题目

Problem

Anton believes that $10\%$ of students at his college are left-handed. Aliya believes that this is an underestimate. She plans to carry out a hypothesis test of the null hypothesis $p=0.1$ against the alternative hypothesis $p>0.1$ , where $p$ is the actual proportion of students at the college that are left-handed. She chooses a random sample of $20$ students from the college. She will reject the null hypothesis if at least $5$ of these students are left-handed.

(a) Explain what is meant by a Type I error in this context.

(1)

(b) Find the probability of a Type I error in the test.

(3)

(c) Given that the true value of $p$ is $0.3$ , find the probability of a Type II error in the test.

(2)

题目中文翻译

Anton 认为，他所在学院有 $10\%$ 的学生是左撇子。Aliya 认为这一比例被低估了。她计划进行假设检验，原假设为 $p=0.1$ ，备择假设为 $p>0.1$ ，其中 $p$ 是该学院左撇子学生的实际比例。她从该学院随机抽取 $20$ 名学生。如果其中至少有 $5$ 名学生是左撇子，她就拒绝原假设。

(a) 解释第一类错误在本题中的含义。

(1)

(b) 求该检验中发生第一类错误的概率。

(3)

(c) 已知 $p$ 的真值为 $0.3$ ，求该检验中发生第二类错误的概率。

(2)

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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第一类错误是在真实比例为10%时，错误推断左撇子比例高于10%。

答题过程

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It means concluding that more than 10% of students at the college are left-handed when the true proportion is 10%.

(b)

解法一

思路

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原假设下人数服从Bin20,0.1，拒绝域为至少5人。

答题过程

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Let $L$ be the number of left-handed students in the sample. Under $H_0$ , $L\sim\mathrm{B}(20,0.1)$ . Thus

\begin{align*} &\,\mathrm{P}(\text{Type I error})\\[4mm] =&\,\mathrm{P}(L\geq5\mid p=0.1)\\[4mm] =&\,1-(0.9)^{20}-20(0.1)(0.9)^{19}\\[4mm] &\,-\binom{20}{2}(0.1)^2(0.9)^{18}\\[4mm] &\,\hspace{2pt}-\binom{20}{3}(0.1)^3(0.9)^{17}\\[4mm] &\,\hspace{4pt}-\binom{20}{4}(0.1)^4(0.9)^{16}\\[4mm] =&\,\boxed{0.0432}\quad\text{(3 s.f.).} \end{align*}

(c)

解法一

思路

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真实比例0.3时，第二类错误为人数不超过4，导致未拒绝原假设。

答题过程

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When $p=0.3$ , $L\sim\mathrm{B}(20,0.3)$ . A Type II error occurs when $L\leq4$ , giving

\begin{align*} &\,\mathrm{P}(\text{Type II error})\\[4mm] =&\,(0.7)^{20}+20(0.3)(0.7)^{19}\\[4mm] &\,+\binom{20}{2}(0.3)^2(0.7)^{18}\\[4mm] &\,\hspace{2pt}+\binom{20}{3}(0.3)^3(0.7)^{17}\\[4mm] &\,\hspace{4pt}+\binom{20}{4}(0.3)^4(0.7)^{16}\\[4mm] =&\,\boxed{0.238}\quad\text{(3 s.f.).} \end{align*}

CIE 9709 2022 June Paper 63 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

(c)

解法一

思路

答题过程