题目

Problem

Chrystal is studying the lengths of pine cones that have fallen from a tree. She believes that the length, $X$ cm, of the pine cones can be modelled by a normal distribution with mean 6 cm and standard deviation 0.75 cm.

She collects a random sample of 80 pine cones and their lengths are recorded in the table below.

Length, x cm	$x<5$	$5 \leq x < 5.5$	$5.5 \leq x < 6$	$6 \leq x < 6.5$	$x \geq 6.5$
Frequency	6	14	24	26	10

(a) Stating your hypotheses clearly and using a 10% level of significance, test Chrystal’s belief. Show your working clearly and state the expected frequencies, the test statistic and the critical value used.

(10)

Chrystal’s friend David asked for more information about the lengths of the 80 pine cones. Chrystal told him that

\begin{align*} \sum x=&\,464,\\ \sum x^2=&\,2722.59. \end{align*}

(b) Calculate unbiased estimates of the mean and variance of the lengths of the pine cones.

(3)

David used the calculations from part (b) to test whether or not the lengths of the pine cones are normally distributed using Chrystal’s sample.

His test statistic was 3.50 (to 3 significant figures) and he did not pool any classes.

(3)

(d) Estimate, to 2 significant figures, the proportion of pine cones from the tree that are longer than 7 cm.

(2)

(Total 18 marks)

题目中文翻译

Chrystal 在研究从树上掉落的松果长度。她认为松果的长度 $X$ cm 可以用均值为 6 cm、标准差为 0.75 cm 的正态分布建模。

她收集了 80 个松果的随机样本，长度记录如下表。

长度，x cm	$x<5$	$5 \leq x < 5.5$	$5.5 \leq x < 6$	$6 \leq x < 6.5$	$x \geq 6.5$
频数	6	14	24	26	10

(a) 清楚写出假设，并在 10% 显著性水平下检验 Chrystal 的看法。清楚写出你的工作过程，并给出期望频数、检验统计量和所用临界值。

Chrystal 的朋友 David 想了解这 80 个松果长度的更多信息。Chrystal 告诉他

\begin{align*} \sum x=&\,464,\\ \sum x^2=&\,2722.59. \end{align*}

(b) 求松果长度均值和方差的无偏估计。

David 用 (b) 中的计算结果，利用 Chrystal 的样本检验松果长度是否服从正态分布。

他的检验统计量为 3.50（保留 3 位有效数字），并且没有合并任何组。

(d) 估计来自这棵树的松果长度超过 7 cm 的比例，答案保留 2 位有效数字。

解答

(a)

解法一

思路

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这是检验指定正态模型 $N(6,0.75^2)$ 是否适合数据的卡方拟合优度检验。先用该模型计算五组的期望频数，再求卡方统计量。由于模型的两个参数都是题目预先给定、并非由样本估计，所以自由度为 $5-1=4$ 。

答题过程

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The hypotheses are:

$H_0$ : $N(6,0.75^2)$ is a suitable model for the lengths of the fallen pine cones.
$H_1$ : $N(6,0.75^2)$ is not a suitable model for the lengths of the fallen pine cones.

Under $H_0$ ,

X\sim N(6,0.75^2).

The class boundaries have standardised values

\begin{align*} \frac{5-6}{0.75}=&\,-\frac{4}{3},\\ \frac{5.5-6}{0.75}=&\,-\frac{2}{3},\\ \frac{6-6}{0.75}=&\,0,\\ \frac{6.5-6}{0.75}=&\,\frac{2}{3}. \end{align*}

For example, the expected frequency in the second class is

\begin{align*} E_2=&\,80P(5\leqslant X<5.5)\\ =&\,80P(-1.3333\ldots\leqslant Z<-0.6667\ldots)\\ =&\,12.9025\ldots \end{align*}

Using the normal distribution, the expected frequencies are:

