I constructed a Welch's t-test in order to examine the difference between the means of two samples (check whether or not one is larger than the other).

$H_{0}: \mu_0 = \mu_1 ~\textrm{vs}~ H_{1}: \mu_0 < \mu_1$

I computed the t-score as follows:

$T = \frac{\bar{x}_{0} - \bar{x}_{1}}{\sqrt{\frac{s_{0}^{2}}{n_{0}} + \frac{s_{1}^{2}}{n_{1}}}}$

My rejection rule:

$T < t_{\alpha, \textrm{df}}$

Now, I would like to compute the power of this test. I know that the power is the probability of rejecting $H_{0}$ when $H_{1}$ is true. So $P(T < t_{\alpha, \textrm{df}} | \mu_0 < \mu_1)$.

Is that correct?

How would I compute that?


  • 1
    $\begingroup$ For a pooled t test, a power computation requires you to specify the difference in population means and sample sizes, also guess the value of common population variance.For the Welch t test you will have specify the difference, sample sizes and make guesses about the two variances. $\endgroup$ – BruceET Mar 17 '19 at 7:53

To compute the power, you need to specify the unknowns under some parameterization; it will depend on the sample sizes, the two variances and the specific difference in means (assuming normality; if you want power under some other situation you have more to specify).

With all those quantities specified, the easiest way to proceed is via simulation, which will give the rejection rate to any desired degree of accuracy (if you're prepared to simulate long enough to attain it).

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