You weigh 30 people who exercise every day, and you weigh 30 people who do not get regular exercise. The exercise group has a mean weight which is 22 pounds less than the non-exercise group's mean. Can you conclude that exercise causes people to weigh less?

Jan 1, 2015

No you cannot.

First of all, it is fairly safe to perform a two-sample independent z-test for this example. Because we are comparing means of a sample with 30 or more people, the central limit theorem states that the resultant distribution of the mean will approximate a normal distribution.

In order to perform a z-test, however, two pieces of information are necessary. First, one needs the difference in means. Second, one needs to know the standard deviations of the samples.

This allows for two important steps. First, one can determine whether it is reasonable to assume equal variance between the two samples. Second, and most importantly, one can calculate the standard error (SE) of the estimate. The z-statistic is equal to $\left(\text{mean difference")/("standard error}\right)$.

Because the standard error can be large if the standard deviation is sufficiently large, consider the following examples. If the SE = 22, then $z = \frac{22}{22} = 1$ and therefore p = 0.32. However, if SE = 10, then $z = \frac{22}{10} = 2.2$ and p = 0.04.

Because we can only state that there is a difference if we reject the null that there is no difference, we cannot come to a conclusion about the hypothesis test without knowing the standard error, which is not given in the question.