ek. A 99% confidence interval for the difference in the population means, without music – with music, is (1.525, 7.275). Does the confidence interval support the claim that employees who listen to music while working are less productive than employees who work without listening to music? yes because the confidence interval does not contain 0 yes because the confidence interval consists entirely of positive numbers no because the confidence interval does not contain 0 no because the confidence interval consists entirely of positive num
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An experiment is conducted to estimate the difference in work productivity for employees under two conditions: with and without music playing in the background. Twenty employees volunteered to be part of the study. Ten of the volunteers were randomly assigned to work for one week with music playing in the background. The other 10 volunteers worked for one week without music playing in the background. Productivity is measured by counting the number of units produced for the week. A 99% confidence interval for the difference in the population means, without music – with music, is (1.525, 7.275). Does the confidence interval support the claim that employees who listen to music while working are less productive than employees who work without listening to music?
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- Acne is a common skin disease that affects most adolescents and can continue into adulthood. A study compared the effectiveness of three acne treatments and a placebo, all in gel form, applied twice daily for 12 weeks. The study's 517 teenage volunteers were randomly assigned to one of the four treatments. Success was assessed as clear or almost clear skin at the end of the 12 week period. The results of the study can be seen in the table below. Using the appropriate statistical test, determine if there is significant evidence that the four treatments perform differently. If so, how do they compare.Computer-mediated communication (CMC) is a form of interaction that heavily involves technology (e.g., instant messaging, email). A study was conducted to compare relational intimacy in people interacting via CMC to people meeting face-to-face (FTF). Participants were 48 undergraduate students, of which half were randomly assigned to the CMC group and half assigned to the FTF group. The variable of interest, relational intimacy score, was measured (on a 7-point scale) for each participant after each of three different meeting sessions. Summary statistics for the first meeting session are given here. The researchers hypothesized that, after the first meeting, the mean relational intimacy score for participants in the CMC group would be lower than the mean relational intimacy score for participants in the FTF group. Test the researchers’ hypothesis using α = .10. CMC FTF Number…An important issue is whether there are racial differences in hypertension among children. We define hypertension as being above the 95th percentile for either systolic blood pressure (SBP) or diastolic blood pressure (DBP) among children of the same age, height, and sex. Since some of the children were observed at multiple visits, a GEE model was run of hypertension on ethnic group. There were three ethnic groups considered: Caucasian, African American, and Hispanic. One issue is that BMI, which may vary among ethnic groups, is positively related to hypertension. Hence, a logistic regression model was run, as shown in Table 13.55. 13.89 What is the estimated OR for hypertension comparing Hispanic boys vs. Caucasian boys based on the results in Table 13.55? (Call this OR2 .) Provide a 95% CI for this estimate. 13.90 Suppose the average BMI of Hispanic boys is higher than that for Caucasian boys. Is BMI a confounder of the association between ethnic group and hypertension? If so, is it…
- A psychologist hypothesizes that the children with type K personality score higher than the children with type R personality on a memory test. The estimated standard error turned out to be 0.9, while the mean score of type K children was 18.8 and the mean score for type R children was 21.3. What is the t-score? a. t = (18.8 - 21.3)/0.9 = 2.78 b. t = (18.8 - 21.3)(0.3) = 8.33 c. t = (18.8 - 21.3)/0.92 = 3.09 d. t = (18.8 - 21.3)(2)(0.9) = 1.39Students in a representative sample of 69 second-year students selected from a large university in England participated in a study of academic procrastination. Each student in the sample completed the Tuckman Procrastination Scale, which measures procrastination tendencies. Scores on this scale can range from 16 to 64, with scores over 40 indicating higher levels of procrastination. For the 69 second-year students in the study at the university, the sample mean procrastination score was 41.00 and the sample standard deviation was 6.83. n USE SALT (a) Construct a 95% confidence interval estimate of u, the population mean procrastination scale for second-year students at this college. (Use technology to calculate the critical value. Round your answers to three decimal places.) (b) How does the confidence interval for the population mean score for second-year students compare to the confidence interval for first-year students of (35.422, 38.618)? What does this tell you about the…According to previous studies, the mean distance each visitor in Greenspan National Park hikes during their visit is 30 kilometers. The park recently closed its shuttle system, which used to transport hikers to many of the park's most popular hiking trails. Because of this, an administrator at the park suspects the mean distance, u, is now less than 30 kilometers. The administrator chooses a random sample of 45 visitors. The mean distance hiked for the sample is 27.2 kilometers. Assume the population standard deviation is 9.9 kilometers. Can the administrator conclude that the mean distance hiked by each visitor is now less than 30 kilometers? Perform a hypothesis test, using the 0.10 level of significance. (a) State the null hypothesis H, and the alternative hYpothesis H,. OFor this example should a correlated or independent sample t test be used? Delayed posttest achievement scores were compared withimmediate posttest scores for all participants.According to previous studies, the mean distance each visitor in Greenspan National Park hikes during their visit is 21 kilometers. The park recently closed its shuttle system, which used to transport hikers to many of the park's most popular hiking trails. Because of this, an administrator at the park suspects the mean distance, u, is now less than 21 kilometers. The administrator chooses a random sample of 85 visitors. The mean distance hiked for the sample is 20.5 kilometers. Assume the population standard deviation is 8.3 kilometers. Can the administrator conclude that the mean distance hiked by each visitor is now less than 21 kilometers? Perform a hypothesis test, using the 0.10 level of significance. (a) State the null hypothesis H and the alternative hypothesis H . H: 0 ORecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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