Currently I'm doing a power calculation using the pwr package. I thought it should be very simple, but I'm not sure if I'm doing it right. I use R to calculate the sample size. I checked my results using other software. The results are really different. The documentation of the other software isn't available, that's why I post this question.

In short. I'm going to perform a two-sample, two-sided t-test. The no. of observations between the two groups are the same. I would like to measure/calculate a difference between the means of 4. The variance of my test is 12. The variance is equal between the groups. sig. level = 0.05 and the power is 0.8.

Using the pwr package I get the following code

d <- 4/12 
pwr.t.test(n=NULL, d=d, sig.level=0.05, power=0.8, type="two.sample", alternative="two.sided") 

With the following results

Two-sample t test power calculation

          n = 142.2462 
          d = 0.3333333 
  sig.level = 0.05 
      power = 0.8 
alternative = two.sided 

NOTE: n is number in each group

However, the other software I use says that I just need 13 replications. Is the other package using another theory about calculating the no. of replications for a t-test? Or am I doing something wrong?

I hope you can help me out.

Best regards, Robin


1 Answer 1


This is from Minitab:

Power and Sample Size 

2-Sample t Test

Testing mean 1 = mean 2 (versus ≠)
Calculating power for mean 1 = mean 2 + difference
α = 0.05  Assumed standard deviation = 3.464

            Sample  Target
Difference    Size   Power  Actual Power
         4      13     0.8      0.806340

The sample size is for each group.

enter image description here

I think you may be confusing the population variance (not 'test' variance) $\sigma^2 = 12$ with the population standard deviation $\sigma = 3.464.$ If I enter $\sigma = 12,$ I get $n = 143.$

I believe the parameter $d$ in your output is intended to be $d=\Delta/\sigma,$ which should be $d = 4/3.464,$ but by using $\sigma = 12,$ you make it $d = 4/12 = 1/3 = 0.333.$

  • $\begingroup$ you're right. This is what I didn't understand. Thank you for this clarification $\endgroup$
    – Soml
    Dec 28, 2018 at 9:57

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