CAIE S2 2018 June — Question 3 5 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2018
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConfidence intervals
TypeFind alpha from CI width
DifficultyStandard +0.3 This is a straightforward confidence interval problem requiring students to work backwards from the interval width to find the confidence level. It involves standard formulas for proportions (width = 2z√(p̂(1-p̂)/n)) and looking up or calculating the z-value, but requires no conceptual insight beyond routine manipulation of the confidence interval formula. Slightly easier than average as it's a direct application with clear numerical values provided.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance

3 A researcher wishes to estimate the proportion, \(p\), of houses in London Road that have only one occupant. He takes a random sample of 64 houses in London Road and finds that 8 houses in the sample have only one occupant. Using this sample, he calculates that an approximate \(\alpha \%\) confidence interval for \(p\) has width 0.130 . Find \(\alpha\) correct to the nearest integer.

Question 3:
AnswerMarks Guidance
AnswerMark Guidance
\(\frac{\frac{8}{64} \times (1 - \frac{8}{64})}{64}\) \(\left(= \frac{7}{4096}\) or \(0.00171\right)\)M1 OE, e.g. \(\frac{\frac{1}{8} \times \frac{7}{8}}{64}\)
\(2 \times z\sqrt{"\frac{7}{4096}"} = 0.130\)M1 Correct equation using their variance
\(z = 1.572\)A1
\(\phi("1.572") = 0.942\), \((0.942 - (1 - 0.942) = 0.884)\)M1 \(2\phi(\text{their } z) - 1\)
\(\alpha = 88\)A1 CAO
Total: 5
## Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{\frac{8}{64} \times (1 - \frac{8}{64})}{64}$ $\left(= \frac{7}{4096}$ or $0.00171\right)$ | M1 | OE, e.g. $\frac{\frac{1}{8} \times \frac{7}{8}}{64}$ |
| $2 \times z\sqrt{"\frac{7}{4096}"} = 0.130$ | M1 | Correct equation using their variance |
| $z = 1.572$ | A1 | |
| $\phi("1.572") = 0.942$, $(0.942 - (1 - 0.942) = 0.884)$ | M1 | $2\phi(\text{their } z) - 1$ |
| $\alpha = 88$ | A1 | CAO |
| **Total: 5** | | |
3 A researcher wishes to estimate the proportion, $p$, of houses in London Road that have only one occupant. He takes a random sample of 64 houses in London Road and finds that 8 houses in the sample have only one occupant. Using this sample, he calculates that an approximate $\alpha \%$ confidence interval for $p$ has width 0.130 . Find $\alpha$ correct to the nearest integer.\\

\hfill \mbox{\textit{CAIE S2 2018 Q3 [5]}}