| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | One-tailed hypothesis test (upper tail, H₁: p > p₀) |
| Difficulty | Moderate -0.3 This is a straightforward one-tailed hypothesis test for a proportion using the binomial distribution with clear parameters (n=300, x=60, p=0.15, α=0.025). It requires standard application of the normal approximation to binomial and comparison with critical value, but involves no conceptual complications or multi-step reasoning beyond the standard test procedure. |
| Spec | 2.04d Normal approximation to binomial2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
**B1** Ho: $p = 0.15$
**B1** H1: $p > 0.15$
**M1** $N(300 \times 0.15, 300 \times 0.15 \times 0.85) = N(45, 38.25)$
**M1** $\frac{59.5-\text{'45'}}{\sqrt{\text{'38.25'}}} = 2.345$
**A1** $z = 1.96$; $2.345 > 1.96$; Evidence prop is higher for new plan
Allow wrong or no cc
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Or: Ho: Approval rate same for new as for old
H1: Approval rate for new $>$ for old
$N\left(0.15, \frac{0.15 \times 0.85}{300}\right) = N(0.15, 0.000425)$
$\frac{\frac{59}{300}+\frac{0.5}{300}-\text{'0.15'}}{\sqrt{\text{'0.000425'}}} = 2.345$
Allow wrong or no cc
comparison (or area comparison)
cwo
---3 When the council published a plan for a new road, only $15 \%$ of local residents approved the plan. The council then published a revised plan and, out of a random sample of 300 local residents, 60 approved the revised plan. Is there evidence, at the $2.5 \%$ significance level, that the proportion of local residents who approve the revised plan is greater than for the original plan?
\hfill \mbox{\textit{CAIE S2 2012 Q3 [5]}}