722. A random sample of 50 accounting graduates at a large university resulted in a mean offer of $49,850 and a standard deviation of $3,300. You would like to determine if there is strong support for the claim that the mean salary offer for accounting graduates of this university is higher than the 2010 national average of $48,722.",no,Free Response,,,,
-LFD,13R,63c,"Indicate whether or not the appropriate hypothesis test would be for a difference in population means for the following scenario. If not, explain why not. Scenario 3: Each person in a random sample of 228 male teenagers and a random sample of 306 female teenagers was asked how many hours he or she spent online in a typical week (Ipsos, January 25, 2006). The sample mean and standard deviation were 15.1 hours and 11.4 hours for males and 14.1 and 11.8 for females. You would like to determine if there is convincing evidence that the mean number of hours spent online in a typical week is greater for male teenagers than for female teenagers.",yes,Free Response,,,,
-LFD,13R,65,"The paper “Sodium content of Lunchtime Fast Food Purchases at Major U.S. Chains” (Archives of Internal Medicine [2010]: 732–734) reported that for a random sample of 850 meal purchases made at Burger King, the mean sodium content was 1,685 mg, and the standard deviation was 828 mg. For a random sample of 2,107 meal purchases made at McDonald’s, the mean sodium content was 1,477 mg, and the standard deviation was 812 mg. Based on these data, is it reasonable to conclude that there is a difference in mean sodium content for meal purchases at Burger King and meal purchases at McDonald’s? Use α = 0.05.","H0: µb - µm = 0 versus Ha: µm - µb ≠ 0, t = 6.22, P ≈ 0.000, reject H0. There is convincing evidence that the mean sodium content is not the same for meal purchases at Burger King and McDonalds.",Free Response,,,,
-OS,5.1,1a,"For the following situation, state whether the parameter of interest is a mean or a proportion: In a survey, one hundred college students are asked how many hours per week they spend on the Internet.",Mean. Each student reports a numerical value: a number of hours.,Free Response,,,,
-OS,5.1,1b,"For the following situation, state whether the parameter of interest is a mean or a proportion: In a survey, one hundred college students are asked: “What percentage of the time you spend on the Internet is part of your course work?”","Mean. Each student reports a number (it is a percentage, and we can average over these percentages).",Free Response,,,,
-OS,5.1,1c,"For the following situation, state whether the parameter of interest is a mean or a proportion: In a survey, one hundred college students are asked whether or not they cited information from Wikipedia in their papers.","Proportion. Each student reports Yes or No, so this is a categorical variable and we use a proportion.",Free Response,,,,
-OS,5.1,1d,"For the following situation, state whether the parameter of interest is a mean or a proportion: In a survey, one hundred college students are asked what percentage of their total weekly spending is on alcoholic beverages.",Mean. Each student reports a number (a percentage).,Free Response,,,,
-OS,5.1,1e,"For the following situation, state whether the parameter of interest is a mean or a proportion: In a sample of one hundred recent college graduates, it is found that 85 percent expect to get a job within one year of their graduation date.","Proportion. Each student reports whether or not he got a job, so this is a categorical variable and we use a proportion.",Free Response,,,,
-OS,5.1,2a,"For the following situation, state whether the parameter of interest is a mean or a proportion: A poll shows that 64% of Americans personally worry a great deal about federal spending and the budget deficit.","Proportion. Each respondent reports whether or not they worry a great deal about federal spending and the budget deficit, so this is a categorical variable and we use a proportion.",Free Response,,,,
-OS,5.1,2b,"For the following situation, state whether the parameter of interest is a mean or a proportion: A survey reports that local TV news has shown a 17% increase in revenue within a two year period while newspaper revenues decreased by 6.4% during this time period.",Mean. Each TV news program and newspaper report a number: revenue.,Free Response,,,,
-OS,5.1,2c,"For the following situation, state whether the parameter of interest is a mean or a proportion: In a survey, high school and college students are asked whether or not they use geolocation services on their smart phones.","Proportion. Each student reports whether or not they use geolocation services on their smart phones, so this is a categorical variable and we use a proportion.",Free Response,,,,
-OS,5.1,2d,"For the following situation, state whether the parameter of interest is a mean or a proportion: In a survey, smart phone users are asked whether or not they use a web-based taxi service.","Proportion. Each user reports whether or not they use a web-based taxi service, so this is a categorical variable and we use a proportion.",Free Response,,,,
-OS,5.1,2e,"For the following situation, state whether the parameter of interest is a mean or a proportion: In a survey, smart phone users are asked how many times they used a web-based taxi service over the last year.",Mean. Each user reports a number: how many times they used a web-based taxi service over the last year.,Free Response,,,,
-OS,5.1,3a,"As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. What population is under consideration in the data set?","The sample is from all computer chips manufactured at the factory during the week of pro- duction. We might be tempted to generalize the population to represent all weeks, but we should exercise caution here since the rate of defects may change over time.",Free Response,,,,
-OS,5.1,3b,"As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. What parameter is being estimated?",The fraction of computer chips manufactured at the factory during the week of production that had defects.,Free Response,,,,
-OS,5.1,3c,"As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. What is the point estimate for the parameter?",We estimate the parameter by computing the observed value in the data: p^ = 27/212 =  0.127,Free Response,,,,
-OS,5.1,3d,"As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. What is the name of the statistic we use to measure the uncertainty of the point estimate?","We quantify this uncertainty using the standard error, which may be abbreviated as SE.",Free Response,,,,
-OS,5.1,3e,"As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. Compute the value of the standard error for this context.",Compute the standard error using the SE formula and plugging in the point estimate p^ = 0:127 for p: SE = sqrt(0.127(1 – 0.127) / 212 = 0.023,Free Response,,,,
-OS,5.1,3f,"As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. The historical rate of defects is 10%. Should the engineer be surprised by the observed rate of defects during the current week?","The standard error is the standard deviation of p^. A value of 0.10 would be about one standard error away from the observed value, which would not represent a very uncommon deviation. (Usually beyond about 2 standard errors is a good rule of thumb.) The engineer should not be surprised.",Free Response,,,,
-OS,5.1,4a,"Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt. What population is under consideration in the data set?","The sample is from all adults in the United States, so US adults is the population under consideration.",Free Response,,,,
-OS,5.1,4b,"Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt. What parameter is being estimated?",The fraction of US adults who could not cover a $400 expense without borrowing money or selling something.,Free Response,,,,
-OS,5.1,4c,"Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt. What is the point estimate for the parameter?",We estimate the parameter by computing the observed value in the data: p^ = 322 / 765 = 0.421,Free Response,,,,
-OS,5.1,4d,"Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt. What is the name of the statistic we use to measure the uncertainty of the point estimate?","We quantify this uncertainty using the standard error, which may be abbreviated as SE.",Free Response,,,,
-OS,5.1,4e,"Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt. Compute the value from part (d) for this context.",We can compute the standard error using the SE formula and plugging in the point estimate p^ = 0:421 for p: SE = sqrt( 0.421(1 – 0.421) / 765) = 0.0179,Free Response,,,,
-OS,5.1,4f,"Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt. A cable news pundit thinks the value is actually 50%. Should she be surprised by the data?","The standard error can be thought of as the standard deviation of ^p. A value of 0.50 would be over 4 standard errors from the observed value, which would represent a very uncommon observation. The news pundit should be surprised by the data.",Free Response,,,,
-OS,5.1,5a,"A nonprofit wants to understand the fraction of households that have elevated levels of lead in their drinking water. They expect at least 5% of homes will have elevated levels of lead, but not more than about 30%. They randomly sample 800 homes and work with the owners to retrieve water samples, and they compute the fraction of these homes with elevated lead levels. They repeat this 1,000 times and build a distribution of sample proportions. What is this distribution called?",Sampling distribution.,Free Response,,,,
-OS,5.1,5b,"A nonprofit wants to understand the fraction of households that have elevated levels of lead in their drinking water. They expect at least 5% of homes will have elevated levels of lead, but not more than about 30%. They randomly sample 800 homes and work with the owners to retrieve water samples, and they compute the fraction of these homes with elevated lead levels. They repeat this 1,000 times and build a distribution of sample proportions. Would you expect the shape of this distribution to be symmetric, right skewed, or left skewed? Explain your reasoning.","To know whether the distribution is skewed, we need to know the proportion. We've been told the proportion is likely above 5% and below 30%, and the success-failure condition would be satisfied for any of these values. If the population proportion is in this range, the sampling distribution will be symmetric.",Free Response,,,,
-OS,5.1,5c,"A nonprofit wants to understand the fraction of households that have elevated levels of lead in their drinking water. They expect at least 5% of homes will have elevated levels of lead, but not more than about 30%. They randomly sample 800 homes and work with the owners to retrieve water samples, and they compute the fraction of these homes with elevated lead levels. They repeat this 1,000 times and build a distribution of sample proportions. If the proportions are distributed around 8%, what is the variability of the distribution?",We use the standard error to describe the variability: SE = sqrt( 0.08(1 – 0.08) / 800) = 0.0096,Free Response,,,,
-OS,5.1,5e,"A nonprofit wants to understand the fraction of households that have elevated levels of lead in their drinking water. They expect at least 5% of homes will have elevated levels of lead, but not more than about 30%. They randomly sample 800 homes and work with the owners to retrieve water samples, and they compute the fraction of these homes with elevated lead levels. They repeat this 1,000 times and build a distribution of sample proportions. Suppose the researchers’ budget is reduced, and they are only able to collect 250 observations per sample, but they can still collect 1,000 samples. They build a new distribution of sample proportions. How will the variability of this new distribution compare to the variability of the distribution when each sample contained 800 observations?",The distribution will tend to be more variable when we have fewer observations per sample.,Free Response,,,,
-OS,5.1,6a,"Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class project, students randomly sample 40 students and check if those students made the list. They repeat this 1,000 times and build a distribution of sample proportions. What is this distribution called?",Sampling distribution,Free Response,,,,
-OS,5.1,6b,"Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class project, students randomly sample 40 students and check if those students made the list. They repeat this 1,000 times and build a distribution of sample proportions. Would you expect the shape of this distribution to be symmetric, right skewed, or left skewed? Explain your reasoning.","Since the proportion is p = 0.16 and n = 40, the success-failure condition is not satis_ed, with the expected number of successes being just 40 x 0.16 = 6.4. When we have too few expected successes, the sampling distribution of p^ is right-skewed.",Free Response,,,,
-OS,5.1,6c,"Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class project, students randomly sample 40 students and check if those students made the list. They repeat this 1,000 times and build a distribution of sample proportions. Calculate the variability of this distribution.","The variability can still be calculated, even if we cannot model ^p using a normal distribution: SE = sqrt( 0.16(1 – 0.16) / 40) = 0.058",Free Response,,,,
-OS,5.1,6e,"Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class project, students randomly sample 40 students and check if those students made the list. They repeat this 1,000 times and build a distribution of sample proportions. Suppose the students decide to sample again, this time collecting 90 students per sample, and they again collect 1,000 samples. They build a new distribution of sample proportions. How will the variability of this new distribution compare to the variability of the distribution when each sample contained 40 observations?","When there are more observations in the sample, the point estimate tends to be less variable. This means the distribution will tend to be less variable when we have more observations per sample. Beyond the required answer: the success-failure condition would be satisfied with this larger sample, so the distribution would also be symmetric in this scenario.",Free Response,,,,
-OS,5.2,7,"In 2013, the Pew Research Foundation reported that “45% of U.S. adults report that they live with one or more chronic conditions” However, this value was based on a sample, so it may not be a perfect estimate for the population parameter of interest on its own. The study reported a standard error of about 1.2%, and a normal model may reasonably be used in this setting. Create a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions. Also interpret the confidence interval in the context of the study.","Recall that the general formula is point estimate ±_z*x SE. First, identify the three different values. The point estimate is 45%, z* = 1.96 for a 95% confidence level, and SE = 1.2%. Then, plug the values into the formula: 45% ± 1:96 x 1.2% = (42.6%; 47.4%) We are 95% confident that the proportion of US adults who live with one or more chronic conditions is between 42.6% and 47.4%.",Free Response,,,,
-OS,5.2,8,"A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter.12. The standard error for this estimate was 2.4%, and a normal distribution may be used to model the sample proportion. Construct a 99% confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter, and interpret the confidence interval in context.","Recall that the general formula is point estimate ± z* x SE. First, identify the three different values. The point estimate is 45%, z* = 2.58 for a 99% confidence level, and SE = 2.4%. Then, plug the values into the formula: 52% ± 2:58 x 2.4% = (45.8%; 58.2%) We are 99% confident that 45.8% to 58.2% of U.S. adult Twitter users get some news on Twitter.",Free Response,,,,
-OS,5.2,9b,"In 2013, the Pew Research Foundation reported that “45% of U.S. adults report that they live with one or more chronic conditions”, and the standard error for this estimate is 1.2%. Identify the following statement as true or false. Provide an explanation to justify your answer: If we repeated this study 1,000 times and constructed a 95% confidence interval for each study, then approximately 950 of those confidence intervals would contain the true fraction of U.S. adults who suffer from chronic illnesses.",True. Notice that the description focuses on the true population value.,Free Response,,,,
