Statistics: The Art and Science of Learning From Data, Books a la Carte Edition (4th Edition)

4th Edition

ISBN: 9780133860825

Author: Alan Agresti, Christine A. Franklin, Bernhard Klingenberg

Publisher: PEARSON

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Textbook Question

Chapter 10, Problem 93CP

Anorexia again Refer to Exercise 10.89, comparing mean weight changes in anorexic girls for cognitive behavioral therapy and a control group. The MINITAB output shows results of doing analyses while assuming equal population standard deviations.

MINITAB output for comparing mean weight changes

Chapter 10, Problem 93CP, Anorexia again Refer to Exercise 10.89, comparing mean weight changes in anorexic girls for , example 1

Difference = μ(cogchange) − μ(controlchange)

Estimate for difference: 3.46

95% CI for difference:(−0.68, 7.59)

T-Test of difference = 0 (vs ≠):

T-Value = 1.68 P-Value = 0.100 DF = 53

Both use Pooled StDev = 7.6369

a. Interpret the reported confidence interval.
b. Interpret the reported P-value.
c. What would be the P-value for H_a: μ₁ > μ₂? Interpret it.
d. What assumptions do these inferences make?

Teenage anorexia Example 8 in Section 9.3 described a study that used a cognitive behavioral therapy to treat a sample of teenage girls who suffered from anorexia. The study observed the mean weight change after a period of treatment. Studies of that type also usually have a control group that receives no treatment or a standard treatment. Then researchers can analyze how the change in weight compares for the treatment group to the control group. In fact, the anorexia study had a control group that received a standard treatment. Teenage girls in the study were randomly assigned to the cognitive behavioral treatment (Group 1) or to the control group (Group 2). The figure shows box plots of the weight changes for the two groups (displayed vertically). The output shows how MINITAB reports inferential comparisons of those two means.

MINITAB Output Comparing Mean Weight Changes

Chapter 10, Problem 93CP, Anorexia again Refer to Exercise 10.89, comparing mean weight changes in anorexic girls for , example 2

Difference = μ(cogchange) − μ(controlchange)

Estimate for difference: 3.46

95% CI for difference: (−0.71, 7.62)

T-Test of difference = 0 (vs ≠):

T-Value = 1.67 P-Value = 0.102 DF = 50

Chapter 10, Problem 93CP, Anorexia again Refer to Exercise 10.89, comparing mean weight changes in anorexic girls for , example 3

Box plots of weight change for anorexia study.

a. Report and interpret the P-value for testing H₀: μ₁ = μ₂ against H_a: μ₁ ≠∙ μ₂.
b. Summarize the assumptions needed for the analysis in part a. Based on the box plots, would you be nervous if you had to perform a one-sided test instead? Why?
c. The reported 95% confidence interval tells us that if the population mean weight change is less for the cognitive behavioral group than for the control group, it is just barely less (less than 1 pound), but if the population mean change is greater, it could be nearly 8 pounds greater. Explain how to get this interpretation from the interval reported.
d. Explain the correspondence between the confidence interval and the decision in the significance test for a 0.05 significance level.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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Chapter 10, Problem 92CP

Chapter 10, Problem 94CP

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Chapter 10 Solutions

Statistics: The Art and Science of Learning From Data, Books a la Carte Edition (4th Edition)

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Solution Summary: The author explains the probability of getting the test statistic as extreme as observed is 0.1 for the true null hypothesis.

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