题目

Problem

Components are manufactured such that their length in mm is normally distributed with mean $\mu$ and variance $\sigma^2$ . Below is a 95% confidence interval for $\mu$ calculated from a random sample of components.

(11.52,\ 13.75)

Using the same random sample,

(a) find a 90% confidence interval for $\mu$ .

(6)

Four 90% confidence intervals are found from independent random samples.

(b) Calculate the probability that only 3 of these 4 intervals will contain $\mu$ .

(2)

(Total for Question 3 is 8 marks)

题目中文翻译

元件的长度（单位 mm）服从均值为 $\mu$ 、方差为 $\sigma^2$ 的正态分布。下面是根据一个元件随机样本算出的 $\mu$ 的 95% 置信区间。

(11.52,\ 13.75)

使用同一个随机样本，

(a) 求 $\mu$ 的 90% 置信区间。

(6)

从相互独立的 4 个随机样本中得到 4 个 90% 置信区间。

(b) 求其中恰好只有 3 个区间包含 $\mu$ 的概率。

(2)

解答

(a)

解法一

思路

展开

同一个样本给出的不同置信水平区间具有相同的样本均值与标准误。先由原 $95\%$ 区间的中点求 $\bar{x}$ ，再用半宽除以 $1.96$ 求标准误；最后改用 $90\%$ 区间的临界值 $1.6449$ 。

答题过程

展开

The sample mean is the midpoint of the $95\%$ confidence interval:

\begin{align*} \bar{x} =&\,\frac{11.52+13.75}{2}\\ =&\,12.635. \end{align*}

The half-width of the $95\%$ confidence interval is

13.75-12.635=1.115.

Equivalently, the full width is $13.75-11.52=2.23$ , so the standard error is $2.23/(2\times1.96)$ . Using the half-width gives

\begin{align*} \operatorname{SE} =&\,\frac{1.115}{1.96}\\ =&\,0.568877\ldots \end{align*}

For a $90\%$ confidence interval, the critical value is $1.6449$ . Hence the interval is

\begin{align*} &\,\bar{x}\pm1.6449\operatorname{SE}\\ =&\,12.635\pm1.6449(0.568877\ldots)\\ =&\,12.635\pm0.935746\ldots. \end{align*}

The endpoints are

\begin{align*} L=&\,12.635-0.935746\ldots=11.699253\ldots,\\ U=&\,12.635+0.935746\ldots=13.570746\ldots. \end{align*}

Therefore, to $3$ significant figures, the $90\%$ confidence interval is

\boxed{(11.7,\ 13.6)\text{ mm}}.

(b)

解法一

思路

展开

每个 $90\%$ 置信区间包含 $\mu$ 的概率为 $0.9$ ，不包含的概率为 $0.1$ 。四个区间来自独立样本，因此包含 $\mu$ 的区间数服从 $\operatorname{Bin}(4,0.9)$ ，计算恰好成功三次的概率。

答题过程

展开

Let $X$ be the number of the four confidence intervals that contain $\mu$ . Then

X\sim\operatorname{Bin}(4,0.9).

Therefore,

\begin{align*} \operatorname{P}(X=3) =&\,\binom{4}{3}(0.9)^3(0.1)\\ =&\,0.2916\\ \approx&\,\boxed{0.292}. \end{align*}

S3 2021 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