-OS,5.2,9c,"In 2013, the Pew Research Foundation reported that “45% of U.S. adults report that they live with one or more chronic conditions”, and the standard error for this estimate is 1.2%. Identify the following statement as true or false. Provide an explanation to justify your answer: The poll provides statistically significant evidence (at the α = 0.05 level) that the percentage of U.S. adults who suffer from chronic illnesses is below 50%.","True. If we examine the 95% confidence interval computed in Exercise ??, we can see that 50% is not included in this interval. This means that in a hypothesis test, we would reject the null hypothesis that the proportion is 0.5.",Free Response,,,,
-OS,5.2,9d,"In 2013, the Pew Research Foundation reported that “45% of U.S. adults report that they live with one or more chronic conditions”, and the standard error for this estimate is 1.2%. Identify the following statement as true or false. Provide an explanation to justify your answer: Since the standard error is 1.2%, only 1.2% of people in the study communicated uncertainty about their answer.","False. The standard error describes the uncertainty in the overall estimate from natural fluctuations due to randomness, not the uncertainty corresponding to individuals' responses.",Free Response,,,,
-OS,5.2,10a,"A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter, and the standard error for this estimate was 2.4%. Identify each of the following statements as true or false. Provide an explanation to justify each of your answers. The data provide statistically significant evidence that more than half of U.S. adult Twitter users get some news through Twitter. Use a significance level of α = 0.01. ","False. 50% is included in the 99% confidence interval, hence a null hypothesis of p = 0.50 would not be rejected at this level.",Free Response,,,,
-OS,5.2,10b,"A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter, and the standard error for this estimate was 2.4%. Identify each of the following statements as true or false. Provide an explanation to justify each of your answers. Since the standard error is 2.4%, we can conclude that 97.6% of all U.S. adult Twitter users were included in the study.","False. The standard error measures the variability of the sample proportion, and is unrelated to the proportion of the population included in the study.",Free Response,,,,
-OS,5.2,10c,"A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter, and the standard error for this estimate was 2.4%. Identify each of the following statements as true or false. Provide an explanation to justify each of your answers. If we want to reduce the standard error of the estimate, we should collect less data.",False. We need to increase the sample size to decrease the standard error.,Free Response,,,,
-OS,5.2,10d,"A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter, and the standard error for this estimate was 2.4%. Identify each of the following statements as true or false. Provide an explanation to justify each of your answers. If we construct a 90% confidence interval for the percentage of U.S. adults Twitter users who get some news through Twitter, this confidence interval will be wider than a corresponding 99% confidence interval.","False. As the confidence level decreases so does the margin of error, and hence the width of the confidence interval.",Free Response,,,,
-OS,5.2,11a,"A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a simple random sample of 64 patients and determines the time (in minutes) between when they checked in to the ER until they were first seen by a doctor. A 95% confidence interval based on this sample is (128 minutes, 147 minutes), which is based on the normal model for the mean. Determine whether the following statement is true or false, and explain your reasoning: We are 95% confident that the average waiting time of these 64 emergency room patients is between 128 and 147 minutes.","False, even if the population distribution is not normal, with a large enough sample size we can assume that the sampling distribution is nearly Normal and calculate a confidence interval. It would however be ideal to see a histogram or Normal probability plot of the population distribution to assess the level of skewness. If the distribution is very strongly skewed, the sample size of 64 may be insufficient for the sample mean to be approximately normally distributed.",Free Response,,,,
-OS,5.2,11b,"A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a simple random sample of 64 patients and determines the time (in minutes) between when they checked in to the ER until they were first seen by a doctor. A 95% confidence interval based on this sample is (128 minutes, 147 minutes), which is based on the normal model for the mean. Determine whether the following statement is true or false, and explain your reasoning: We are 95% confident that the average waiting time of all patients at this hospital’s emergency room is between 128 and 147 minutes.","True, this is the correct interpretation of the confidence interval.",Free Response,,,,
-OS,5.2,11c,"A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a simple random sample of 64 patients and determines the time (in minutes) between when they checked in to the ER until they were first seen by a doctor. A 95% confidence interval based on this sample is (128 minutes, 147 minutes), which is based on the normal model for the mean. Determine whether the following statement is true or false, and explain your reasoning: 95% of random samples have a sample mean between 128 and 147 minutes.","False, the confidence interval is not about a sample mean. The true interpretation of the confidence level would be that 95% of random samples produce confidence intervals that include the true population mean.",Free Response,,,,
-OS,5.2,11d,"A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a simple random sample of 64 patients and determines the time (in minutes) between when they checked in to the ER until they were first seen by a doctor. A 95% confidence interval based on this sample is (128 minutes, 147 minutes), which is based on the normal model for the mean. Determine whether the following statement is true or false, and explain your reasoning: A 99% confidence interval would be narrower than the 95% confidence interval since we need to be more sure of our estimate.","False. To be more confident that we capture the parameter, we need a wider interval.",Free Response,,,,
-OS,5.2,11e,"A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a simple random sample of 64 patients and determines the time (in minutes) between when they checked in to the ER until they were first seen by a doctor. A 95% confidence interval based on this sample is (128 minutes, 147 minutes), which is based on the normal model for the mean. Determine whether the following statement is true or false, and explain your reasoning: The margin of error is 9.5 and the sample mean is 137.5.","True, since the normal model was used to model the sample mean. The margin of error can be calculated as half the width of the interval, and the sample mean is the midpoint of the interval. ME = (146 – 128) / 2 = 9.5; x-bar = 128 + 9.5 = 137.5",Free Response,,,,
-OS,5.2,11f,"A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a simple random sample of 64 patients and determines the time (in minutes) between when they checked in to the ER until they were first seen by a doctor. A 95% confidence interval based on this sample is (128 minutes, 147 minutes), which is based on the normal model for the mean. Determine whether the following statement is true or false, and explain your reasoning: In order to decrease the margin of error of a 95% confidence interval to half of what it is now, we would need to double the sample size. (Hint: the margin of error for a mean scales in the same way with sample size as the margin of error for a proportion.)","False, since in calculation of the standard error we divide the standard deviation by square root of the sample size. In order to cut the standard error in half (and hence the margin of error) we would need to sample 22 = 4 times the number of people in the initial sample.",Free Response,,,,
-OS,5.2,12a,"The General Social Survey asked the question: “For how many days during the past 30 days was your mental health, which includes stress, depression, and problems with emotions, not good?” Based on responses from 1,151 US residents, the survey reported a 95% confidence interval of 3.40 to 4.24 days in 2010. Interpret this interval in context of the data.",We are 90% confident that US residents experience poor mental health 3.40 to 4.24 days per month.,Free Response,,,,
-OS,5.2,12b,"The General Social Survey asked the question: “For how many days during the past 30 days was your mental health, which includes stress, depression, and problems with emotions, not good?” Based on responses from 1,151 US residents, the survey reported a 95% confidence interval of 3.40 to 4.24 days in 2010. What does “95% confident” mean? Explain in the context of the application.","90% of random samples of size 1,151 will yield a confidence interval that contains the true average number of bad mental health days that US residents experience per month.",Free Response,,,,
-OS,5.2,12c,"The General Social Survey asked the question: “For how many days during the past 30 days was your mental health, which includes stress, depression, and problems with emotions, not good?” Based on responses from 1,151 US residents, the survey reported a 95% confidence interval of 3.40 to 4.24 days in 2010. Suppose the researchers think a 99% confidence level would be more appropriate for this interval. Will this new interval be smaller or wider than the 95% confidence interval?","To be more sure they capture the actual mean, they require a wider interval, unless they collect more data.",Free Response,,,,
-OS,5.2,12d,"The General Social Survey asked the question: “For how many days during the past 30 days was your mental health, which includes stress, depression, and problems with emotions, not good?” Based on responses from 1,151 US residents, the survey reported a 95% confidence interval of 3.40 to 4.24 days in 2010. If a new survey were to be done with 500 Americans, do you think the standard error of the estimate be larger, smaller, or about the same.","Less data means less precision. The estimate will probably be less accurate with less data, so the interval will be larger.",Free Response,,,,
-OS,5.2,13a,"A website is trying to increase registration for first-time visitors, exposing 1% of these visitors to a new site design. Of 752 randomly sampled visitors over a month who saw the new design, 64 registered. Check any conditions required for constructing a confidence interval.","The visitors are from a simple random sample, so independence is satisfied. The success-failure condition is also satisfied, with both 64 and 752 - 64 = 688 above 10. Therefore, we can use a normal distribution to model p^ and construct a confidence interval.",Free Response,,,,
-OS,5.2,13b,"A website is trying to increase registration for first-time visitors, exposing 1% of these visitors to a new site design. Of 752 randomly sampled visitors over a month who saw the new design, 64 registered. Compute the standard error.",The sample proportion is p^ = 64 / 752 = 0.085. The standard error is SE = sqrt(0.085(1 – 0.085) / 752) = 0.010,Free Response,,,,
-OS,5.2,13c,"A website is trying to increase registration for first-time visitors, exposing 1% of these visitors to a new site design. Of 752 randomly sampled visitors over a month who saw the new design, 64 registered. Construct and interpret a 90% confidence interval for the fraction of first-time visitors of the site who would register under the new design (assuming stable behaviors by new visitors over time).","For a 90% confidence interval, use z* = 1.65. The confidence interval is therefore 0.085 ± 1:65 x 0.010 = (0.0685; 0.1015). We are 90% confident that 6.85% to 10.15% of first-time site visitors will register using the new design.",Free Response,,,,
-OS,5.2,14, A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they’d received in the mail. Construct a 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they’d received in the mail.,"First, we check conditions: Independence. Since the data come from a random sample, independence is satisfied. Success-failure. We also have at least 10 successes (142) and 10 failures (603 - 142 = 461), so the success-failure condition is satisfied. With the conditions satisfied, p^ = 142 / 603 = 0.235 can be modeled using a normal distribution. Next, we compute the standard error: SE = sqrt(0.235(1 – 0.235) / 603) = 0.0173. For a 95% confidence level, z* = 1.96, and the confidence interval is 0.235 ± 1.96 x 0.0173 = (0:201; 0:269). We are 95% confident that 20.1% to 26.9% of the store's shoppers during the year made their visit because they had received a coupon in the mail.",Free Response,,,,
-OS,5.2,15a,Write the null and alternative hypotheses in words and then symbols for the following situation: A tutoring company would like to understand if most students tend to improve their grades (or not) after they use their services. They sample 200 of the students who used their service in the past year and ask them if their grades have improved or declined from the previous year.,H0: p = 0.5 (The number of students with grades that improve after the program is equal to the number of students who do not have their grades improve) HA: p ≠ 0.5 (Either a majority or a minority of students' grades improved),Free Response,,,,
-OS,5.2,15b,"Write the null and alternative hypotheses in words and then symbols for the following situation: Employers at a firm are worried about the effect of March Madness, a basketball championship held each spring in the US, on employee productivity. They estimate that on a regular business day employees spend on average 15 minutes of company time checking personal email, making personal phone calls, etc. They also collect data on how much company time employees spend on such non-business activities during March Madness. They want to determine if these data provide convincing evidence that employee productivity changed during March Madness.",H0: µ = 15 (The average amount of company time spent not working is 15 minutes) HA: µ ≠ 15 (The average amount of company time spent not working is different than 15 minutes),Free Response,,,,
-OS,5.2,16a,"Write the null and alternative hypotheses in words and using symbols for the following situations: Since 2008, chain restaurants in California have been required to display calorie counts of each menu item. Prior to menus displaying calorie counts, the average calorie intake of diners at a restaurant was 1100 calories. After calorie counts started to be displayed on menus, a nutritionist collected data on the number of calories consumed at this restaurant from a random sample of diners. Do these data provide convincing evidence of a difference in the average calorie intake of a diners at this restaurant?",H0: µ = 1100 (The current average calorie intake is 1100 calories) HA: µ ≠ 1100 (The current average calorie intake is different than 1100 calories.),Free Response,,,,
-OS,5.2,16b,"Write the null and alternative hypotheses in words and using symbols for the following situations: The state of Wisconsin would like to understand the fraction of its adult residents that consumed alcohol in the last year, specifically if the rate is different from the national rate of 70%. To help them answer this question, they conduct a random sample of 852 residents and ask them about their alcohol consumption.",H0: p = 0.7 (The fraction of Wisconsin adults who consume alcohol is 0.7) HA: p ≠ 0.7 (The fraction of Wisconsin adults who consume alcohol is different from 0.7),Free Response,,,,
-OS,5.2,17,"A study suggests that 60% of college student spend 10 or more hours per week communicating with others online. You believe that this is incorrect and decide to collect your own sample for a hypothesis test. You randomly sample 160 students from your dorm and find that 70% spent 10 or more hours a week communicating with others online. A friend of yours, who offers to help you with the hypothesis test, comes up with the following set of hypotheses. Indicate any errors you see. H0: p^ < 0.6; HA: p^ > 0.7","First, the hypotheses should be about the population proportion (p), not the sample proportion. Second, the null hypothesis should have an equal sign. Third, the alternative hypothesis should have a not-equals sign and reference the null value, p0 = 0.6, not the observed sample proportion. The correct way to set up these hypotheses is: H0: p = 0.6; HA: p ≠ 0.6",Free Response,,,,
