一所大型学院正在举办钢琴比赛。每位学生演奏了一首指定乐曲，两位评委分别给出满分 80 分的评分。下表给出了随机抽取的 14 名学生所得的分数。

(a) 其中一名学生声称，平均而言评委 1 给出的分数高于评委 2。使用 Wilcoxon 配对符号秩检验，在 5% 显著性水平下检验数据是否支持该学生的说法。

(b) 说明为什么在这种情况下使用 Wilcoxon 配对符号秩检验比使用配对样本 $t$ 检验更合适。

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。

同一位学生的两份评分组成一对。取第一位减第二位，用差值的绝对值排名并恢复符号。

Let $D$ be Judge 1’s mark minus Judge 2’s mark for the same performance. Test $H_0$ that the population median difference is zero against $H_1$ that it is positive. The differences in student order are

Their signed ranks are

For the positive alternative, use the negative rank sum $W_-=7+6+12+8=33$. The lower one-sided $5\%$ critical value for $14$ pairs is $25$. Since $33>25$, we fail to reject $H_0$. There is insufficient evidence at the $5\%$ level that Judge 1 tends to award higher marks.

小样本配对 $t$ 检验需要差值总体正态。这里的评分差值来自有上下限的整数评分，正态假设没有保证。

There is no basis for assuming that the population distribution of the paired mark differences is normal. A signed-rank test can be used under a symmetry assumption without requiring normality of those differences.