题目

Problem

Krishi owns a farm on which he keeps chickens.

He selects, at random, 10 of the eggs produced and weighs each of them.

You may assume that these weights are a random sample from a normal distribution with standard deviation 1.9 g

The total weight of these 10 eggs is 537.2 g

(a) Find a 95% confidence interval for the mean weight of the eggs produced by Krishi’s chickens.

(4)

Krishi was hoping to obtain a 99% confidence interval of width at most 1.5 g

(b) Calculate the minimum sample size necessary to achieve this.

(4)

(Total 8 marks)

题目中文翻译

Krishi 在自己的农场里养鸡。

他随机选取了产出的 10 枚鸡蛋，并称量每一枚。

你可以假设这些重量是来自一个标准差为 1.9 g 的正态分布的随机样本。

这 10 枚鸡蛋的总重量为 537.2 g。

(a) 求 Krishi 的鸡所产鸡蛋平均重量的 95% 置信区间。

(b) 计算达到“宽度至多为 1.5 g”的 99% 置信区间所需的最小样本量。

解答

(a)

解法一

思路

展开

先用总重量除以 $10$ 求样本均值。总体标准差已知为 $1.9$ g，且总体为正态分布，因此 $95\%$ 置信区间使用标准正态临界值 $1.96$ 。

答题过程

展开

The sample mean is

\bar{x}=\frac{537.2}{10}=53.72\text{ g}.

For a $95\%$ confidence interval, $z=1.96$ . Therefore,

\begin{align*} \bar{x}\pm z\frac{\sigma}{\sqrt n} =&\,53.72\pm1.96\frac{1.9}{\sqrt{10}}\\ =&\,53.72\pm1.17763\ldots \end{align*}

Hence the $95\%$ confidence interval is

\boxed{(52.54,54.90)}\text{ g}.

(b)

解法一

思路

展开

$99\%$ 置信区间的临界值为 $2.5758$ 。区间宽度是误差界的两倍，把它设为不超过 $1.5$ g，解关于 $n$ 的不等式，最后向上取最小整数。

答题过程

展开

For a $99\%$ confidence interval, $z=2.5758$ . The required width satisfies

2(2.5758)\frac{1.9}{\sqrt n}\leq1.5.

Therefore,

\begin{align*} \sqrt n \geq&\,\frac{2(2.5758)(1.9)}{1.5}\\ \sqrt n \geq&\,6.52536\ldots\\ n \geq&\,42.5803\ldots \end{align*}

Since $n$ must be a whole number, the minimum sample size is

\boxed{43}.

IAL 2022 Jan S3 UNUSED Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程