The NIH wants to compare the mean weight loss for five different diets. They mea
ID: 3225556 • Letter: T
Question
The NIH wants to compare the mean weight loss for five different diets. They measure the weight loss (in pounds) of 5 men assigned to each of the diets for one month. The resulting data set is below.
a) Which of the following is NOT an assumption that should be checked before performing an ANOVA?
The average weight loss should be roughly the same for each diet.
The variability in weight loss should be roughly the same for each diet.
Weight loss should be roughly normally distributed for each diet.
b) What is the F statistic for the ANOVA? Give your answer to at least three decimal places.
c) Using a 0.01 level of significance, what is the critical point that one would compare to the F statistic in order to make a conclusion? Give your answer to three decimal places.
d) What is the P-value from the ANOVA? Give your answer to four decimal places.
e) What is the proper conclusion for NIH in this case?
Reject the claim that all diets are equivalent in terms of mean weight loss because the F statistic is larger than the critical point.
Fail to reject the claim that all diets are equivalent in terms of mean weight loss because the F statistic is smaller than the critical point.
Diet 4 has the largest mean weight loss because the F statistic is larger than the critical point.
Fail to reject the claim that all diets are equivalent in terms of mean weight loss because the F statistic is larger than the critical point.
Diet 5 has the largest mean weight loss because the F statistic is larger than the critical point.
Explanation / Answer
(a)
The variability in weight loss shoud be roughly the same for each diet
Null Hypothesis (Ho): There is no significance difference between the mean weight loss for five different diets
Alternative Hypothesis (H1): There is significance difference between the mean weight loss for five different diets
(b)
From the above ANOVA table F statistic value = 78. 237
(c)
At alpha = 0.01 level of significance with (1, 48) degrees of freedom F critical value = 7.194
(d)
From the ANOVA table the P = 1.2E-11
(e) F caluculated value > Fcritical value
i.e 78.23694 > 7.194218
Then Reject Ho
Fail to reject the claim that all diets are equivalent in terms of mean weight loss because the F statistic is larger than the critical point
Anova: Single Factor SUMMARY Groups Count Sum Average Variance Pounds 25 -16.3 -0.652 2.178433 Diet 25 75 3 2.083333 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 166.7138 1 166.7138 78.23694 1.2E-11 7.194218 Within Groups 102.2824 48 2.130883 Total 268.9962 49