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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.
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c
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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)
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b
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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
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c
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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.
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d
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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.
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a
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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.
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d
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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.
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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.
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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.
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The point estimate is (317.64 + 394.56) / 2 = 356.1. The margin of error is 394.56 - 356.1 = 38.46.
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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.
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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.
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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.
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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.
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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.
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Random: Random sample of students. 10%: 45 < 10% of 1000. Normal/Large Sample: n= 45 ≥ 30.
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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.
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Random: Random sample of bags. 10%: Assume 75 < 10% of population. Normal/Large Sample: n= 75 ≥ 30.
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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.
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H0: µ = 11.5, Ha: µ < 11.5, where µ = the true mean battery life when playing videos for all tablets.
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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.
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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.
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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.
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H0: µ = 50, Ha: µ > 50, where µ = the true mean percent of purchases for which an alternative supplier offered lower prices.
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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.
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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.
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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.
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x¯= 62.8 ≠ 64
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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.
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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
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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.
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x¯= 4.7 < 5
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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.
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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
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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.
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t= (125.7 – 115) / (29.8 / sqrt(45) = 2.41
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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.
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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.
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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?
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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.
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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.
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t= (19.28 - 19.2) / (0.81 / sqrt(75) = 0.86
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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.
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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.
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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?
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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.
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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?
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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.
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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.
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Type I error: Finding convincing evidence that the true mean speed is greater than 25 mph when it really isn’t.
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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?
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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.
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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.
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Type I error: Finding convincing evidence that the true mean sleep increase is greater than 0 when it really isn’t.
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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?
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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.
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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?
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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.
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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?
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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.
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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?
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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.
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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.
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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.
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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.
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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.
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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.
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1 - 0.61 = 0.39
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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.
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Increase the sample size or use a larger significance level.
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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.
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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.
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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.
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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.
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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.
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1 - 0.49 = 0.51
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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.
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Increase the sample size or use a larger significance level.
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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.
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H0: µ = 200, Ha: α ≠ 200, where µ = the true mean response time of European servers (in milliseconds).
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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.
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Random: Random sample. 10%: 14 < 10% of population. Normal/Large Sample: A graph of the data reveals no strong skewness or outliers.
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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?
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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.
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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.
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No! We cannot draw any conclusions about the United States because we only collected information from a random sample of servers in Europe.
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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.
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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.
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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.
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Random: Random sample. 10%: 24 < 10% of population. Normal/Large Sample: A graph shows a roughly symmetric shape with no outliers.
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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?
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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.
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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.
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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.
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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?
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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.
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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?
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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.
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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?
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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.
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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?
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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.
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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.
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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.
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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?
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The researcher should repeat the procedure on these four people to see if they again perform well.
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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?
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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b
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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.
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a
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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
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d
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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.
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c
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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.
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a
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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.
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a
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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?
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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.
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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.
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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.
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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?
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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
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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.
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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.
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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?
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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.
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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?
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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
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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t = 0.98, P-value = 0.1675 using df = 26.96.
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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?
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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.
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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.
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To increase the power of this test, we could increase the sample sizes or the significance level.
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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.
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t = 0.80, P-value = 0.2131 using df = 48.46
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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?
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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.
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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.
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To increase the power of this test, we could increase the sample sizes or the significance level.
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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?
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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.
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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?
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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.
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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.
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Because we rejected H0, it is possible that we made a Type I error.
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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?
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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.
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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?
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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.
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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.
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Because we rejected Ha, it is possible that we made a Type I error.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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Two-sample t test; the data are being produced using two distinct groups of cars in a randomized experiment.
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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.
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Paired t test; this is a matched pairs experimental design in which both treatments are applied to each subject in a random order.
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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.
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Paired t test; the data were collected from the male and female partners in 40 couples.
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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.
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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.
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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.
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Two-sample t test; the data come from independent random samples of male and female college professors.
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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.
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Two-sample t test; the data are being produced using two distinct groups of plots in a randomized experiment.
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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
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d
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