题目

Problem

Amir believes that $20\%$ of the students at his college are left-handed. His friend believes that the true proportion, $p$ , is less than $20\%$ . Amir plans to use the binomial distribution to test the null hypothesis, $\mathrm{H}_0\colon p=0.2$ , against the alternative hypothesis, $\mathrm{H}_1\colon p<0.2$ .

He decides to choose $35$ students at random. If $3$ or fewer of these students are left-handed, Amir will reject his belief.

(a) Find the significance level of the test.

(3)

(b) State the probability of a Type I error.

(1)

It is now given that the true value of $p$ is $0.05$ .

(c) Find the probability of a Type II error.

(3)

题目中文翻译

Amir 认为，他所在学院有 $20\%$ 的学生是左撇子。他的朋友认为，真实比例 $p$ 低于 $20\%$ 。Amir 打算用二项分布检验原假设 $\mathrm{H}_0\colon p=0.2$ ，备择假设为 $\mathrm{H}_1\colon p<0.2$ 。

他决定随机选取 $35$ 名学生。如果其中至多有 $3$ 名学生是左撇子，Amir 就否定自己的看法。

(a) 求该检验的显著性水平。

(3)

(b) 写出发生第一类错误的概率。

(1)

现已知 $p$ 的真值为 $0.05$ 。

(c) 求发生第二类错误的概率。

(3)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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检验的实际显著性水平是原假设成立时落入拒绝域的概率，计算二项分布不超过3的概率。

答题过程

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Under $H_0$ , let $L$ be the number of left-handed students in the sample. Then $L\sim\mathrm{B}(35,0.2)$ , and the rejection region is $L\leq3$ .

\begin{align*} &\,\mathrm{P}(L\leq3\mid H_0)\\[4mm] =&\,\sum_{r=0}^3\binom{35}{r}(0.2)^r(0.8)^{35-r}\\[4mm] =&\,(0.8)^{35}+35(0.2)(0.8)^{34}\\[4mm] &\,+\binom{35}{2}(0.2)^2(0.8)^{33}\\[4mm] &\,\hspace{2pt}+\binom{35}{3}(0.2)^3(0.8)^{32}\\[4mm] =&\,0.06052398\ldots. \end{align*}

The actual significance level is therefore $\boxed{6.05\%}$ (3 s.f.).

(b)

解法一

思路

展开

第一类错误为原假设正确却被拒绝，其概率就是原假设下拒绝域的概率。

答题过程

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\begin{align*} &\,\mathrm{P}(\text{Type I error})\\[4mm] =&\,\mathrm{P}(L\leq3\mid p=0.2)\\[4mm] \approx&\,\boxed{0.0605}\quad\text{(3 s.f.).} \end{align*}

(c)

解法一

思路

展开

真实比例为0.05时，第二类错误是未拒绝原假设，即样本中至少4人左撇子。

答题过程

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When the true value is $p=0.05$ , $L\sim\mathrm{B}(35,0.05)$ . A Type II error occurs when $H_0$ is not rejected, that is, when $L\geq4$ .

\begin{align*} &\,\mathrm{P}(\text{Type II error})\\[4mm] =&\,1-\sum_{r=0}^3\binom{35}{r}(0.05)^r(0.95)^{35-r}\\[4mm] =&\,1-(0.95)^{35}-35(0.05)(0.95)^{34}\\[4mm] &\,-\binom{35}{2}(0.05)^2(0.95)^{33}\\[4mm] &\,\hspace{2pt}-\binom{35}{3}(0.05)^3(0.95)^{32}\\[4mm] =&\,1-0.90424521\ldots\\[4mm] =&\,\boxed{0.0958}\quad\text{(3 s.f.).} \end{align*}

CIE 9709 2025 March Paper 62 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