一所学院在面试研究职位申请者时使用两项评估 X 和 Y。今年，这两项评估已经用于大量申请者。

参加评估 X 的 9 名申请者的随机样本分数如下。

参加评估 Y 的 10 名申请者的随机样本分数如下。

面试官认为评估 X 的总体中位数分数低于评估 Y 的总体中位数分数。

在 1% 显著性水平下进行 Wilcoxon 秩和检验，检验数据是否支持面试官的看法。

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。

将两组分数合在一起从小到大排名。若测评 X 的总体中位数较低，X 组的秩和应偏小。

Under the common-shape location model, test $H_0$ that the population medians are equal against $H_1$ that the median for assessment X is lower. The ranks for the X scores in the order given are

The ranks for the Y scores in the order given are $2,11,16,6,10,13,12,14,17,19$. Thus the rank sum for X is $W_X=70$; the Y rank sum is $120$, and the two sums total $19(20)/2=190$. For sample sizes $9$ and $10$, the lower one-sided $1\%$ critical value for the size-$9$ rank sum is $61$. Since $70>61$, we fail to reject $H_0$. There is insufficient evidence at the $1\%$ level that the population median score is lower for assessment X.