Answer to Question #184627 in Statistics and Probability for jeykey

Question

. An investigator wants to compare two diet programs in children who are obese. One diet is a low-fat diet, and the other is a low carbohydrate diet. The plan is to enroll children and weigh them at the start of the study. Each child will then be randomly assigned to either the low fat or the low carbohydrate diet. Each child will follow the assigned diet for 8 weeks, at which time they will again be weighed. The number of pounds lost will be computed for each child. Based on data reported from diet trials in adults, the investigator expects that 20% of all children will not complete the study. A 95% confidence interval will be estimated to quantify the difference in weight loss between the two diets and the investigator would like the margin of error to be no more than 3 pounds. How many children should be recruited into the study? 

Answer 1

The sample sizes are computed as follows:

$n_i = 2 (\dfrac{Z\sigma}{E})^2$

Suppose one such study compared the same diets in adults and involved 100 participants in each diet group. The study reported a standard deviation in weight lost over 8 weeks on a low fat diet of 8.4 pounds and a standard deviation in weight lost over 8 weeks on a low carbohydrate diet of 7.7 pounds. These data can be used to estimate the common standard deviation in weight lost as follows:

$s_p = \sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n+n-2}}$

$= \sqrt{\dfrac{(100-1)8.4^2+(100-1)7.7^2}{100+100-2}}$

$= 8.1$

We now use this value and the other inputs to compute the sample sizes:

$n_i = 2 (\dfrac{Z\sigma}{E})^2$

$= 2(\dfrac{1.96 \times 8.1}{3})^2$

$= 56.0$

Samples of size n₁=56 and n₂=56 will ensure that the 95% confidence interval for the difference in weight lost between diets will have a margin of error of no more than 3 pounds. Again, these sample sizes refer to the numbers of children with complete data. The investigators anticipate a 20% attrition rate. In order to ensure that the total sample size of 112 is available at 8 weeks, the investigator needs to recruit more participants to allow for attrition.

N (number to enroll) * (% retained) = desired sample size

Therefore N (number to enroll) = desired sample size/(% retained)

$N = \dfrac{112}{0.80} = 140$