-OS,5.2,18,"A study suggests that the 25% of 25 year olds have gotten married. You believe that this is incorrect and decide to collect your own sample for a hypothesis test. From a random sample of 25 year olds in census data with size 776, you find that 24% of them are married. A friend of yours offers to help you with setting up the hypothesis test and comes up with the following hypotheses. Indicate any errors you see. H0: p^ = 0.24; HA: p^ ≠ 0.24","First, the hypotheses should be about the population proportion (p), not the sample pro- portion. Second, the null value should be what we are testing (0.25), not the observed value (0.24). The correct way to set up these hypotheses is: H0: p = 0.25; HA: p ≠ 0.25",Free Response,,,,
-OS,5.2,19a,"Teens were surveyed about cyberbullying, and 54% to 64% reported experiencing cyberbullying (95% confidence interval). Answer the following questions based on this interval. A newspaper claims that a majority of teens have experienced cyberbullying. Is this claim supported by the confidence interval? Explain your reasoning.","This claim is reasonable, since the entire interval lies above 50%.",Free Response,,,,
-OS,5.2,19b,"Teens were surveyed about cyberbullying, and 54% to 64% reported experiencing cyberbullying (95% confidence interval). Answer the following questions based on this interval. A researcher conjectured that 70% of teens have experienced cyberbullying. Is this claim supported by the confidence interval? Explain your reasoning.","The value of 70% lies outside of the interval, so we have convincing evidence that the re- searcher's conjecture is wrong.",Free Response,,,,
-OS,5.2,19c,"Teens were surveyed about cyberbullying, and 54% to 64% reported experiencing cyberbullying (95% confidence interval). Answer the following questions based on this interval. Without actually calculating the interval, determine if the claim of the researcher that 70% of teens have experienced cyberbullying would be supported based on a 90% confidence interval?","A 90% confidence interval will be narrower than a 95% confidence interval. Even without calculating the interval, we can tell that 70% would not fall in the interval, and we would reject the researcher's conjecture based on a 90% confidence level as well.",Free Response,,,,
-OS,5.2,20a,"Exercise 5.11 provides a 95% confidence interval for the mean waiting time at an emergency room (ER) of (128 minutes, 147 minutes). Answer the following questions based on this interval. A local newspaper claims that the average waiting time at this ER exceeds 3 hours. Is this claim supported by the confidence interval? Explain your reasoning.",This claim does is not supported since 3 hours (180 minutes) is not in the interval.,Free Response,,,,
-OS,5.2,20b,"Exercise 5.11 provides a 95% confidence interval for the mean waiting time at an emergency room (ER) of (128 minutes, 147 minutes). Answer the following questions based on this interval. The Dean of Medicine at this hospital claims the average wait time is 2.2 hours. Is this claim supported by the confidence interval? Explain your reasoning.","2.2 hours (132 minutes) is in the 95% confidence interval, so we do not have evidence to say she is wrong.",Free Response,,,,
-OS,5.2,20c,"Exercise 5.11 provides a 95% confidence interval for the mean waiting time at an emergency room (ER) of (128 minutes, 147 minutes). Answer the following questions based on this interval. Without actually calculating the interval, determine if the claim of the Dean that the average wait time is 2.2 hours would be supported based on a 99% confidence interval?","A 99% confidence interval will be wider than a 95% confidence interval. Even without calculating the interval, we can tell that 132 minutes would be in it, and we would not reject her claim based on a 99% confidence level as well.",Free Response,,,,
-OS,5.2,21,"Do a majority of US adults believe raising the minimum wage will help the economy, or is there a majority who do not believe this? A Rasmussen Reports survey of a random sample of 1,000 US adults found that 42% believe it will help the economy. Conduct an appropriate hypothesis test to help answer the research question.","Here's the answer in the Prepare, Check, Calculate, and Conclude framework. Prepare. Set up hypotheses. H0: p = 0.5; HA: p ≠ 0.5 We will use a significance level of α = 0:05. Check. Simple random sample gets us independence, and the success-failure conditions is satisfied since 0.42 x 1000 = 420 and (1 – 0.42) x 1000 = 580 are both at least 10. Calculate. SE = sqrt(0.5(1 – 0.5) / 1000) = 0.016. Z = (0.42-0.5) / 0.016 = -5, which has a one-tail area of about 0.0000003, so the p-value is twice this one-tail area at 0.0000006. Conclude. Because the p-value is less than α = 0:05, we reject the null hypothesis and conclude that the fraction of US adults who believe raising the minimum wage will help the economy is not 50%. Because the observed value is less than 50% and we have rejected the null hypothesis, we can conclude that this belief is held by fewer than 50% of US adults. For reference, the survey also explores support for changing the minimum wage, which is itself a different question.",Free Response,,,,
-OS,5.2,22,"Getting enough sleep. 400 students were randomly sampled from a large university, and 289 said they did not get enough sleep. Conduct a hypothesis test to check whether this represents a statistically significant difference from 50%, and use a significance level of 0.01.","Here's the answer in the Prepare, Check, Calculate, and Conclude framework. Prepare. Set up hypotheses. H0: p = 0.5, HA: p ≠ 0.5. We are told to use a significance level of α = 0:01. Check. Simple random sample gets us independence, and the success-failure conditions is satisfied since 289 and 400 - 289 = 111 are both at least 10. Calculate. p^ = 289/400 = 0.7225. SE = sqrt(0.5(1 – 0.5) / 400) = 0.025. Z = (0.7225 – 0.5) / 0.025 = 8.9, which has a one-tail area of about 3e-19 (0.0000000000000000003). So the p-value is twice this one-tail area at 6e-19 (0.0000000000000000006). Conclude. Because the p-value is less than _ = 0:01, we reject the null hypothesis and conclude that the fraction of students who don't get enough sleep is different than 50%. Because the observed value is greater than 50% and we have rejected the null hypothesis, we can conclude that a majority of students at the surveyed university don't get enough sleep.",Free Response,,,,
-OS,5.2,23,You are given the following hypotheses: H0: p = 0.3; HA: p ≠ 0.3. We know the sample size is 90. For what sample proportion would the p-value be equal to 0.05? Assume that all conditions necessary for inference are satisfied.,"If the p-value is 0.05, this means the test statistic would be either Z = -1.96 or Z = 1.96. We'll show the calculations for Z = 1.96. The standard error would be SE = sqrt(0.3(1 – 0.3) / 90) = 0.048. Finally, we set up the test statistic formula and solve for p^: Z = (p^ - 0.3) / SE; 1.96 = (p^ - 0.3) / 0.048; p^ = 0.394. Alternatively, if Z = -1.96 was used: p^ = 0.206.",Free Response,,,,
-OS,5.2,24,"You are given the following hypotheses: H0: p = 0.9; HA: p ≠ 0.9. We know that the sample size is 1,429. For what sample proportion would the p-value be equal to 0.01? Assume that all conditions necessary for inference are satisfied.","If the p-value is 0.01, this means the test statistic would be either Z = -2.58 or Z = 2.58. If either of these Zs is chosen, it is okay. We'll use Z = 2.58 for our calculations. The standard error would be SE = sqrt(0.9(1 – 0.9) / 90) = 00079. Finally, we set up the test statistic formula and solve for p^: Z = (p^ - 0.9) / SE; 2.58 = (p^ - 0.9) / 0.0079; p^ = 0.92 Alternatively, if Z = -2.58 was used: p^ = 0.88.",Free Response,,,,
-OS,5.2,25a,"A patient named Diana was diagnosed with Fibromyalgia, a long-term syndrome of body pain, and was prescribed anti-depressants. Being the skeptic that she is, Diana didn’t initially believe that anti-depressants would help her symptoms. However after a couple months of being on the medication she decides that the anti-depressants are working, because she feels like her symptoms are in fact getting better. Write the hypotheses in words for Diana’s skeptical position when she started taking the anti-depressants.",H0: Anti-depressants do not affect the symptoms of Fibromyalgia. HA: Anti-depressants do affect the symptoms of Fibromyalgia (either helping or harming).,Free Response,,,,
-OS,5.2,25b,"A patient named Diana was diagnosed with Fibromyalgia, a long-term syndrome of body pain, and was prescribed anti-depressants. Being the skeptic that she is, Diana didn’t initially believe that anti-depressants would help her symptoms. However after a couple months of being on the medication she decides that the anti-depressants are working, because she feels like her symptoms are in fact getting better. What is a Type 1 Error in this context?",Concluding that anti-depressants either help or worsen Fibromyalgia symptoms when they actually do neither.,Free Response,,,,
-OS,5.2,25c,"A patient named Diana was diagnosed with Fibromyalgia, a long-term syndrome of body pain, and was prescribed anti-depressants. Being the skeptic that she is, Diana didn’t initially believe that anti-depressants would help her symptoms. However after a couple months of being on the medication she decides that the anti-depressants are working, because she feels like her symptoms are in fact getting better. What is a Type 2 Error in this context?",Concluding that anti-depressants do not affect Fibromyalgia symptoms when they actually do.,Free Response,,,,
-OS,5.2,26a,"In the scenario, there is a value of interest and two scenarios (I and II). For each part, report if the value of interest is larger under scenario I, scenario II, or whether the value is equal under the scenarios: The standard error of p^ when (I) n = 125 or (II) n = 500.",Scenario I is higher. Recall that a sample mean based on less data tends to be less accurate and have larger standard errors.,Free Response,,,,
-OS,5.2,26b,"In the scenario, there is a value of interest and two scenarios (I and II). For each part, report if the value of interest is larger under scenario I, scenario II, or whether the value is equal under the scenarios: The margin of error of a confidence interval when the confidence level is (I) 90% or (II) 80%.","Scenario I is higher. The higher the confidence level, the higher the corresponding margin of error.",Free Response,,,,
-OS,5.2,26c,"In the scenario, there is a value of interest and two scenarios (I and II). For each part, report if the value of interest is larger under scenario I, scenario II, or whether the value is equal under the scenarios: The p-value for a Z-statistic of 2.5 calculated based on a (I) sample with n = 500 or based on a (II) sample with n = 1000.",They are equal. The sample size does not affect the calculation of the p-value for a given Z-score.,Free Response,,,,
-OS,5.2,26d,"In the scenario, there is a value of interest and two scenarios (I and II). For each part, report if the value of interest is larger under scenario I, scenario II, or whether the value is equal under the scenarios: The probability of making a Type 2 Error when the alternative hypothesis is true and the significance level is (I) 0.05 or (II) 0.10.","Scenario I is higher. If the null hypothesis is harder to reject (lower α), then we are more likely to make a Type 2 Error when the alternative hypothesis is true.",Free Response,,,,
-OS,5R,27a,"The General Social Survey asked the question: “After an average work day, about how many hours do you have to relax or pursue activities that you enjoy?” to a random sample of 1,155 Americans. A 95% confidence interval for the mean number of hours spent relaxing or pursuing activities they enjoy was (1.38, 1.92). Interpret this interval in context of the data.",We are 95% confident that Americans spend an average of 1.38 to 1.92 hours per day relaxing or pursuing activities they enjoy.,Free Response,,,,
-OS,5R,27b,"The General Social Survey asked the question: “After an average work day, about how many hours do you have to relax or pursue activities that you enjoy?” to a random sample of 1,155 Americans. A 95% confidence interval for the mean number of hours spent relaxing or pursuing activities they enjoy was (1.38, 1.92). Suppose another set of researchers reported a confidence interval with a larger margin of error based on the same sample of 1,155 Americans. How does their confidence level compare to the confidence level of the interval stated above?",Their confidence level must be higher as the width of the confidence interval increases as the confidence level increases.,Free Response,,,,
-OS,5R,27c,"The General Social Survey asked the question: “After an average work day, about how many hours do you have to relax or pursue activities that you enjoy?” to a random sample of 1,155 Americans. A 95% confidence interval for the mean number of hours spent relaxing or pursuing activities they enjoy was (1.38, 1.92). Suppose next year a new survey asking the same question is conducted, and this time the sample size is 2,500. Assuming that the population characteristics, with respect to how much time people spend relaxing after work, have not changed much within a year. How will the margin of error of the 95% confidence interval constructed based on data from the new survey compare to the margin of error of the interval stated above?","The new margin of error will be smaller, since as the sample size increases, the standard error decreases, which will decrease the margin of error.",Free Response,,,,
-OS,5R,28,"We learned that a Rasmussen Reports survey of 1,000 US adults found that 42% believe raising the minimum wage will help the economy. Construct a 99% confidence interval for the true proportion of US adults who believe this.","We could note the data and context provided: p^ = 0.42, n = 1000, and we're using a 99% confidence level. Checking conditions. While it doesn't explicitly say here whether the sample is random, we will assume the survey by this company was in fact random, which would satisfy the independence condition. The success-failure condition is also satisfied since 0.42 x 1000 and 0.58 x 1000 are both at least 10. Next, we compute the standard error for the confidence interval using p^ = 0.42: SE = sqrt(0.42(1 – 0.42) / 1000 = 0.016 Next, we can compute the confidence interval itself, where we use z* = 2.58 for a 99% confidence level: 2.42 ± 2.58 x 0:016 = (0.379; 0.461) We are 99% confident that 37.9% to 46.1% of US adults believe that increasing the minimum wage would help the economy.",Free Response,,,,
-OS,5R,29a,"A food safety inspector is called upon to investigate a restaurant with a few customer reports of poor sanitation practices. The food safety inspector uses a hypothesis testing framework to evaluate whether regulations are not being met. If he decides the restaurant is in gross violation, its license to serve food will be revoked. Write the hypotheses in words.",H0: The restaurant meets food safety and sanitation regulations. HA: The restaurant does not meet food safety and sanitation regulations.,Free Response,,,,
-OS,5R,29b,"A food safety inspector is called upon to investigate a restaurant with a few customer reports of poor sanitation practices. The food safety inspector uses a hypothesis testing framework to evaluate whether regulations are not being met. If he decides the restaurant is in gross violation, its license to serve food will be revoked. What is a Type 1 Error in this context?",The food safety inspector concludes that the restaurant does not meet food safety and sanitation regulations and shuts down the restaurant when the restaurant is actually safe.,Free Response,,,,
-OS,5R,29c,"A food safety inspector is called upon to investigate a restaurant with a few customer reports of poor sanitation practices. The food safety inspector uses a hypothesis testing framework to evaluate whether regulations are not being met. If he decides the restaurant is in gross violation, its license to serve food will be revoked. What is a Type 2 Error in this context?",The food safety inspector concludes that the restaurant meets food safety and sanitation regulations and the restaurant stays open when the restaurant is actually not safe.,Free Response,,,,
