题目

Problem

A greengrocer is investigating the weights of two types of orange, type A and type B. She believes that on average type A oranges weigh greater than 5 grams more than type B oranges. She collects a random sample of 40 type A oranges and 32 type B oranges and records the weight, $x$ grams, of each orange.

The table shows a summary of her data.

	$n$	$\bar{x}$	$\sum x^2$
Type A oranges	40	140.4	790 258
Type B oranges	32	134.7	581 430

(a) Calculate unbiased estimates for the variance of the weights of the population of type A oranges and the variance of the weights of the population of type B oranges.

(3)

(b) Test, at the 5% level of significance, the greengrocer’s belief. You should state the hypotheses and the critical value used for this test.

(7)

(2)

(Total 12 marks)

题目中文翻译

一位蔬果商正在研究两种橙子的重量，A 型和 B 型。她认为平均而言，A 型橙子的重量比 B 型橙子重超过 5 克。她随机抽取了 40 个 A 型橙子和 32 个 B 型橙子，并记录每个橙子的重量 $x$ （克）。

下表给出了她的数据摘要。

	$n$	$\bar{x}$	$\sum x^2$
A 型橙子	40	140.4	790 258
B 型橙子	32	134.7	581 430

(a) 计算 A 型橙子总体重量方差和 B 型橙子总体重量方差的无偏估计。

(b) 在 5% 显著性水平下，检验这位蔬果商的看法。你应写出假设以及该检验所用的临界值。

解答

(a)

解法一

思路

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题目给出每组的样本量、样本均值与 $\sum x^2$ 。总体方差的无偏估计要先用 $\sum x^2-n\bar{x}^2$ 求修正平方和，再除以 $n-1$ ；A、B 两组分别计算。

答题过程

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For type A oranges,

\begin{align*} s_A^2=&\,\frac{\sum x^2-n\bar{x}^2}{n-1}\\ =&\,\frac{790258-40(140.4)^2}{39}\\ =&\,45.4256\ldots \end{align*}

For type B oranges,

\begin{align*} s_B^2=&\,\frac{\sum x^2-n\bar{x}^2}{n-1}\\ =&\,\frac{581430-32(134.7)^2}{31}\\ =&\,26.4232\ldots \end{align*}

Therefore, the unbiased variance estimates are

\boxed{s_A^2=45.4\text{ g}^2, \qquad s_B^2=26.4\text{ g}^2}

to 3 significant figures.

(b)

解法一

思路

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蔬果商声称 A 型橙子的总体均重比 B 型高超过 $5$ 克，所以检验 $\mu_A-\mu_B>5$ 。用 (a) 的无偏方差估计构造两个独立样本均值之差的标准误，再把检验统计量与 $5\%$ 右尾临界值比较。

答题过程

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Let $\mu_A$ and $\mu_B$ be the population mean weights of type A and type B oranges respectively.

\begin{align*} H_0:&\ \mu_A-\mu_B=5,\\ H_1:&\ \mu_A-\mu_B>5. \end{align*}

Using the estimates from part (a),

\begin{align*} \widehat{\operatorname{Var}}(\bar X_A)=&\,\frac{45.4256\ldots}{40}\\ =&\,1.13564\ldots, \end{align*}

and

\begin{align*} \widehat{\operatorname{Var}}(\bar X_B)=&\,\frac{26.4232\ldots}{32}\\ =&\,0.825726\ldots \end{align*}

Therefore, the standard error is

\begin{align*} \operatorname{SE}=&\,\sqrt{1.13564\ldots+0.825726\ldots}\\ =&\,1.40049\ldots \end{align*}

The test statistic is

\begin{align*} z=&\,\frac{(140.4-134.7)-5}{1.40049\ldots}\\ =&\,0.4998\ldots \end{align*}

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

z_{0.95}=1.6449.

Since

0.4998<1.6449,

the result is not significant, so $H_0$ is not rejected. There is insufficient evidence to support the greengrocer’s belief that type A oranges weigh, on average, more than 5 grams more than type B oranges.

解法二

思路

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官方评分资料也接受 p-value 判定。沿用相同的右尾检验统计量 $z=0.4998\ldots$ ，求其右尾概率；若 p-value 大于 $0.05$ ，结论同样是不拒绝原假设。

答题过程

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Using the same hypotheses and test statistic,

z=0.4998\ldots

The one-tailed p-value is

\begin{align*} p=&\,P(Z>0.4998\ldots)\\ =&\,0.3086\ldots \end{align*}

Since

0.3086>0.05,

$H_0$ is not rejected. There is insufficient evidence to support the greengrocer’s belief that the difference between the population mean weights exceeds 5 grams.

(c)

解法一

思路

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题目没有说明单个橙子的重量服从正态分布。由于两个样本量都较大，可用中心极限定理把两个样本均值视为近似正态；同时可用样本方差 $s_A^2$ 、 $s_B^2$ 分别近似代替未知总体方差。

答题过程

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Because both sample sizes are large, the central limit theorem allows both sample means, $\bar X_A$ and $\bar X_B$ , to be treated as approximately normally distributed.

The large sample sizes also allow the unknown population variances to be replaced by their sample estimates, so $s_A^2\approx\sigma_A^2$ and $s_B^2\approx\sigma_B^2$ in the standard error.

IAL 2020 Oct S3 Q5

题目

解答

(a)

解法一

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答题过程

(b)

解法一

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解法二

思路

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(c)

解法一

思路

答题过程