一家公司正在判断两台机器 $X$ 和 $Y$ 中哪一台能更快地生产某种电气元件。分别从机器 $X$ 生产的 8 个元件和机器 $Y$ 生产的 9 个元件中随机抽样，记录生产一个这种元件所需的时间，单位为分钟。数据如下。

经理声称，平均而言，机器 $X$ 生产一个元件所需的时间少于机器 $Y$。

(a) 在 5% 显著性水平下进行 Wilcoxon 秩和检验，检验数据是否支持经理的说法。

(b) 假设两台机器生产元件所需时间均服从正态分布且方差相等，在 5% 显著性水平下进行 $t$ 检验，检验数据是否支持经理的说法。

(c) 一般来说，你是否预期 (a) 和 (b) 两部分检验的结论相同？给出理由。

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。

把两台机器的时间合并从小到大排名。若机器 $X$ 较快，它的秩和会偏小，所以做左尾检验。

Under the common-shape location model, test $H_0\colon m_X=m_Y$ against $H_1\colon m_X<m_Y$, where $m_X$ and $m_Y$ are the population median times. The ranks are

For sample sizes $8$ and $9$, the lower one-sided $5\%$ critical value for the rank sum of the size-$8$ sample is $54$. Since $55>54$, we fail to reject $H_0$. There is insufficient evidence at the $5\%$ level that machine $X$ has a lower median component time than machine $Y$.

按题设的正态、等方差假设做独立两样本 $t$ 检验。保留原始时间的数值差异，并合并组内方差。

Test $H_0\colon\mu_X=\mu_Y$ against $H_1\colon\mu_X<\mu_Y$, where the parameters are population mean production times. The sample means are $39.7/8$ and $49.5/9$. The individual unbiased variance estimates are

The pooled variance estimate is

The statistic is

For $15$ degrees of freedom, the lower $5\%$ critical value is $-1.753$. Since $-1.82037\ldots<-1.753$, we reject $H_0$. Under the stated assumptions, there is sufficient evidence at the $5\%$ level that machine $X$ has a lower population mean production time.

比较两种检验实际使用的信息。一个看数值差距，另一个只看排序。

The conclusions need not agree. The $t$-test uses the actual times and assumes normal populations with equal variances. The rank-sum test uses only the ordering. Changing an extreme observation without changing its rank can alter the $t$ statistic while leaving the rank-sum statistic unchanged.