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IAL 2023 June S2 Q4

A Level / Edexcel / S2

IAL 2023 June Paper · Question 4

题目

Problem

A manufacturer produces candles. Those candles that pass a quality inspection are suitable for sale.

It is known that 2% of the candles produced by the manufacturer are not suitable for sale.

A random sample of 125 candles produced by the manufacturer is taken.

(a) Given nn is large, state a condition for which the binomial distribution B(n,p)B(n, p) can be reasonably approximated by a Poisson distribution.

(1)

(b) Use a suitable approximation to find the probability that no more than 6 of the candles are not suitable for sale.

(4)

The manufacturer also produces candle holders.

Charlie believes that 5% of candle holders produced by the factory have minor defects. The manufacturer claims that the true proportion is less than 5% To test the manufacturer’s claim, a random sample of 30 candle holders is taken and none of them are found to contain minor defects.

(c) (i) Carry out a test of the manufacturer’s claim using a 5% level of significance. You should state your hypotheses clearly.

(5)

(ii) Give a reason why this is not an appropriate test.

(1)

Ashley suggests changing the sample size to 50

(d) Comment on whether or not this change would make the test appropriate. Give a reason for your answer.

(2)

(Total for Question 4 is 13 marks)

题目中文翻译

一家制造商生产蜡烛。通过质量检验的蜡烛可供出售。

已知该制造商生产的蜡烛中有 2% 不适合出售。

随机抽取了 125 支该制造商生产的蜡烛。

(a) 已知 nn 很大,说明在什么条件下二项分布 B(n,p)B(n,p) 可以合理地用泊松分布近似。

(b) 使用适当的近似方法,求这些蜡烛中不超过 6 支不适合出售的概率。

该制造商还生产烛台。

Charlie 认为工厂生产的烛台中有 5% 存在轻微缺陷。 制造商声称真实比例小于 5%。 为检验制造商的说法,随机抽取 30 个烛台,其中没有一个被发现有轻微缺陷。

(c) (i) 使用 5% 显著性水平,对制造商的说法进行检验。 你应清楚陈述你的假设。

(ii) 说明为什么这不是一个合适的检验。

Ashley 建议将样本量改为 50。

(d) 评论这一改变是否会使该检验变得合适。 给出理由。

(第 4 题共 13 分)

解答

(a)

解法一

思路

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当试验次数 nn 很大而成功概率 pp 很小时,二项分布中的成功事件属于稀有事件,此时可以使用参数为 npnp 的泊松分布作近似。题目只要求给出关于 pp 的条件。

答题过程

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The probability pp must be small, for example

p<0.1.p<0.1.

(b)

解法一

思路

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设不适合出售的蜡烛数为 XX,则 XB(125,0.02)X\sim B(125,0.02)。由于 nn 较大且 pp 很小,可以用均值为 np=2.5np=2.5 的泊松分布近似,再求不超过 6 支的累积概率。

答题过程

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Let XX be the number of candles that are not suitable for sale. Then

XB(125,0.02).X\sim B(125,0.02).

Since nn is large and pp is small,

XY,YPo(125×0.02)=Po(2.5).X\approx Y,\qquad Y\sim \operatorname{Po}(125\times 0.02) =\operatorname{Po}(2.5).

Therefore,

P(X6)P(Y6)0.98580.986.\begin{align*} P(X\leqslant 6)\approx&\, P(Y\leqslant 6)\\ \approx&\, 0.9858\\ \approx&\, \boxed{0.986}. \end{align*}

(c)(i)

解法一

思路

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制造商声称缺陷率低于 5%5\%,因此作左尾检验。以 5%5\% 为原假设中的比例,计算在 H0H_0 下观察到 0 个缺陷品的概率。这个概率就是左尾 pp 值,再与显著性水平 0.050.05 比较。

答题过程

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Let pp be the true proportion of candle holders with minor defects.

H0:p=0.05,H1:p<0.05.H_0:p=0.05, \qquad H_1:p<0.05.

Under H0H_0, if DD is the number with minor defects in the sample, then

DB(30,0.05).D\sim B(30,0.05).

The observed value is D=0D=0, so

p-value=P(D0)=P(D=0)=(0.95)30=0.2146.\begin{align*} \text{$p$-value}=&\, P(D\leqslant 0)\\ =&\, P(D=0)\\ =&\, (0.95)^{30}\\ =&\, 0.2146. \end{align*}

Since 0.2146>0.050.2146>0.05, there is insufficient evidence to reject H0H_0.

There is insufficient evidence at the 5%5\% significance level to support the manufacturer’s claim that the true proportion of candle holders with minor defects is less than 5%5\%.

(c)(ii)

解法一

思路

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这是左尾检验,而样本中的缺陷品数量不可能低于 0。即使取得最极端的结果 D=0D=0,其概率仍大于 0.050.05,所以不存在能使原假设在 5%5\% 水平下被拒绝的临界域。

答题过程

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The most extreme possible result in the lower tail is D=0D=0, but

P(D=0)=0.9530=0.2146>0.05.P(D=0)=0.95^{30}=0.2146>0.05.

Therefore the critical region is empty, so H0H_0 can never be rejected at the 5%5\% significance level. Hence this is not an appropriate test.

(d)

解法一

思路

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样本量改成 50 后,仍检查左尾最极端结果 0 个缺陷品的概率。如果这个概率依然高于 0.050.05,临界域仍为空,检验仍不合适。

答题过程

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For a sample of size 50, the probability of the most extreme possible lower-tail result is

P(D=0)=0.9550=0.0769>0.05.P(D=0)=0.95^{50}=0.0769>0.05.

Therefore the critical region would still be empty and H0H_0 could not be rejected. Increasing the sample size to 50 would not make the test appropriate.