Solution to Exercise 8.28.

  First, we must formulate the null and alternative hypotheses. The question posed is, ``Does this data (sic) suggest the the true average lateral recumbency time under these conditions is less than 20 min.?'' These data do certainly ``suggest'' that the true average lateral recumbency time under these conditions is less than 20 min. since the sample average of 18.86 min. is less than 20 min. I am criticizing the author's wording here. A more appropriate question for motivating a statistical test of hypotheses is, ``Do these data support (or strongly support) the claim that the true average lateral recumbency time under these conditions is less than 20 min.?'' Note the difference between the rather weak ``suggest'' and the much stronger ``support the claim.'' In my opinion, one doesn't need to perform a test of hypotheses to ``suggest.'' So much for philosophy. Anyway, letting $\mu$ denote the ``average lateral recumbency time under these conditions,'' we want to test

$$H_0 : \mu \ge 20 , \quad vs. \quadH_1 : \mu < 20 .$$

The sample size n = 73 is large enough that we can use a z-test:

$$z \; = \; \frac{\bar{x}-\mu_0}{s/\sqrt{n}} \; = \; \frac{18.86 - 20}{8.6/sqrt{73}} \; = \; -1.132577 .$$

Small (negative) values of z are evidence against H₀. We reject at the .05 level of significance if

$$z \; \le \; -z_{.05} \; = \; -1.645 .$$

Clearly z doesn't satisfy this, so we cannot reject H₀. So, although the data ``suggest'' that the mean is less than 20, they do not offer strong evidence for that hypothesis.

Well, I screwed up and didn't read that the author asked for a test at the 0.1 level of significance. So, the critical value should be -z-sub-.1 = -1.282, and -1.13 is not less than this amount so we still cannot reject. Note that this is clearly a 1-sided testing situation since we are asked if there is evidence if the true mean is LESS than 20 min.

3/31/2001