CAIE S2 2010 June — Question 1 5 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2010
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of binomial distributions
TypeOne-tailed hypothesis test (lower tail, H₁: p < p₀)
DifficultyModerate -0.3 This is a straightforward one-tailed binomial hypothesis test with clearly stated hypotheses (p = 1/3 vs p < 1/3), small sample size (n=20) allowing direct probability calculation, and standard significance level. The question requires routine application of the binomial test procedure with no conceptual complications or novel problem-solving, making it slightly easier than average for A-level.
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

1 At the 2009 election, \(\frac { 1 } { 3 }\) of the voters in Chington voted for the Citizens Party. One year later, a researcher questioned 20 randomly selected voters in Chington. Exactly 3 of these 20 voters said that if there were an election next week they would vote for the Citizens Party. Test at the \(2.5 \%\) significance level whether there is evidence of a decrease in support for the Citizens Party in Chington, since the 2009 election.

AnswerMarks Guidance
\(H_0\): Pop prop = \(\frac{1}{3}\) (or unchanged)B1 Accept p
\(H_1\): Pop prop < \(\frac{1}{3}\) (or decreased)M1 Attempt Bin(20, \(\frac{1}{3}\)) P(≤ 3)
\(\binom{20}{0}\left(\frac{2}{3}\right)^{20} + 20\binom{}{}\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^{19} + \binom{20}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^{18} + \binom{20}{3}\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^{17}\) Allow one term omitted
\(= 0.0604/0.0605\)A1
comp "0.0604" with 0.025M1 For comparison of their 0.0604
No evidence that support decreased or support probably not decreasedA1n Correct conclusion no contradictions
[5]
SC Use Of Normal Standardising with or without ccM1
Obtains \(z = -1.502\)A1
Valid Comparison with \(z = -1.96\)M1
Correct conclusionA1n
$H_0$: Pop prop = $\frac{1}{3}$ (or unchanged) | B1 | Accept p
$H_1$: Pop prop < $\frac{1}{3}$ (or decreased) | M1 | Attempt Bin(20, $\frac{1}{3}$) P(≤ 3)
$\binom{20}{0}\left(\frac{2}{3}\right)^{20} + 20\binom{}{}\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^{19} + \binom{20}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^{18} + \binom{20}{3}\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^{17}$ | | Allow one term omitted
$= 0.0604/0.0605$ | A1 |
comp "0.0604" with 0.025 | M1 | For comparison of their 0.0604
No evidence that support decreased or support probably not decreased | A1n | Correct conclusion no contradictions
 | [5] |
SC Use Of Normal Standardising with or without cc | M1 |
Obtains $z = -1.502$ | A1 |
Valid Comparison with $z = -1.96$ | M1 |
Correct conclusion | A1n |
1 At the 2009 election, $\frac { 1 } { 3 }$ of the voters in Chington voted for the Citizens Party. One year later, a researcher questioned 20 randomly selected voters in Chington. Exactly 3 of these 20 voters said that if there were an election next week they would vote for the Citizens Party. Test at the $2.5 \%$ significance level whether there is evidence of a decrease in support for the Citizens Party in Chington, since the 2009 election.

\hfill \mbox{\textit{CAIE S2 2010 Q1 [5]}}