| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2007 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Interpret test conclusion or error |
| Difficulty | Standard +0.8 This question requires understanding of two-sample proportion hypothesis testing and, crucially, a conceptual understanding of Type I error. Part (a) is standard S3 material, but part (b) tests whether students understand that Type I error probability is fixed by the significance level regardless of the true parameter value (a common misconception). This conceptual subtlety elevates it above routine application. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: p_K = p_S\); \(H_1: p_K \neq p_S\) | B1 | Both; OE; allow A&B or 1&2 |
| SL \(\alpha = 0.05\); CV \(\ | z\ | = 1.96\) |
| \(\hat{p} = \dfrac{(150\times0.28)+(250\times0.34)}{400}\) | M1 | Used |
| \(= \dfrac{127}{400}\) or \(0.317\) to \(0.318\) | A1 | CAO/AWFW \((0.3175)\) |
| \(z = \dfrac{(\hat{p}_K - \hat{p}_S)-0}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_K}+\frac{1}{n_S}\right)}}\) | M1 | Used; accept unpooled denominator |
| \(\ | z\ | = \dfrac{\ |
| \(= \ | 1.24\ | \) to \(\ |
| Thus accept \(H_0\) as \(\ | z\ | < 1.96\) |
| Thus no evidence, at 5% level, of a difference between two proportions of male customers in two salons | E1\(\checkmark\) | \(\checkmark\) on \(z\) and CV with same sign; in context and qualified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Zero | B1 | CAO |
| since cannot make a Type I error when \(H_0\) is false | B1 | OE |
## Question 3:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: p_K = p_S$; $H_1: p_K \neq p_S$ | B1 | Both; OE; allow A&B or 1&2 |
| SL $\alpha = 0.05$; CV $\|z\| = 1.96$ | B1 | CAO |
| $\hat{p} = \dfrac{(150\times0.28)+(250\times0.34)}{400}$ | M1 | Used |
| $= \dfrac{127}{400}$ or $0.317$ to $0.318$ | A1 | CAO/AWFW $(0.3175)$ |
| $z = \dfrac{(\hat{p}_K - \hat{p}_S)-0}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_K}+\frac{1}{n_S}\right)}}$ | M1 | Used; accept unpooled denominator |
| $\|z\| = \dfrac{\|0.28-0.34\|}{\sqrt{0.3175\times0.6825\left(\frac{1}{150}+\frac{1}{250}\right)}}$ | A1$\checkmark$ | $\checkmark$ on $\hat{p}$; accept no pooling |
| $= \|1.24\|$ to $\|1.25\|$ | A1 | AWFW; $\|1.26\|$ to $\|1.27\|$ |
| Thus accept $H_0$ as $\|z\| < 1.96$ | A1$\checkmark$ | $\checkmark$ on $z$ and CV with same sign |
| Thus no evidence, at 5% level, of a difference between two proportions of male customers in two salons | E1$\checkmark$ | $\checkmark$ on $z$ and CV with same sign; in context and qualified |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Zero | B1 | CAO |
| since cannot make a Type I error when $H_0$ is false | B1 | OE |3 Kutz and Styler are two unisex hair salons. An analysis of a random sample of 150 customers at Kutz shows that 28 per cent are male. An analysis of an independent random sample of 250 customers at Styler shows that 34 per cent are male.
\begin{enumerate}[label=(\alph*)]
\item Test, at the $5 \%$ level of significance, the hypothesis that there is no difference between the proportion of male customers at Kutz and that at Styler.
\item State, with a reason, the probability of making a Type I error in the test in part (a) if, in fact, the actual difference between the two proportions is 0.05 .
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2007 Q3 [11]}}