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Many Americans complain about being sleep deprived. A team of psychologists researched the reasons for this sleep deprivation. After determining that one reason was poor time management, they devised a program to help people manage their time better. They recruited a random sample of 30 people at a major shopping center and randomly split them into three groups of size 10. Group 1 (the controls) received a logbook asking them to record the number of hours slept for a week but nothing else. Group 2 (the informed group) was shown a video to help improve time management. Group 3 (the intervention) participated in a two-day training course on improving time management. Groups 2 and 3 were also asked to keep records for one week on the number of hours slept. The psychologists were interested in learning if providing information or an intervention affects sleep time. Below are 95% Tukey simultaneous confidence intervals for the pairwise comparisons of all three pairs of means. Many Americans complain about being sleep deprived. A team of psychologists researched the reasons for this sleep deprivation. After determining that one reason was poor time management, they devised a program to help people manage their time better. They recruited a random sample of 30 people at a major shopping center and randomly split them into three groups of size 10. Group 1 (the controls) received a logbook asking them to record the number of hours slept for a week but nothing else. Group 2 (the informed group) was shown a video to help improve time management. Group 3 (the intervention) participated in a two-day training course on improving time management. Groups 2 and 3 were also asked to keep records for one week on the number of hours slept. The psychologists were interested in learning if providing information or an intervention affects sleep time. Below are 95% Tukey simultaneous confidence intervals for the pairwise comparisons of all three pairs of means. Based on these results, which of the following statements is true? A) Clearly, all three means are the same, since all confidence intervals include zero or nearly do so. B) We can be 95% confident that the intervention mean is different from both the control mean and the information mean, and the control mean and the intervention mean are not different. C) We can be sure that the intervention group is different from the control and information groups, and the information group is not different from the control group. D) We can be confident that the intervention mean is different from the control mean and information mean, while the control mean and information mean are not different, but we do not know the confidence level. Based on these results, which of the following statements is true?


A) Clearly, all three means are the same, since all confidence intervals include zero or nearly do so.
B) We can be 95% confident that the intervention mean is different from both the control mean and the information mean, and the control mean and the intervention mean are not different.
C) We can be sure that the intervention group is different from the control and information groups, and the information group is not different from the control group.
D) We can be confident that the intervention mean is different from the control mean and information mean, while the control mean and information mean are not different, but we do not know the confidence level.

E) A) and B)
F) A) and C)

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A company runs a three-day workshop on strategies for working effectively in teams. On each day, a different strategy is presented. Forty-eight employees of the company attend the workshop. At the outset, all 48 are divided into 12 teams of four. The teams remain the same for the entire workshop. Strategies are presented in the morning. In the afternoon, the teams are presented with a series of small tasks. The number of these tasks completed successfully, using the strategy taught that morning, is recorded for each team. The mean number of tasks completed successfully by all teams each day and the standard deviation are computed. The results follow. A company runs a three-day workshop on strategies for working effectively in teams. On each day, a different strategy is presented. Forty-eight employees of the company attend the workshop. At the outset, all 48 are divided into 12 teams of four. The teams remain the same for the entire workshop. Strategies are presented in the morning. In the afternoon, the teams are presented with a series of small tasks. The number of these tasks completed successfully, using the strategy taught that morning, is recorded for each team. The mean number of tasks completed successfully by all teams each day and the standard deviation are computed. The results follow. The researchers did an ANOVA F test of the data and obtained the following results. Which of the following conclusions is most reasonable? A) There is moderate evidence that the strategies taught are effective in increasing the number of tasks completed successfully for the first two days, but the effect appears to wear off. B) An ANOVA F test is not appropriate for these data. Instead, the company should have done several tests to see if the number of tasks completed successfully differed for the three days. This analysis would have shown that the treatment was effective. C) The data provide strong evidence that the mean number of tasks completed successfully differs for the three strategies taught. D) The data appear to provide little or no evidence that the strategies taught differ in their effectiveness in helping teams complete tasks successfully. The researchers did an ANOVA F test of the data and obtained the following results. A company runs a three-day workshop on strategies for working effectively in teams. On each day, a different strategy is presented. Forty-eight employees of the company attend the workshop. At the outset, all 48 are divided into 12 teams of four. The teams remain the same for the entire workshop. Strategies are presented in the morning. In the afternoon, the teams are presented with a series of small tasks. The number of these tasks completed successfully, using the strategy taught that morning, is recorded for each team. The mean number of tasks completed successfully by all teams each day and the standard deviation are computed. The results follow. The researchers did an ANOVA F test of the data and obtained the following results. Which of the following conclusions is most reasonable? A) There is moderate evidence that the strategies taught are effective in increasing the number of tasks completed successfully for the first two days, but the effect appears to wear off. B) An ANOVA F test is not appropriate for these data. Instead, the company should have done several tests to see if the number of tasks completed successfully differed for the three days. This analysis would have shown that the treatment was effective. C) The data provide strong evidence that the mean number of tasks completed successfully differs for the three strategies taught. D) The data appear to provide little or no evidence that the strategies taught differ in their effectiveness in helping teams complete tasks successfully. Which of the following conclusions is most reasonable?


