Statistical Power - Example


We study the effect of a treatment on some quantity, and compare research subjects by measuring the quantity before and after the treatment, analyzing the data using a paired t-test. Let, denote the pre-treatment and post-treatment measures on subject i. The possible effect of the treatment should be visible in the differences, which we assume to be independently distributed, all with the same expected value and variance.

We proceed by analyzing D as in a one-sided t-test. The null hypothesis will be: (no effect), and the alternative: (positive effect). The test statistic is:

where n is the sample size, is the average of the and is the sample variance. The null hypothesis is rejected when

with 1.64 the approximate decision threshold for a level 0.05 test based on a normal approximation to the test statistic.

Now suppose that the alternative hypothesis is true and . Then the power is

\pi(\tau)&=&P(\sqrt{n}\bar{D}/\hat{\sigma}_D > 1.64|\tau) \\ &=&P\left(\sqrt{n}(\bar{D}-\tau+\tau)/\hat{\sigma}_D > 1.64\right|\tau)\\
&=& P\left(\sqrt{n}(\bar{D}-\tau)/\hat{\sigma}_D > 1.64-\sqrt{n}\tau/\hat{\sigma}_D\right|\tau)\\

Since approximately follows a standard normal distribution when the alternative hypothesis is true, the approximate power can be calculated as

Note that according to this formula the power increases with the values of the parameter . For a specific value of a higher power may be obtained by increasing the sample size n.

It is, of course, not possible to guarantee a sufficient large power for all values of, as may be very close to 0. In fact the minimum (infimum) value of the power is equal to the size of the test, in this example 0.05. However it is of no importance to distinguish between and small positive values. If it is desirable to have enough power, say at least 0.90, to detect values of, the required sample size can be calculated approximately:

\pi(1)\approx 1-\Phi(1.64-\sqrt{n}/\hat{\sigma}_D) >0{.}90\ ,

from which it follows that



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