某俱乐部的 9 名运动员有了一位新教练。教练采用了一个新的训练计划，他认为该计划会减少这些运动员的比赛用时。每名运动员在完成新训练计划前后各完成一次 1500 m 计时测试。他们的用时（秒）如下。

(a) 在 5% 显著性水平下进行配对 t 检验，以检验教练的看法。

进一步研究表明，该训练计划的效果倾向于使较慢运动员的用时减少得比较快运动员更多。

(b) 说明为什么(a)中使用的配对 t 检验在这种情况下可能不是合适的检验。

(c) 提出一种可以替代配对 t 检验的合适检验。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

同一位运动员的训练前后成绩组成一对。取训练前减训练后，成绩改善对应正差值。

Let $\mu_D$ be the population mean of the paired differences, defined as before minus after. Test $H_0\colon\mu_D=0$ against $H_1\colon\mu_D>0$.

The differences are $5,0,-1,6,1,-2,8,7,15$, with sum $39$ and sum of squares $405$. Thus

For $8$ degrees of freedom the upper $5\%$ critical value is $1.860$. Since $2.39349\ldots>1.860$, we reject $H_0$. There is sufficient evidence at the $5\%$ level that the training programme reduces mean race times.

检验要求配对差值的总体近似正态。较慢运动员的改善幅度更大，可能使差值的分布偏斜。

Larger reductions among slower athletes could produce an asymmetric distribution of before-minus-after differences. The population distribution of the differences may therefore fail the normality assumption of the paired $t$-test.

如果差值分布可能偏斜，可只检验配对差值的正负。这样无需对差值分布作正态或对称假设。

A one-sided paired sign test is suitable. Omit any zero differences and test whether a positive difference occurs more often than a negative one.