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CIE 9231 2023 June Paper 41 Q4

A Level / CIE / FS

CIE 9231 2023 June Paper 41 Paper · Question 4

题目

Problem

A random sample of 13 technology companies is chosen and the numbers of employees in 2018 and in 2022 are recorded.

CompanyABCDEFGHIJKLM
Number in 2018104191262349705143514942912863041104
Number in 20221062412722810125253215644924782941154

A researcher claims that there has been an increase in the median number of employees at technology companies between 2018 and 2022.

(a) Carry out a Wilcoxon matched-pairs signed-rank test, at the 5% significance level, to test whether the data supports this claim.

[7]

The researcher notices that the figures for company G have been recorded incorrectly. In fact, the number of employees in 2018 was 32 and the number of employees in 2022 was 35.

(b) Explain, with numerical justification, whether or not the conclusion of the test in part (a) remains the same.

[2]
题目中文翻译

随机抽取 13 家科技公司,并记录它们在 2018 年和 2022 年的员工人数。

公司ABCDEFGHIJKLM
2018 年人数104191262349705143514942912863041104
2022 年人数1062412722810125253215644924782941154

一名研究者声称,2018 年到 2022 年间,科技公司的员工人数中位数有所增加。

(a) 在 5% 显著性水平下进行 Wilcoxon 配对符号秩检验,检验这些数据是否支持该声称。

(7)

研究者注意到,公司 G 的数据记录有误。事实上,2018 年的员工人数是 32,2022 年的员工人数是 35。

(b) 结合数值说明,解释 (a) 中检验的结论是否保持不变。

(2)
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

按公司配对,用 2022 年员工数减去 2018 年员工数。将绝对差排名后,计算负秩和以检验增加。

答题过程

展开

Let mDm_D be the population median of the within-company changes, defined as 20222022 minus 20182018. Test H0 ⁣:mD=0H_0\colon m_D=0 against H1 ⁣:mD>0H_1\colon m_D>0. For A through G the changes and signed ranks, respectively, are

 2, 5, 1, −6, 42, 11, −3, 2, 4, 1, −5, 12, 9, −3.\begin{align*} &\,2,\ 5,\ 1,\ -6,\ 42,\ 11,\ -3,\\ &\,2,\ 4,\ 1,\ -5,\ 12,\ 9,\ -3. \end{align*}

For H through M they are

 7, 20, 12, −8, −10, 50, 6, 11, 10, −7, −8, 13.\begin{align*} &\,7,\ 20,\ 12,\ -8,\ -10,\ 50,\\ &\,6,\ 11,\ 10,\ -7,\ -8,\ 13. \end{align*}

The negative rank sum is W−=5+3+7+8=23W_-=5+3+7+8=23. The lower one-sided 5%5\% critical value for 1313 pairs is 2121. Since 23>2123>21, we fail to reject H0H_0. There is insufficient evidence at the 5%5\% level of a positive median change in employee numbers.

(b)

解法一

思路

展开

更正只使 G 的差值从负变正,绝对差不变。把 G 的秩从负秩和移走,再与临界值比较。

答题过程

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For G the change becomes 35−32=335-32=3 instead of −3-3. Its absolute rank is still 33. The negative rank sum becomes 23−3=2023-3=20. Since 20≤2120\le21, the corrected data lead to rejection of H0H_0, so the conclusion changes. There is now sufficient evidence at the 5%5\% level of a positive median change.