Sample size calculation example (2004-05-20)

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I received a question in Hong Kong about how to double check a power calculation in a paper by Tugwell et all in the 1995 NEJM [Medline]. In the paper, they state that

With the tender-joint count used as the primary outcome, a sample of 75 patients per group was needed in order to have a 5 percent probability of a Type I error and a power of 80 percent to detect a difference of 5 tender joints between groups, with a standard deviation of 9.5, and to allow for a 25 percent dropout rate.

and the formula for estimating the sample size would be

where D represents the minimum clinically relevant difference. This formula assumes that you have two parallel groups and you are comparing them with a continuous outcome variable using a two-sided t-test. There are several variations on this formula, but they all give the same answer. Extracting the information from the text shown above, you would get

Using these numbers, I get an estimated sample size of 56.60. In order to allow for dropouts, divide this number by 0.75 to get 75.47. I would round this number up, but could not criticize someone who rounded it downward.

This page was written by Steve Simon while working at Children's Mercy Hospital. Although I do not hold the copyright for this material, I am reproducing it here as a service, as it is no longer available on the Children's Mercy Hospital website. Need more information? I have a page with general help resources. You can also browse for pages similar to this one at Category: Sample size justification.