# Power of the t-test under unequal sample sizes

I heard that the power of a t-test used with unequal samples is limited by the size of the smaller sample. Can I take this to mean that the power of a t-test with unequal sample sizes is equal to the power of a t-test used on equal sample sizes where $n$ equals the size of the smaller sample?

(Note, by $n$, I usually mean the total sample size, so I interpret your last sentence to be 'where $\bf{.5}$$n equals the size of the smaller sample'.) No, not quite. Consider this simulation (conducted with R): set.seed(9) power1010 = vector(length=10000) power9010 = vector(length=10000) for(i in 1:10000){ n1a = rnorm(10, mean=0, sd=1) n2a = rnorm(10, mean=.5, sd=1) n1c = rnorm(90, mean=0, sd=1) n2c = rnorm(10, mean=.5, sd=1) power1010[i] = t.test(n1a, n2a, var.equal=T)p.value power9010[i] = t.test(n1c, n2c, var.equal=T)p.value } mean(power1010<.05)  0.184 mean(power9010<.05)  0.323  What we see here is that when the total sample size is 20, with equal group sizes, n_1=n_2=10, power is 18\%; but when the total sample size is 100, but the smaller group has n_2=10, power is 32\%. Thus power can increase when the size of the larger group goes up even though the smaller sample size stays the same. This answer is adapted from my answer here: How should one interpret the comparison of means from different sample sizes?, which you will probably want to read for more on this topic. • You're welcome, @Jimj. You might want to check out the other answer as well. – gung - Reinstate Monica Jul 10 '13 at 2:02 • That's right, @John, I discuss that in the other answer that I link to. My only point here is that: the power of a t-test w/ unequal sample sizes is not equal to the power of a t-test w/ equal sample sizes where .5n equals the size of the smaller sample. – gung - Reinstate Monica Jul 10 '13 at 2:59 To understand the comment, consider the effect of letting the second sample get larger and larger while the first stays constant in size. Eventually, the sample mean for the second sample converges to the population mean it was drawn from, and the standard error of the mean becomes zero. If you examine the ordinary two-sample test statistic$$t = \frac{\bar {X}_1 - \bar{X}_2}{S_p \cdot \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$$then as n_2 gets large, \bar{X}_2 goes to \mu_2, and the term in the square root goes to 1/n_1. Now look at S_p:$$S_p = \sqrt{\frac{(n_1-1)S_{X_1}^2+(n_2-1)S_{X_2}^2}{n_1+n_2-2}}$$Let w_1 = \frac{n_1-1}{(n_1-1)+(n_2-1)} and similarly for w_2$$S_p = \sqrt{w_1 S_{X_1}^2+w_2 S_{X_2}^2}$$As n_2 gets large, w_2 goes to 1, while S_{X_2} goes to \sigma_2; S_p becomes \sigma_2. So what are we left with? The statistic now looks like this:$$ \frac{\bar {X}_1 - \mu_2}{\sigma_2 / \sqrt{n_1}}$$An ordinary one-sample z-test ... whose power is a function of n_1. You might find it instructive to consider the same calculation in the case of the Welch t-test (I believe you get a one-sample t statistic). • In the equation for the ordinary two-sample test statistic, is it a typo to put the pooled standard deviation Sp as part of the denominator? – vtshen Mar 25 at 5:14 • No. \!$$\!$$\!$$\!$ – Glen_b Mar 25 at 6:24