CAIE S2 2022 November — Question 4 8 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2022
SessionNovember
Marks8
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TopicHypothesis test of a Poisson distribution
TypeFind Type I error probability
DifficultyStandard +0.3 This is a standard hypothesis testing question on Poisson distributions with straightforward application of definitions. Parts (a) and (c) are routine bookwork, while parts (b) and (d) require understanding Type I/II errors but involve direct calculation from tables with clear parameters (λ=7.2 for the test). The scaling from 10m² to 30m² is simple multiplication, and all steps follow standard procedures taught in S2 with no novel problem-solving required.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail
The number of faults in cloth made on a certain machine has a Poisson distribution with mean 2.4 per 10 m\(^2\). An adjustment is made to the machine. It is required to test at the 5% significance level whether the mean number of faults has decreased. A randomly selected 30 m\(^2\) of cloth is checked and the number of faults is found.
  1. State suitable null and alternative hypotheses for the test. [1]
  2. Find the probability of a Type I error. [3]
Exactly 3 faults are found in the randomly selected 30 m\(^2\) of cloth.
  1. Carry out the test at the 5% significance level. [2]
Later a similar test was carried out at the 5% significance level, using another randomly selected 30 m\(^2\) of cloth.
  1. Given that the number of faults actually has a Poisson distribution with mean 0.5 per 10 m\(^2\), find the probability of a Type II error. [2]
Question 4:
AnswerMarks
4Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
AnswerMarks
4(a)H : Population mean = 7.2 or 2.4
0
H : Population mean < 7.2 or 2.4
AnswerMarks Guidance
1B1 or λ or μ = 7.2 or 2.4 (Not just ‘mean’).
or λ or μ < 7.2 or 2.4
1
AnswerMarks Guidance
4(b)λ = 7.2 B1
 7.22 
 P(X 2)  =e−7.2 1+7.2+  or e–7.2(1 + 7.2 + 25.92)
 2 
or 0.0007465 + 0.0053754 + 0.01935 [= 0.0255]
7.23
 P(X 3)  ='0.0255'+e−7.2 or ‘0.0255’ +e–7.2 (62.21)
3!
AnswerMarks Guidance
or ‘0.0255’ +0.04644 [= 0.0719]M1 Both expressions needed, allow any λ
If λ ≠ 7.2 allow P(X ⩽ n) for 2 consecutive values of n
with P(X ⩽ n) < 0.05 and P(X ⩽ n + 1) > 0.05.
AnswerMarks
P(Type I) = 0.02547 or 0.0255 (3 sf)B1
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
4(c)3 > 2 or P(X ⩽ 3) > 0.05 or ‘0.0719’ > 0.05 M1
or 3 outside critical region. FT their CR in (b).
[Not reject H ]
0
AnswerMarks Guidance
No evidence that [mean] number of faults has decreasedA1 FT No contradictions. In context, not definite.
2
AnswerMarks Guidance
4(d)1 – e–1.5(1 + 1.5 + 1.52 / 2) or 1 – e–1.5(1 + 1.5 + 1.125)
or 1 – (0.2231 + 0.3347 + 0.2510)M1 Must see expression. FT their CR in (b).
= 0.191 (3 sf)A1
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
--- 4(a) ---
4(a) | H : Population mean = 7.2 or 2.4
0
H : Population mean < 7.2 or 2.4
1 | B1 | or λ or μ = 7.2 or 2.4 (Not just ‘mean’).
or λ or μ < 7.2 or 2.4
1
--- 4(b) ---
4(b) | λ = 7.2 | B1 | SOI
 7.22 
 P(X 2)  =e−7.2 1+7.2+  or e–7.2(1 + 7.2 + 25.92)
 2 
or 0.0007465 + 0.0053754 + 0.01935 [= 0.0255]
7.23
 P(X 3)  ='0.0255'+e−7.2 or ‘0.0255’ +e–7.2 (62.21)
3!
or ‘0.0255’ +0.04644 [= 0.0719] | M1 | Both expressions needed, allow any λ
If λ ≠ 7.2 allow P(X ⩽ n) for 2 consecutive values of n
with P(X ⩽ n) < 0.05 and P(X ⩽ n + 1) > 0.05.
P(Type I) = 0.02547 or 0.0255 (3 sf) | B1
3
Question | Answer | Marks | Guidance
--- 4(c) ---
4(c) | 3 > 2 or P(X ⩽ 3) > 0.05 or ‘0.0719’ > 0.05 | M1 | For a valid comparison
or 3 outside critical region. FT their CR in (b).
[Not reject H ]
0
No evidence that [mean] number of faults has decreased | A1 FT | No contradictions. In context, not definite.
2
--- 4(d) ---
4(d) | 1 – e–1.5(1 + 1.5 + 1.52 / 2) or 1 – e–1.5(1 + 1.5 + 1.125)
or 1 – (0.2231 + 0.3347 + 0.2510) | M1 | Must see expression. FT their CR in (b).
= 0.191 (3 sf) | A1
2
Question | Answer | Marks | Guidance
The number of faults in cloth made on a certain machine has a Poisson distribution with mean 2.4 per 10 m$^2$. An adjustment is made to the machine. It is required to test at the 5% significance level whether the mean number of faults has decreased. A randomly selected 30 m$^2$ of cloth is checked and the number of faults is found.

\begin{enumerate}[label=(\alph*)]
\item State suitable null and alternative hypotheses for the test. [1]

\item Find the probability of a Type I error. [3]
\end{enumerate}

Exactly 3 faults are found in the randomly selected 30 m$^2$ of cloth.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Carry out the test at the 5% significance level. [2]
\end{enumerate}

Later a similar test was carried out at the 5% significance level, using another randomly selected 30 m$^2$ of cloth.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Given that the number of faults actually has a Poisson distribution with mean 0.5 per 10 m$^2$, find the probability of a Type II error. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE S2 2022 Q4 [8]}}
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