题目

Problem

A dog breeder claims that the mean weight of male Great Dane dogs is 20 kg more than the mean weight of female Great Dane dogs.

Tammy believes that the mean weight of male Great Dane dogs is more than 20 kg more than the mean weight of female Great Dane dogs.

She takes random samples of 50 male and 50 female Great Dane dogs and records their weights.

The results are summarised below, where x denotes the weight, in kg, of a male Great Dane dog and y denotes the weight, in kg, of a female Great Dane dog.

∑x = 3610 ∑x² = 260 955.6 ∑y = 2585 ∑y² = 133 757.2

(a) Find unbiased estimates for the mean and variance of the weights of

(i) the male Great Dane dogs,

(ii) the female Great Dane dogs.

(5)

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

(7)

(c) For the test in part (b), state whether or not it is necessary to assume that the weights of the Great Dane dogs are normally distributed. Give a reason for your answer.

(2)

(d) State an assumption you have made in carrying out the test in part (b).

(1)

(Total 15 marks)

题目中文翻译

一名狗繁育者声称，雄性大丹犬的平均体重比雌性大丹犬的平均体重多 20 kg。

Tammy 认为，雄性大丹犬的平均体重比雌性大丹犬的平均体重多于 20 kg。

她随机抽取了 50 只雄性大丹犬和 50 只雌性大丹犬，并记录它们的体重。

结果汇总如下，其中 x 表示雄性大丹犬的体重（单位：kg），y 表示雌性大丹犬的体重（单位：kg）。

∑x = 3610 ∑x² = 260 955.6 ∑y = 2585 ∑y² = 133 757.2

(a) 求以下体重的均值和方差的无偏估计：

(i) 雄性大丹犬，

(ii) 雌性大丹犬。

(b) 清楚写出假设，进行一个适当的检验来评估 Tammy 的看法。使用 5% 显著性水平，并写出临界值。

(d) 写出你在进行 (b) 中检验时所作的一个假设。

解答

(a)(i)

解法一

思路

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雄性样本量为 $50$ 。样本均值是总体均值的无偏估计；总体方差的无偏估计用分母 $n-1=49$ ，并由 $\sum x^2-n\bar{x}^2$ 求离均差平方和。

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For the male dogs,

\bar{x}=\frac{3610}{50}=\boxed{72.2}\text{ kg}.

The unbiased estimate of the variance is

\begin{align*} s_x^2 =&\,\frac{\sum x^2-n\bar{x}^2}{n-1}\\ =&\,\frac{260955.6-50(72.2)^2}{49}\\ =&\,\boxed{6.4}\text{ kg}^2. \end{align*}

(a)(ii)

解法一

思路

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对雌性样本重复同一无偏估计过程，样本量同样为 $50$ 。

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For the female dogs,

\bar{y}=\frac{2585}{50}=\boxed{51.7}\text{ kg}.

The unbiased estimate of the variance is

\begin{align*} s_y^2 =&\,\frac{\sum y^2-n\bar{y}^2}{n-1}\\ =&\,\frac{133757.2-50(51.7)^2}{49}\\ =&\,\boxed{2.3}\text{ kg}^2. \end{align*}

(b)

解法一

思路

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Tammy 认为雄性总体均值比雌性总体均值多出的部分大于 $20$ kg，因此进行右尾检验：原假设取差为 $20$ ，备择假设取差大于 $20$ 。两个样本都很大，可用样本方差估计总体方差并作标准正态检验。

答题过程

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Let $\mu_x$ and $\mu_y$ be the population mean weights of male and female Great Dane dogs respectively.

\begin{aligned} H_0:&\ \mu_x-\mu_y=20,\\ H_1:&\ \mu_x-\mu_y>20. \end{aligned}

Under $H_0$ , the estimated standard error is

\begin{align*} \operatorname{SE} =&\,\sqrt{\frac{s_x^2}{50}+\frac{s_y^2}{50}}\\ =&\,\sqrt{\frac{6.4}{50}+\frac{2.3}{50}}\\ =&\,0.41713\ldots \end{align*}

The test statistic is

\begin{align*} Z =&\,\frac{\bar{x}-\bar{y}-20} {\operatorname{SE}}\\ =&\,\frac{72.2-51.7-20}{0.41713\ldots}\\ =&\,1.1986\ldots\\ \approx&\,1.20. \end{align*}

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

z=1.6449.

Since

1.1986\ldots<1.6449,

the test statistic does not lie in the critical region, so $H_0$ is not rejected. There is insufficient evidence at the $5\%$ significance level to support Tammy’s belief that the mean weight of male Great Dane dogs is more than $20$ kg greater than that of female Great Dane dogs.

(c)

解法一

思路

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两个样本量都是 $50$ ，属于大样本。中心极限定理使两个样本均值近似服从正态分布，因此不需要假设原始体重本身服从正态分布。

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It is not necessary to assume that the dog weights are normally distributed. Both samples are large, so by the Central Limit Theorem the two sample means are approximately normally distributed.

(d)

解法一

思路

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(b) 用两个样本方差代替了未知的总体方差，因此所作假设是样本方差可视为相应总体方差的值。

答题过程

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It is assumed that the sample variances are equal to the corresponding population variances, so that

\begin{aligned} \sigma_x^2=&\,s_x^2,\\ \sigma_y^2=&\,s_y^2. \end{aligned}

IAL 2022 Jan S3 UNUSED Q5

题目

解答

(a)(i)

解法一

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(a)(ii)

解法一

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(b)

解法一

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(c)

解法一

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(d)

解法一

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