A) There is moderate evidence that the strategies taught are effective in increasing the number of tasks completed successfully for the first two days, but the effect appears to wear off.
B) An ANOVA F test is not appropriate for these data. Instead, the company should have done several tests to see if the number of tasks completed successfully differed for the three days. This analysis would have shown that the treatment was effective.
C) The data provide strong evidence that the mean number of tasks completed successfully differs for the three strategies taught.
D) The data appear to provide little or no evidence that the strategies taught differ in their effectiveness in helping teams complete tasks successfully.

E) B) and C)
F) A) and D)

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Investigators gave caffeine to fruit flies to see if it affected their rest. The four treatments were a control, a low caffeine dose of 1 mg/ml of blood, a medium dose of 3 mg/ml of blood, and a higher caffeine dose of 5 mg/ml of blood. Twelve fruit flies were assigned at random to the four treatments, three to each treatment, and the minutes of rest measured over a 24-hour period were recorded. Assume the data that follow are four independent SRSs (one from each of the four populations of caffeine levels) and that the distribution of the yields is Normal. Investigators gave caffeine to fruit flies to see if it affected their rest. The four treatments were a control, a low caffeine dose of 1 mg/ml of blood, a medium dose of 3 mg/ml of blood, and a higher caffeine dose of 5 mg/ml of blood. Twelve fruit flies were assigned at random to the four treatments, three to each treatment, and the minutes of rest measured over a 24-hour period were recorded. Assume the data that follow are four independent SRSs (one from each of the four populations of caffeine levels) and that the distribution of the yields is Normal. A partial ANOVA table produced by Minitab follows, along with the means and standard deviation of the yields for the four groups. The value of the ANOVA F statistic for testing equality of the population means of the average rest time for the four caffeine levels is: A) 2.78. B) 4.73. C) 4.82. D) 7.41. A partial ANOVA table produced by Minitab follows, along with the means and standard deviation of the yields for the four groups. Investigators gave caffeine to fruit flies to see if it affected their rest. The four treatments were a control, a low caffeine dose of 1 mg/ml of blood, a medium dose of 3 mg/ml of blood, and a higher caffeine dose of 5 mg/ml of blood. Twelve fruit flies were assigned at random to the four treatments, three to each treatment, and the minutes of rest measured over a 24-hour period were recorded. Assume the data that follow are four independent SRSs (one from each of the four populations of caffeine levels) and that the distribution of the yields is Normal. A partial ANOVA table produced by Minitab follows, along with the means and standard deviation of the yields for the four groups. The value of the ANOVA F statistic for testing equality of the population means of the average rest time for the four caffeine levels is: A) 2.78. B) 4.73. C) 4.82. D) 7.41. The value of the ANOVA F statistic for testing equality of the population means of the average rest time for the four caffeine levels is:


A) 2.78.
B) 4.73.
C) 4.82.
D) 7.41.

