题目

Problem

The volume of water in a bottle has a normal distribution with unknown mean, $\mu$ millilitres, and known standard deviation, $\sigma$ millilitres.

A random sample of 150 of the bottles of water gave a 95% confidence interval for $\mu$ of $(327.84,329.76)$

(a) Using the confidence interval given, test whether or not $\mu=328$ State your hypotheses clearly and write down the significance level you have used.

(3)

A second random sample, of 200 of these bottles of water, had a mean volume of 328 millilitres.

(b) Calculate a 98% confidence interval for $\mu$ based on this second sample. You must show all steps in your working. (Solutions relying entirely on calculator technology are not acceptable.)

(6)

Using five different random samples of 200 of these bottles of water, five 98% confidence intervals for $\mu$ are to be found.

(3)

(Total for Question 3 is 12 marks)

题目中文翻译

瓶中水的容量服从均值未知、标准差已知的正态分布，均值为 $\mu$ 毫升，标准差为 $\sigma$ 毫升。

对 150 个水瓶的随机样本给出的 $\mu$ 的 95% 置信区间为 $(327.84,329.76)$

(a) 利用给出的置信区间，检验 $\mu=328$ 是否成立。请清楚写出你的原假设，并写明所用的显著性水平。

(3)

另随机抽取 200 瓶水，所得第二个样本的平均容量为 328 毫升。

(b) 基于这个第二个样本，求 $\mu$ 的 98% 置信区间。你必须写出所有步骤。 （完全依赖计算器技术的解法不予接受。）

(6)

抽取五个不同的随机样本，每个样本各含 200 瓶水，将得到五个 $\mu$ 的 98% 置信区间。

(3)

解答

(a)

解法一

思路

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$95\%$ 置信区间对应显著性水平 $5\%$ 的双尾检验。检验值 328 位于给定区间内，所以不能拒绝 $H_0$ 。

答题过程

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The hypotheses are

H_0:\mu=328, \qquad H_1:\mu\neq328.

The significance level is $5\%$ . Since

328\in(327.84,329.76),

$H_0$ is not rejected. There is insufficient evidence at the $5\%$ significance level that $\mu$ differs from 328 ml.

(b)

解法一

思路

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先由第一个 $95\%$ 置信区间的中点和半宽恢复已知总体标准差 $\sigma$ 。再对第二个样本使用 $98\%$ 区间的临界值 $2.3263$ 及新样本量 200。

答题过程

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The midpoint and half-width of the first confidence interval are

\begin{align*} \overline x_1 =&\,\frac{327.84+329.76}{2}\\ =&\,328.8, \end{align*}

and

\frac{329.76-327.84}{2}=0.96.

For the $95\%$ confidence interval,

\begin{align*} 1.96\frac{\sigma}{\sqrt{150}}=&\,0.96,\\ \sigma =&\,\frac{0.96\sqrt{150}}{1.96}\\ =&\,5.9987\ldots. \end{align*}

For a $98\%$ confidence interval, $z=2.3263$ . Using the second sample, the confidence limits are

\begin{align*} &\,328\pm2.3263 \big(\frac{5.9987\ldots}{\sqrt{200}}\big)\\ =&\,328\pm0.9867\ldots. \end{align*}

Therefore, the $98\%$ confidence interval is

\boxed{(327.013240\ldots,\ 328.986759\ldots)}.

Equivalently, to the nearest millilitre, it is $(327,329)$ .

(c)

解法一

思路

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每个 $98\%$ 置信区间包含 $\mu$ 的概率为 $0.98$ 。五个区间来自不同随机样本，可视为独立；令 $Y$ 表示包含 $\mu$ 的区间数，则求 $P(Y>3)=P(Y=4)+P(Y=5)$ 。

答题过程

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Let $Y$ be the number of the five confidence intervals that contain $\mu$ . Then

Y\sim\operatorname{B}(5,0.98).

Therefore,

\begin{align*} P(Y>3) =&\,P(Y=4)+P(Y=5)\\ =&\,\binom54(0.98)^4(0.02)\\ &\qquad{}+(0.98)^5\\ =&\,0.996157\ldots\\ =&\,\boxed{0.996}\quad\text{(3 s.f.)}. \end{align*}

也可以令 $U$ 为不包含 $\mu$ 的区间数，则 $U\sim\operatorname{B}(5,0.02)$ ，而 $Y>3$ 等价于 $U\leqslant1$ ，因此

P(U\leqslant1)=(0.98)^5+5(0.02)(0.98)^4=0.996157\ldots.

S3 2024 June Q3

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