CAIE S2 2011 June — Question 3 6 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2011
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of binomial distributions
TypeExplain Type I or II error
DifficultyStandard +0.3 This is a straightforward one-tailed binomial hypothesis test with standard structure: define hypotheses, calculate probability under H₀, compare to significance level, and conclude. Part (ii) tests understanding of Type I/II errors but requires only textbook definitions applied to the context. Slightly easier than average due to small sample size making calculations simple and no unusual complications.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail

3 At an election in 2010, 15\% of voters in Bratfield voted for the Renewal Party. One year later, a researcher asked 30 randomly selected voters in Bratfield whether they would vote for the Renewal Party if there were an election next week. 2 of these 30 voters said that they would.
  1. Use a binomial distribution to test, at the \(4 \%\) significance level, the null hypothesis that there has been no change in the support for the Renewal Party in Bratfield against the alternative hypothesis that there has been a decrease in support since the 2010 election.
  2. (a) Explain why the conclusion in part (i) cannot involve a Type I error.
    (b) State the circumstances in which the conclusion in part (i) would involve a Type II error.

Question 3:
Part (i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(0.85^{30} + 30 \times 0.85^{29} \times 0.15 + {}^{30}C_2 \times 0.85^{28} \times 0.15^2 = 0.151\)M1, A1 Allow just \(0.85^{30} + 30 \times 0.85^{29} \times 0.15\) (or critical region \(X=0\) or \(X=2\) not in CR)
\(> 0.04\)M1 Comp with 0.04 (can be implied by diagram)
No evidence decrease or Accept no decreaseA1\(\sqrt{}\)[4] Correct Conclusion (ft); Use of \(P(X=2)\) only: max M0A0M1A1
Part (ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
(a) Not rejected H₀B1[1] Both independent marks
(b) Has been decrease, or \(\pi\) (or \(p\)) \(< 0.15\)B1[1] Must be in context
## Question 3:

### Part (i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.85^{30} + 30 \times 0.85^{29} \times 0.15 + {}^{30}C_2 \times 0.85^{28} \times 0.15^2 = 0.151$ | M1, A1 | Allow just $0.85^{30} + 30 \times 0.85^{29} \times 0.15$ (or critical region $X=0$ or $X=2$ not in CR) |
| $> 0.04$ | M1 | Comp with 0.04 (can be implied by diagram) |
| No evidence decrease or Accept no decrease | A1$\sqrt{}$[4] | Correct Conclusion (ft); Use of $P(X=2)$ only: max M0A0M1A1 |

### Part (ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** Not rejected H₀ | B1[1] | Both independent marks |
| **(b)** Has been decrease, or $\pi$ (or $p$) $< 0.15$ | B1[1] | Must be in context |

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3 At an election in 2010, 15\% of voters in Bratfield voted for the Renewal Party. One year later, a researcher asked 30 randomly selected voters in Bratfield whether they would vote for the Renewal Party if there were an election next week. 2 of these 30 voters said that they would.
\begin{enumerate}[label=(\roman*)]
\item Use a binomial distribution to test, at the $4 \%$ significance level, the null hypothesis that there has been no change in the support for the Renewal Party in Bratfield against the alternative hypothesis that there has been a decrease in support since the 2010 election.
\item (a) Explain why the conclusion in part (i) cannot involve a Type I error.\\
(b) State the circumstances in which the conclusion in part (i) would involve a Type II error.
\end{enumerate}

\hfill \mbox{\textit{CAIE S2 2011 Q3 [6]}}