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CIE 9231 2022 June Paper 41 Q1

A Level / CIE / FS

CIE 9231 2022 June Paper 41 Paper · Question 1

题目

Problem

A manager is investigating the times taken by employees to complete a particular task as a result of the introduction of new technology. He claims that the mean time taken to complete the task is reduced by more than 0.40.4 minutes. He chooses a random sample of 1010 employees. The times taken, in minutes, before and after the introduction of the new technology are recorded in the table.

EmployeeAABBCCDDEEFFGGHHIIJJ
Time before new technology10.29.812.411.610.811.214.610.612.311.0
Time after new technology9.68.512.410.910.210.612.810.812.510.6

(a) Test at the 10%10\% significance level whether the manager’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

一位经理正在调查引入新技术后,员工完成某项任务所需的时间。他声称,完成该任务的平均时间减少了超过 0.40.4 分钟。他随机抽取 1010 名员工,记录引入新技术前后每人完成任务所需的时间,单位为分钟,结果如下表。

员工AABBCCDDEEFFGGHHIIJJ
引入新技术前的时间10.29.812.411.610.811.214.610.612.311.0
引入新技术后的时间9.68.512.410.910.210.612.810.812.510.6

(a) 在 10%10\% 的显著性水平下,检验经理的声称是否有依据。

(7)

(b) 写出使这一检验有效所必需的一项假设。

(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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取同一员工引入技术前后的时间差。检验平均减少量是否超过 0.40.4,所以用配对差值做右尾检验。

答题过程

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Let μD\mu_D be the population mean of before-minus-after times. Test H0 ⁣:μD=0.4H_0\colon\mu_D=0.4 against H1 ⁣:μD>0.4H_1\colon\mu_D>0.4. The differences are

 0.6, 1.3, 0, 0.7, 0.6, 0.6, 1.8, −0.2, −0.2, 0.4.\begin{align*} &\,0.6,\ 1.3,\ 0,\ 0.7,\ 0.6,\\ &\,0.6,\ 1.8,\ -0.2,\ -0.2,\ 0.4. \end{align*}

Hence

 dˉ=5.6/10=0.56, sD2=6.74−5.62/109=0.400444….\begin{align*} &\,\bar d=5.6/10=0.56,\\ &\,s_D^2=\frac{6.74-5.6^2/10}{9}=0.400444\ldots. \end{align*}

The statistic is

t=0.56−0.4sD2/10=0.799555….t=\frac{0.56-0.4}{\sqrt{s_D^2/10}}=0.799555\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 time was reduced by more than 0.40.4 minutes.

(b)

解法一

思路

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配对检验把每对观察化成一个差值,正态性假设也针对差值。

答题过程

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The population distribution of paired differences in task completion times is assumed to be normal.