题目

Problem

The time, $T$ minutes, for a certain daily bus journey is normally distributed. The bus company claims that the mean of $T$ is $45$ . A passenger believes that the mean of $T$ is actually greater than $45$ . She notes the times taken for this journey on a random sample of $60$ days. The results are summarised below.

n=60\qquad \sum t=2750\qquad \sum t^2=127\,000

(a) Calculate unbiased estimates of the population mean and variance.

(3)

(b) Test the passenger’s belief at the $5\%$ significance level.

(5)

题目中文翻译

某条每天运行的公交线路的行程用时为 $T$ 分钟，服从正态分布。公交公司声称 $T$ 的均值为 $45$ 。一名乘客认为 $T$ 的均值实际上大于 $45$ 。她随机选取 $60$ 天，记录这些天的行程用时。结果汇总如下。

n=60\qquad \sum t=2750\qquad \sum t^2=127\,000

(a) 计算总体均值和总体方差的无偏估计值。

(3)

(b) 在 $5\%$ 的显著性水平下检验该乘客的看法。

(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

(a)

解法一

思路

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用汇总值计算样本均值与无偏方差，保留精确分数供下一问检验使用。

答题过程

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An unbiased estimate of the population mean is

\begin{align*} \overline t=&\,\frac{2750}{60}=\frac{275}{6}\\[4mm] \approx&\,\boxed{45.8\text{ minutes}}. \end{align*}

An unbiased estimate of the population variance is

\begin{align*} s^2=&\,\frac1{59}\bigg(127000-\frac{2750^2}{60}\bigg)\\[4mm] =&\,\frac1{59}\bigg(127000-\frac{378125}{3}\bigg)\\[4mm] =&\,\frac{2875}{177}\approx\boxed{16.2\text{ minutes}^2}. \end{align*}

(b)

解法一

思路

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乘客怀疑均值变大，所以是右尾检验。使用(a)未舍入的均值和无偏方差，避免把接近临界值的统计量算偏。

答题过程

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Let $\mu$ be the population mean journey time. Test

H_0:\mu=45,\qquad H_1:\mu>45.

For the large random sample, use the normal distribution of the sample mean with the population variance estimated by $s^2$ from (a). The test statistic is

z=\frac{275/6-45}{\sqrt{(2875/177)/60}}\approx1.602.

The critical value for a $5\%$ upper-tail test is $1.645$ . Since $1.602<1.645$ , do not reject $H_0$ .

There is insufficient evidence at the $5\%$ significance level that the population mean journey time exceeds $45$ minutes, so the passenger’s belief is not supported.

CIE 9709 2025 June Paper 61 Q3

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程