8.8 The observed difference is substantial. Of those who took vitamin C, 143/321 = 0.445 had no sick days; of those taking the placebo, 92/320 = 0.288 had no sick days. Using the pooled sample success proportion (143 + 92)/(321 + 320) = 0.367, the z value is

far larger than the 2.575 value required for a two-tailed test at the 1 percent level. (The two-sided P value is 0.00004.) These data strongly reject the null hypothesis that the chances of having no sick days is not affected by taking vitamin C.

8.18 A difference-in-means test is supposed to use two independent random samples. This is not the case here, because the 26 students in the front group are also included in the total sample of 50 students. He should have compared the 26 in front with the 13 in back.
   There are 26 + 13 - 2 = 37 degrees of freedom. The pooled variance is

and the t value is

which is statistically significant at the 5 percent level, since it exceeds the 2.0 value required for statistical significance at the 5 percent level using a two-tailed t test with 37 degrees of freedom. (Statistical software shows P[t > 286] = 0.014, giving a two-sided P value of 0.028.) The observed difference is substantial, though not overwhelming, representing roughly the difference between a B and B+ average. (On a 4-point scale, 9.68 and 8.79 convert to 3.23 and 2.93.)
   Even though the difference is statistically significant and substantial, perhaps causation runs in the opposite direction; maybe better students like to sit in the front where they can see the blackboard and hear the teacher, while those who want to daydream or are unprepared to answer questions try to hide in the back rows.

8.34 The pooled sample proportion is 0.365:

giving a z value of 14.69:

The P value is virtually zero, decisively rejecting the null hypothesis that the noun proportions are identical. Yet, the difference between 36 and 37 percent does not seem substantial.

8.36 For a difference-in-means test, with possibly unequal population variances,

Computer software shows there to be 138.2 degrees of freedom and a two-sided P value of 0.274, not nearly enough to reject the null hypothesis at the 5 percent level.
   A matched-pair test is appropriate for these data because it is the very same 75 people in each sample; we have night and day data for the same person. For a matched-pairs test, the t value is

With 75 - 1 = 74 degrees of freedom, the two-sided P value is 0.230, still not nearly large enough to reject the null hypothesis at the 5 percent level.

8.48 Here are the sample means and standard deviations:

For the female data, the t value for a difference-in-means test is only 0.585; with 17.2 degrees of freedom, P[t > 0.585] = 0.2830, giving a two-sided P value of 2(0.2830) = 0.566, not nearly statistically significant at the 5 percent level:

For the male data, the t value for a difference-in-means test is 9.591; with 16.4 degrees of freedom, the two-sided P value is 0.00000004, decisively significant at the 5 percent level:

The improvement in the average female time is less than 1 percent, which does not seem substantial. The improvement in the average male time is a substantial 5.8 percent.