6. Suppose you have a new program created that can help people with dyslipidemia improve their lipid profile. You would like to design a randomized clinical trial to evaluate the program's effectiveness by comparing the average change in HDL in patients treated with your program to the average change in patients in a control group.Use data from the control group of a different study (problem #6, HW#6) to help. The data are displayed in table 2 below. Table 2: HDL levels (Mean SD)Study Group Study entry After exercise Change Control(n=26) 42.7 2.7 42.1 2.2 -0.6 2.4

a. *(1 point)We will consider the new program a success if we can improve average HDL by at least 4 mg/dl (approximately 10% of control group baseline level).Using the standard deviation of the change in control group in the previous study as your estimate of the standard deviation of change (in both groups), calculate the minimum number of patients you will need to have 80% power to indicate a difference between groups.Recommended sample size calculator:Go to powerandsamplesize.com, select "Go straight to the Calculators", and use the calculator for Compare 2 Means: 2-sample, 2-sided Equality. (When determining the sample size do not clear the sample size box.Just enter your values for the mean, standard deviation, etc. and select calculate.If you clear the sample size box, the calculator will not work. If this happens, go back to the home page and restart the calculator.) IF YOU DO NOT USE THE RECOMMENDED CALCULATOR, CITE THE CALCULATOR YOU USED.

b. Describe your size sample calculation for part a. Your description should contain enough information so that a reader could reproduce your sample size calculation.

c. If you expect that 20% of the patients will drop out of the study before follow-up, how many subjects do you need to enroll in each group at pre-randomization?

d. *(1 point)Using the sample size you calculated in part A, what would be the power to demonstrate a 4 mg/dl in the change in HDL between groups, if the standard deviation for change in HDL is 25% greater than assumed in part a (i.e., 3.0 instead of 2.4 mg/dl)?