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 0.4 minutes. He chooses a random sample of 10 employees. The times taken, in minutes, before and after the introduction of the new technology are recorded in the table.

Employee	A	B	C	D	E	F	G	H	I	J
Time before new technology	10.2	9.8	12.4	11.6	10.8	11.2	14.6	10.6	12.3	11.0
Time after new technology	9.6	8.5	12.4	10.9	10.2	10.6	12.8	10.8	12.5	10.6

(a) Test at the 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

一名经理正在研究引入新技术后，员工完成某项特定任务所需时间的情况。他声称，完成该任务所需的平均时间减少了超过 0.4 分钟。他随机抽取 10 名员工。引入新技术前后所需的时间（单位：分钟）记录在表中。

员工	A	B	C	D	E	F	G	H	I	J
新技术前用时	10.2	9.8	12.4	11.6	10.8	11.2	14.6	10.6	12.3	11.0
新技术后用时	9.6	8.5	12.4	10.9	10.2	10.6	12.8	10.8	12.5	10.6

(a) 在 10% 显著性水平下，检验这名经理的声称是否成立。

(7)

(b) 说明该检验有效所需的一个假设。

(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

展开

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 $0.4$ ，所以用配对差值做右尾检验。

答题过程

展开

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

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

Hence

\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=\frac{0.56-0.4}{\sqrt{s_D^2/10}}=0.799555\ldots.

With $9$ degrees of freedom, the upper $10\%$ critical value is $1.383$ . Since the statistic is smaller, we fail to reject $H_0$ . There is insufficient evidence at the $10\%$ level that the population mean time was reduced by more than $0.4$ minutes.

(b)

解法一

思路

展开

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

展开

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

CIE 9231 2022 June Paper 42 Q1

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程