一所拥有大量学生的学校正在更新其校徽。每名学生都设计了一个新校徽，两名教师分别给每个校徽打出满分 50 分的分数。随机抽取 12 名学生，所得分数如下表所示。

其中一名学生声称教师 2 给出的分数高于教师 1。

(a) 在 5% 显著性水平下进行 Wilcoxon 配对符号秩检验，以检验数据是否支持该声称。

后来发现，教师 1 错误地输入了学生 C 的分数。她原本想输入的分数是 24，而不是 40。这个错误已被更正。

(b) 判断这一更正是否会影响(a)中检验的结论。

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。

同一份标志设计的两位老师评分构成一对。对差值绝对值排名，再恢复符号，检验第二位老师的评分是否偏高。

Let $D$ be Teacher 2’s mark minus Teacher 1’s mark for the same design. Test $H_0$ that the population median difference is zero against $H_1$ that it is positive. The differences and signed ranks are

For the positive alternative, a small negative rank sum supports the claim. Here $T_-=7+2+11=20$. The one-sided $5\%$ lower critical value for $12$ nonzero differences is $17$. Since $20>17$, we fail to reject $H_0$. The data provide insufficient evidence that Teacher 2 tends to award higher marks.

学生 C 的差值只改变正负，绝对值和排序都不变。将其秩从负秩和移到正秩和，再比较同一临界值。

The corrected difference for C is $32-24=8$, replacing $-8$. Its absolute rank remains $7$, so $T_-=20-7=13$. Since $13\le17$, we now reject $H_0$. After the correction there is sufficient evidence at the $5\%$ level that Teacher 2 tends to award higher marks.