Let\'s carry out an F test to see whether true mean change in body fat mass diff
ID: 3219402 • Letter: L
Question
Let's carry out an F test to see whether true mean change in body fat mass differs for the four treatments.
Appendix Table 7 gives the 95% Studentized range critical value q = 3.74 (using error df = 60, the closest tabled value to df = N - k = 53). The first two T-K intervals are
The remaining intervals are
We would conclude that 1 is not significantly different from 2 and that 3 is not significantly different from 4. We would also conclude that 1 and 2 are significantly different than 3 and 4. Note that Treatments 1 and 2 were treatments that administered a placebo in place of the growth hormone and Treatments 3 and 4 were treatments that included the growth hormone. This analysis was the basis of the researchers' conclusion that growth hormone, with or without sex steroids, decreased body fat mass.
The G+S treatment resulted in the largest (negative) change in body fat mass. According to the T-K intervals, which treatments were significantly different from G+S?
a. P+P and G+P
b. P+P and P+S
c. P+S and G+P
According to the T-K intervals, which treatments were significantly different from P+P?
a. none
b. G+S
c. G+P and G+S
1. Let 1, 2, 3, and 4 denote the true mean change in body fat for treatments P + P, P + S, G + P, and G + S, respectively. 2. H0: 1 = 2 = 3 = 4 3. Ha: At least two among 1, 2, 3, and 4 are different. 4. Significance level: = .01 5. Test statistic: 6. Assumptions: figure below shows boxplots for the four samples. The boxplots are roughly symmetric, and there are no outliers. The largest standard deviation (s1 = 1.545) is not more than twice as big as the smallest (s2 = 1.264). The subjects were randomly assigned to treatments. The assumptions of ANOVA are reasonable. 7. Computation: SSTr = n1(x1 - )2 + n2(x2 - )2 + ... + nk(xk - )2 = 14(0.064 - (-1.15))2 + 14(-0.286 - (-1.15))2 + 13(-2.023 - (-1.15))2 + 16(-2.250 - (-1.15))2 = 60.37 treatment df = k - 1 = 3 SSE = (n1 - 1)s12 + (n2 - 1)s22 + ... + (nk - 1)sk2 = 13(2.387) + 13(1.484) + 12(1.598) + 15(2.155) = 101.81 error df = N - k = 57 - 4 = 53 Thus, 8. P-value: Appendix Table 6 shows that for df1 = 3 and df2 = 60 (the closest tabled df to df = 53), the value 6.17 captures upper-tail area .001. Because F = 10.48 > 6.17, it follows P-value < .001. 9. Conclusion: Since P-value , we reject H0. The mean change in body fat mass is not the same for all four treatments. Treatment P P 0.1 0.6 2.2 0.7 2.0 0.7 0.0 2.6 1.4 1.5 2.8 0.3 1.0 1.0 14 0.064 1.545 2.387 Change in Body Fat Mass (kg) P S G P 0.1 1.6 0.2 0.4 0.0 0.4 0.4 2.0 -0.9 3.4 1.1 2.8 1.2 2.2 0.1 1.8 0.7 3.3 2.0 2.1 -0.9 3.6 3.0 0.4 1.0 3.1 1.2 14 13 0.286 2.023 1.264 1.218 1.598 1.484 G S 3.1 3.2 2.0 2.0 3.3 0.5 4.5 0.7 1.8 2.3 1.3 1.0 5.6 2.9 1.6 0.2 16 2.250 1.468 2.155Explanation / Answer
The G+S treatment resulted in the largest (negative) change in body fat mass. According to the T-K intervals, which treatments were significantly different from G+S?
a. P+P and G+P
b. P+P and P+S
c. P+S and G+P
Ans: Correct answer is b. P+P and P+S.
We know that the confidence intervals of 1 - 3 and 2 - 3 do not include zero. Hence, according to the T-K intervals, treatments P+P and P+S were significantly different from G+S.
According to the T-K intervals, which treatments were significantly different from P+P?
a. none
b. G+S
c. G+P and G+S
Ans: The correct answer is c. G+P and G+S.
These two treatments do not have zero in the confidence interval with P+P.
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