Question

“The US 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 six 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, giving the number of observations and the sample mean and variance of the subjects’ weight loss in each program.

Treatment Number of Observations, n Sample Mean, M (Pounds) Program A 17.9 Sample Variance, s? (Pounds squared) 99.7 95.7 Program B Program C Program D 14 19.5 10.2 22.1 99.7 100.6 The physician selects a significance level of a = 0.05. Why should he perform an analysis of variance (ANOVA) with a = 0.05 instead of simply using multiple t tests with a = 0.05 to compare the six pairs of group means? O Multiple t tests at a = 0.05 result in an experimentwise alpha level smaller than a = 0.05. O The probability of a Type I error when performing multiple t tests at a = 0.05 will be greater than 0.05. O An ANOVA avoids the problem of an inflated testwise alpha level. The probability of a Type I error when performing multiple t tests at a = 0.05 will be less than 0.05.

Answer 1

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Here' a problem with using multiple t-test:

Every time you conduct a t-test there is a chance that you will make a Type I error. This error is usually 5%. By running two t-tests on the same data you will have increased your chance of "making a mistake" to 10%. The formula for determining the new error rate for multiple t-tests is not as simple as multiplying 5% by the number of tests. However, if you are only making a few multiple comparisons, the results are very similar if you do. As such, three t-tests would be 15% (actually, 14.3%) and so on. These are unacceptable errors. An ANOVA controls for these errors so that the Type I error remains at 5% and you can be more confident that any statistically significant result you find is not just running lots of tests.

Answer is 3rd. ANOVA avoids the problem of inflated alpha level due to multiple t-testa