| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | One-tailed hypothesis test (upper tail, H₁: p > p₀) |
| Difficulty | Standard +0.3 This is a standard S3 hypothesis testing question covering binomial tests and normal approximation. Part (a) requires routine binomial hypothesis test calculations at 10% significance. Part (b)(i) applies normal approximation to proportions (standard bookwork). Part (b)(ii) on power calculation is slightly less routine but still follows a standard procedure taught in S3. The question is methodical rather than conceptually challenging, making it slightly easier than average for A-level. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: P = 0.5\), \(H_1: P > 0.5\) | B1 | Both hypotheses correct |
| \(X \sim B(50, 0.5)\) under \(H_0\) | ||
| \(P(X \geq 29) = 1 - P(X \leq 28)\) | M1 | Attempting correct tail probability |
| \(= 1 - 0.8389 = 0.1611\) | A1 | Correct probability |
| \(0.1611 > 0.10\) | M1 | Comparing with 0.10 |
| Insufficient evidence to reject \(H_0\); claim not supported | A1 | Correct conclusion |
| Conclusion in context | B1 | Must be in context of peas |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: P = 0.5\), \(H_1: P > 0.5\) | (already stated) | |
| \(\hat{P} \sim N\left(0.5, \frac{0.5 \times 0.5}{500}\right) = N(0.5, 0.0005)\) | M1 | Correct Normal approximation |
| \(z = \frac{0.542 - 0.5}{\sqrt{0.0005}} = \frac{0.042}{0.02236} = 1.879\) | M1 A1 | Correct standardisation and value |
| \(1.879 > 1.2816\) | M1 | Comparing with \(z = 1.2816\) |
| Sufficient evidence to reject \(H_0\); claim is supported | A1 | Correct conclusion in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| If \(P = 0.55\), \(\hat{P} \sim N\left(0.55, \frac{0.55 \times 0.45}{500}\right)\) | M1 | Using \(P = 0.55\) for distribution |
| \(= N(0.55, 0.000495)\) | A1 | Correct variance |
| Power \(= P(\hat{P} > 0.529 \mid P = 0.55)\) | M1 | Correct probability statement using critical value |
| \(z = \frac{0.529 - 0.55}{\sqrt{0.000495}} = \frac{-0.021}{0.02225} = -0.944\) | M1 | Correct standardisation |
| \(P(Z > -0.944) = \Phi(0.944) = 0.8274\) | A1 | Correct value (accept awrt 0.827) |
# Question 7:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: P = 0.5$, $H_1: P > 0.5$ | B1 | Both hypotheses correct |
| $X \sim B(50, 0.5)$ under $H_0$ | | |
| $P(X \geq 29) = 1 - P(X \leq 28)$ | M1 | Attempting correct tail probability |
| $= 1 - 0.8389 = 0.1611$ | A1 | Correct probability |
| $0.1611 > 0.10$ | M1 | Comparing with 0.10 |
| Insufficient evidence to reject $H_0$; claim not supported | A1 | Correct conclusion |
| Conclusion in context | B1 | Must be in context of peas |
## Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: P = 0.5$, $H_1: P > 0.5$ | | (already stated) |
| $\hat{P} \sim N\left(0.5, \frac{0.5 \times 0.5}{500}\right) = N(0.5, 0.0005)$ | M1 | Correct Normal approximation |
| $z = \frac{0.542 - 0.5}{\sqrt{0.0005}} = \frac{0.042}{0.02236} = 1.879$ | M1 A1 | Correct standardisation and value |
| $1.879 > 1.2816$ | M1 | Comparing with $z = 1.2816$ |
| Sufficient evidence to reject $H_0$; claim is supported | A1 | Correct conclusion in context |
## Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| If $P = 0.55$, $\hat{P} \sim N\left(0.55, \frac{0.55 \times 0.45}{500}\right)$ | M1 | Using $P = 0.55$ for distribution |
| $= N(0.55, 0.000495)$ | A1 | Correct variance |
| Power $= P(\hat{P} > 0.529 \mid P = 0.55)$ | M1 | Correct probability statement using critical value |
| $z = \frac{0.529 - 0.55}{\sqrt{0.000495}} = \frac{-0.021}{0.02225} = -0.944$ | M1 | Correct standardisation |
| $P(Z > -0.944) = \Phi(0.944) = 0.8274$ | A1 | Correct value (accept awrt 0.827) |
These pages (22, 23, and 24) are **answer space pages** from the AQA exam paper P59726/Jun13/MS03. They contain:
- Blank lined answer spaces for **Question 7**
- An "END OF QUESTIONS" notice
- A blank page with "DO NOT WRITE ON THIS PAGE" notice
**There is no mark scheme content on these pages.** These are student answer booklet pages only — no questions, model answers, mark allocations, or examiner guidance are printed here.
To find the mark scheme content for Question 7, you would need the separate **AQA Mark Scheme document** for this paper, not the question paper/answer booklet itself.7 It is claimed that the proportion, $P$, of people who prefer cooked fresh garden peas to cooked frozen garden peas is greater than 0.50 .
\begin{enumerate}[label=(\alph*)]
\item In an attempt to investigate this claim, a sample of 50 people were each given an unlabelled portion of cooked fresh garden peas and an unlabelled portion of cooked frozen garden peas to taste. After tasting each portion, the people were each asked to state which of the two portions they preferred.
Of the 50 people sampled, 29 preferred the cooked fresh garden peas.
Assuming that the 50 people may be considered to constitute a random sample, use a binomial distribution and the $10 \%$ level of significance to investigate the claim.\\
(6 marks)
\item It was then decided to repeat the tasting in part (a) but to involve a sample of 500 , rather than 50, people.
Of the 500 people sampled, 271 preferred the cooked fresh garden peas.
\begin{enumerate}[label=(\roman*)]
\item Assuming that the 500 people may be considered to constitute a random sample, use an approximation to the distribution of the sample proportion, $\widehat { P }$, and the $10 \%$ level of significance to again investigate the claim.
\item The critical value of $\widehat { P }$ for the test in part (b)(i) is 0.529 , correct to three significant figures. It is also given that, in fact, 55 per cent of people prefer cooked fresh garden peas.
Estimate the power for a test of the claim that $P > 0.50$ based on a random sample of 500 people and using the $10 \%$ level of significance.\\
(5 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S3 2013 Q7 [15]}}