题目

Problem

A manager claims that the lengths of the rubber tubes that his company produces have a median of 5.50 cm. The lengths, in cm, of a random sample of 11 tubes produced by this company are as follows.

\begin{align*} &\,5.56,\ 5.45,\ 5.47,\ 5.58,\ 5.54,\ 5.52,\\ &\,5.60,\ 5.35,\ 5.59,\ 5.51,\ 5.62. \end{align*}

It is required to test at the 10% significance level the null hypothesis that the population median length is 5.50 cm against the alternative hypothesis that the population median length is not equal to 5.50 cm.

Show that both a sign test and a Wilcoxon signed-rank test give the same conclusion and state this conclusion.

[9]

题目中文翻译

一名经理声称，他的公司生产的橡胶管长度的中位数为 5.50 cm。该公司生产的 11 根橡胶管组成一个随机样本，其长度（单位 cm）如下。

\begin{align*} &\,5.56,\ 5.45,\ 5.47,\ 5.58,\ 5.54,\ 5.52,\\ &\,5.60,\ 5.35,\ 5.59,\ 5.51,\ 5.62. \end{align*}

要求在 10% 显著性水平下，检验总体长度中位数为 5.50 cm 的原假设，对立假设为总体长度中位数不等于 5.50 cm。

证明符号检验和 Wilcoxon 符号秩检验会给出相同结论，并说明该结论。

(9)

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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符号检验用低于基准的个数，符号秩检验还用绝对差的排名。因为要检验中位数是否不同，两种方法都取双侧。

答题过程

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Let $m$ be the population median tube length. Both tests use $H_0\colon m=5.50$ and $H_1\colon m\ne5.50$ .

Sign test. Three lengths are below $5.50$ and eight are above it, with no ties. Under $H_0$ the number $S$ below the benchmark has distribution $\operatorname{Binomial}(11,1/2)$ . The exact two-sided probability is

\begin{align*} &\,2\Pr(S\le3)=\frac{2}{2^{11}}\sum_{j=0}^3\binom{11}{j}\\ =&\,0.2265625>0.10. \end{align*}

So the sign test does not reject $H_0$ .

Wilcoxon signed-rank test. Subtracting $5.50$ gives the differences

\begin{align*} &\,0.06,\ -0.05,\ -0.03,\ 0.08,\ 0.04,\ 0.02,\\ &\,0.10,\ -0.15,\ 0.09,\ 0.01,\ 0.12. \end{align*}

Ranking their absolute values gives the signed ranks in observation order

\begin{align*} &\,6,\ -5,\ -3,\ 7,\ 4,\ 2,\\ &\,9,\ -11,\ 8,\ 1,\ 10. \end{align*}

The smaller rank sum is $5+3+11=19$ . The two-sided $10\%$ critical value for $11$ observations is $13$ . Since $19>13$ , the signed-rank test also does not reject $H_0$ .

Both tests give insufficient evidence at the $10\%$ level that the population median tube length differs from $5.50$ cm.

CIE 9231 2022 June Paper 43 Q5

题目

解答

解法一

思路

答题过程