Question

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 (#6, HW #6) to help. The data are displayed in table 2 below. Table 2: HDL levels (Mean SD) Study Group Control (n-26) Study entry Change After exercise 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 the 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 an (i.e., 3.0 instead of 2.4 mg/dl)?

Answer 1

This is a simple problem related to calculation of sample size and its related concepts of derivation .

For the calculation purposes ,i have assumed the confidence limit of significence as 95 %

a & b).

We need to find the minimum sample size n required in order to achieve the aimed statistical power of 0.80, for this two-tailed test, for the given significance level of a 0.95. The critical value in this - 0.06 case is Traditionally, this analysis of computing the minimum sample size to achieve a target statistical power is called A Priori Power Analys is In this case, the effect size passed is d 4. The minimum required sample size to reached the target power of 0.80 will be computed using the given value of the effect size d 4, a 0.95, and the type of tail specified For all the purposes, we will assume that we are testing for one population mean, and we are using the z-statistics. This is a simplification that makes the calculations less cumbersome, while not losing too much precision First we give the formula for commputing the probability of making a type II error (8). Let us recall that (6) is the probability of rejecting a false null hypothesis. We will use the following notation: Ho is the hypothesized value of the population mean, and u is the true population value. Also, we assume that the effect size d 4 we have provided is Cohen's d, which is defined as: The probability of making a type Il error (3) is computed as folllows: BPr(Fail to Reject a False Null Hypothesis Ho) = Pr-2e <tu) X Pr-e 7/Vi X =Pr-e +

X - Pr <H) But, if the true population mean is u, then has a standard normal distribution. Therefore, the following is obtained: μー く) B Pr + Z - /Vn =Pr-e =Pr-e-dyn Z (uAp Observe that we have assumed without loss of generality that Ho 0 by symmetry, because if it is not the case, we would use the other tail to get the exact same result. From the previous equation can be observed that the value of B depends on n. Indeed, the value of B will decrease as n increases. In order to find the minimum n to reach the target power, we need to construct the following table: Lower (a dVn) Upper (b dVn) B P(a sZs b) Sample Size (n) -zC- zC- Power 1-B -4.06 -3.94 1 0 1 -4.06 -3.94 C -5.5969 0 -5.7169 1 -6.9882 3 6.8682 C 1 -7.94 4 -8.06 0 1 Therefore, looking at the table above, we find the minimum value of n that exceeds the minimum required target statistical power is n = 1. This completes the calculation.

c). We will have to increase the sample size by 20 % to take care of 20 % drop out after follow up.

Final expected size =26

Thus initial size required

= (100%+20%)*26

~31

d). Observe the power table attached in solution a and b

For n =1, the power is 99.2 % ~100 %