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IAL 2021 Oct S3 Q1

A Level / Edexcel / S3

IAL 2021 Oct Paper · Question 1

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。

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 3030 mm，因此进行左尾检验。总体标准差已知为 2.52.5 mm，且样本量为 8080，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 μ\mu be the population mean length, in mm, of screws made by the machine after the repairs.

H0: μ=30,H1: μ<30.\begin{aligned} H_0:&\ \mu=30,\\ H_1:&\ \mu<30. \end{aligned}

Under H0H_0, the test statistic is

z= xˉ−μσ/n= 29.5−302.5/80= −1.7888…\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%5\% significance level, the critical value is −1.6449-1.6449.

Since

−1.7888…<−1.6449,-1.7888\ldots<-1.6449,

the test statistic lies in the critical region, so H0H_0 is rejected. There is sufficient evidence at the 5%5\% significance level to support the manager’s claim that the mean length of the screws is less than 3030 mm.