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CIE 9231 2024 June Paper 41 Q6

A Level / CIE / FS

CIE 9231 2024 June Paper 41 Paper · Question 6

题目

Problem

Jade is a swimming instructor at a sports college. She claims that, as a result of an intensive training course, the mean time taken by students to swim 50 metres has reduced by more than 1 second. She chooses a random sample of 10 students. The times taken, in seconds, before and after the training course are recorded in the table.

StudentABCDEFGHIJ
Time before course54.247.452.159.055.351.048.952.258.451.4
Time after course50.146.352.558.851.448.449.548.758.351.4

(a) Test, at the 10% significance level, whether Jade’s claim is justified.

[7]

(b) State an assumption that is necessary for this test to be valid.

[1]
题目中文翻译

Jade 是一所体育学院的游泳教练。她声称,经过一次强化训练课程后,学生游 50 米所用的平均时间减少了超过 1 秒。她随机抽取 10 名学生。训练课程前后所用的时间如下表,单位为秒。

学生ABCDEFGHIJ
训练前时间54.247.452.159.055.351.048.952.258.451.4
训练后时间50.146.352.558.851.448.449.548.758.351.4

(a) 在 10% 显著性水平下,检验 Jade 的声称是否合理。

(7)

(b) 说明使该检验有效所需要的一个假设。

(1)
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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取每位学生训练前减训练后的时间。检验平均减少是否超过 11 秒,所以分子是样本平均减少量减 11。

答题过程

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Let μD\mu_D be the population mean reduction in swim time. Test H0 ⁣:μD=1H_0\colon\mu_D=1 against H1 ⁣:μD>1H_1\colon\mu_D>1. The paired reductions are 4.1,1.1,−0.4,0.2,3.9,2.6,−0.6,3.5,0.1,04.1,1.1,-0.4,0.2,3.9,2.6,-0.6,3.5,0.1,0, giving

 dˉ=14.5/10=1.45, sD2=52.81−14.52/109=3.53166….\begin{align*} &\,\bar d=14.5/10=1.45,\\ &\,s_D^2=\frac{52.81-14.5^2/10}{9}=3.53166\ldots. \end{align*}

The test statistic is

t=1.45−1sD2/10=0.75722….t=\frac{1.45-1}{\sqrt{s_D^2/10}}=0.75722\ldots.

With 99 degrees of freedom, the upper 10%10\% critical value is 1.3831.383. Since the statistic is smaller, we fail to reject H0H_0. There is insufficient evidence at the 10%10\% level that the population mean swimming time was reduced by more than one second.

(b)

解法一

思路

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配对 tt 检验的正态假设针对训练前后时间之差。

答题过程

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The population distribution of the paired differences in swimming times is assumed to be normal.