题目

Problem

A machine makes screws with a mean length of 30 mm and a standard deviation of 2.5 mm.

A manager claims that, following some repairs, the machine is now making screws with a mean length of less than 30 mm. The manager takes a random sample of 80 screws and finds that they have a mean length of 29.5 mm.

Use a suitable test, at the 5% level of significance, to determine whether there is evidence to support the manager’s claim. State your hypotheses clearly.

(5)

(Total 5 marks)

题目中文翻译

一台机器制造螺钉，其平均长度为 30 mm，标准差为 2.5 mm。

一位经理声称，在经过一些维修后，这台机器现在制造的螺钉平均长度小于 30 mm。经理随机抽取了 80 个螺钉，发现它们的平均长度为 29.5 mm。

使用合适的检验，在 5% 显著性水平下判断是否有证据支持经理的说法。清楚写出假设。

(5)

解答

解法一

思路

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经理声称维修后的总体平均长度小于 $30$ mm，因此进行左尾检验。总体标准差已知为 $2.5$ mm，且样本量为 $80$ ，所以使用单样本均值的标准正态检验。

答题过程

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Let $\mu$ be the population mean length, in mm, of screws made by the machine after the repairs.

\begin{align*} H_0:&\ \mu=30,\\ H_1:&\ \mu<30. \end{align*}

Under $H_0$ , the test statistic is

\begin{align*} z =&\,\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\\ =&\,\frac{29.5-30}{2.5/\sqrt{80}}\\ =&\,-1.7888\ldots \end{align*}

For a one-tailed test at the $5\%$ significance level, the critical value is $-1.6449$ .

Since

-1.7888\ldots<-1.6449,

the test statistic lies in the critical region, so $H_0$ is rejected. There is sufficient evidence at the $5\%$ significance level to support the manager’s claim that the mean length of the screws is less than $30$ mm.

Equivalently, the critical region for the sample mean is

\begin{align*} \bar X<&\,30-1.6449\frac{2.5}{\sqrt{80}}\\ =&\,29.540236\ldots. \end{align*}

The observed mean $29.5$ is in this region. The corresponding one-tailed $p$ -value is $0.036819\ldots<0.05$ , giving the same conclusion.

S3 2021 Oct Q1

题目

解答

解法一

思路

答题过程