题目

Problem

When a child completes an online exercise called a Mathlit, they might be awarded a medal. The publishers claim that the probability that a randomly chosen child who completes a Mathlit will be awarded a medal is $\dfrac13$ . Asha wishes to test this claim. She decides that if she is awarded no medals while completing $10$ Mathlits, she will conclude that the true probability is less than $\dfrac13$ .

(a) Use a binomial distribution to find the probability of a Type I error.

(2)

The true probability of being awarded a medal is denoted by $p$ .

(b) Given that the probability of a Type II error is $0.8926$ , find the value of $p$ .

(3)

题目中文翻译

孩子完成一项名为 Mathlit 的在线练习时，有可能获得一枚奖牌。发布者声称，从完成一项 Mathlit 的孩子中随机选取一人，该孩子获得奖牌的概率为 $\dfrac13$ 。Asha 想检验这一说法。她决定，如果自己完成 $10$ 项 Mathlit 却没有获得任何奖牌，就认定真实概率小于 $\dfrac13$ 。

(a) 用二项分布求发生第一类错误的概率。

(2)

获得奖牌的真实概率记为 $p$ 。

(b) 已知发生第二类错误的概率为 $0.8926$ ，求 $p$ 的值。

(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

展开

第一类错误为中奖概率确实为1/3时，因10次全无奖而拒绝原假设。

答题过程

展开

Let $M$ be the number of medals in the 10 Mathlits. Under $H_0:p=\frac13$ ,

M\sim\mathrm{B}\bigg(10,\frac13\bigg).

The rejection region is $M=0$ . Thus

\begin{align*} &\,\mathrm{P}(\text{Type I error})\\[4mm] =&\,\bigg(\frac23\bigg)^{10}\\[4mm] \approx&\,\boxed{0.0173}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

展开

真实概率为p时，第二类错误是至少得到一枚奖牌而未拒绝原假设，其概率为1减去零枚的概率。

答题过程

展开

A Type II error occurs when the null hypothesis is not rejected despite the true medal probability being less than $\frac13$ . Thus

1-(1-p)^{10}=0.8926.

Hence

(1-p)^{10}=0.1074.

Therefore

\begin{align*} p=&\,1-(0.1074)^{1/10}\\[4mm] =&\,0.1999808\ldots. \end{align*}

Therefore $\boxed{p=0.200}$ (3 s.f.).

CIE 9709 2023 June Paper 62 Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程