用餐愉快 | 宽数题库

第 1 小问

题目详情

一家大型米其林餐厅的经理对员工表现不满意，并认为厨师们每天做出的菜不超过 13 份。为检验这一假设，经理随机观察了 $49$ 位厨师在某一天做出的菜品数量。他发现这 $40$ 位厨师做出的菜品数量均值和方差分别为 15 和 4。在这些信息下，经理进行统计检验，以判断厨师们平均每天做出的菜是否显著多于 13 份。

The manager of a large Michelin star restaurant is disappointed with his staff and believes that his chefs are cooking no more than 13 dishes a day. To evaluate this hypothesis, the manager randomly observes the number of dishes that $49$ chefs cook on a random day. He finds that the mean and variance of the number of meals the $40$ chefs cooked is 15 and 4. With these values, the manager runs a statistical test to see if his chefs are making significantly more than 13 dishes a day.

解析

在这个检验中，经理检验的是如下假设：

H_0: \mu = 13 \ \ \ \ \ \ H_a : \mu \gt 13

由于样本量相当大（ $n \gt 30$ ），可以使用 Z 统计量。

Z = \frac{\bar{x}- \mu_0}{\sqrt{\sigma^2/n}} = \frac{15 - 13}{2 / 7} = \boxed{7}

Original Explanation

For this test, the manager is testing the following hypotheses:

H_0: \mu = 13 \ \ \ \ \ \ H_a : \mu \gt 13

Since we have a reasonably large amount of samples ( $n \gt 30$ ), we can use the Z-statistic.

Z = \frac{\bar{x}- \mu_0}{\sqrt{\sigma^2/n}} = \frac{15 - 13}{2 / 7} = \boxed{7}

第 2 小问

题目详情

一家大型米其林餐厅的经理希望检测厨师平均做菜数量是否相差 1 份。为此，他记录了 $49$ 位厨师在某一周内做出的菜品数量，并发现每位厨师做菜数量的方差为 4。他以 $\alpha=0.05$ 检验原假设 $H_0: \mu = 13$ 对备择假设 $H_a: \mu \gt 13$ 。II 类错误的概率是多少？

The manager of a large Michelin star restaurant is seeking to detect a difference equal to one dish in the average number of meals cooked by his chefs. To check this, he records the number of dishes that $49$ make on a given week. He finds the variance of the number of dishes cooked per chef to be 4. He runs a statistical test with $\alpha=0.05$ to test the null hypothesis $H_0: \mu = 13$ against his alternative hypothesis $H_a: \mu \gt 13$ . What is the probability of a Type II error?

解析

要解这道题，先求拒绝域。由于样本量较大（ $n \gt 30$ ），可以使用 z 统计量。

z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{\bar{x} - 13}{2 / \sqrt{49}} > 1.645 \Rightarrow \bar{x} > 13.47

接下来计算 II 类错误概率（通常记作 $\beta$ ）。回忆一下， $\beta$ 就是在备择假设为真时错误地未拒绝原假设的概率。

\beta = P(\textrm{Fail to reject } H_0 \ \vert \ H_a \ \textrm{is true}) = P(\bar{x} \leq 13.47 \ \vert \ \mu = 14) \\\ \\ = P(Z \leq \frac{13.47 - 14}{2 / \sqrt{49}}) = P(Z \leq -1.855)

Original Explanation

To solve this question, we'll begin by computing the rejection region. Since the sample size is reasonably large ( $n \gt 30$ ) we can use the z-statistic.

z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{\bar{x} - 13}{2 / \sqrt{49}} > 1.645 \Rightarrow \bar{x} > 13.47

Next we want to calculate the probability of a type 2 error (commonly known as $\beta$ ). Recall that $\beta$ is simply the probability that we incorrectly fail to reject the null hypothesis.

\beta = P(\textrm{Fail to reject } H_0 \ \vert \ H_a \ \textrm{is true}) = P(\bar{x} \leq 13.47 \ \vert \ \mu = 14) \\\ \\ = P(Z \leq \frac{13.47 - 14}{2 / \sqrt{49}}) = P(Z \leq -1.855)

第 3 小问

题目详情

经理进行统计检验，原假设为 $H_0: \mu = 13$ ，备择假设为 $H_a: \mu = 14$ 。若 $\sigma^2=4$ ，要使 $\alpha=\beta=0.05$ ，样本量应为多少？

The manager of a large Michelin star restaurant is seeking to detect a difference equal to one dish in the average number of meals cooked by his chefs. He runs a statistical test with the null hypothesis $H_0: \mu = 13$ against his alternative hypothesis $H_a: \mu = 14$ . Assuming $\sigma^2=4$ , what sample size will ensure that $\alpha=\beta=0.05$ ?

解析

本题中，单侧上尾、显著性水平为 $\alpha$ 的检验所需样本量公式为：

n = \frac{(Z_{\alpha} + Z_\beta)^2 \cdot \sigma^2}{(\mu_0 - \mu_a )^2}

因为 $\alpha=\beta = 0.05$ ，所以 $Z_{\alpha} = Z_{\beta} = 1.645$ 。代入上式得到：

n = \frac{(1.645 + 1.645)^2 \times 4}{(13-14)^2} \approx 43.29

因此，经理需要 $n=44$ 个观测值，才能进行功效为 95% 的检验。

Original Explanation

For this question, recall the equation for calculating the sample size for an upper-tail $\alpha$ -level test is:

n = \frac{(Z_{\alpha} + Z_\beta)^2 \cdot \sigma^2}{(\mu_0 - \mu_a )^2}

Since we know that $\alpha=\beta = 0.05$ , it is also the case that $Z_{\alpha} = Z_{\beta} = 1.645$ . Plugging this into the equation above gives us:

n = \frac{(1.645 + 1.645)^2 \times 4}{(13-14)^2} \approx 43.29

The manager will need $n=44$ observations in order to conduct a test with 95% power.