题目

Problem

A biologist wishes to test whether the mean concentration $\mu$ , in suitable units, of a certain pollutant in a river is below the permitted level of $0.5$ . She measures the concentration, $x$ , of the pollutant at $50$ randomly chosen locations in the river. The results are summarised below.

$n=50\qquad\sum x=23.0\qquad\sum x^2=13.02$

(a) Carry out a test at the $5\%$ significance level of the null hypothesis $\mu=0.5$ against the alternative hypothesis $\mu<0.5$ .

(7)

Later, a similar test is carried out at the $5\%$ significance level using another sample of size $50$ and the same hypotheses as before. You should assume that the standard deviation is unchanged.

(b) Given that, in fact, the value of $\mu$ is $0.4$ , find the probability of a Type II error.

(5)

题目中文翻译

一名生物学家想检验某条河中一种污染物的平均浓度 $\mu$ （以适当的单位计）是否低于许可限值 $0.5$ 。她在河中随机选取 $50$ 处地点，测量该污染物的浓度 $x$ 。结果汇总如下。

$n=50\qquad\sum x=23.0\qquad\sum x^2=13.02$

(a) 在 $5\%$ 的显著性水平下检验原假设 $\mu=0.5$ ，备择假设为 $\mu<0.5$ 。

(7)

后来，另取一个容量为 $50$ 的样本，采用与之前相同的假设，在 $5\%$ 的显著性水平下进行了一次类似的检验。假设标准差保持不变。

(b) 已知 $\mu$ 的实际值为 $0.4$ ，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

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先求样本均值与无偏方差，利用中心极限定理作左尾正态检验，并给出非确定性的情境结论。

答题过程

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The hypotheses are

H_0:\mu=0.5,\qquad H_1:\mu<0.5.

The sample estimates are

\bar x=\frac{23}{50}=0.46,

and

\begin{align*} s^2=&\,\frac{13.02-23^2/50}{49}=\frac{61}{1225}\\[4mm] =&\,0.049795918\ldots. \end{align*}

By the Central Limit Theorem, the sample mean is approximately normal. The test statistic is

z=\frac{0.46-0.5}{\sqrt{(61/1225)/50}} =-1.26750\ldots.

The lower-tail 5% critical value is $-1.645$ . Since $-1.26750\ldots>-1.645$ , we do not reject $H_0$ .

There is insufficient evidence at the 5% significance level that the population mean pollutant concentration is below 0.5.

(b)

解法一

思路

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用第一问未舍入方差估计不变的总体方差。先求原假设下的临界样本均值，再以真实均值0.4计算未拒绝概率。

答题过程

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Using the same estimated standard deviation, the standard error is

s_{\overline X}=\sqrt{\frac{61/1225}{50}}=0.03155817\ldots.

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

c=0.5-1.645s_{\overline X}=0.4480868\ldots.

We reject $H_0$ when $\overline X<c$ . If the true mean is 0.4, a Type II error occurs when $\overline X\geq c$ , giving

\begin{align*} &\,\mathrm{P}(\text{Type II error})\\[4mm] \approx&\,1-\Phi\bigg(\frac{0.4480868\ldots-0.4}{0.03155817\ldots}\bigg)\\[4mm] =&\,1-\Phi(1.523751\ldots)\\[4mm] =&\,\boxed{0.0638}\quad\text{(3 s.f.).} \end{align*}

CIE 9709 2023 November Paper 61 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