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CIE 9231 2025 Nov Paper 41 Q4

A Level / CIE / FS

CIE 9231 2025 Nov Paper 41 Paper · Question 4

题目

Problem

A researcher believes that the median mm of a population has changed from its known previous value m0m_0. The researcher collects a random sample of size 28. She ranks the data and calculates a test statistic TT using the Wilcoxon signed-rank test. The conclusion of the test carried out at a 1% significance level is that there is not sufficient evidence to support her belief.

Using a normal approximation, find the least possible value of TT.

[5]
题目中文翻译

一名研究者认为某总体的中位数 mm 已经从其已知的先前值 m0m_0 发生了改变。该研究者收集了一个容量为 28 的随机样本。她对数据进行排序,并使用 Wilcoxon 符号秩检验计算检验统计量 TT。在 1% 显著性水平下进行检验后的结论是,没有足够证据支持她的看法。

使用正态近似,求 TT 的最小可能值。

(5)
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

解法一

思路

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备择假设是中位数发生变化,所以显著性水平平分到两尾。较小秩和的临界点用下尾,并加 0.50.5 作连续性修正。

答题过程

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Under the null hypothesis, the sum of the positive ranks has mean and variance

 28(29)4=203, 28(29)(57)24=1928.5.\begin{align*} &\,\frac{28(29)}4=203,\\ &\,\frac{28(29)(57)}{24}=1928.5. \end{align*}

The test statistic TT is the smaller of the positive and negative rank sums. By symmetry, a lower-tail probability for one signed sum is compared with 0.0050.005 for this two-sided test. To be outside the rejection region, the corrected statistic must satisfy

T+0.5−2031928.5>−2.576.\frac{T+0.5-203}{\sqrt{1928.5}}>-2.576.

This gives T>89.375…T>89.375\ldots. Since TT is an integer, the least possible value is 9090.