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So, I am new here.

I need to perform a sample size calculation for a clinical trial. The study sample will be select according to criteria of person's height. Persons within the particular height range (female, 1.6m to 1.7m) will be invited to participate in trial. We know the expected sample standard deviation from previous trial. But my concern is that sample is not from a normal distribution. Usual power/sample size calculation need the assumption of normal distribution of test statistic under $H_0$ and $H_1$, but here I believe we have truncated normal distribution. So how may I modify power.t.test, or make some other calculation in R, to accommodate this? My colleague says to just rely on central limit theory and assume normal with 1.65 mean and known standard deviation, but I believe this is wrong due to the truncation. Any advice would be appreciated.

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    $\begingroup$ I would use simulation to calculate the power. Anyway, given the truncated nature of your data, the use of a t-test may be questionable (I'm assuming that you're going to carry out the t-test on your truncated variable. This is what I understand from your question). $\endgroup$
    – boscovich
    Commented Apr 10, 2012 at 9:33
  • $\begingroup$ I am fairly sure the CLT is not applicable here. $\endgroup$
    – LeelaSella
    Commented Apr 10, 2012 at 10:16
  • $\begingroup$ Thank you @andrea . I mention t-test because that is what I do /if/ the sample is normal distribution. I would like to make a calculation using truncation distribution, or I also think uniform distribution is good in this case because sample near the mean of the population and range is small. Can I make sample size calculation with truncated normal distribution or uniform distribution ? I prefer calculation instead of the simulation. Please to know your further comment, and also to confirm the central limit theorem is wrong to use for here. $\endgroup$
    – Joseph King
    Commented Apr 10, 2012 at 10:17
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    $\begingroup$ What I would use, assuming you want to compare the mean of the height in two groups, is a permutation test en.wikipedia.org/wiki/Resampling_(statistics) $\endgroup$
    – boscovich
    Commented Apr 10, 2012 at 13:34
  • $\begingroup$ Andrea, truncation (within the center of the distribution) actually makes the t-test more applicable, not less applicable: the sampling distribution of the mean will be remarkably close to normal. P Sellaz, because the distribution will likely be close to uniform (and not very skewed), the CLT gives good insight even with samples as small as $5$ or so. Joseph, your thinking is good. $\endgroup$
    – whuber ♦
    Commented Apr 10, 2012 at 14:31

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Your colleague is correct.

In the US, 1.6 to 1.7 m is near the middle of the range of adult female heights. According to Wolfram Alpha, which summarizes NHANES 2006 data, the height distribution in this range should look close to this:

Female heights between 1.6 and 1.7 m

This is extremely close to uniform: its mean is 1.649 m and its standard deviation is 0.0287 m (whereas a uniform distribution in this range would have a mean of 1.650 m and SD of 0.0289 m). Its skewness coefficient is only 0.054.

Accordingly, independent samples drawn from this distribution will have means that are close to normally distributed. Here, for instance, is a histogram of means of 10,000 samples of just four heights drawn (independently) from this distribution:

Histogram of means

It is only very, very slightly non-normal (a Kolmogorov-Smirnov test rejects normality at p=0.94%, which is amazingly large given there are 10,000 data points). For the purpose of planning comparisons of mean heights among random groups of women, the normal approximation will work well. Standard power calculations ought to give good guidance.

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  • $\begingroup$ thank you again. So I can use 0.0287 as SD in standard power calculation ? With example of following R code to detect 0.01 difference from control to intervention group ? power.t.test(delta = 0.01, sd = 0.0287 , sig.level = 0.05 , power = 0.9 , type = "two.sample", alternative = "two.sided") thank you more $\endgroup$
    – Joseph King
    Commented Apr 10, 2012 at 18:53

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