-OS,5R,29d,"A food safety inspector is called upon to investigate a restaurant with a few customer reports of poor sanitation practices. The food safety inspector uses a hypothesis testing framework to evaluate whether regulations are not being met. If he decides the restaurant is in gross violation, its license to serve food will be revoked. Between a Tupe 1 and Type 2 error, which error is more problematic for the restaurant owner? Why?",A Type 1 Error may be more problematic for the restaurant owner since his restaurant gets shut down even though it meets the food safety and sanitation regulations.,Free Response,,,,
-OS,5R,29e,"A food safety inspector is called upon to investigate a restaurant with a few customer reports of poor sanitation practices. The food safety inspector uses a hypothesis testing framework to evaluate whether regulations are not being met. If he decides the restaurant is in gross violation, its license to serve food will be revoked. Between a Tupe 1 and Type 2 error, which error is more problematic for the diners? Why?",A Type 2 Error may be more problematic for diners since the restaurant deemed safe by the inspector is actually not.,Free Response,,,,
-OS,5R,29f,"A food safety inspector is called upon to investigate a restaurant with a few customer reports of poor sanitation practices. The food safety inspector uses a hypothesis testing framework to evaluate whether regulations are not being met. If he decides the restaurant is in gross violation, its license to serve food will be revoked. As a diner, would you prefer that the food safety inspector requires strong evidence or very strong evidence of health concerns before revoking a restaurant’s license? Explain your reasoning.","A diner would prefer strong evidence as any indication of evidence might mean there may be an issue with the restaurant meeting food safety and sanitation regulations, and diners would rather a restaurant that meet the regulations get shut down than a restaurant that doesn't meet the regulations not get shut down.",Free Response,,,,
-OS,5R,30a,"Determine if the following statement is true or false, and explain your reasoning. If false, state how it could be corrected: If a given value (for example, the null hypothesized value of a parameter) is within a 95% confidence interval, it will also be within a 99% confidence interval.",True.,Free Response,,,,
-OS,5R,30b,"Determine if the following statement is true or false, and explain your reasoning. If false, state how it could be corrected: Decreasing the significance level (α) will increase the probability of making a Type 1 Error.",False. The significance level is the probability of the Type 1 Error.,Free Response,,,,
-OS,5R,30c,"Determine if the following statement is true or false, and explain your reasoning. If false, state how it could be corrected: Suppose the null hypothesis is p = 0.5 and we fail to reject H0. Under this scenario, the true population proportion is 0.5.","False. Failure to reject H0 only means there wasn't sufficient evidence to reject it, not that it has been confirmed.",Free Response,,,,
-OS,5R,30d,"Determine if the following statement is true or false, and explain your reasoning. If false, state how it could be corrected: With large sample sizes, even small differences between the null value and the observed point estimate, a difference often called the effect size, will be identified as statistically significant.",True.,Free Response,,,,
-OS,5R,31a,"A USA Today/Gallup poll asked a group of unemployed and underemployed Americans if they have had major problems in their relationships with their spouse or another close family member as a result of not having a job (if unemployed) or not having a full-time job (if underemployed). 27% of the 1,145 unemployed respondents and 25% of the 675 underemployed respondents said they had major problems in relationships as a result of their employment status. What are the hypotheses for evaluating if the proportions of unemployed and underemployed people who had relationship problems were different?",H0: punemp = punderemp: The proportions of unemployed and underemployed people who are having relationship problems are equal. HA: punemp ≠ punderemp: The proportions of unemployed and underemployed people who are having relationship problems are different.,Free Response,,,,
-OS,5R,31b,"A USA Today/Gallup poll asked a group of unemployed and underemployed Americans if they have had major problems in their relationships with their spouse or another close family member as a result of not having a job (if unemployed) or not having a full-time job (if underemployed). 27% of the 1,145 unemployed respondents and 25% of the 675 underemployed respondents said they had major problems in relationships as a result of their employment status. The p-value for this hypothesis test is approximately 0.35. Explain what this means in context of the hypothesis test and the data.","If in fact the two population proportions are equal, the probability of observing at least a 2% difference between the sample proportions is approximately 0.35. Since this is a high probability we fail to reject the null hypothesis. The data do not provide convincing evidence that the proportion of unemployed and underemployed people who are having relationship problems are different.",Free Response,,,,
-OS,5R,32,"It is believed that nearsightedness affects about 8% of all children. In a random sample of 194 children, 21 are nearsighted. Conduct a hypothesis test for the following question: do these data provide evidence that the 8% value is inaccurate?","Here's the answer in the Prepare, Check, Calculate, and Conclude framework. Prepare. Set up hypotheses. H0: p = 0.08, HA: p ≠ 0.08. We will use a significance level of α = 0:05. Check. Simple random sample gets us independence, and the success-failure condition is satisfied since 21 and 194 - 21 = 173 are both at least 10. Calculate. Several calculations: p^ = 21=194 = 0.108. SE = sqrt(0.08(1 – 0.08) / 194) = 0.0195. Z = (0.108 – 0.08) / 0.0195 = 1.44 which has a one-tail area of about 0.075, so the p-value is twice this one-tail area at 0.15. Conclude. Because the p-value is bigger than α = 0:05, we do not reject the null hypothesis. The sample does not provide convincing evidence that the fraction of children who are nearsighted is different from 0.08.",Free Response,,,,
-OS,5R,33,"The nutrition label on a bag of potato chips says that a one ounce (28 gram) serving of potato chips has 130 calories and contains ten grams of fat, with three grams of saturated fat. A random sample of 35 bags yielded a confidence interval for the number of calories per bag of 128.2 to 139.8 calories. Is there evidence that the nutrition label does not provide an accurate measure of calories in the bags of potato chips?","Because 130 is inside the confidence interval, we do not have convincing evidence that the true average is any different than what the nutrition label suggests.",Free Response,,,,
-OS,5R,34,"Define the term “sampling distribution” of the sample proportion, and describe how the shape, center, and spread of the sampling distribution change as the sample size increases when p = 0.1.","The sampling distribution is the distribution of sample proportions from samples of the same size randomly sampled from the same population. As the same size increases, the shape of the sampling distribution (when p = 0.1) will go from being right-skewed to being more symmetric and resembling the normal distribution. With larger sample sizes, the spread of the sampling distribution gets smaller. Regardless of the sample size, the center of the sampling distribution is equal to the true mean of that population, provided the sampling isn't biased.",Free Response,,,,
-OS,5R,35,"Determine whether the following statement is true or false, and explain your reasoning: “With large sample sizes, even small differences between the null value and the observed point estimate can be statistically significant.”","True. If the sample size gets ever larger, then the standard error will become ever smaller. Eventually, when the sample size is large enough and the standard error is tiny, we can find statistically significant yet very small differences between the null value and point estimate (assuming they are not exactly equal).",Free Response,,,,
-OS,5R,36,"Same observation, different sample size. Suppose you conduct a hypothesis test based on a sample where the sample size is n = 50, and arrive at a p-value of 0.08. You then refer back to your notes and discover that you made a careless mistake, the sample size should have been n = 500. Will your p-value increase, decrease, or stay the same? Explain.","As the sample size increases the standard error will decrease, the sample statistic will increase, and the p-value will decrease.",Free Response,,,,
-OS,5R,37a,"A study examined the average pay for men and women entering the workforce as doctors for 21 different positions. If each gender was equally paid, then we would expect about half of those positions to have men paid more than women and women would be paid more than men in the other half of positions. Write appropriate hypotheses to test this scenario.","In effect, we're checking whether men are paid more than women (or vice-versa), and we'd expect these outcomes with either chance under the null hypothesis: H0:  p = 0.5; HA: p ≠ 0.5 We'll use p to represent the fraction of cases where men are paid more than women.",Free Response,,,,
-OS,5R,37b,"A study examined the average pay for men and women entering the workforce as doctors for 21 different positions. Men were, on average, paid more in 19 of those 21 positions. Supposing these 21 positions represent a simple random sample, complete a hypothesis test based on men and women being equally paid.","In effect, we're checking whether men are paid more than women (or vice-versa), and we'd expect these outcomes with either chance under the null hypothesis: H0:  p = 0.5; HA: p ≠ 0.5 We'll use p to represent the fraction of cases where men are paid more than women. There isn't a good way to check independence here since the jobs are not a simple random sample. However, independence doesn't seem unreasonable, since the individuals in each job are different from each other. The success-failure condition is met since we check it using the null proportion: p0n = (1 - p0)n = 10.5 is greater than 10. We can compute the sample proportion, SE, and test statistic: p^ = 19 / 21 = 0.905; SE = sqrt(0.5 x (1 – 0.5) / 21 = 0.109; Z = (0.905 – 0.5) / 0.109 = 3.72. The test statistic Z corresponds to an upper tail area of about 0.0001, so the p-value is 2 times this value: 0.0002. Because the p-value is smaller than 0.05, we reject the notion that all these gender pay disparities are due to chance. Because we observe that men are paid more in a higher proportion of cases and we have rejected H0, we can conclude that men are being paid higher amounts in ways not explainable by chance alone.",Free Response,,,,
-OS,6.1,1a,"Suppose that 8% of college students are vegetarians. Determine if the following statement is true or false, and explain your reasoning The distribution of the sample proportions of vegetarians in random samples of size 60 is approximately normal since n ≥ 30.","False. For the distribution of p^ to be approximately normal, we need to have at least 10 successes and 10 failures in our sample (on the average).",Free Response,,,,
-OS,6.1,1b,"Suppose that 8% of college students are vegetarians. Determine if the following statement is true or false, and explain your reasoning The distribution of the sample proportions of vegetarian college students in random samples of size 50 is right skewed.","True. The success-failure condition is not satisfied np = 50 x 0.08 = 4 and n(1 x p) = 50 x 0.92 = 46; therefore we know that the distribution of ^p is not approximately normal. In most samples we would expect p^ to be close to 0.08, the true population proportion. While p^ can be as high as 1 (though we would expect this to effectively never happen), it can only go as low as 0. Therefore the distribution would probably take on a right-skewed shape. Plotting the sampling distribution would confirm this suspicion.",Free Response,,,,
-OS,6.1,1c,"Suppose that 8% of college students are vegetarians. Determine if the following statement is true or false, and explain your reasoning A random sample of 125 college students where 12% are vegetarians would be considered unusual.","False. Standard error of p^ in samples with n = 125 can be calculated as: SEp^ = sqrt(0.08 x 0.92 / 125) = 0.0243 A p^ of 0.12 is only (0.12 – 0.08) / 0.0243 = 1.65 standard errors away from the mean, which would not be considered unusual.",Free Response,,,,
-OS,6.1,1d,"Suppose that 8% of college students are vegetarians. Determine if the following statement is true or false, and explain your reasoning A random sample of 250 college students where 12% are vegetarians would be considered unusual.","True. Standard error of p^ in samples with n = 125 can be calculated as: SEp^ = sqrt(0.08 x 0.92 / 250) = 0.0172 A p^ of 0.12 is 0: (12 x 0:08) / 0.0172 = 2.32 standard errors away from the mean, which might be considered unusual.",Free Response,,,,
-OS,6.1,1e,"Suppose that 8% of college students are vegetarians. Determine if the following statement is true or false, and explain your reasoning The standard error would be reduced by one-half if we increased the sample size from 125 to 250.","False. Since n appears under the square root sign in the formula for the standard error, increasing the sample size from 125 to 250 would decrease the standard error of the sample proportion only by a factor of sqrt(2).",Free Response,,,,
-OS,6.1,2a,"About 77% of young adults think they can achieve the American dream. Determine if the following statement is true or false, and explain your reasoning: The distribution of sample proportions of young Americans who think they can achieve the American dream in samples of size 20 is left skewed.","True. The success-failure condition is not satisfied np = 20 x 0.77 = 15.4 and n(1 x p) = 20 x 0.23 = 4.6; therefore we know that the distribution of ^p is not approximately normal. In most samples we would expect p^ to be close to 0.77, the true population proportion. While p^ can be as low as 0 (though we would expect this to happen very rarely), it can only go as high as 1. Therefore, since 0.77 is closer to 1, the distribution would probably take on a left skewed shape. Plotting the sampling distribution would confirm this suspicion.",Free Response,,,,
-OS,6.1,2b,"About 77% of young adults think they can achieve the American dream. Determine if the following statement is true or false, and explain your reasoning: The distribution of sample proportions of young Americans who think they can achieve the American dream in random samples of size 40 is approximately normal since n ≥ 30.","False. Unlike with means, for the sampling distribution of proportions to be approximately normal, we need to have at least 10 successes and 10 failures in our sample. We do not use n ≥ 30 as a condition to check for the normality of the distribution of p^.",Free Response,,,,
-OS,6.1,2c,"About 77% of young adults think they can achieve the American dream. Determine if the following statement is true or false, and explain your reasoning: A random sample of 60 young Americans where 85% think they can achieve the American dream would be considered unusual.","False. Standard error of p^ in samples with n = 60 can be calculated as: SEp^ = sqrt( 0.77 x 0.23 / 60) = 0.384. A p^ of 0.85 is only Z = (0.85-0.77) / 0.0384 = 2.08 standard errors away from the mean, which would be considered unusual.",Free Response,,,,
-OS,6.1,2d,"About 77% of young adults think they can achieve the American dream. Determine if the following statement is true or false, and explain your reasoning: A random sample of 120 young Americans where 85% think they can achieve the American dream would be considered unusual.","True. Standard error of p^ in samples with n = 120 can be calculated as: SEp^ = sqrt( 0.77 x 0.23 / 120) = 0.046. A p^ of 0.85 is Z = (0.85 – 0.77) / 0.046 = 1.73 standard errors away from the mean, which would not be considered unusual.",Free Response,,,,
-OS,6.1,3a,"Suppose that 90% of orange tabby cats are male. Determine if the following statement is true or false, and explain your reasoning: The distribution of sample proportions of random samples of size 30 is left skewed.","True. The success-failure condition is not satisfied np = 30 x 0.90 = 27 and n(1 - p) = 30 x 0.10 = 3; therefore we know that the distribution of ^p is not nearly normal. In most samples we would expect p^ to be close to 0.90, the true population proportion. While p^ can be as low as 0 (thought we would expect this to happen very rarely), it can only go as high as 1. Therefore the distribution would probably take on a left-skewed shape. Plotting the sampling distribution would confirm this suspicion.",Free Response,,,,