E) B) and D)
F) A) and C)

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Which of the following is a true statement about analysis of variance?


A) The null hypothesis test states that all population means i are the same.
B) The alternative is many-sided.
C) The F test assesses evidence for some differences among the population means i.
D) All of the answer options are correct.

E) All of the above
F) A) and B)

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Investigators gave caffeine to fruit flies to see if it affected their rest. The four treatments were a control, a low caffeine dose of 1 mg/ml of blood, a medium dose of 3 mg/ml of blood, and a higher caffeine dose of 5 mg/ml of blood. Twelve fruit flies were assigned at random to the four treatments, three to each treatment, and the minutes of rest measured over a 24-hour period were recorded. Assume the data that follow are four independent SRSs (one from each of the four populations of caffeine levels) and that the distribution of the yields is Normal. Investigators gave caffeine to fruit flies to see if it affected their rest. The four treatments were a control, a low caffeine dose of 1 mg/ml of blood, a medium dose of 3 mg/ml of blood, and a higher caffeine dose of 5 mg/ml of blood. Twelve fruit flies were assigned at random to the four treatments, three to each treatment, and the minutes of rest measured over a 24-hour period were recorded. Assume the data that follow are four independent SRSs (one from each of the four populations of caffeine levels) and that the distribution of the yields is Normal. A partial ANOVA table produced by Minitab follows, along with the means and standard deviation of the yields for the four groups. The pooled standard deviation is: A) 22.84. B) 23.21. C) 91.37. D) 2154.82. A partial ANOVA table produced by Minitab follows, along with the means and standard deviation of the yields for the four groups. Investigators gave caffeine to fruit flies to see if it affected their rest. The four treatments were a control, a low caffeine dose of 1 mg/ml of blood, a medium dose of 3 mg/ml of blood, and a higher caffeine dose of 5 mg/ml of blood. Twelve fruit flies were assigned at random to the four treatments, three to each treatment, and the minutes of rest measured over a 24-hour period were recorded. Assume the data that follow are four independent SRSs (one from each of the four populations of caffeine levels) and that the distribution of the yields is Normal. A partial ANOVA table produced by Minitab follows, along with the means and standard deviation of the yields for the four groups. The pooled standard deviation is: A) 22.84. B) 23.21. C) 91.37. D) 2154.82. The pooled standard deviation is:


A) 22.84.
B) 23.21.
C) 91.37.
D) 2154.82.

E) All of the above
F) B) and D)

Correct Answer

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Investigators gave caffeine to fruit flies to see if it affected their rest. The four treatments were a control, a low caffeine dose of 1 mg/ml of blood, a medium dose of 3 mg/ml of blood, and a higher caffeine dose of 5 mg/ml of blood. Twelve fruit flies were assigned at random to the four treatments, three to each treatment, and the minutes of rest measured over a 24-hour period were recorded. Assume the data that follow are four independent SRSs (one from each of the four populations of caffeine levels) and that the distribution of the yields is Normal. Investigators gave caffeine to fruit flies to see if it affected their rest. The four treatments were a control, a low caffeine dose of 1 mg/ml of blood, a medium dose of 3 mg/ml of blood, and a higher caffeine dose of 5 mg/ml of blood. Twelve fruit flies were assigned at random to the four treatments, three to each treatment, and the minutes of rest measured over a 24-hour period were recorded. Assume the data that follow are four independent SRSs (one from each of the four populations of caffeine levels) and that the distribution of the yields is Normal. A partial ANOVA table produced by Minitab follows, along with the means and standard deviation of the yields for the four groups. The conclusion that you would draw from this test is: A) accept the null hypothesis; all of the groups have the same mean. B) fail to reject the null hypothesis; there is no evidence to suggest that the groups have different means. C) reject the null hypothesis; all of the group means are different. D) reject the null hypothesis; at least one of the groups has a mean that is different. A partial ANOVA table produced by Minitab follows, along with the means and standard deviation of the yields for the four groups. Investigators gave caffeine to fruit flies to see if it affected their rest. The four treatments were a control, a low caffeine dose of 1 mg/ml of blood, a medium dose of 3 mg/ml of blood, and a higher caffeine dose of 5 mg/ml of blood. Twelve fruit flies were assigned at random to the four treatments, three to each treatment, and the minutes of rest measured over a 24-hour period were recorded. Assume the data that follow are four independent SRSs (one from each of the four populations of caffeine levels) and that the distribution of the yields is Normal. A partial ANOVA table produced by Minitab follows, along with the means and standard deviation of the yields for the four groups. The conclusion that you would draw from this test is: A) accept the null hypothesis; all of the groups have the same mean. B) fail to reject the null hypothesis; there is no evidence to suggest that the groups have different means. C) reject the null hypothesis; all of the group means are different. D) reject the null hypothesis; at least one of the groups has a mean that is different. The conclusion that you would draw from this test is:


