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IAL 2022 Jan S2 Q3

A Level / Edexcel / S2

IAL 2022 Jan Paper · Question 3

题目

Problem

A photocopier in a school is known to break down at random at a mean rate of 8 times per week.

(a) Give a reason why a Poisson distribution could be used to model the number of breakdowns.

(1)

The headteacher of the school replaces the photocopier with a refurbished one and wants to find out if the rate of breakdowns has increased or decreased.

(b) Write down suitable null and alternative hypotheses that the headteacher should use.

(1)

The refurbished photocopier was monitored for the first week after it was installed.

(c) Using a 5% level of significance, find the critical region to test whether the rate of breakdowns has now changed.

(3)

(d) Find the actual significance level of a test based on the critical region from part (c).

(2)

During the first week after it was installed there were 4 breakdowns.

(e) Comment on this finding in the light of the critical region found in part (c).

(2)

(Total for Question 3 is 9 marks)

题目中文翻译

学校里的一台复印机已知会随机出故障,平均每周 8 次。

(a) 给出一个理由说明为什么可以用泊松分布来建模故障次数。

校长把复印机换成了一台翻新的,并想知道故障率是增加还是减少了。

(b) 写出校长应使用的合适原假设和备择假设。

翻新的复印机在安装后的第一周被监测。

(c) 使用 5% 显著性水平,求用于检验故障率是否已经改变的临界区域。

(d) 求基于 (c) 中临界区域的检验实际显著性水平。

安装后的第一周出现了 4 次故障。

(e) 结合 (c) 中的临界区域,评论这一结果。

(第 3 题共 9 分)

解答

(a)

解法一

思路

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泊松模型适用于固定时间内随机发生的计数事件,通常要求事件相互独立并以恒定平均速率发生。给出其中一个与复印机故障情境相符的理由即可。

答题过程

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The photocopier breakdowns can be assumed to occur independently at a constant mean rate.

(b)

解法一

思路

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原有平均故障率为每周 88 次。校长要检验故障率是否“增加或减少”,所以备择假设必须是双侧的 λ8\lambda\ne8

答题过程

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Let λ\lambda be the mean number of breakdowns per week for the refurbished photocopier.

H0:λ=8,H1:λ8.H_0:\lambda=8, \qquad H_1:\lambda\ne8.

(c)

解法一

思路

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H0H_0 下,一周故障次数服从 Po(8)\operatorname{Po}(8)。这是 5% 双侧检验,所以从分布两端分别寻找概率不超过 0.0250.025 的最大临界区域。下尾比较 P(X2)P(X\leqslant2)P(X3)P(X\leqslant3);上尾比较 P(X14)P(X\geqslant14)P(X15)P(X\geqslant15),确保选出的每一尾都不超过 0.0250.025

答题过程

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Under H0H_0,

XPo(8).X\sim\operatorname{Po}(8).

For the lower tail,

P(X2)=0.0138,P(X3)=0.0424.P(X\leqslant2)=0.0138, \qquad P(X\leqslant3)=0.0424.

Therefore the lower critical region is X2X\leqslant2.

For the upper tail,

P(X14)=0.0342,P(X15)=0.0173.P(X\geqslant14)=0.0342, \qquad P(X\geqslant15)=0.0173.

Therefore the upper critical region is X15X\geqslant15.

Hence the critical region is

X2orX15.\boxed{X\leqslant2\quad\text{or}\quad X\geqslant15}.

(d)

解法一

思路

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实际显著性水平就是在原假设为真时落入临界区域的总概率,因此把 (c) 的两个尾部概率相加。

答题过程

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The actual significance level is

P(X2)+P(X15)=0.0138+0.0173=0.0311.\begin{align*} P(X\leqslant2)+P(X\geqslant15) =&\,0.0138+0.0173\\[2mm] =&\,0.0311. \end{align*}

Thus the actual significance level is

0.0311, or 3.11%.\boxed{0.0311\text{, or }3.11\%}.

(e)

解法一

思路

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观察值 44 不在 (c) 的临界区域内,因此不能拒绝原假设。结论必须回到题目情境:证据不足以说明翻新后复印机的平均故障率发生了变化。

答题过程

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The observed value 44 is not in the critical region, so we do not reject H0H_0.

There is insufficient evidence at the 5% significance level to suggest that refurbishment has changed the mean number of photocopier breakdowns per week.