来自两所学校（school A 和 school B）的同龄学生全年参加大量小测，每名学生获得一个满分为 1000 的分数。school A 的 123 名学生和 school B 的 147 名学生的分数从低到高排名（最低为 rank 1，最高为 rank 270）。school A 学生的秩和为 15 355。

(a) 在 $5\%$ 显著性水平下进行 Wilcoxon 秩和检验，研究 school A 和 school B 学生的平均分是否存在差异。

(b) 写出 Wilcoxon 秩和检验有效所需的一个假设。

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。

这是两组独立样本的秩和检验。用两组样本量求零假设下的秩和均值和方差，再做连续性修正并比较双侧临界值。

Under the common-shape location model, let $H_0$ be that the two population median marks are equal, and $H_1$ that they are different. For school A’s rank sum $W$, under $H_0$ with no ties,

The sample sizes are large, so we use a normal approximation. For the observed lower-tail value $15355$, the continuity-corrected boundary is $15355.5$. Thus

The two-sided $5\%$ critical values are $\pm1.96$. Since $|z|>1.96$, we reject $H_0$. There is sufficient evidence at the $5\%$ level that typical marks differ between the two schools.

要把秩和差异解释成总体位置的差异，需要对两组总体分布的形状作假设。

The population distributions of marks in the two schools are assumed to have the same shape, so that any systematic difference can be interpreted as a location shift.