A) accept the null hypothesis; all of the groups have the same mean.
B) fail to reject the null hypothesis; there is no evidence to suggest that the groups have different means.
C) reject the null hypothesis; all of the group means are different.
D) reject the null hypothesis; at least one of the groups has a mean that is different.

E) A) and C)
F) C) and D)

Correct Answer

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Investigators gave caffeine to fruit flies to see if it affected their rest. The four treatments were a control, a low caffeine dose of 1 mg/ml of blood, a medium dose of 3 mg/ml of blood, and a higher caffeine dose of 5 mg/ml of blood. Twelve fruit flies were assigned at random to the four treatments, three to each treatment, and the minutes of rest measured over a 24-hour period were recorded. Assume the data that follow are four independent SRSs (one from each of the four populations of caffeine levels) and that the distribution of the yields is Normal. Investigators gave caffeine to fruit flies to see if it affected their rest. The four treatments were a control, a low caffeine dose of 1 mg/ml of blood, a medium dose of 3 mg/ml of blood, and a higher caffeine dose of 5 mg/ml of blood. Twelve fruit flies were assigned at random to the four treatments, three to each treatment, and the minutes of rest measured over a 24-hour period were recorded. Assume the data that follow are four independent SRSs (one from each of the four populations of caffeine levels) and that the distribution of the yields is Normal. A partial ANOVA table produced by Minitab follows, along with the means and standard deviation of the yields for the four groups. The P-value of this test is: A) greater than 0.1. B) between 0.05 and 0.1. C) less than 0.05. D) It is not possible to determine the P-value from the information provided. A partial ANOVA table produced by Minitab follows, along with the means and standard deviation of the yields for the four groups. Investigators gave caffeine to fruit flies to see if it affected their rest. The four treatments were a control, a low caffeine dose of 1 mg/ml of blood, a medium dose of 3 mg/ml of blood, and a higher caffeine dose of 5 mg/ml of blood. Twelve fruit flies were assigned at random to the four treatments, three to each treatment, and the minutes of rest measured over a 24-hour period were recorded. Assume the data that follow are four independent SRSs (one from each of the four populations of caffeine levels) and that the distribution of the yields is Normal. A partial ANOVA table produced by Minitab follows, along with the means and standard deviation of the yields for the four groups. The P-value of this test is: A) greater than 0.1. B) between 0.05 and 0.1. C) less than 0.05. D) It is not possible to determine the P-value from the information provided. The P-value of this test is:


A) greater than 0.1.
B) between 0.05 and 0.1.
C) less than 0.05.
D) It is not possible to determine the P-value from the information provided.

E) B) and C)
F) B) and D)

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The one-way ANOVA is a generalization of what other statistical test?


A) ANOVA is not related to any other tests.
B) the two-sample t test for independent means
C) the paired t test for independent means
D) Levene's test for differences between variations

E) A) and C)
F) B) and D)

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