题目

Problem

A biologist claims that the mean weight of grey squirrels is 200 grams more than the mean weight of red squirrels.

To test this, the research student collects and weighs a random sample of 80 grey squirrels and 80 red squirrels. The results are summarised below, where

x denotes the weights, in grams, of the grey squirrels
y denotes the weights, in grams, of the red squirrels

x̄ = 500

s_x^2 = 1156

∑y = 23 200

∑y^2 = 6 741 351

(a) Find an unbiased estimate of the mean and an unbiased estimate of the variance for the weights of the red squirrels.

(3)

(b) Stating your hypotheses clearly, carry out a suitable test to assess the research student’s belief. Use a 5% level of significance and state your critical value.

(7)

(1)

(Total for Question 4 is 11 marks)

题目中文翻译

一位生物学家声称，灰松鼠的平均体重比红松鼠的平均体重多 200 克。

为检验这一点，研究学生随机测量了 80 只灰松鼠和 80 只红松鼠的重量。结果汇总如下，其中：

x 表示灰松鼠的重量（单位：克）
y 表示红松鼠的重量（单位：克）

x̄ = 500

s_x^2 = 1156

∑y = 23 200

∑y^2 = 6 741 351

(a) 求红松鼠体重的均值无偏估计和方差无偏估计。

(b) 清楚写出假设，进行一个适当的检验来检验研究学生的看法。使用 5% 的显著性水平，并写出临界值。

解答

(a)

解法一

思路

展开

红松鼠样本量为 80。用总重量除以样本量求样本均值，再以分母 $n-1$ 的公式计算总体方差的无偏估计。

答题过程

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For the red squirrels, the unbiased estimate of the mean is

\begin{align*} \overline y =&\,\frac{\sum y}{80}\\ =&\,\frac{23200}{80}\\ =&\,\boxed{290}. \end{align*}

The unbiased estimate of the variance is

\begin{align*} s_y^2 =&\,\frac{\sum y^2-80\overline y^{,2}}{80-1}\\ =&\,\frac{6741351-80(290)^2}{79}\\ =&\,\boxed{169}. \end{align*}

(b)

解法一

思路

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题目要检验灰松鼠总体均值是否比红松鼠总体均值高出超过 200 克，因此使用右尾两样本均值差检验。两组样本量均为 80，以样本方差估计总体方差后计算标准误与 $z$ 统计量。

答题过程

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Let $\mu_x$ and $\mu_y$ be the population mean weights of grey and red squirrels respectively. The hypotheses are

\begin{aligned} H_0:&\ \mu_x-\mu_y=200,\\ H_1:&\ \mu_x-\mu_y>200. \end{aligned}

The estimated standard error of the difference in sample means is

\begin{align*} \operatorname{SE}(\overline X-\overline Y) =&\,\sqrt{\frac{1156}{80}+\frac{169}{80}}\\ =&\,4.0697\ldots. \end{align*}

Therefore, the test statistic is

\begin{align*} z =&\,\frac{(500-290)-200} {\sqrt{\frac{1156}{80}+\frac{169}{80}}}\\ =&\,2.4571\ldots\\ =&\,2.46. \end{align*}

For a one-tailed test at the $5\%$ significance level, the critical value is $1.6449$ , giving the rejection region

z\geqslant1.6449.

Since $2.46>1.6449$ , $H_0$ is rejected. There is sufficient evidence at the $5\%$ significance level to support the belief that the mean weight of grey squirrels is more than 200 grams greater than the mean weight of red squirrels.

(c)

解法一

思路

展开

检验中用两组样本方差代替了相应的总体方差，因此需要假设两组样本方差分别等于对应的总体方差。

答题过程

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For both types of squirrel, assume that the sample variance is equal to the corresponding population variance.

IAL 2025 Jun S3 Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程