题目

Problem

A farmer sells strawberries in baskets. The contents of each of 100 randomly selected baskets were weighed and the results, given to the nearest gram, are shown below.

Weight of strawberries (grams)	Number of baskets
302 – 303	5
304 – 305	13
306 – 307	10
308 – 309	18
310 – 311	25
312 – 313	20
314 – 315	5
316 – 317	4

The farmer proposes that the weight of strawberries per basket, in grams, should be modelled by a normal distribution with a mean of 310 g and standard deviation 4 g. Using his model, the farmer obtains the following expected frequencies.

Weight of strawberries (s, grams)	Expected frequency
$s \le 303.5$	a
$303.5 < s \le 305.5$	7.8
$305.5 < s \le 307.5$	13.6
$307.5 < s \le 309.5$	18.4
$309.5 < s \le 311.5$	19.6
$311.5 < s \le 313.5$	16.3
$313.5 < s \le 315.5$	10.6
$s > 315.5$	b

(a) Find the value of a and the value of b. Give your answers correct to one decimal place.

(5)

Before $s \le 303.5$ and $s > 315.5$ are included, for the remaining cells,

\sum \frac{(O-E)^2}{E}=9.71

(b) Using a 5% significance level, test whether the data are consistent with the model. You should state your hypotheses, the test statistic and the critical value used.

(7)

An alternative model uses estimates for the population mean and standard deviation from the data given. Using these estimated values no expected frequency is below 5 Another test is to be carried out, using a 5% significance level, to assess whether the data are consistent with this alternative model.

(2)

(Total for Question 6 is 14 marks)

题目中文翻译

一位农民按篮出售草莓。随机抽取的 100 个篮子中的草莓重量（按克取整）如下。

草莓重量（克）	篮子数
302 – 303	5
304 – 305	13
306 – 307	10
308 – 309	18
310 – 311	25
312 – 313	20
314 – 315	5
316 – 317	4

农民提出，每篮草莓重量（克）可由均值 310 g、标准差 4 g 的正态分布建模。按照该模型，农民得到以下期望频数。

草莓重量（s, 克）	期望频数
$s \le 303.5$	a
$303.5 < s \le 305.5$	7.8
$305.5 < s \le 307.5$	13.6
$307.5 < s \le 309.5$	18.4
$309.5 < s \le 311.5$	19.6
$311.5 < s \le 313.5$	16.3
$313.5 < s \le 315.5$	10.6
$s > 315.5$	b

(a) 求 a 和 b 的值，答案保留 1 位小数。

在把 $s \le 303.5$ 和 $s > 315.5$ 包含进去之前，其余格子的

\sum \frac{(O-E)^2}{E}=9.71

(b) 在 5% 显著性水平下检验数据是否与该模型一致。你应写出原假设、检验统计量和临界值。

另一个模型使用了由所给数据估计出的总体均值和标准差。使用这些估计值时，没有期望频数低于 5。现在要进行另一项 5% 显著性水平的检验，以评估数据是否与该替代模型一致。

解答

(a)

解法一

思路

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模型为 $N(310,4^2)$ ，而期望频数等于样本量 $100$ 乘相应区间概率。分别把 $303.5$ 和 $315.5$ 标准化，求左右两端的尾概率。

答题过程

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Let $W$ be the weight, in grams, of strawberries in a basket. Under the proposed model,

W\sim N(310,4^2).

For the first class,

\begin{align*} a =&\,100P(W\leq303.5)\\ =&\,100P\bigg( Z\leq\frac{303.5-310}{4} \bigg)\\ =&\,100P(Z\leq-1.625)\\ =&\,5.2081\ldots\\ \approx&\,\boxed{5.2}. \end{align*}

For the final class,

\begin{align*} b =&\,100P(W>315.5)\\ =&\,100P\bigg( Z>\frac{315.5-310}{4} \bigg)\\ =&\,100P(Z>1.375)\\ =&\,8.4565\ldots\\ \approx&\,\boxed{8.5}. \end{align*}

(b)

解法一

思路

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这是卡方拟合优度检验。题目已经给出中间六组对统计量的总贡献，只需加上首尾两组的贡献。模型参数 $310$ 和 $4$ 是预先指定而非由数据估计，因此自由度为类别数减一，即 $8-1=7$ 。

答题过程

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The hypotheses are

\begin{aligned} H_0:&\ \text{The data are consistent with }N(310,16),\\ H_1:&\ \text{The data are not consistent with }N(310,16). \end{aligned}

Using the expected frequencies from part (a),

\begin{align*} \chi^2 =&\,\sum\frac{(O-E)^2}{E}\\ =&\,9.71+\frac{(5-5.2)^2}{5.2}\\ &\,\hspace{2pt}+\frac{(4-8.5)^2}{8.5}\\ =&\,12.100\ldots\\ \approx&\,12.1. \end{align*}

Since no parameters have been estimated from the sample, the number of degrees of freedom is

\nu=8-1=7.

At the $5\%$ significance level, the critical value is

\chi^2_7(0.05)=14.067.

Since

12.100\ldots<14.067,

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 suggest that the data are inconsistent with the proposed $N(310,16)$ model.

(c)

解法一

思路

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替代模型从本组数据估计了均值和标准差，相当于额外估计两个参数，因此自由度还要减 $2$ 。类别数仍为 $8$ ，所以自由度由 $7$ 降至 $5$ ，相应的 $5\%$ 卡方临界值也会降低。

答题过程

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Two population parameters, the mean and the standard deviation, are now estimated from the data. Therefore, two additional degrees of freedom must be subtracted:

\nu=8-1-2=5.

Hence the $5\%$ critical value decreases from $14.067$ to

\boxed{\chi^2_5(0.05)=11.070}.

IAL 2022 Jan S3 Q6

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