某公司的每周利润以美元为单位，服从正态分布。过去，每周利润服从分布 $\mathrm{N}(736,26^2)$。管理层变更后，随机选取 $35$ 周，所得的平均每周利润为 $725$ 美元。

(a) 写出一个必要的假设，并在 $2\%$ 的显著性水平下检验每周平均利润是否下降。

再随机选取 $35$ 周，求出样本的平均每周利润，并在 $2\%$ 的显著性水平下进行类似的检验。

(b) 写出发生第一类错误的概率。

(c) 已知目前每周平均利润实际为 $718$ 美元，求发生第二类错误的概率。

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。

假设改变管理后标准差仍为 $26$。备择是均值降低，使用左尾 $2\%$ 临界值。

Assume that the population standard deviation of weekly profit is unchanged at $26$ dollars. Let $\mu$ be the population mean weekly profit after the management change. Test

Under $H_0$, the mean of the $35$ independent normal observations has distribution $\mathrm N(736,26^2/35)$. The test statistic is

For a lower-tailed test at the $2\%$ level, the critical value is $\Phi^{-1}(0.02)\approx-2.054$. Since $-2.503<-2.054$, reject $H_0$.

There is sufficient evidence at the $2\%$ significance level that the population mean weekly profit has decreased.

连续正态检验可精确选择拒绝域概率，因此第一类错误概率等于显著性水平。

A Type I error is rejecting a true $H_0$. For this continuous normal test, its probability equals the significance level:

先在原假设下找固定的临界样本均值，再在真实均值 $718$ 下求未拒绝的右尾概率。

Let $c$ be the critical sample mean. Under $H_0$,

The rejection region is $\overline X<c$. When the true mean profit is $718$, $\overline X\sim\mathrm N(718,26^2/35)$. A Type II error occurs when $\overline X\ge c$, so