题目

Problem

In the past, $2\%$ of talking dolls made by a certain firm have been faulty. Following a change in the manufacturing process, a quality control officer suspects that the percentage of faulty talking dolls has increased. In order to test his suspicion, he selects $50$ talking dolls at random. If the number of faulty talking dolls is more than $3$ , he will conclude that the percentage has increased.

(a) State suitable null and alternative hypotheses for the test.

(1)

(b) State what is meant by a Type II error in this context.

(1)

(c) Given that the true percentage of faulty talking dolls is now $12\%$ , use a binomial distribution to find the probability of a Type II error.

(3)

题目中文翻译

过去，某公司生产的会说话的玩偶中有 $2\%$ 是次品。生产工艺改变后，一名质量检验员怀疑次品所占的百分比有所上升。为了检验他的猜测，他随机抽取了 $50$ 个会说话的玩偶。如果次品数超过 $3$ ，他就断定该百分比有所上升。

(a) 写出该检验适用的原假设与备择假设。

(1)

(b) 说明在此情境中第二类错误的含义。

(1)

(c) 已知现在次品所占的真实百分比为 $12\%$ ，用二项分布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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用总体次品率 $p$ 表示假设；怀疑的是“增加”，所以备择假设是单侧。

答题过程

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Let $p$ be the population proportion of faulty talking dolls after the change in manufacturing process. Suitable hypotheses are

\boxed{H_0:p=0.02,\qquad H_1:p>0.02}.

(b)

解法一

思路

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第二类错误是在比例实际上已经增加时，样本未能使我们作出增加的结论。

答题过程

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A Type II error is concluding that the percentage of faulty talking dolls has not increased when it actually has increased.

(c)

解法一

思路

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真实比例换成 $0.12$ ，但检验规则不变：只有样本次品数大于 $3$ 才拒绝原假设，所以第二类错误对应至多 $3$ 件次品。

答题过程

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Let $F$ be the number of faulty dolls in the sample. When the true faulty proportion is $0.12$ ,

F\sim\mathrm B(50,0.12).

A Type II error occurs when $F\le3$ . Therefore

\begin{align*} \mathrm P(\text{Type II error}) =&\,\sum_{r=0}^3\binom{50}{r}(0.12)^r(0.88)^{50-r}\\[4mm] =&\,(0.88)^{50}+50(0.12)(0.88)^{49}\\[4mm] &\,\hspace{2pt}+\binom{50}{2}(0.12)^2(0.88)^{48}\\[4mm] &\,\hspace{4pt}+\binom{50}{3}(0.12)^3(0.88)^{47}\\[4mm] \approx&\,0.00167546+0.0114236\\[4mm] &\,+0.0381651+0.0832694\\[4mm] \approx&\,\boxed{0.135}. \end{align*}

CIE 9709 2026 June Paper 61 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