bnolton/AP_Stat_Inference_Helper
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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
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Sample mean, x_bar = 30.35
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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
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Sample standard deviation, sx = 14.24
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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.
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Sample Proportion, p^ = 36/50 = 0.72
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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.”
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Sample porportion, p^ = 19/172 = 0.11
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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.
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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.
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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?
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The point estimate = p^ = (0.63 + 0.69) / 2 = 0.66; the margin of error is = 0.69-0.66 = 0.03
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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.
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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.
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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.
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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.
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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?
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The point estimate = p^ = (0.56 + 0.62) / 2 = 0.59; the margin of error is = 0.62-0.59 = 0.03
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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About 90% of the intervals from the random samples would capture the true median household income for all households in Arizona in 2015.
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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.
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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.
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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.
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About 90% of the intervals from the random samples would capture the true median household income for all households in New Jersey in 2015.
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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.
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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.
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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.
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Incorrect; we shouldn’t use the results of one sample to predict the results for future samples.
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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.
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Correct; a confidence interval provides an interval of plausible values for a parameter.
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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.
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Incorrect; the population mean always stays the same, regardless of the number of samples taken.
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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.
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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.
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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.
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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).
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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.
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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.
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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.
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Correct; this is the meaning of 95% confidence.
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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).
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Incorrect; it doesn’t make sense to say that a sample contains an interval.
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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.
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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.
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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%.
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The length would increase.
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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?
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The confidence interval would be half as wide.
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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.
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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.
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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%.
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The length would decrease.
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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?
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The confidence interval would be half as wide.
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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?
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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.
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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.
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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.
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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?
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Decrease confidence level. Drawback: less confidence that our interval captures the true average. Increase sample size. Drawback: larger samples cost more time and money.
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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?
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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.
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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.
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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.
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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?
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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.
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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?
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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.
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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.
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b
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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.
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e
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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
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e
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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.
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c
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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.
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Random: Met—SRS. 10%: Not met because 50 > 10% of seniors in the dormitory. Large Counts: Met because 14 ≥ 10 and 36 ≥ 10.
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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.
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Random: Met—SRS. 10%: Met because 50 < 10% of students at his college. Large Counts: Met because 38 ≥ 10 and 12 ≥ 10.
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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.
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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.
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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.
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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.
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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?
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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^.
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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?
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The capture rate will usually be greater than the specified confidence level.
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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?
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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*.
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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?
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The capture rate will almost always be less than the specified confidence level.
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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.
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Table A: z* 5 2.33; Tech: z* 5 2.326
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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.
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0.1996±2.326 sqrt((0.1996(1-0.1996))/4579) → (0.1859, 0.2133)
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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.
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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.
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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.
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Table A: z* 5 2.05; Tech: z* 5 2.054
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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.
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0.1105±2.054 sqrt((0.1105(1-0.1105))/172) → (0.0614, 0.1596)
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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All seniors at Tonya’s high school; the true proportion of all seniors who plan to attend the prom.
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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.
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Random: Random sample. 10%: 50<, 10% of the population. Large Counts: 36 ≥ 10 and 14 ≥ 10.
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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^.
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0.72±1.645 sqrt((0.72(1-0.72))/50) = (0.616, 0.824)
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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.
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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.
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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.
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All students at this high school; the true proportion of all students that the student body president knows by name.
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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.
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Random: Random sample. 10%: 100 < 10% of the population. Large Counts: 46 ≥ 10 and 54 ≥ 10.
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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^.
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0.46±2.576 sqrt((0.46(1-0.46))/100) = (0.332, 0.588)
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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
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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.
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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.
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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.
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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.
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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.
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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.
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We do not know the size of the sample that was taken from each age group.
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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.
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Larger, because this is a smaller sample size than the original group.
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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.
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We do not know the size of the sample that was taken from each group.
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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.
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Larger, because this is a smaller sample size than the original group.
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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.
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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.”
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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.
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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.
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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.
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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.
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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.
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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.
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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^.
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Solving 2.576 sqrt((0.44(0.56))/n) ≤ 0.03 gives n ≥ 1817.
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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?
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Solving 2.576 sqrt((0.50(0.50))/n) ≤ 0.03 gives n ≥ 1844. The conservative approach requires 27 more adults.
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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^.
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Solving 1.645 sqrt((0.75(0.25))/n) ≤ 0.04 gives n ≥ 318.
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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?
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Solving 1.645 sqrt((0.50(0.50))/n) ≤ 0.04 gives n ≥ 423. The conservative approach requires 105 more people.
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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?
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Solving 1.645 sqrt((0.50(0.50))/n) ≤ 0.04 gives n ≥ 423.
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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?
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Solving 1.96 sqrt((0.50(0.50))/n) ≤ 0.03 gives n ≥ 1068.
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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.
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Solving 0.03 = z* sqrt((0.64(0.36))/1028) gives z* = 2.00. The confidence level is likely 95%.
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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.
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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.
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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.
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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%.
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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?
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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.
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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.
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a
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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.
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a
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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
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d
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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
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a
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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
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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.
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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.
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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.
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The dataset used in training consists of 1014 question-answer pairs on the topic of inference for the AP Statistics exam created by the owner of this model. Specifically the problems involve 1 and 2 sample means and proportions confidence intervals and significance tests (no chi-sqaure or inference for slope). These problems were taken from three textbooks (two paid for copies, 1 open source). There are 928 free response questions and 86 multiple choice questions.
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