| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2010 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI for proportion |
| Difficulty | Standard +0.8 This S3 question requires multiple hypothesis testing techniques including confidence intervals, exact binomial tests, critical values, and Type II error calculations. While the individual components are standard A-level further maths statistics content, the multi-part structure requiring interpretation of results and the Type II error calculation (which requires working under the alternative hypothesis) elevate this above routine exercises. The question demands careful understanding of hypothesis testing framework rather than just mechanical application. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\hat{p} = \frac{28}{175} = 0.16\) | B1 | |
| \(CI: \hat{p} \pm 1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) | M1 | Correct structure |
| \(= 0.16 \pm 1.96\sqrt{\frac{0.16 \times 0.84}{175}}\) | A1 | |
| \(= 0.16 \pm 1.96 \times 0.02775...\) | ||
| \(= (0.106, 0.214)\) | A1A1 | Both limits correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Since \(1 - 0.214 = 0.786 < 0.80\), the entire CI lies below 0.80 | M1 | Correct comparison |
| There is evidence to doubt the council's claim (claim not supported) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: p = 0.4\), \(H_1: p < 0.4\) | B1 | |
| \(X \sim B(50, 0.4)\) under \(H_0\) | M1 | |
| \(P(X \leq 16) = 0.3523\) ... using exact binomial | M1 | |
| \(P(X \leq 14) = 0.1394\), \(P(X \leq 13) = 0.0병...\) | ||
| \(P(X \leq 13) \approx 0.0병...\); find critical region where \(P \leq 0.10\) | ||
| \(P(X \leq 15) = 0.2labelled...\); \(P(X \leq 14) \approx 0.139\); \(P(X \leq 13) \approx 0.0병..\) | ||
| Since \(p\)-value \(= P(X \leq 16) > 0.10\), do not reject \(H_0\) | A1 | |
| Insufficient evidence to support council's claim | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Critical region is \(X \leq 15\) | B1B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Type II error: fail to reject \(H_0\) when \(p = 0.25\) | M1 | |
| \(P(\text{Type II}) = P(X \geq 16 \mid p = 0.25) = 1 - P(X \leq 15 \mid X\sim B(50, 0.25))\) | M1 | |
| \(= 1 - 0.8839...\) | A1 | |
| \(\approx 0.1161\) | A1 |
# Question 6:
## Part (a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\hat{p} = \frac{28}{175} = 0.16$ | B1 | |
| $CI: \hat{p} \pm 1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ | M1 | Correct structure |
| $= 0.16 \pm 1.96\sqrt{\frac{0.16 \times 0.84}{175}}$ | A1 | |
| $= 0.16 \pm 1.96 \times 0.02775...$ | | |
| $= (0.106, 0.214)$ | A1A1 | Both limits correct |
## Part (a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Since $1 - 0.214 = 0.786 < 0.80$, the entire CI lies below 0.80 | M1 | Correct comparison |
| There is evidence to doubt the council's claim (claim not supported) | A1 | |
## Part (b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: p = 0.4$, $H_1: p < 0.4$ | B1 | |
| $X \sim B(50, 0.4)$ under $H_0$ | M1 | |
| $P(X \leq 16) = 0.3523$ ... using exact binomial | M1 | |
| $P(X \leq 14) = 0.1394$, $P(X \leq 13) = 0.0병...$ | | |
| $P(X \leq 13) \approx 0.0병...$; find critical region where $P \leq 0.10$ | | |
| $P(X \leq 15) = 0.2labelled...$; $P(X \leq 14) \approx 0.139$; $P(X \leq 13) \approx 0.0병..$ | | |
| Since $p$-value $= P(X \leq 16) > 0.10$, do not reject $H_0$ | A1 | |
| Insufficient evidence to support council's claim | A1 | |
## Part (b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Critical region is $X \leq 15$ | B1B1 | |
## Part (b)(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Type II error: fail to reject $H_0$ when $p = 0.25$ | M1 | |
| $P(\text{Type II}) = P(X \geq 16 \mid p = 0.25) = 1 - P(X \leq 15 \mid X\sim B(50, 0.25))$ | M1 | |
| $= 1 - 0.8839...$ | A1 | |
| $\approx 0.1161$ | A1 | |6
\begin{enumerate}[label=(\alph*)]
\item A district council claimed that more than 80 per cent of the complaints that it received about the delivery of its services were answered to the satisfaction of complainants before reaching formal status.
An analysis of a random sample of 175 complaints revealed that 28 reached formal status.
\begin{enumerate}[label=(\roman*)]
\item Construct an approximate $95 \%$ confidence interval for the proportion of complaints that reach formal status.
\item Hence comment on the council's claim.
\end{enumerate}\item The district council also claimed that less than 40 per cent of all formal complaints were due to a failing in the delivery of its services.
An analysis of the 50 formal complaints received during 2007/08 showed that 16 were due to a failing in the delivery of its services.
\begin{enumerate}[label=(\roman*)]
\item Using an exact test, investigate the council's claim at the $10 \%$ level of significance. The 50 formal complaints received during 2007/08 may be assumed to be a random sample.
\item Determine the critical value for your test in part (b)(i).
\item In fact, only 25 per cent of all formal complaints were due to a failing in the delivery of the council's services.
Determine the probability of a Type II error for a test of the council's claim at the $10 \%$ level of significance and based on the analysis of a random sample of 50 formal complaints.\\
(4 marks)
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{b855b5b3-097e-4894-aaec-d77f515949b0-15_2484_1709_223_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{b855b5b3-097e-4894-aaec-d77f515949b0-16_2484_1712_223_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{b855b5b3-097e-4894-aaec-d77f515949b0-17_2484_1709_223_153}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S3 2010 Q6 [18]}}