题目

Problem

Past evidence shows that 7% of pears grown by a farmer are unfit for sale. This season it is believed that the proportion of pears that are unfit for sale has decreased. To test this belief a random sample of $n$ pears is taken. The random variable $Y$ represents the number of pears in the sample that are unfit for sale.

(a) Find the smallest value of $n$ such that $Y = 0$ lies in the critical region for this test at a 5% level of significance.

(3)

In the past, 8% of the pears grown by the farmer weigh more than 180 g. This season the farmer believes the proportion of pears weighing more than 180 g has changed. She takes a random sample of 75 pears and finds that 11 of them weigh more than 180 g.

(b) Test, using a suitable approximation, whether there is evidence of a change in the proportion of pears weighing more than 180 g. You should use a 5% level of significance and state your hypotheses clearly.

(6)

(Total for Question 4 is 9 marks)

题目中文翻译

过去证据表明，某农场种植的梨中有 7% 不适合出售。本季认为不适合出售的梨的比例下降了。为检验这一判断，随机抽取 $n$ 个梨。随机变量 $Y$ 表示样本中不适合出售的梨的数量。

(a) 求最小的 $n$ 值，使得在 5% 显著性水平下， $Y = 0$ 落入该检验的临界区域。

过去，农场种植的梨中有 8% 重量超过 180 g。本季农场主认为重量超过 180 g 的梨的比例发生了变化。她随机抽取 75 个梨，其中 11 个重量超过 180 g。

(b) 使用适当的近似方法，检验是否有证据表明重量超过 180 g 的梨的比例发生了变化。你应使用 5% 显著性水平并清楚陈述你的假设。

（第 4 题共 9 分）

解答

(a)

解法一

思路

展开

在原假设下，每个梨不适合出售的概率为 $0.07$ ，所以 $Y$ 服从二项分布。要使 $Y=0$ 落入 5% 显著性水平的下侧临界域，必须有 $P(Y=0)<0.05$ 。解这个指数不等式，并检查相邻整数，便可确定最小样本量。

答题过程

展开

Under $H_0$ ,

Y\sim\operatorname{B}(n,0.07).

For $Y=0$ to lie in the critical region,

\begin{align*} P(Y=0)=0.93^n<&\,0.05 \\ n\ln(0.93)<&\,\ln(0.05). \end{align*}

Since $\ln(0.93)<0$ , division reverses the inequality:

\begin{align*} n>&\,\frac{\ln(0.05)}{\ln(0.93)} \\ >&\,41.28\ldots \end{align*}

Also,

0.93^{41}=0.0510\ldots>0.05, \qquad 0.93^{42}=0.0475\ldots<0.05.

Therefore, the smallest possible value is

\boxed{n=42}.

(b)

解法一

思路

展开

“比例发生变化”表示双侧检验。令 $p$ 为重量超过 180 g 的梨所占比例，在原假设下样本计数服从 $\operatorname{B}(75,0.08)$ 。由于 $n$ 较大而 $p$ 较小，可用均值为 $75\times0.08=6$ 的泊松分布近似。观测值 11 位于上尾，所以计算上尾概率并与双侧检验每一尾的 $0.025$ 比较。

答题过程

展开

Let $p$ be the proportion of pears weighing more than 180 g. The hypotheses are

H_0:p=0.08, \qquad H_1:p\neq0.08.

Let $X$ be the number of pears in the sample that weigh more than 180 g. Under $H_0$ ,

X\sim\operatorname{B}(75,0.08) \approx\operatorname{Po}(75\times0.08) =\operatorname{Po}(6).

Using the Statistical Tables,

\begin{align*} P(X\geqslant11) =&\,1-P(X\leqslant10) \\ =&\,1-0.9574 \\ =&\,0.0426. \end{align*}

Since this is a two-tailed test, the upper-tail significance level is $0.025$ . As

0.0426>0.025,

we do not reject $H_0$ . There is insufficient evidence at the 5% significance level to suggest that the proportion of pears weighing more than 180 g has changed.

解法二

思路

展开

也可以用临界域作判断。仍采用 $\operatorname{Po}(6)$ 近似，分别确定概率不超过 $0.025$ 的下尾与上尾边界，再判断观测值 11 是否落入临界域。

答题过程

展开

Under $H_0$ ,

X\approx\operatorname{Po}(6).

For the lower tail,

P(X\leqslant1)=0.0174<0.025, \qquad P(X\leqslant2)=0.0620>0.025.

For the upper tail,

P(X\geqslant12)=1-P(X\leqslant11) =1-0.9799=0.0201<0.025,

whereas

P(X\geqslant11)=0.0426>0.025.

Therefore, the critical region is

\boxed{X\leqslant1\quad\text{or}\quad X\geqslant12}.

The observed value $X=11$ does not lie in the critical region, so we do not reject $H_0$ . There is insufficient evidence at the 5% significance level to suggest that the proportion of pears weighing more than 180 g has changed.

IAL 2022 June S2 Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

解法二

思路

答题过程