-OS,6.1,3b,"Suppose that 90% of orange tabby cats are male. Determine if the following statement is true or false, and explain your reasoning: Using a sample size that is 4 times as large will reduce the standard error of the sample proportion by one-half.","True. Since n appears in a square root for SE, using a sample size that is 4 times as large will reduce the SE by half.",Free Response,,,,
-OS,6.1,3c,"Suppose that 90% of orange tabby cats are male. Determine if the following statement is true or false, and explain your reasoning: The distribution of sample proportions of random samples of size 140 is approximately normal.",True. The success-failure condition is satisfied np = 140 x 0.90 = 126 and n(1 - p) = 140 x 0.10 = 14; therefore the distribution of p^ is nearly normal.,Free Response,,,,
-OS,6.1,3d,"Suppose that 90% of orange tabby cats are male. Determine if the following statement is true or false, and explain your reasoning: The distribution of sample proportions of random samples of size 280 is approximately normal.",True. The success-failure condition is satisfied np = 280 x 0.90 = 252 and n(1 - p) = 70 x 0.10 = 28; therefore the distribution of p^ is nearly normal.,Free Response,,,,
-OS,6.1,4a,"About 25% of young Americans have delayed starting a family due to the continued economic slump. Determine if the following statement is true or false, and explain your reasoning: The distribution of sample proportions of young Americans who have delayed starting a family due to the continued economic slump in random samples of size 12 is right skewed.","True. The success-failure condition is not satisfied np = 12 x 0.25 = 4 and n(1 - p) = 12 x 0.75 = 8; therefore we know that the distribution of p^ is not approximately normal. In most samples we would expect p^ to be close to 0.25, the true population proportion. While p^ can be as high as 1 (though we would expect this to happen very rarely), it can only go as low as 0. Therefore, since 0.25 is closer to 0, the distribution would probably take on a right skewed shape. Plotting the sampling distribution would confirm this suspicion.",Free Response,,,,
-OS,6.1,4b,"About 25% of young Americans have delayed starting a family due to the continued economic slump. Determine if the following statement is true or false, and explain your reasoning: In order for the distribution of sample proportions of young Americans who have delayed starting a family due to the continued economic slump to be approximately normal, we need random samples where the sample size is at least 40.","True. The minimum sample size that yields at least 10 successes and 10 failures can be calculated as n = max(10 / 0.25, 10 / 0.75) = max(40, 13.3) = 40. We need a sample of at least n = 40 to meet the success failure condition.",Free Response,,,,
-OS,6.1,4c,"About 25% of young Americans have delayed starting a family due to the continued economic slump. Determine if the following statement is true or false, and explain your reasoning: A random sample of 50 young Americans where 20% have delayed starting a family due to the continued economic slump would be considered unusual.","False. Standard error of ^p in samples with n = 50 can be calculated as: SEp^ = sqrt(0.25 x 0.75 / 50) = 0.0612. A p^ of 0.20 is only Z = (0.20 – 0.25) / 0.0612 = -0.82 standard errors away from the mean, which would not be considered unusual.",Free Response,,,,
-OS,6.1,4d,"About 25% of young Americans have delayed starting a family due to the continued economic slump. Determine if the following statement is true or false, and explain your reasoning: A random sample of 150 young Americans where 20% have delayed starting a family due to the continued economic slump would be considered unusual.","False. Standard error of p^ in samples with n = 150 can be calculated as: SE^p = sqrt(0.25 x 0.75 / 150) = 0.0354 A p^ of 0.20 is Z = (0.20 – 0.25) / 0.0354 = -1.41 standard errors away from the mean, which would not be considered unusual.",Free Response,,,,
-OS,6.1,4e,"About 25% of young Americans have delayed starting a family due to the continued economic slump. Determine if the following statement is true or false, and explain your reasoning: Tripling the sample size will reduce the standard error of the sample proportion by one-third.","False. Since n appears under the square root sign in the formula for the standard error, doubling the sample size would decrease the standard error of the sample proportion only by a factor of sqrt(3).",Free Response,,,,
-OS,6.1,5a,"The General Social Survey asked a random sample of 1,390 Americans the following question: “On the whole, do you think it should or should not be the government’s responsibility to promote equality between men and women?” 82% of the respondents said it “should be”. At a 95% confidence level, this sample has 2% margin of error. Based on this information, determine if the following statements are true or false, and explain your reasoning: We are 95% confident that between 80% and 84% of Americans in this sample think it’s the government’s responsibility to promote equality between men and women.","False. A confidence interval is constructed to estimate the population proportion, not the sample proportion.",Free Response,,,,
-OS,6.1,5b,"The General Social Survey asked a random sample of 1,390 Americans the following question: “On the whole, do you think it should or should not be the government’s responsibility to promote equality between men and women?” 82% of the respondents said it “should be”. At a 95% confidence level, this sample has 2% margin of error. Based on this information, determine if the following statements are true or false, and explain your reasoning: We are 95% confident that between 80% and 84% of all Americans think it’s the government’s responsibility to promote equality between men and women.","True. This is the correct interpretation of the confidence interval, which can be calculated as 0.82 ± 0.02 = (0.80; 0.84).",Free Response,,,,
-OS,6.1,5c,"The General Social Survey asked a random sample of 1,390 Americans the following question: “On the whole, do you think it should or should not be the government’s responsibility to promote equality between men and women?” 82% of the respondents said it “should be”. At a 95% confidence level, this sample has 2% margin of error. Based on this information, determine if the following statements are true or false, and explain your reasoning: If we considered many random samples of 1,390 Americans, and we calculated 95% confidence intervals for each, 95% of these intervals would include the true population proportion of Americans who think it’s the government’s responsibility to promote equality between men and women.",True. This is the correct interpretation of the confidence level.,Free Response,,,,
-OS,6.1,5d,"The General Social Survey asked a random sample of 1,390 Americans the following question: “On the whole, do you think it should or should not be the government’s responsibility to promote equality between men and women?” 82% of the respondents said it “should be”. At a 95% confidence level, this sample has 2% margin of error. Based on this information, determine if the following statements are true or false, and explain your reasoning: In order to decrease the margin of error to 1%, we would need to quadruple (multiply by 4) the sample size.","True. Since the sample size appears under the square root sign in calculation of the standard error, in order to halve the margin of error we would need to quadruple the sample size.",Free Response,,,,
-OS,6.1,5e,"The General Social Survey asked a random sample of 1,390 Americans the following question: “On the whole, do you think it should or should not be the government’s responsibility to promote equality between men and women?” 82% of the respondents said it “should be”. At a 95% confidence level, this sample has 2% margin of error. Based on this information, determine if the following statements are true or false, and explain your reasoning: Based on this confidence interval, there is sufficient evidence to conclude that a majority of Americans think it’s the government’s responsibility to promote equality between men and women.",True. The confidence interval lies entirely above 50%.,Free Response,,,,
-OS,6.1,6a,"The Marist Poll published a report stating that 66% of adults nationally think licensed drivers should be required to retake their road test once they reach 65 years of age. It was also reported that interviews were conducted on 1,018 American adults, and that the margin of error was 3% using a 95% confidence level. Verify the margin of error reported by The Marist Poll.","With a random sample from < 10% of the population, independence is satisfied. The success- failure condition is also satisfied. Hence, the margin of error can be calculated as follows: ME = 1.96 x sqrt(0.66 x 0.34 / 1018) = 0.029 ≈ 3%",Free Response,,,,
-OS,6.1,6b,"The Marist Poll published a report stating that 66% of adults nationally think licensed drivers should be required to retake their road test once they reach 65 years of age. It was also reported that interviews were conducted on 1,018 American adults, and that the margin of error was 3% using a 95% confidence level. Based on a 95% confidence interval, does the poll provide convincing evidence that more than 70% of the population think that licensed drivers should be required to retake their road test once they turn 65?","A 95% confidence interval for the proportion of adults who think that licensed drivers should be required to re-take their road test once they reach 65 years of age can be calculated as 0.66 ± 0.03 = (0.63; 0.69). Since two thirds (roughly 67%) is contained in the interval we wouldn't reject a null hypothesis where p = 0.67. Therefore, the data do not provide evidence that more than two thirds of the population think that drivers over the age of 65 should re-take their road test.",Free Response,,,,
-OS,6.1,7,A local news outlet reported that 56% of 600 randomly sampled Kansas residents planned to set off fireworks on July 4th. Determine the margin of error for the 56% point estimate using a 95% confidence level.,"With a random sample, independence is satisfied. The success-failure condition is also satisfied. Hence, the margin of error can be calculated as follows: ME = 1.96 x sqrt(0.56 x 0.44 / 600) = 0.0397 ≈ 4%",Free Response,,,,
-OS,6.1,8a,"Greece has faced a severe economic crisis since the end of 2009. A Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that 25% of them said they would rate their lives poorly enough to be considered “suffering”. Describe the population parameter of interest. What is the value of the point estimate of this parameter?","The population parameter of interest is the proportion of all Greeks who would rate their lives poorly enough to be considered “suffering"", p. The point estimate for this parameter is the proportion of Greeks in this sample who would rate their lives as such, p^ = 0:25.",Free Response,,,,
-OS,6.1,8b,"Greece has faced a severe economic crisis since the end of 2009. A Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that 25% of them said they would rate their lives poorly enough to be considered “suffering”. Check if the conditions required for constructing a confidence interval based on these data are met.","1. Independence: The sample is random, and 1,000 < 10% of all Greeks, therefore the life rating of one Greek in this sample is independent of another. 2. Success-failure: 1,000 x 0.25 = 250 > 10 and 1,000 x 0.75 = 750 > 10. Since the observations are independent and the success-failure condition is met, p^ is expected to be approximately normal.",Free Response,,,,
-OS,6.1,8c,"Greece has faced a severe economic crisis since the end of 2009. A Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that 25% of them said they would rate their lives poorly enough to be considered “suffering”. Construct a 95% confidence interval for the proportion of Greeks who are “suffering”.","A 95% confidence interval can be calculated as follows: 0.25 ± 1.96 sqrt( 0.25 x 0.75 / 1000) = (0.22312, 0.2769) We are 95% confident that approximately 22% to 28% of Greeks would rate their lives poorly enough to be considered “suffering"".",Free Response,,,,
-OS,6.1,8d,"Greece has faced a severe economic crisis since the end of 2009. A Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that 25% of them said they would rate their lives poorly enough to be considered “suffering”. Without doing any calculations, describe what would happen to the confidence interval if we decided to use a higher confidence level.",Increasing the confidence level would increase the margin of error hence widen the interval.,Free Response,,,,
-OS,6.1,8e,"Greece has faced a severe economic crisis since the end of 2009. A Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that 25% of them said they would rate their lives poorly enough to be considered “suffering”. Without doing any calculations, describe what would happen to the confidence interval if we used a larger sample.",Increasing the sample size would decrease the margin of error hence make the interval narrower.,Free Response,,,,
-OS,6.1,9a,"A survey on 1,509 high school seniors who took the SAT and who completed an optional web survey shows that 55% of high school seniors are fairly certain that they will participate in a study abroad program in college. Is this sample a representative sample from the population of all high school seniors in the US? Explain your reasoning.","No. The sample only represents students who took the SAT, and this was also an online survey.",Free Response,,,,
-OS,6.1,9b,"A survey on 1,509 high school seniors who took the SAT and who completed an optional web survey shows that 55% of high school seniors are fairly certain that they will participate in a study abroad program in college. Let’s suppose the conditions for inference are met. Even if your answer to part (a) indicated that this approach would not be reliable, this analysis may still be interesting to carry out (though not report). Construct a 90% confidence interval for the proportion of high school seniors (of those who took the SAT) who are fairly certain they will participate in a study abroad program in college, and interpret this interval in context.","A 90% confidence interval can be calculated as follows: 0.55 ± 1.65 x sqrt(0.55 x 0.45 / 1509) = (0.5289, 0.5711) We are 95% confident that 53% to 57% of high school seniors are fairly certain that they will participate in a study abroad program in college.",Free Response,,,,
-OS,6.1,9c,"A survey on 1,509 high school seniors who took the SAT and who completed an optional web survey shows that 55% of high school seniors are fairly certain that they will participate in a study abroad program in college. What does “90% confidence” mean?","90% of random samples of 1,509 high school seniors would produce a 90% confidence interval that includes the true proportion of high school seniors who took the SAT are fairly and who certain that they will participate in a study abroad program in college.",Free Response,,,,
-OS,6.1,9d,"A survey on 1,509 high school seniors who took the SAT and who completed an optional web survey shows that 55% of high school seniors are fairly certain that they will participate in a study abroad program in college. Based on this interval, would it be appropriate to claim that the majority of high school seniors are fairly certain that they will participate in a study abroad program in college?","Yes. The interval lies entirely above 50%. Therefore, at 90% confidence level, it would be appropriate to claim the majority of high school seniors who took the SAT who are fairly certain they will participate in a study abroad program in college.",Free Response,,,,
-OS,6.1,10a,"The General Social Survey asked 1,578 US residents: “Do you think the use of marijuana should be made legal, or not?” 61% of the respondents said it should be made legal. Is 61% a sample statistic or a population parameter? Explain.","61% is a sample statistic, it's the observed sample proportion.",Free Response,,,,