Length class	$x<5$	$5\leqslant x<5.5$	$5.5\leqslant x<6$	$6\leqslant x<6.5$	$x\geqslant6.5$
Observed, $O_i$	6	14	24	26	10
Expected, $E_i$	7.2969	12.9025	19.8006	19.8006	20.1994

The contributions to the test statistic are:

Length class	$x<5$	$5\leqslant x<5.5$	$5.5\leqslant x<6$	$6\leqslant x<6.5$	$x\geqslant6.5$
$\dfrac{(O_i-E_i)^2}{E_i}$	0.2305	0.0934	0.8906	1.9410	5.1500

Therefore,

\begin{align*} \chi^2=&\,\sum\frac{(O_i-E_i)^2}{E_i}\\ =&\,8.3055\ldots \end{align*}

Equivalently, the accepted computational form

\chi^2=\sum\frac{O_i^2}{E_i}-80

gives the same value.

Since no parameters have been estimated from the sample,

\nu=5-1=4.

At the $10\%$ significance level, the critical value is

\chi^2_4(0.10)=7.779.

Since

8.3055>7.779,

the result is significant, so $H_0$ is rejected. There is sufficient evidence that $N(6,0.75^2)$ is not a suitable model for the lengths of the fallen pine cones; therefore, the data do not support Chrystal’s belief.

(b)

解法一

思路

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样本均值本身就是总体均值的无偏估计。总体方差的无偏估计要用分母 $n-1=79$ ，并由 $\sum x$ 与 $\sum x^2$ 计算修正平方和。

答题过程

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The unbiased estimate of the population mean is

\begin{align*} \hat{\mu}=\bar{x}=&\,\frac{\sum x}{n}\\ =&\,\frac{464}{80}\\ =&\,\boxed{5.8\text{ cm}}. \end{align*}

First calculate the corrected sum of squares:

\begin{align*} S_{xx}=&\,\sum x^2-\frac{(\sum x)^2}{n}\\ =&\,2722.59-\frac{464^2}{80}\\ =&\,31.39. \end{align*}

The unbiased estimate of the population variance is therefore

\begin{align*} s^2=&\,\frac{S_{xx}}{n-1}\\ =&\,\frac{31.39}{79}\\ =&\,0.397341\ldots \end{align*}

Therefore,

\boxed{s^2=0.397\text{ cm}^2}

to 3 significant figures.

(c)

解法一

思路

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David 用样本估计了正态分布的均值和方差，因此除了五组频数总和带来的一个限制，还要扣除两个已估计参数，自由度为 $5-1-2=2$ 。再把给定的检验统计量与 $10\%$ 临界值比较。

答题过程

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David estimated two parameters, the mean and the variance, from the sample. Hence the number of degrees of freedom is

\nu=5-1-2=2.

At the $10\%$ significance level, the critical value is

\chi^2_2(0.10)=4.605.

Since

3.50<4.605,

the result is not significant, so the null hypothesis is not rejected. There is insufficient evidence that a normal distribution is unsuitable; hence a normal distribution is a plausible model for the lengths of the pine cones.

(d)

解法一

思路

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用 (b) 的无偏估计建立正态模型，均值为 $5.8$ 、方差为 $0.39734\ldots$ ，所以标准差为其平方根。把 $7$ 标准化后求标准正态分布的右尾概率；题目要的是比例，不需要乘以样本量 $80$ 。

答题过程

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Using the estimates from part (b),

X\sim N(5.8,0.397341\ldots).

The estimated standard deviation is

s=\sqrt{0.397341\ldots}=0.63035\ldots

Therefore,

\begin{align*} P(X>7)=&\,P\bigg(Z>\frac{7-5.8}{0.63035\ldots}\bigg)\\ =&\,P(Z>1.9037\ldots)\\ =&\,0.02847\ldots \end{align*}

Hence the estimated proportion, to 2 significant figures, is

\boxed{0.028}.

IAL 2021 Jan S3 Q5

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

(d)

解法一

思路

答题过程