Suppose a team of doctors wants to study the effect of different types of exercise on reducing body fat percentage in adult women. The 48 participants in the study consist of women between the ages of 40 and 49 with body fat percentages ranging from 36%–38%. The participants were each randomly assigned to one of four exercise regimens.
All participants were asked to adhere to their assigned exercise regimens for eight weeks. To control for the effect of diet on weight loss, the researchers provided the participants with all meals for the duration of the study. After eight weeks, the doctors recorded the change in body fat percentage for each of the participants.
The doctors plan to use this data in an analysis of variance (ANOVA) F‑test. They calculate the variance between samples as 16.760000 and the variance within samples as 1.191277.
Choose all correct facts about the doctors' ANOVA F‑test statistic.
The F‑test statistic is 14.0689.
The F‑test statistic has 4 degrees of freedom in the numerator and 47 degrees of freedom in the denominator.
The F‑test statistic is 0.0711.
The F‑test statistic indicates which treatment groups, if any, are significantly different from each other.
The F‑test statistic increases as the differences among the treatment groups' sample means increase.
The F‑test statistic has 3 degrees of freedom in the numerator and 44 degrees of freedom in the denominator.
We have k= 4 groups
N= 12*4 = 48 total participants
The between group variance or MSb is the variation by its df ie SSb/df
The within group variance or MSw is the variation by its df ie SSw/df
F-ratio = MSb/MSw
Df numerator = k-1= 3
Df denominatior = N-k = 44
The ANOVA table is as follows:
So F-statistic is 14.0689
F-statistic increases if difference among the sample means of the treatment groups increases.
Df numerator is 3 and df denominatior is 44
So options 1, 5 and 6 are correct.
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