-OS,6.1,10b,"The General Social Survey asked 1,578 US residents: “Do you think the use of marijuana should be made legal, or not?” 61% of the respondents said it should be made legal. Construct a 95% confidence interval for the proportion of US residents who think marijuana should be made legal, and interpret it in the context of the data.","A 95% confidence interval can be calculated as follows: 0.61 ± 1.96 x sqrt(0.61(1 – 0.61) / 1578) = (0.586, 0.634) We are 95% confident that approximately 58.6% to 63.4% of Americans think marijuana should be legalized.",Free Response,,,,
-OS,6.1,10c,"The General Social Survey asked 1,578 US residents: “Do you think the use of marijuana should be made legal, or not?” 61% of the respondents said it should be made legal. A critic points out that this 95% confidence interval is only accurate if the statistic follows a normal distribution, or if the normal model is a good approximation. Is this true for these data? Explain.","1. Independence: The sample is random, and comprises less than 10% of the American population, therefore we can assume that the individuals in this sample are independent of each other 2. Success-failure: The number of successes (people who said marijuana should be legalized: 1578 x 0.61 = 962.58) and failures (people who said it shouldn't be: 1578 x 0.39 = 615.42) are both greater than 10, therefore the success-failure condition is met as well. Therefore the distribution of the sample proportion is expected to be approximately normal.",Free Response,,,,
-OS,6.1,10d,"The General Social Survey asked 1,578 US residents: “Do you think the use of marijuana should be made legal, or not?” 61% of the respondents said it should be made legal. A news piece on this survey’s findings states, “Majority of Americans think marijuana should be legalized.” Based on your confidence interval, is this news piece’s statement justified?","Yes, the interval is above 50%, suggesting, with 95% confidence, that the true population proportion of Americans who think marijuana should be legalized is greater than 50%.",Free Response,,,,
-OS,6.1,11a,"A Kaiser Family Foundation poll for US adults in 2019 found that 79% of Democrats, 55% of Independents, and 24% of Republicans supported a generic “National Health Plan”. There were 347 Democrats, 298 Republicans, and 617 Independents surveyed. A political pundit on TV claims that a majority of Independents support a National Health Plan. Do these data provide strong evidence to support this type of statement?","(i) Let's prepare for running a hypothesis test. We want to check for a majority (or minority), so we use the following hypotheses: H0: p = 0.5 HA: p ≠ 0.5 We have a sample proportion of p^ = 0.55 and a sample size of n = 617 independents. (ii) Next, we check whether the conditions are met to proceed. Since this is a random sample, independence is satisfied. The success-failure condition is also satisfied: 617 x 0.5 and 617 x (1 – 0.5) are both at least 10 (we use the null proportion p0 = 0.5 for this check in a one- proportion hypothesis test). (iii) Next, we can start performing calculations. We can model ^p using a normal distribution with a standard error of SE = 0.02 (We use the null proportion, p0 = 0.5, to compute the standard error for a one-proportion hypothesis test.) Next, we compute the test statistic: Z = (0.55 - 0:5) / 0.02 = 2.5 This yields a one-tail area of 0.0062, and a p-value of 2 x 0.0062 = 0.0124. (iv) Lastly, we make a conclusion. Because the p-value is smaller than 0.05, we reject the null hypothesis. We have strong evidence that the support is different from 0.5, and since the data provide a point estimate above 0.5, we have strong evidence to support this claim by the TV pundit.",Free Response,,,,
-OS,6.1,11b,"A Kaiser Family Foundation poll for US adults in 2019 found that 79% of Democrats, 55% of Independents, and 24% of Republicans supported a generic “National Health Plan”. There were 347 Democrats, 298 Republicans, and 617 Independents surveyed. Would you expect a confidence interval for the proportion of Independents who oppose the public option plan to include 0.5? Explain.","No. Generally we expect a hypothesis test and a confidence interval to align, so we would expect the confidence interval to show a range of plausible values entirely above 0.5. However, if the confidence level is misaligned (e.g. a 99% confidence level and α = 0.05 significance level), then this is no longer generally true.",Free Response,,,,
-OS,6.1,12a,"Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school. A newspaper article states that only a minority of the Americans who decide not to go to college do so because they cannot afford it and uses the point estimate from this survey as evidence. Conduct a hypothesis test to determine if these data provide strong evidence supporting this statement.","The hypotheses are as follows: H0: p = 0:5 (50% of Americas who decide not to go to college because they cannot afford it do so because they cannot afford it) HA: p < 0:5 (Less than 50% of Americas who decide not to go to college because they cannot afford it do so because they cannot afford it) Before calculating the test statistic we should check that the conditions are satisfied. 1. Independence: The sample is representative and we can safely assume that 331 < 10% of all American adults who decide not to go to college, therefore whether or not one person in the sample decided not to go to college because they can't afford it is independent of another. 2. Success-failure: 331 x 0.5 = 165.5 > 10 and 331 x 0.5 = 165.5 > 10. Since the observations are independent and the success-failure condition is met, ^p is expected to be approximately normal. The test statistic can be calculated as follows: Z = (0.48 – 0.5) / (sqrt(0.5 x 0.5 / 331)) = -0.73 p-value = 0.2327 Since the p-value is large, we fail to reject H0. The data do not provide strong evidence that less than half of American adults who decide not to go to college make this decision because they cannot afford college.",Free Response,,,,
-OS,6.1,13a,"Some people claim that they can tell the difference between a diet soda and a regular soda in the first sip. A researcher wanting to test this claim randomly sampled 80 such people. He then filled 80 plain white cups with soda, half diet and half regular through random assignment, and asked each person to take one sip from their cup and identify the soda as diet or regular. 53 participants correctly identified the soda. Do these data provide strong evidence that these people are any better or worse than random guessing at telling the difference between diet and regular soda?","The hypotheses are as follows: H0: p = 0:5 (Results are equivalent to randomly guessing) HA: p ≠ 0:5 (Results are different than just randomly guessing) Before conducting the hypothesis test, we must first check that the conditions for inference are satisfied. 1. Independence: The sample is random, therefore whether or not one person in the sample can identify a soda correctly in independent of another. 2. Success-failure: 80 x 0.5 = 40 > 10 and 80 x 0.5 = 40 > 10. Since the observations are independent and the success-failure condition is met, p^ is expected to be approximately normal. The test statistic and the p-value can be calculated as follows: p^ = 53 / 80 = 0.6625 Z = (0.6625 – 0.5) / sqrt(0.5 x 0.5 / 80)) = 2.91 p-value = 0.0036 Since the p-value < α (use α = 0.05 since not given), we reject the null hypothesis. Since we rejected H0 and the point estimate suggests people are better than random guessing, we can conclude the rate of correctly identifying a soda for these people is significantly better than just by random guessing.",Free Response,,,,
-OS,6.1,14a,"Exercise 6.12 presents the results of a poll where 48% of 331 Americans who decide to not go to college do so because they cannot afford it. Calculate a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it, and interpret the interval in context.","We have previously confirmed that the independence condition is satisfied. We need to recheck the success-failure condition using the sample proportion: 331 x 0.48 = 158.88 > 10 and 331 x 0.52 = 172.12 > 10. An 80% confidence interval can be calculated as follows: 0.48 ± 1.65 x sqrt(0.48 x 0.52 / 331) = (0.435, 0.525) We are 90% confident that the 43.5% to 52.5% of all Americans who decide not to go to college do so because they cannot afford it. This agrees with the conclusion of the earlier hypothesis test since the interval includes 50%.",Free Response,,,,
-OS,6.1,14b,Exercise 6.12 presents the results of a poll where 48% of 331 Americans who decide to not go to college do so because they cannot afford it. Suppose we wanted the margin of error for the 90% confidence level to be about 1.5%. How large of a survey would you recommend?,"We are asked to solve for the sample size required to achieve a 1.5% margin of error for a 90% confidence interval and the point estimate is p^ = 0:48. 0.01 ≥ 1.65 x sqrt(0.48 x 0.52 / n) n ≥ 3020.16 ≈  3121 The sample size n should be at least 3,121.",Free Response,,,,
-OS,6.1,15,"Exercise 6.11 presents the results of a poll evaluating support for a generic “National Health Plan” in the US in 2019, reporting that 55% of Independents are supportive. If we wanted to estimate this number to within 1% with 90% confidence, what would be an appropriate sample size?","Since a sample proportion (p^ = 0.55) is available, we use this for the sample size calculations. The margin of error for a 90% confidence interval is 1.65 x SE  We want this to be less than 0.01, where we use p^ in place of p: 1.65 x sqrt( 0.55(1 – 0.55) / n) ≤ 0.01 From this, we get that n must be at least 6739.",Free Response,,,,
-OS,6.1,16,"As discussed in Exercise 6.10, the General Social Survey reported a sample where about 61% of US residents thought marijuana should be made legal. If we wanted to limit the margin of error of a 95% confidence interval to 2%, about how many Americans would we need to survey?","We are asked to solve for the sample size required to achieve a 2% margin of error for a 95% confidence interval and the point estimate is p^ = 0.61. 1.96 x sqrt( 0.61(1 – 0.61) / n) ≤ 0.02 n ≥ 2284.792. The sample size n should be at least 2,285.",Free Response,,,,
-OS,6.2,19,,,Free Response,,,,
-OS,6.2,19a,"A study asked 1,924 male and 3,666 female undergraduate college students their favorite color. A 95% confidence interval for the difference between the proportions of males and females whose favorite color is black (pmale − pfemale) was calculated to be (0.02, 0.06). Based on this information, determine if the following statement about undergraduate college students is true or false, and explain your reasoning for each statement you identify as false: We are 95% confident that the true proportion of males whose favorite color is black is 2% lower to 6% higher than the true proportion of females whose favorite color is black.","False. Since (pmale - pfemale) is positive, the proportion of males whose favorite color is black is higher than the proportion of females.",Free Response,,,,
-OS,6.2,19b,"A study asked 1,924 male and 3,666 female undergraduate college students their favorite color. A 95% confidence interval for the difference between the proportions of males and females whose favorite color is black (pmale − pfemale) was calculated to be (0.02, 0.06). Based on this information, determine if the following statement about undergraduate college students is true or false, and explain your reasoning for each statement you identify as false: We are 95% confident that the true proportion of males whose favorite color is black is 2% to 6% higher than the true proportion of females whose favorite color is black.",True.,Free Response,,,,
-OS,6.2,19c,"A study asked 1,924 male and 3,666 female undergraduate college students their favorite color. A 95% confidence interval for the difference between the proportions of males and females whose favorite color is black (pmale − pfemale) was calculated to be (0.02, 0.06). Based on this information, determine if the following statement about undergraduate college students is true or false, and explain your reasoning for each statement you identify as false: 95% of random samples will produce 95% confidence intervals that include the true difference between the population proportions of males and females whose favorite color is black.",True.,Free Response,,,,
-OS,6.2,19d,"A study asked 1,924 male and 3,666 female undergraduate college students their favorite color. A 95% confidence interval for the difference between the proportions of males and females whose favorite color is black (pmale − pfemale) was calculated to be (0.02, 0.06). Based on this information, determine if the following statement about undergraduate college students is true or false, and explain your reasoning for each statement you identify as false: We can conclude that there is a significant difference between the proportions of males and females whose favorite color is black and that the difference between the two sample proportions is too large to plausibly be due to chance.",True.,Free Response,,,,
-OS,6.2,19e,"A study asked 1,924 male and 3,666 female undergraduate college students their favorite color. A 95% confidence interval for the difference between the proportions of males and females whose favorite color is black (pmale − pfemale) was calculated to be (0.02, 0.06). Based on this information, determine if the following statement about undergraduate college students is true or false, and explain your reasoning for each statement you identify as false: The 95% confidence interval for (pfemale −pmale) cannot be calculated with only the information given in this exercise.","False. To get the 95% confidence interval for (pfemale - pmale), all we have to do is to swap the bounds of the confidence interval for (pmale - pfemale) and take their negatives: (-0.06,-0.02).",Free Response,,,,
-OS,6.2,20a,"The United States federal government shutdown of 2018–2019 occurred from December 22, 2018 until January 25, 2019, a span of 35 days. A Survey USA poll of 614 randomly sampled Americans during this time period reported that 48% of those who make less than $40,000 per year and 55% of those who make $40,000 or more per year said the government shutdown has not at all affected them personally. A 95% confidence interval for (p<40K −p≥40K), where p is the proportion of those who said the government shutdown has not at all affected them personally, is (-0.16, 0.02). Based on this information, determine if the following statements are true or false, and explain your reasoning if you identify the statement as false. At the 5% significance level, the data provide convincing evidence of a real difference in the proportion who are not affected personally between Americans who make less than $40,000 annually and Americans who make $40,000 annually.",False. The confidence interval includes 0.,Free Response,,,,
-OS,6.2,20b,"The United States federal government shutdown of 2018–2019 occurred from December 22, 2018 until January 25, 2019, a span of 35 days. A Survey USA poll of 614 randomly sampled Americans during this time period reported that 48% of those who make less than $40,000 per year and 55% of those who make $40,000 or more per year said the government shutdown has not at all affected them personally. A 95% confidence interval for (p<40K −p≥40K), where p is the proportion of those who said the government shutdown has not at all affected them personally, is (-0.16, 0.02). Based on this information, determine if the following statements are true or false, and explain your reasoning if you identify the statement as false. We are 95% confident that 16% more to 2% fewer Americans who make less than $40,000 per year are not at all personally affected by the government shutdown compared to those who make $40,000 or more per year.","False. We are 95% confident that 16% fewer to 2% Americans who make less than $40,000 per year are not at all personally affected by the government shutdown compared to those who make $40,000 or more per year.",Free Response,,,,
-OS,6.2,20c,"The United States federal government shutdown of 2018–2019 occurred from December 22, 2018 until January 25, 2019, a span of 35 days. A Survey USA poll of 614 randomly sampled Americans during this time period reported that 48% of those who make less than $40,000 per year and 55% of those who make $40,000 or more per year said the government shutdown has not at all affected them personally. A 95% confidence interval for (p<40K −p≥40K), where p is the proportion of those who said the government shutdown has not at all affected them personally, is (-0.16, 0.02). Based on this information, determine if the following statements are true or false, and explain your reasoning if you identify the statement as false. A 90% confidence interval for (p<40K − p≥40K) would be wider than the (−0.16, 0.02) interval.",False. As the confidence level decreases the width of the confidence level decreases as well.,Free Response,,,,
