D.27 Answers: Tests for paired means

Answers to exercises in Sect. 29.11.

Answer to Exercise 29.1: $H_0$: ${\mu }_d=0$ and $H_1$: ${\mu }_d>0$: differences are positive when the dip rating is better than the raw rating. $t=(5.2-0)/3.06=1.699$; $P$ larger than 0.05: the evidence doesn’t support the alternative hypothesis.

Answer to Exercise 29.2: 1. Because it is the bloood pressure reduction, and a reduction is what the drug is meant to produce, so expect the reductions to be positive nunmbers. 2. Differences shown below. 3. Histogram of differences: Fig. D.7. 4. $H_0$: ${\mu }_d=0$ and $H_1$: ${\mu }_d>0$ (because the differences are reductions). 5. $t=8.12$. 6. $P=0.001÷2=0.0005$ (one-tailed test). 7. Very strong evidence ($P=0.0005$) that the drug reduces the average systolic blood pressure (mean reduction: 8.6 mm Hg) in the population.

FIGURE D.7: A histogram of the systolic blood pressure reductions (in mm Hg)

Answer to Exercise 29.3: $H_0$: ${\mu }_d=0$ and $H_1$: ${\mu }_d>0$, where differences are positive when the intention to smoke is reduced after exercise. $t=(0.66-0)/0.37=1.78$; $P$-value larger than 0.05: the evidence doesn’t support the alternative hypothesis. No evidence ($P>0.05$) that the mean ‘intention to smoke’ reduced after exercise in women (mean change in intention to smoke: -0.66; std. error: 0.37).

Answer to Exercise 29.4: $H_0$: ${\mu }_d=0$ and $H_1$: ${\mu }_d>0$, where differences refer to the reduction in ferritin. $\overline{d}=-424.25$ and $s=2092.693$ and $n=20$, so $t=-0.90663$. $t$ is ‘small’; $P>0.05$ (actually $P=0.376$): the evidence doesn’t support the alternative hypothesis. Since $n<25$, the test may not be statistically valid (the histogram of data (Fig. D.8) suggests that the population might have a normal distribution), though the $P$-value is very large so it probably makes little difference.

FIGURE D.8: A histogram of the change in ferritin concentration