题目

Problem

A teacher at a large college gave a mathematical puzzle to all the students. The median time taken by a random sample of 24 students to complete the puzzle was 18.0 minutes. The students were then given practice in solving puzzles. Two weeks later, the students were given another mathematical puzzle of the same type as the first. The times, in minutes, taken by the random sample of 24 students to complete this puzzle are as follows.

\begin{align*} &\,18.2,\ 17.5,\ 16.4,\ 15.1,\\ &\,20.5,\ 26.5,\ 19.2,\ 23.2. \end{align*}

\begin{align*} &\,17.9,\ 18.8,\ 25.8,\ 19.9,\\ &\,17.7,\ 16.2,\ 17.3,\ 16.6. \end{align*}

\begin{align*} &\,17.1,\ 20.1,\ 20.3,\ 12.6,\\ &\,16.0,\ 21.4,\ 22.7,\ 18.4. \end{align*}

The teacher claims that the practice has not made any difference to the average time taken to complete a puzzle of this type.

Carry out a Wilcoxon signed-rank test, at the 10% significance level, to test whether there is sufficient evidence to reject the teacher’s claim.

[10]

题目中文翻译

一所大型学院的一名教师给所有学生出了一道数学谜题。随机抽取的 24 名学生完成该谜题所需时间的中位数为 18.0 分钟。随后，学生接受了解谜训练。两周后，学生又得到一道与第一道同类型的数学谜题。该随机样本中的 24 名学生完成这道谜题所需时间（单位：分钟）如下。

\begin{align*} &\,18.2,\ 17.5,\ 16.4,\ 15.1,\\ &\,20.5,\ 26.5,\ 19.2,\ 23.2. \end{align*}

\begin{align*} &\,17.9,\ 18.8,\ 25.8,\ 19.9,\\ &\,17.7,\ 16.2,\ 17.3,\ 16.6. \end{align*}

\begin{align*} &\,17.1,\ 20.1,\ 20.3,\ 12.6,\\ &\,16.0,\ 21.4,\ 22.7,\ 18.4. \end{align*}

这名教师声称，训练并没有使完成此类谜题所需的平均时间发生任何变化。

在 10% 显著性水平下进行 Wilcoxon 符号秩检验，检验是否有足够证据拒绝这名教师的声称。

(10)

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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以题给的 $18.0$ 分钟为比较基准。对练习后时间的差值排名，用较小秩和做双侧检验。

答题过程

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Using $18.0$ minutes as the reference, test $H_0\colon m=18.0$ against $H_1\colon m\ne18.0$ , where $m$ is the population median time after practice. The differences from $18.0$ , in data order, are

\begin{align*} &\,0.2,\ -0.5,\ -1.6,\ -2.9,\ 2.5,\ 8.5,\\ &\,1.2,\ 5.2,\ -0.1,\ 0.8,\ 7.8,\ 1.9,\\ &\,-0.3,\ -1.8,\ -0.7,\ -1.4,\ -0.9,\ 2.1,\\ &\,2.3,\ -5.4,\ -2.0,\ 3.4,\ 4.7,\ 0.4. \end{align*}

Ranking their absolute values gives the signed ranks, in the same order,

\begin{align*} &\,2,\ -5,\ -11,\ -18,\ 17,\ 24,\ 9,\ 21,\\ &\,-1,\ 7,\ 23,\ 13,\ -3,\ -12,\ -6,\ -10,\\ &\,-8,\ 15,\ 16,\ -22,\ -14,\ 19,\ 20,\ 4. \end{align*}

The negative rank sum is $110$ and the positive rank sum is $190$ . Under $H_0$ the sum of positive ranks has mean $24(25)/4=150$ and variance $24(25)(49)/24=1225$ . For the smaller observed rank sum, use the lower-tail continuity correction, so

z=\frac{110.5-150}{35}=-1.12857\ldots.

By symmetry, the lower critical value for the two-sided $10\%$ test is $-1.645$ . Since $z>-1.645$ , we fail to reject $H_0$ . Using the stated benchmark, there is insufficient evidence at the $10\%$ level to reject the teacher’s claim. A test of individual before-and-after changes would require the original times for each student.

CIE 9231 2022 June Paper 42 Q6

题目

解答

解法一

思路

答题过程