Calculate the power to detect a change of -3 mmHg when using a sample size of 500 per group.27
(a) Determine the standard error (recall that the standard deviation for patients was expected to be about 12 mmHg).
(b) Identify the null distribution and rejection regions.
(c) Identify the alternative distribution when µtrmt − µctrl = −3.
(d) Compute the probability we reject the null hypothesis.
The researchers decided 3 mmHg was the minimum difference that was practically important, and with a sample size of 500, we can be very certain (97.7% or better) that we will detect any such difference. We now have moved to another extreme where we are exposing an unnecessary number of patients to the new drug in the clinical trial. Not only is this ethically questionable, but it would also cost a lot more money than is necessary to be quite sure we’d detect any important effects.
The most common practice is to identify the sample size where the power is around 80%, and sometimes 90%. Other values may be reasonable for a specific context, but 80% and 90% are most commonly targeted as a good balance between high power and not exposing too many patients to a new treatment (or wasting too much money).
We could compute the power of the test at several other possible sample sizes until we find one that’s close to 80%, but there’s a better way. We should solve the problem backwards.
27(a) The standard error is given as SE = √
500 + 122
500 = 0.76.
(b) & (c) The null distribution, rejection boundaries, and alternative distribution are shown below:
−9 −6 −3 0 3 6 9 xtrmt − xctrl
Null distributionDistribution with µtrmt − µctrl = −3
The rejection regions are the areas on the outside of the two dotted lines and are at ±0.76× 1.96 = ±1.49. (d) The area of the alternative distribution where µtrmt − µctrl = −3 has been shaded. We compute the Z-score and
find the tail area: Z = −1.49−(−3)
0.76 = 1.99 → 0.977. With 500 patients per group, we would be about 97.7% sure
(or more) that we’d detect any effects that are at least 3 mmHg in size.