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CIE 9231 2025 Nov Paper 43 Q6

A Level / CIE / FS

CIE 9231 2025 Nov Paper 43 Paper · Question 6

题目

Problem

Nine athletes in a club have a new coach. The coach adopts a new training programme which he believes will reduce the race times of these athletes. Each athlete completes a 1500 m1500\text{ m} time trial before and after completing the new training programme. Their times, in seconds (s), are recorded.

AthleteABCDEFGHI
Time before training (s)250251252267276291310320335
Time after training (s)245251253261275293302313320

(a) Carry out a paired t-test at the 5%5\% significance level to test the coach’s belief.

[7]

Further research suggests that the effects of the training programme tend to reduce the times of the slower athletes by more than those of the faster athletes.

(b) Suggest a reason why the paired t-test used in part (a) may not have been an appropriate test in this case.

[1]

(c) Suggest a suitable alternative test that could have been used instead of a paired t-test.

[1]
题目中文翻译

某俱乐部的 9 名运动员有了一位新教练。该教练采用了一项新的训练计划,并认为这会减少这些运动员的比赛用时。每名运动员在完成新训练计划前后各完成一次 1500 m1500\text{ m} 计时测试。他们的用时(单位为秒,s)记录如下。

运动员ABCDEFGHI
训练前用时(秒)250251252267276291310320335
训练后用时(秒)245251253261275293302313320

(a) 在 5%5\% 显著性水平下进行 paired t-test,检验教练的看法。

(7)

进一步研究表明,该训练计划的效果往往使较慢运动员的用时减少得比较快运动员更多。

(b) 说明为什么 (a) 中使用的 paired t-test 在此情境中可能并不合适。

(1)

(c) 提出一个可代替 paired t-test 使用的合适检验。

(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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同一位运动员的训练前后成绩组成一对。取训练前减训练后,成绩改善对应正差值。

答题过程

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Let μD\mu_D be the population mean of the paired differences, defined as before minus after. Test H0 ⁣:μD=0H_0\colon\mu_D=0 against H1 ⁣:μD>0H_1\colon\mu_D>0.

The differences are 5,0,−1,6,1,−2,8,7,155,0,-1,6,1,-2,8,7,15, with sum 3939 and sum of squares 405405. Thus

 dˉ=399, sD2=405−392/98=29.5, t=39/929.5/9=2.39349….\begin{align*} &\,\bar d=\frac{39}{9},\\ &\,s_D^2=\frac{405-39^2/9}{8}=29.5,\\ &\,t=\frac{39/9}{\sqrt{29.5/9}}=2.39349\ldots. \end{align*}

For 88 degrees of freedom the upper 5%5\% critical value is 1.8601.860. Since 2.39349…>1.8602.39349\ldots>1.860, we reject H0H_0. There is sufficient evidence at the 5%5\% level that the training programme reduces mean race times.

(b)

解法一

思路

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检验要求配对差值的总体近似正态。较慢运动员的改善幅度更大,可能使差值的分布偏斜。

答题过程

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Larger reductions among slower athletes could produce an asymmetric distribution of before-minus-after differences. The population distribution of the differences may therefore fail the normality assumption of the paired tt-test.

(c)

解法一

思路

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如果差值分布可能偏斜,可只检验配对差值的正负。这样无需对差值分布作正态或对称假设。

答题过程

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A one-sided paired sign test is suitable. Omit any zero differences and test whether a positive difference occurs more often than a negative one.