-OS,6.2,20d,"The United States federal government shutdown of 2018–2019 occurred from December 22, 2018 until January 25, 2019, a span of 35 days. A Survey USA poll of 614 randomly sampled Americans during this time period reported that 48% of those who make less than $40,000 per year and 55% of those who make $40,000 or more per year said the government shutdown has not at all affected them personally. A 95% confidence interval for (p<40K −p≥40K), where p is the proportion of those who said the government shutdown has not at all affected them personally, is (-0.16, 0.02). Based on this information, determine if the following statements are true or false, and explain your reasoning if you identify the statement as false. A 95% confidence interval for (p≥40K − p<40K) is (-0.02, 0.16).",True.,Free Response,,,,
-OS,6.2,21a,"Exercise 6.11 presents the results of a poll evaluating support for a generically branded “National Health Plan” in the United States. 79% of 347 Democrats and 55% of 617 Independents support a National Health Plan. Calculate a 95% confidence interval for the difference between the proportion of Democrats and Independents who support a National Health Plan (pD – pI), and interpret it in this context. We have already checked conditions for you.","Standard error: SE = sqrt((0.79(1-0.79)/347) + (0.55(1-0.55)/617)) = 0.03 So 0.79 – 0.55 ± 1.96 x 0.03 = (0.181, 0.299) We are 95% confident that the proportion of Democrats who support the plan is 18.1% to 29.9% higher than the proportion of Independents who support the plan.",Free Response,,,,
-OS,6.2,21b,"Exercise 6.11 presents the results of a poll evaluating support for a generically branded “National Health Plan” in the United States. 79% of 347 Democrats and 55% of 617 Independents support a National Health Plan. True or false: If we had picked a random Democrat and a random Independent at the time of this poll, it is more likely that the Democrat would support the National Health Plan than the Independent.",True.,Free Response,,,,
-OS,6.2,22,"According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Calculate a 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data.","Before calculating the confidence interval we should check that the conditions are satisfied. 1. Independence: Both samples are random, and 11,545 < 10% of all Californians and 4,691 < 10% of all Oregonians, therefore how much one Californian sleeps is independent of how much another Californian sleeps and how much one Oregonian sleeps is independent of how much another Oregonian sleeps. In addition, the two samples are independent of each other. 2. Success-failure: 11,545 x 0.08 = 923.6 > 10; 11,545 x 0.92 = 10621.4 > 10; 4,691 x 0.088 = 412.8 > 10; 4,691 x 0.912 = 4278.2 > 10. Since the observations are independent and the success-failure condition is met, p^CA - p^OR is expected to be approximately normal. A 95% confidence interval for the difference between the population proportions can be calculated as follows: (0.08 – 0.088) x 1.96 x sqrt((0.08 x 0.92 / 11,545) + (0.088 x 0.912 / 4,691)) = (-0.017, 0.001) We are 95% confident that the difference between the proportions of Californians and Oregonians who are sleep deprived is between -1.7% and 0.1%. In other words, we are 95% confident that 1.7% less to 0.1% more Californians than Oregonians are sleep deprived.",Free Response,,,,
-OS,6.2,24,"Exercise 6.22 provides data on sleep deprivation rates of Californians and Oregonians. The proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Conduct a hypothesis test to determine if these data provide strong evidence the rate of sleep deprivation is different for the two states. (Reminder: Check conditions)","The hypotheses are: H0: pCA = pOR, HA: pCA ≠ pOR We have confirmed in Exercise ?? that the independence condition is satisfied but we need to recheck the success-failure condition using p^pool and expected counts. successCA = nCA x pCA = 11,545 x 0.08 = 923.6 ≈ 924; successOR = nOR x pOR = 4,691 x 0.088 = 412.8 ≈ 413; p^pool = (924 + 413) / (11,545 + 4,691) = 0.0821; 11,545 x 0.082 = 946.69 > 10; 11,545 x 0.918 = 10598.31 > 10; 4,691 x 0.082 = 384.662 > 10; 4,691 x 0.918 = 4306.338 > 10. Since the observations are independent and the success-failure condition is met, p^CA - p^OR is expected to be approximately normal. Next we calculate the test statistic and the p-value: Z = ((0.08 – 0.088) – 0) / sqrt((0.082 x 0.918 / 11,545) + (0.082 x 0.918 / 4,691)) = -1.68; p-value = 0.093. Since the p-value > α (use α = 0:05 since not given), we fail to reject H0 and conclude that the data do not provide strong evidence that the rate of sleep deprivation is different for the two states.",Free Response,,,,
-OS,6.2,29,"In July 2008 the US National Institutes of Health announced that it was stopping a clinical study early because of unexpected results. The study population consisted of HIV-infected women in sub-Saharan Africa who had been given single dose Nevaripine (a treatment for HIV) while giving birth, to prevent transmission of HIV to the infant. The study was a randomized comparison of continued treatment of a woman (after successful childbirth) with Nevaripine vs Lopinavir, a second drug used to treat HIV. 240 women participated in the study; 120 were randomized to each of the two treatments. Twentyfour weeks after starting the study treatment, each woman was tested to determine if the HIV infection was becoming worse (an outcome called virologic failure). Twenty-six of the 120 women treated with Nevaripine experienced virologic failure, while 10 of the 120 women treated with the other drug experienced virologic failure. State appropriate hypotheses to test for difference in virologic failure rates between treatment groups, complete the hypothesis test and state an appropriate conclusion. ","The hypotheses are as follows: H0: pN = pL: There is no difference in virologic failure rates between the Nevaripine and Lopinavir groups. HA: pN ≠ pL: There is some difference in virologic failure rates between the Nevaripine and Lopinavir groups. First, check conditions. 1. Independence: Random assignment was used, so the observations in each group are independent. If the patients in the study are representative of those in the general population (something impossible to check with the given information), then we can also confidently generalize the findings to the population. 2. Success-failure: First we need to find p^pool and then use that to calculate the numbers of expected successes and failures in each group. p^pool = (26 + 10) / (120 + 120) = 0.15; 120 x 0.15 = 18 > 10; 120 x 0.85 = 102 > 10. Since the observations are independent and the success-failure condition is met, (p^C – p^T ) is approximately normal. We should keep in mind that it is unclear if the findings generalize to the population. Next we calculate the test statistic and the p-value: p^N = 26 / 120 = 0.2167; p^L = 10 / 120 = 0.0833 Z = (0.2167 – 0.0833) / sqrt((0.15 x 0.85 / 120) + (0.15 x 0.85 / 120)) = 2.89 p-value = 0.0039. Since the p-value is low, we reject H0. There is strong evidence of a difference in virologic failure rates between the Nevaripine and Lopinavir groups. Treatment and virologic failure do not appear to be independent.",Free Response,,,,
-OS,6.2,30,"An apple a day keeps the doctor away. A physical education teacher at a high school wanting to increase awareness on issues of nutrition and health asked her students at the beginning of the semester whether they believed the expression “an apple a day keeps the doctor away”, and 40% of the students responded yes. Throughout the semester she started each class with a brief discussion of a study highlighting positive effects of eating more fruits and vegetables. She conducted the same apple-a-day survey at the end of the semester, and this time 60% of the students responded yes. Can she used a two proportion method from this section for this analysis? Explain your reasoning.",No. The samples at the beginning and at the end of the semester are not independent since the survey is conducted on the same students.,Free Response,,,,
-OS,7.1,1a,"An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical t-value (t*) for the given sample size and confidence level: n = 6, CL = 90%","n = 6, CL = 90%, df = 6 - 1 = 5, t* = 2.02",Free Response,,,,
-OS,7.1,1b,"An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical t-value (t*) for the given sample size and confidence level: n = 21, CL = 98%","n = 21, CL = 98%, df = 21 - 1 = 20, t* = 2.53",Free Response,,,,
-OS,7.1,1c,"An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical t-value (t*) for the given sample size and confidence level: n = 29, CL = 95%","n = 29, CL = 95%, df = 29 - 1 = 28, t* = 2.05",Free Response,,,,
-OS,7.1,1d,"An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical t-value (t*) for the given sample size and confidence level: n = 12, CL = 99%","n = 12, CL = 99%, df = 12 - 1 = 11, t* = 3.11",Free Response,,,,
-OS,7.1,3a,"An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at α = 0.05: n = 11, T = 1.91",n = 11 T = 1:91 df = 11 - 1 = 10 p-value = 0.085 Do not reject H0,Free Response,,,,
-OS,7.1,3b,"An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at α = 0.05: n = 17, T = −3.45",n = 17 T = -3:45 df = 17 - 1 = 16 p-value = 0.003 Reject H0,Free Response,,,,
-OS,7.1,3c,"An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at α = 0.05: n = 7, T = 0.83",n = 7 T = 0:83 df = 7 - 1 = 6 p-value = 0.438 Do not reject H0,Free Response,,,,
-OS,7.1,3d,"An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at α = 0.05: n = 28, T = 2.13",n = 28 T = 2:13 df = 28 - 1 = 27 p-value = 0.042 Reject H0,Free Response,,,,
-OS,7.1,4a,"An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at α = 0.01: n = 26, T = 2.485",n = 26 T = 2.485 df = 26 - 1 = 25 p-value = 0.020 Do not reject H0,Free Response,,,,
-OS,7.1,4b,"An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at α = 0.01: n = 18, T = 0.5",n = 18 T = 0.5 df = 18 - 1 = 17 p-value = 0.623 Do not reject H0,Free Response,,,,
-OS,7.1,5,"A 95% confidence interval for a population mean, µ, is given as (18.985, 21.015). This confidence interval is based on a simple random sample of 36 observations. Calculate the sample mean and standard deviation. Assume that all conditions necessary for inference are satisfied. Use the t-distribution in any calculations.","The sample mean is the mid-point of the confidence interval, i.e. the average of the upper and lower bounds: x-bar = (18.985 + 21.015) / 2 = 20. The margin of error is 21.015 - 20 = 1.015. Since n = 36, df = 35, and the critical t-score is T* = 2.03. Then, 1.015 = 2.03 x s / sqrt(36); so s = 3.",Free Response,,,,
-OS,7.1,6,"A 90% confidence interval for a population mean is (65, 77). The population distribution is approximately normal and the population standard deviation is unknown. This confidence interval is based on a simple random sample of 25 observations. Calculate the sample mean, the margin of error, and the sample standard deviation.","The sample mean is the mid-point of the confidence interval, i.e. the average of the upper and lower bounds: x-bar = (65 + 77) / 2 = 71. The margin of error is 77-71 = 6. Since n = 25, df = 24, and the critical t-score is T* = 1.71. Then, 6 = 1.71 x s / sqrt(25); so s = 17.54.",Free Response,,,,
-OS,7.1,9,You are given the following hypotheses: H0: µ = 60; HA: µ ≠ 60; We know that the sample standard deviation is 8 and the sample size is 20. For what sample mean would the p-value be equal to 0.05? Assume that all conditions necessary for inference are satisfied.,"For the single tails to each be 0.025 at n - 1 = 20 - 1 = 19 degrees of freedom, T score must equal to be either -2.09 or +2.09. Then, either: -2:09 = (x-bar – 60) / (8 / sqrt(20)); x = 56.26 or 2:09 = (x-bar – 60) / (8 / sqrt(20)); x = 63.74",Free Response,,,,
-OS,7.1,10,"For a given confidence level, t* is larger than z*. Explain how t* being slightly larger than z* affects the width of the confidence interval.","With a larger critical value, the confidence interval ends up being wider. This makes intuitive sense as when we have a small sample size and the population standard deviation is unknown, we should have a wider interval than if we knew the population standard deviation, or if we had a large enough sample size.",Free Response,,,,
-OS,7.1,11a,"Georgianna claims that in a small city renowned for its music school, the average child takes less than 5 years of piano lessons. We have a random sample of 20 children from the city, with a mean of 4.6 years of piano lessons and a standard deviation of 2.2 years. Evaluate Georgianna’s claim (or that the opposite might be true) using a hypothesis test.","We will conduct a 1-sample t-test. H0: µ = 5. HA: µ ≠ 5. We'll use α = 0:05. This is a random sample, so the observations are independent. To proceed, we assume the distribution of years of piano lessons is approximately normal. SE = 2.2= / sqrt(20) = 0.4919. The test statistic is T = (4.6 - 5)=SE = -0:81. df = 20 - 1 = 19. The one-tail area is about 0.21, so the p-value is about 0.42, which is bigger than α = 0:05 and we do not reject H0. That is, we do not have sufficiently strong evidence to reject the notion that the average is 5 years.",Free Response,,,,
-OS,7.1,11b,"Georgianna claims that in a small city renowned for its music school, the average child takes less than 5 years of piano lessons. We have a random sample of 20 children from the city, with a mean of 4.6 years of piano lessons and a standard deviation of 2.2 years. Construct a 95% confidence interval for the number of years students in this city take piano lessons, and interpret it in context of the data.","Using SE = 0.4919 and t* =19 = 2.093, the confidence interval is (3.57, 5.63). We are 95% confident that the average number of years a child takes piano lessons in this city is 3.57 to 5.63 years.",Free Response,,,,
-OS,7.1,11c,"Georgianna claims that in a small city renowned for its music school, the average child takes less than 5 years of piano lessons. We have a random sample of 20 children from the city, with a mean of 4.6 years of piano lessons and a standard deviation of 2.2 years. Do your results from the hypothesis test and the confidence interval agree? Explain your reasoning.","They agree, since we did not reject the null hypothesis and the null value of 5 was in then t-interval.",Free Response,,,,
-OS,7.1,12a,"Researchers interested in lead exposure due to car exhaust sampled the blood of 52 police officers subjected to constant inhalation of automobile exhaust fumes while working traffic enforcement in a primarily urban environment. The blood samples of these officers had an average lead concentration of 124.32 µg/l and a SD of 37.74 µg/l; a previous study of individuals from a nearby suburb, with no history of exposure, found an average blood level concentration of 35 µg/l. Write down the hypotheses that would be appropriate for testing if the police officers appear to have been exposed to a different concentration of lead.","H0: µ = 35, HA: µ ≠ 35.",Free Response,,,,
-OS,7.1,12b,"Researchers interested in lead exposure due to car exhaust sampled the blood of 52 police officers subjected to constant inhalation of automobile exhaust fumes while working traffic enforcement in a primarily urban environment. The blood samples of these officers had an average lead concentration of 124.32 µg/l and a SD of 37.74 µg/l; a previous study of individuals from a nearby suburb, with no history of exposure, found an average blood level concentration of 35 µg/l. Explicitly state and check all conditions necessary for inference on these data.","1. Independence: if we can assume that these 52 officers represent a random sample (big assumption), then independence would be satisfied, but we cannot check this. 2. Normality: We don't have a plot of the distribution that we can use to check this condition. We at least have more than 30 observations, so the distribution would have to be extremely skewed to be an issue. We again cannot check this, but this seems like a less concerning issue than the independence consideration.",Free Response,,,,
