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CIE 9709 2022 June Paper 61 Q7

A Level / CIE / S2

CIE 9709 2022 June Paper 61 Paper · Question 7

题目

Problem

In the past, the mean time for Jenny’s morning run was 28.228.2 minutes. She does some extra training and she wishes to test whether her mean time has been reduced. After the training Jenny takes a random sample of 4040 morning runs. She decides that if the sample mean run time is less than 2727 minutes she will conclude that the training has been effective. You may assume that, after the training, Jenny’s run time has a standard deviation of 4.04.0 minutes.

(a) State suitable null and alternative hypotheses for Jenny’s test.

(1)

(b) Find the probability that Jenny will make a Type I error.

(3)

(c) Jenny found that the sample mean run time was 27.227.2 minutes.

Explain briefly whether it is possible for her to make a Type I error or a Type II error or both.

(2)
题目中文翻译

过去,Jenny 晨跑的平均用时为 28.228.2 分钟。她做了一些额外训练,想检验自己的平均用时是否缩短。训练后,Jenny 随机抽取 4040 次晨跑。她决定,如果样本平均用时少于 2727 分钟,就认为训练有效。可以假设训练后 Jenny 的晨跑用时的标准差为 4.04.0 分钟。

(a) 写出适合 Jenny 的检验的原假设和备择假设。

(1)

(b) 求 Jenny 犯第一类错误的概率。

(3)

(c) Jenny 测得样本平均用时为 27.227.2 分钟。

简要说明她可能犯第一类错误、第二类错误,还是两类错误都有可能。

(2)
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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以训练后的总体平均跑步时间为参数,备择为减少。

答题过程

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Let μ\mu be Jenny’s population mean morning-run time after training, in minutes.

H0:μ=28.2,H1:μ<28.2.\boxed{H_0:\mu=28.2,\qquad H_1:\mu<28.2}.

(b)

解法一

思路

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第一类错误是在原假设正确时样本均值小于指定拒绝边界27。利用样本量40的正态近似计算。

答题过程

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Under H0H_0, the Central Limit Theorem gives

X‾≈N(28.2,4240).\overline X\approx\mathrm{N}\bigg(28.2,\frac{4^2}{40}\bigg).

A Type I error occurs when X‾<27\overline X<27 despite H0H_0 being true. Hence

 P(Type I error)≈ Φ(27−28.24/40)= 0.0288898…= 0.0289(3 s.f.).\begin{align*} &\,\mathrm{P}(\text{Type I error})\\[4mm] \approx&\,\Phi\bigg(\frac{27-28.2}{4/\sqrt{40}}\bigg)\\[4mm] =&\,0.0288898\ldots\\[4mm] =&\,\boxed{0.0289}\quad\text{(3 s.f.).} \end{align*}

(c)

解法一

思路

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27.2没有落入小于27的拒绝域,故没有拒绝原假设。第一类错误不可能,若真实均数已减少则可能发生第二类错误。

答题过程

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Since 27.2>2727.2>27, H0H_0 is not rejected. A Type I error is therefore impossible. A Type II error is possible if Jenny’s true population mean time has decreased but the test has failed to detect it.