题目

Problem

In the past Laxmi’s time, in minutes, for her journey to college had mean $32.5$ and standard deviation $3.1$ . After a change in her route, Laxmi wishes to test whether the mean time has decreased. She notes her journey times for a random sample of $50$ journeys and she finds that the sample mean is $31.8$ minutes. You should assume that the standard deviation is unchanged.

(a) Carry out a hypothesis test, at the $8\%$ significance level, of whether Laxmi’s mean journey time has decreased.

(5)

Later Laxmi carries out a similar test with the same hypotheses, at the $8\%$ significance level, using another random sample of size $50$ .

(b) Given that the population mean is now $31.5$ , find the probability of a Type II error.

(5)

题目中文翻译

过去，Laxmi 到学院的路上用时（以分钟为单位）的均值为 $32.5$ ，标准差为 $3.1$ 。改走另一条路线后，Laxmi 想检验平均用时是否缩短。她随机选取 $50$ 次行程并记录用时，发现样本均值为 $31.8$ 分钟。假设标准差保持不变。

(a) 在 $8\%$ 的显著性水平下进行假设检验，判断 Laxmi 的平均行程用时是否缩短。

(5)

后来，Laxmi 采用相同的假设和另一个容量为 $50$ 的随机样本，在 $8\%$ 的显著性水平下进行了类似的检验。

(b) 已知现在的总体均值为 $31.5$ ，求发生第二类错误的概率。

(5)

原卷通用作答要求

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。

解答

(a)

解法一

思路

展开

以总体平均旅程时间作左尾8%检验，样本量50可用中心极限定理，标准差按题意保持3.1。

答题过程

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Let $\mu$ be the population mean journey time after the route change, in minutes.

H_0:\mu=32.5,\qquad H_1:\mu<32.5.

Using the large-sample normal approximation,

z=\frac{31.8-32.5}{3.1/\sqrt{50}} =-1.59669\ldots.

The lower-tail 8% critical value is $\Phi^{-1}(0.08)=-1.40507\ldots$ . Since the test statistic is smaller, we reject $H_0$ .

There is sufficient evidence at the 8% significance level that Laxmi’s population mean journey time has decreased.

(b)

解法一

思路

展开

先以原假设均值32.5确定临界样本均值，再在真实均值31.5下求未拒绝的上尾概率。

答题过程

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

\begin{align*} c=&\,32.5+\Phi^{-1}(0.08)\frac{3.1}{\sqrt{50}}\\[4mm] =&\,31.8840079\ldots. \end{align*}

We reject $H_0$ when $\overline X<c$ . When the true mean is 31.5,

\overline X\approx\mathrm{N}\bigg(31.5,\frac{3.1^2}{50}\bigg).

Thus

\begin{align*} &\,\mathrm{P}(\text{Type II error})\\[4mm] \approx&\,1-\Phi\bigg(\frac{31.8840079\ldots-31.5}{3.1/\sqrt{50}}\bigg)\\[4mm] =&\,0.190537\ldots\\[4mm] =&\,\boxed{0.191}\quad\text{(3 s.f.).} \end{align*}

CIE 9709 2022 November Paper 63 Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程