-OS,7.1,12c,"Researchers interested in lead exposure due to car exhaust sampled the blood of 52 police officers subjected to constant inhalation of automobile exhaust fumes while working traffic enforcement in a primarily urban environment. The blood samples of these officers had an average lead concentration of 124.32 µg/l and a SD of 37.74 µg/l; a previous study of individuals from a nearby suburb, with no history of exposure, found an average blood level concentration of 35 µg/l. Test the hypothesis that the downtown police officers have a higher lead exposure than the group in the previous study. Interpret your results in context.","The test statistic and the p-value can be calculated as follows: T = (124.32 – 35) / (7.74 / sqrt(52)  = 17.07; df = 52 - 1 = 51 p-value = 2 x P(T > 17.07) < 0.001. The hypothesis test yields a very small p-value, so we reject H0. Given the direction of the data, there is very convincing evidence that the police officers have been exposed to a higher concentration of lead than individuals living in a suburban area.",Free Response,,,,
-OS,7.1,13,A market researcher wants to evaluate car insurance savings at a competing company. Based on past studies he is assuming that the standard deviation of savings is $100. He wants to collect data such that he can get a margin of error of no more than $10 at a 95% confidence level. How large of a sample should he collect?,10 ≥ 1:96 x 100 / sqrt(n); n = 384.16 He should survey at least 385 customers. Note that we need to round up the calculated sample size.,Free Response,,,,
-OS,7.1,14a,"The standard deviation of SAT scores for students at a particular Ivy League college is 250 points. Two statistics students, Raina and Luke, want to estimate the average SAT score of students at this college as part of a class project. They want their margin of error to be no more than 25 points. Raina wants to use a 90% confidence interval. How large a sample should she collect?",25 ≥ 1.65 x 250 / sqrt(n); n = 272.25. Raina should collect a sample of at least 273 students.,Free Response,,,,
-OS,7.1,14b,"The standard deviation of SAT scores for students at a particular Ivy League college is 250 points. Two statistics students, Raina and Luke, want to estimate the average SAT score of students at this college as part of a class project. They want their margin of error to be no more than 25 points. Luke wants to use a 99% confidence interval. Without calculating the actual sample size, determine whether his sample should be larger or smaller than Raina’s, and explain your reasoning.","If Luke had the same sample size as Raina but used a higher confidence level, he would end up with wider interval. To keep the width of his confidence interval the same as Raina's Luke will need a higher sample size.",Free Response,,,,
-OS,7.1,14c,"The standard deviation of SAT scores for students at a particular Ivy League college is 250 points. Two statistics students, Raina and Luke, want to estimate the average SAT score of students at this college as part of a class project. They want their margin of error to be no more than 25 points. Calculate the minimum required sample size for Luke.",25 ≥ 2.58 x 250 / sqrt(n); n = 665.64. Luke should collect a sample of at least 666 students.,Free Response,,,,
-OS,7.2,15,"Air quality measurements were collected in a random sample of 25 country capitals in 2013, and then again in the same cities in 2014. We would like to use these data to compare average air quality between the two years. Should we use a paired or non-paired test? Explain your reasoning.","Paired, data are recorded in the same cities at two different time points. The temperature in a city at one point is not independent of the temperature in the same city at another time point.",Free Response,,,,
-OS,7.2,16a,"Determine if the following statement is true or false. If false, explain: In a paired analysis we first take the difference of each pair of observations, and then we do inference on these differences.",True.,Free Response,,,,
-OS,7.2,16b,"Determine if the following statement is true or false. If false, explain: Two data sets of different sizes cannot be analyzed as paired data.",True.,Free Response,,,,
-OS,7.2,16c,"Determine if the following statement is true or false. If false, explain: Consider two sets of data that are paired with each other. Each observation in one data set has a natural correspondence with exactly one observation from the other data set.",True.,Free Response,,,,
-OS,7.2,16d,"Determine if the following statement is true or false. If false, explain: Consider two sets of data that are paired with each other. Each observation in one data set is subtracted from the average of the other data set’s observations.","False. We find the deference of each pair of observations, and then we do inference on these divergences.",Free Response,,,,
-OS,7.2,17a,"In the following scenario, determine if the data are paired: Compare pre- (beginning of semester) and post-test (end of semester) scores of students.","Since it's the same students at the beginning and the end of the semester, there is a pairing between the datasets, for a given student their beginning and end of semester grades are dependent.",Free Response,,,,
-OS,7.2,17b,"In the following scenario, determine if the data are paired: Assess gender-related salary gap by comparing salaries of randomly sampled men and women.","Since the subjects were sampled randomly, each observation in the men's group does not have a special correspondence with exactly one observation in the other (women's) group.",Free Response,,,,
-OS,7.2,17c,"In the following scenario, determine if the data are paired: Compare artery thicknesses at the beginning of a study and after 2 years of taking Vitamin E for the same group of patients. ","Since it's the same subjects at the beginning and the end of the study, there is a pairing between the datasets, for a subject student their beginning and end of semester artery thickness are dependent.",Free Response,,,,
-OS,7.2,17d,"In the following scenario, determine if the data are paired: Assess effectiveness of a diet regimen by comparing the before and after weights of subjects.","Since it's the same subjects at the beginning and the end of the study, there is a pairing between the datasets, for a subject student their beginning and end of semester weights are dependent.",Free Response,,,,
-OS,7.2,18a,"In the following scenario, determine if the data are paired: We would like to know if Intel’s stock and Southwest Airlines’ stock have similar rates of return. To find out, we take a random sample of 50 days, and record Intel’s and Southwest’s stock on those same days.","Paired, on the same day the stock prices may be dependent on external factors that affect the price of both stocks.",Free Response,,,,
-OS,7.2,18b,"In the following scenario, determine if the data are paired: We randomly sample 50 items from Target stores and note the price for each. Then we visit Walmart and collect the price for each of those same 50 items.","Paired, the prices are for the same items.",Free Response,,,,
-OS,7.2,18c,"In the following scenario, determine if the data are paired: A school board would like to determine whether there is a difference in average SAT scores for students at one high school versus another high school in the district. To check, they take a simple random sample of 100 students from each high school.","Not paired, these are two independent random samples, individual students are not matched.",Free Response,,,,
-OS,7.2,21a,We considered the change in the number of days exceeding 90°F from 1948 and 2018 at 197 randomly sampled locations from the NOAA database in Exercise 7.19. The mean and standard deviation of the reported differences are 2.9 days and 17.2 days. Calculate a 90% confidence interval for the average difference between number of days exceeding 90°F between 1948 and 2018. We’ve already checked the conditions for you.,"We checked conditions in an earlier exercise, so we'll jump right to the calculations. First we compute the standard error and find z*: SE = 17.2 / sqrt(197) = 1.23; Z* = 1.65. Next, we can compute the confidence interval: 2.9 ± 1.65 x 1:23 = (0.87; 4.93)",Free Response,,,,
-OS,7.2,21c,We considered the change in the number of days exceeding 90°F from 1948 and 2018 at 197 randomly sampled locations from the NOAA database in Exercise 7.19. The mean and standard deviation of the reported differences are 2.9 days and 17.2 days. Interpret the 90% interval in context.,We are 90% confident that there was an increase of 0.87 to 4.93 in the average number of days that hit 90°F in 2018 relative to 1948 for NOAA stations.,Free Response,,,,
-OS,7.2,21c,We considered the change in the number of days exceeding 90°F from 1948 and 2018 at 197 randomly sampled locations from the NOAA database in Exercise 7.19. The mean and standard deviation of the reported differences are 2.9 days and 17.2 days. Does the 90% confidence interval provide convincing evidence that there were more days exceeding 90°F in 2018 than in 1948 at NOAA stations? Explain.,"Yes, since the interval lies entirely above 0.",Free Response,,,,
-OS,7.2,22a,We considered the differences between the reading and writing scores of a random sample of 200 students who took the High School and Beyond Survey in Exercise 7.20. The mean and standard deviation of the differences are x(read−write) = −0.545 and 8.887 points. Calculate a 95% confidence interval for the average difference between the reading and writing scores of all students.,"A 95% confidence interval can be calculated using xdiff as follows: -0.545 ± 1:98 x 8.887 / sqrt(200) = (-1.79, 0.70)",Free Response,,,,
-OS,7.2,22b,We considered the differences between the reading and writing scores of a random sample of 200 students who took the High School and Beyond Survey in Exercise 7.20. The mean and standard deviation of the differences are x(read−write) = −0.545 and 8.887 points. Interpret the 95% confidence interval in context.,"We are 95% confident that on the reading test students score, on average, 1.79 points lower to 0.70 points higher than they do on the writing test.",Free Response,,,,
-OS,7.2,22c,We considered the differences between the reading and writing scores of a random sample of 200 students who took the High School and Beyond Survey in Exercise 7.20. The mean and standard deviation of the differences are x(read−write) = −0.545 and 8.887 points. Does the 95% confidence interval provide convincing evidence that there is a real difference in the average scores? Explain.,"No, since 0 is included in the interval.",Free Response,,,,
-OS,7.3,29,"Casein is a common weight gain supplement for humans. Does it have an effect on chickens? Using data provided in Exercise 7.27, test the hypothesis that the average weight of chickens that were fed casein is different than the average weight of chickens that were fed soybean. If your hypothesis test yields a statistically significant result, discuss whether or not the higher average weight of chickens can be attributed to the casein diet. Assume that conditions for inference are satisfied.","The hypotheses are H0: µc = µs and HA: µc ≠ µs. We are told to assume that conditions for inference are satisfied. T = ((323.58 – 246.43) – 0) / sqrt((64.4322 / 12) + (54.1322 / 14)) = 3.48; df = 11; p-value = P(|T| > 3.27) < 0.01. Since p-value < 0:05, reject H0. The data provide strong evidence that the average weight of chickens that were fed casein is different than the average weight of chickens that were fed soybean. Since this is a randomized experiment the type of diet is the only difference between the various groups of chicken, and hence the observed differences are can be attributed to differences in diet.",Free Response,,,,
-OS,7.3,31,"Subjects from Central Prison in Raleigh, NC, volunteered for an experiment involving an “isolation” experience. The goal of the experiment was to find a treatment that reduces subjects’ psychopathic deviant T scores. This score measures a person’s need for control or their rebellion against control, and it is part of a commonly used mental health test called the Minnesota Multiphasic Personality Inventory (MMPI) test. The experiment had three treatment groups: (1) Four hours of sensory restriction plus a 15 minute “therapeutic” tape advising that professional help is available. (2) Four hours of sensory restriction plus a 15 minute “emotionally neutral” tape on training hunting dogs. (3) Four hours of sensory restriction but no taped message.","Let µdiff = µpre - µpost. Then, for each treatment the hypotheses are: H0: µdiff = 0: Treatment has no effect. HA: µdiff > 0: Treatment has an effect on P.D.T. scores, either positive or negative. The conditions that need to be satisfied are: 1. Independence: The subjects are randomly assigned to treatments, so independence within and between groups is satisfied. 2. Normality: All three sample sizes are smaller than 30, so we look for clear outliers. There is a borderline outlier in the first treatment group. Since it is borderline, we will proceed, but we should report this caveat with any results. The calculation of the test statistics and p-values are shown below. Treatment 1: T = (6.21 – 0) / (12.3 / sqrt(14) = 1.89; df = 13; p-value = 2 x P(T > 1.89) = 0.081. Treatment 2: T = (2.86 – 0) / (7.94 / sqrt(14) = 1.35; df = 13; p-value = 2 x P(T > 1.35) = 0.200. Treatment 3: T = (-3.21 – 0) / (8.57 / sqrt(14) = -1.40; df = 13; p-value = 2 x P(T > -1.40) = 0.185. We do not reject the null hypothesis for any of these groups. As earlier noted, there is some uncertainty about if the method applied is reasonable for the first group.",Free Response,,,,
-OS,7.3,32a,"Determine if the following statement is true or false, and explain your reasoning for statements you identify as false: When comparing means of two samples where n1 = 20 and n2 = 40, we can use the normal model for the difference in means since n2 ≥ 30.","False, in order to be able to use a Z test both sample sizes need to be above 30.",Free Response,,,,
-OS,7.3,32b,"Determine if the following statement is true or false, and explain your reasoning for statements you identify as false: As the degrees of freedom increases, the t-distribution approaches normality.",True.,Free Response,,,,
-OS,7.3,32c,"Determine if the following statement is true or false, and explain your reasoning for statements you identify as false: We use a pooled standard error for calculating the standard error of the difference between means when sample sizes of groups are equal to each other.","False, we use the pooled standard deviation when the variability in groups is constant.",Free Response,,,,
-OS,7.3,33,"A large farm wants to try out a new type of fertilizer to evaluate whether it will improve the farm’s corn production. The land is broken into plots that produce an average of 1,215 pounds of corn with a standard deviation of 94 pounds per plot. The owner is interested in detecting any average difference of at least 40 pounds per plot. How many plots of land would be needed for the experiment if the desired power level is 90%? Use α = 0.05. Assume each plot of land gets treated with either the current fertilizer or the new fertilizer.",Difference we care about: 40. Single tail of 90%: 1.28 x SE. At the 5% significance level the rejection region bounds span ±1.96 x SE. 40 = 1.28 x SE + 1.96 x SE = 3.24 x SE; SE = 12.35 = sqrt(942 / n + 942 / n). n = 115.86. We will need 116 plots of land for each fertilizer.,Free Response,,,,
-OS,7.3,34,"A medical research group is recruiting people to complete short surveys about their medical history. For example, one survey asks for information on a person’s family history in regards to cancer. Another survey asks about what topics were discussed during the person’s last visit to a hospital. So far, as people sign up, they complete an average of just 4 surveys, and the standard deviation of the number of surveys is about 2.2. The research group wants to try a new interface that they think will encourage new enrollees to complete more surveys, where they will randomize each enrollee to either get the new interface or the current interface. How many new enrollees do they need for each interface to detect an effect size of 0.5 surveys per enrollee, if the desired power level is 80%? Use α = 0.05.",Difference we care about: 0.5. Single tail of 90%: 1.28 x SE. At the 5% significance level the rejection region bounds span ±1.96 x SE. 0.5 = 0.84 x SE + 1.96 x SE = 32.8 x SE; SE = 0.1786 = sqrt(2.22 / n + 2.22 / n). n = 303.47. We will need 304 plots of land for each fertilizer.,Free Response,,,,