Subjects in a study by Dabonneville et al. included a sample of 40 men who claim
ID: 3049468 • Letter: S
Question
Subjects in a study by Dabonneville et al. included a sample of 40 men who claimed to engage in a variety of sports activities (multisport). The mean Body Mass Index (BMI) for these men was 22.41 with a standard deviation of 1.27. A sample of 24 male rugby players has a mean BMI of 27.75 with a standard deviation of 2.64.
a. Is there sufficient evidence for one to claim that in general, rugby players have a higher BMI than the multisport men? Perform the test at the 0.01 level of significance.
b. Suppose you were to do a new study. Assume that the higher standard deviation of 2.64 applies to the whole population. Your hypothesis is that the multisport athletes and rugby players have a difference in mean BMI of about 5.0. How many individuals would you need to detect that difference? Use 0.01 significance, 0.90 power.
c. Now suppose who were less sure about how different multisport athletes and rugby players differ in BMI. Using the same standard deviation (2.64), you hypothesize that the mean difference in BMI is only 2.0. How many individuals would you need for that study? Use both 0.05 significance and 0.80 power as well as 0.01 significance and 0.90 power.
Explanation / Answer
a)
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 40 22.41 1.27 0.20
2 24 27.75 2.64 0.54
Difference = (1) - (2)
Estimate for difference: -5.340
99% upper bound for difference: -4.170
T-Test of difference = 0 (vs <): T-Value = -10.90 P-Value = 0.000 DF = 62
Both use Pooled StDev = 1.8974
p-value = 0.000 < 0.01
there is sufficient evidence for one to claim that in general, rugby players have a higher BMI than the multisport men
b)
Power and Sample Size
1-Sample Z Test
Testing mean = null (versus null)
Calculating power for mean = null + difference
= 0.01 Assumed standard deviation = 2.64
Sample Target
Difference Size Power Actual Power
5 5 0.9 0.951457
sample size = 5
c)
Power and Sample Size
1-Sample Z Test
Testing mean = null (versus null)
Calculating power for mean = null + difference
= 0.01 Assumed standard deviation = 2.64
Sample Target
Difference Size Power Actual Power
2 21 0.8 0.814825
sample size = 21 when power = 0.80
Power and Sample Size
1-Sample Z Test
Testing mean = null (versus null)
Calculating power for mean = null + difference
= 0.01 Assumed standard deviation = 2.64
Sample Target
Difference Size Power Actual Power
2 26 0.9 0.900964
sample size = 26 when power = 0.90