U.S. Centers for Disease Control and Prevention said about 72 million Americans—nearly one out of every four of the country’s 304 million residents—are considered obese. Millions more are designated as overweight.” The American trend toward obesity has prompted many physicians to offer weight-loss programs to their
Multiple comparison procedures #2
“The U.S. Centers for Disease Control and Prevention said about 72 million Americans—nearly one out of every four of the country’s 304 million residents—are considered obese. Millions more are designated as overweight.” The American trend toward obesity has prompted many physicians to offer weight-loss programs to their patients. [Source: Price, W. T. “Physicians Get into the Weight-Loss Business.” Florida Today, July 7, 2008.]
A physician conducted an experimental study to compare the effectiveness of four different weight-loss programs. In the study, 56 obese adults were randomly assigned to the four programs so that each program had 14 adults. The programs lasted for 6 months. The weights of the subjects were measured before and after the programs, and each subject’s weight loss was computed in pounds.
The following table summarizes the results of the study.
|
Treatment, j |
Number of Observations, njj |
Sample Mean, x̄ jx̄j(Pounds) |
|---|---|---|
| Program A | 14 | 22.1 |
| Program B | 14 | 17.9 |
| Program C | 14 | 19.5 |
| Program D | 14 | 10.2 |
The four treatments define four populations of interest. ANOVA was performed to test the hypothesis that the means of the four populations are equal. The results are presented in the following ANOVA table.
|
ANOVA Table |
|||||
|---|---|---|---|---|---|
|
Source of Variation |
Sum of Squares |
Degrees of Freedom |
Mean Square |
F |
P-Value |
| Treatments | 2,008.85 | 3 | 669.62 | 8.40 | ≤ 0.0001 |
| Error | 4,144.00 | 52 | 79.69 | ||
| Total | 6,152.85 | 55 |
At a significance level of α = 0.05, the null hypothesis that the population means are equal is rejected.
Suppose that you would like to perform multiple comparisons of the sample means to determine where the differences in the population means occur. In this question, you will perform only one of the six possible pairwise comparisons.
Let μ22 denote the population mean for the second treatment (Program B) and μ44 the population mean for the fourth treatment (Program D). You want to test whether the two population means are equal.
What is the null hypothesis?
H₀: μ22 = μ44
H₀: μ22 < μ44
H₀: μ22 ≠ μ44
H₀: μ22 > μ44
What is the alternative hypothesis?
Haa: μ22 > μ44
Haa: μ22 ≠ μ44
Haa: μ22 = μ44
Haa: μ22 < μ44
In this experiment, the estimated standard error of the difference between any two sample means is3.3741_______ .
Use the tool to help you answer the following question.
Select a Distribution
Use Fisher’s least significant difference (LSD) procedure to test the hypothesis that the two population means are equal at a comparison wise error rate of α = 0.05.
The value of the t-test statistic is1.27______ , and its critical values are _____ .
Another possible test statistic is the difference between the two sample means, which is_____ . Their least significant difference is _____ .
The confidence
You _____ reject the null hypothesis that the two population means are equal, because the t-test statistic is ______ extreme than its critical values.
Alternatively, you can make this conclusion because the magnitude of the difference between the two sample means is_____ than their LSD.
Also, you can make this conclusion because the confidence interval estimate of the difference between the two population means ____ 0.
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