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IAL 2024 June S3 Q5

A Level / Edexcel / S3

IAL 2024 June Paper · Question 5

题目

Problem

A manager of a large company is investigating the time it takes the company’s employees to complete a task.

The manager believes that the mean time for full-time employees to complete the task is more than a minute quicker than the mean time for part-time employees to complete the task.

The manager collects a random sample of 605 full-time employees and 45 part-time employees and records the times, t minutes, it takes each employee to complete the task. The results are summarised in the table below.

nnt‾\overline ts2s^2
Full-time employees6055.69
Part-time employees457.04

(a) Test, at the 5% level of significance, the manager’s claim. You should state your hypotheses, test statistic, critical value and conclusion clearly.

(8)

(b) State two assumptions you have made in carrying out the test in part (a)

(2)

The company increases the size of the sample of part-time employees to 46 The time taken to complete the task by the extra employee is 8 minutes.

(c) Find an unbiased estimate of the variance for the sample of 46 part-time employees.

(4)
(Total for Question 5 is 14 marks)
题目中文翻译

一家大公司的经理正在调查公司员工完成一项任务所需的时间。

经理认为，全职员工完成该任务的平均时间比兼职员工完成该任务的平均时间快超过 1 分钟。

经理随机抽取了 605 名全职员工和 45 名兼职员工，并记录每位员工完成该任务所用的时间 t 分钟。 结果汇总如下表。

nnt‾\overline ts2s^2
全职员工6055.69
兼职员工457.04

(a) 在 5% 显著性水平下检验经理的说法。 你应清楚写出原假设、检验统计量、临界值和结论。

(b) 写出在完成 (a) 小题的检验时所作的两个假设。

公司将兼职员工的样本量增加到 46。 新增的那位员工完成任务所用时间为 8 分钟。

(c) 求 46 名兼职员工样本的无偏方差估计。

解答

(a)

解法一

思路

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“全职员工快超过 1 分钟”可写成兼职总体均值减全职总体均值大于 1。以两组样本方差估计总体方差，建立右尾两样本均值差检验，再将 zz 统计量与 5%5\% 单尾临界值比较。

答题过程

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Let μp\mu_p and μf\mu_f be the population mean completion times for part-time and full-time employees respectively. The hypotheses are

H0: μp−μf=1,H1: μp−μf>1.\begin{aligned} H_0:&\ \mu_p-\mu_f=1,\\ H_1:&\ \mu_p-\mu_f>1. \end{aligned}

The estimated standard error is

SE⁡(T‾p−T‾f)= 445+9605= 0.3221….\begin{align*} \operatorname{SE}(\overline T_p-\overline T_f) =&\,\sqrt{\frac{4}{45}+\frac{9}{605}}\\ =&\,0.3221\ldots. \end{align*}

Therefore, the test statistic is

z= (7.0−5.6)−1445+9605= 1.2417…= 1.24.\begin{align*} z =&\,\frac{(7.0-5.6)-1} {\sqrt{\frac{4}{45}+\frac{9}{605}}}\\ =&\,1.2417\ldots\\ =&\,1.24. \end{align*}

For a one-tailed test at the 5%5\% significance level, the critical value is 1.64491.6449, giving the rejection region

z⩾1.6449.z\geqslant1.6449.

Since 1.24<1.64491.24<1.6449, H0H_0 is not rejected. There is insufficient evidence at the 5%5\% significance level that full-time employees are more than one minute quicker than part-time employees, so the manager’s claim is not supported.

(b)

解法一

思路

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检验中依靠两组样本均值的正态近似，并以样本方差代替对应总体方差。因此分别写出这两个针对两组员工的假设。

答题过程

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The assumptions are:

  1. The completion times for both full-time and part-time employees are normally distributed, or both samples are sufficiently large for the central limit theorem to apply.
  2. For both groups, the sample variance is equal to the corresponding population variance.

(c)

解法一

思路

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先由原样本的均值和无偏方差恢复 ∑t\sum t 与 ∑t2\sum t^2。加入新的观测值 8 后更新这两个总和，再用样本量 46 的无偏方差公式计算。

答题过程

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For the original sample of 45 part-time employees,

∑t=45(7)=315.\sum t=45(7)=315.

Using

s2=∑t2−nt‾,2n−1,s^2=\frac{\sum t^2-n\overline t^{,2}}{n-1},

the original sum of squares is

∑t2= (45−1)(4)+45(72)= 2381.\begin{align*} \sum t^2 =&\,(45-1)(4)+45(7^2)\\ =&\,2381. \end{align*}

After including the extra employee,

∑t= 315+8=323,∑t2= 2381+82=2445,\begin{align*} \sum t=&\,315+8=323,\\ \sum t^2=&\,2381+8^2=2445, \end{align*}

so the new sample mean is

t‾=32346=7.0217….\overline t=\frac{323}{46}=7.0217\ldots.

Therefore, the new unbiased estimate of the variance is

s2= 2445−46(7.0217…)246−1= 3.9328…= 3.93(3 s.f.).\begin{align*} s^2 =&\,\frac{2445-46(7.0217\ldots)^2}{46-1}\\ =&\,3.9328\ldots\\ =&\,\boxed{3.93}\quad\text{(3 s.f.)}. \end{align*}