题目

Problem

A scientist is investigating the masses of birds of a certain species in country $X$ and country $Y$ . She takes a random sample of 50 birds of this species from country $X$ and a random sample of 80 birds of this species from country $Y$ . She records their masses in kg, $x$ and $y$ , respectively. Her results are summarised as follows.

\begin{align*} &\,\sum x=75.5,\qquad \sum x^2=115.2,\\ &\,\sum y=116.8,\qquad \sum y^2=172.6. \end{align*}

The population mean masses of these birds in countries $X$ and $Y$ are $\mu_x$ kg and $\mu_y$ kg respectively.

Test, at the 5% significance level, the null hypothesis $\mu_x = \mu_y$ against the alternative hypothesis $\mu_x > \mu_y$ . State your conclusion in the context of the question.

[8]

题目中文翻译

一位科学家正在研究某种鸟在国家 $X$ 和国家 $Y$ 中的体重。她从国家 $X$ 随机抽取 50 只这种鸟，并从国家 $Y$ 随机抽取 80 只这种鸟。她分别记录它们的体重（单位 kg），记为 $x$ 和 $y$ 。结果汇总如下。

\begin{align*} &\,\sum x=75.5,\qquad \sum x^2=115.2,\\ &\,\sum y=116.8,\qquad \sum y^2=172.6. \end{align*}

国家 $X$ 和国家 $Y$ 中这种鸟的总体平均体重分别为 $\mu_x$ kg 和 $\mu_y$ kg。

在 5% 显著性水平下，检验原假设 $\mu_x = \mu_y$ ，备择假设为 $\mu_x > \mu_y$ 。结合题目背景说明你的结论。

(8)

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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两组样本量都较大。分别估计方差并用均值差的正态近似，按题给备择假设做右尾检验。

答题过程

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For $X$ , $\bar x=75.5/50=1.51$ and $s_X^2=(115.2-75.5^2/50)/49=0.0243877\ldots$ . For $Y$ , $\bar y=116.8/80=1.46$ and $s_Y^2=(172.6-116.8^2/80)/79=0.0262278\ldots$ . The estimated variance of the difference of sample means is $s_X^2/50+s_Y^2/80=0.000815603\ldots$ . The large-sample statistic for $H_0\colon\mu_X=\mu_Y$ against $H_1\colon\mu_X>\mu_Y$ is

z=\frac{1.51-1.46}{\sqrt{s_X^2/50+s_Y^2/80}}=1.75077\ldots.

The upper $5\%$ critical value is $1.645$ . Since $z>1.645$ , we reject $H_0$ . There is sufficient evidence at the $5\%$ level that the mean mass of birds of this species is greater in country $X$ than in country $Y$ .

CIE 9231 2022 Nov Paper 42 Q3

题目

解答

解法一

思路

答题过程