| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| 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 binomial hypothesis test with a large sample requiring normal approximation. Part (a) involves standard procedure: stating hypotheses, calculating test statistic, comparing to critical value. Part (b) asks for a basic assumption (independence/random sampling). While it requires correct application of the normal approximation and continuity correction, it's a routine S3 question with no conceptual surprises, making it slightly easier than average. |
| Spec | 5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: p = 0.25\), \(H_1: p > 0.25\) | B1 | Both hypotheses correct |
| \(X \sim B(375, 0.25)\) under \(H_0\) | ||
| Since \(n\) large, \(np = 93.75\), \(npq = 70.3125\), use Normal approximation | M1 | Normal approximation stated or implied |
| \(X \sim N(93.75, 70.3125)\) approximately | A1 | Both parameters correct |
| \(P(X \geq 108) = P\left(Z \geq \frac{107.5 - 93.75}{\sqrt{70.3125}}\right)\) | M1 | Continuity correction, standardising |
| \(= P(Z \geq 1.639...)\) | A1 | Correct \(z\)-value |
| \(= 1 - \Phi(1.64) = 0.0505\) | ||
| Critical value at 2% is \(z = 2.054\) | M1 | Comparison with correct critical value |
| \(1.64 < 2.054\), insufficient evidence to reject \(H_0\) | A1 | Correct conclusion in context |
| The consumer report's claim is not supported at the 2% significance level |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The 375 visitors form a random sample (of all visitors who used the catering facilities) | B1 | Must mention random sample |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p = 0.25$, $H_1: p > 0.25$ | B1 | Both hypotheses correct |
| $X \sim B(375, 0.25)$ under $H_0$ | | |
| Since $n$ large, $np = 93.75$, $npq = 70.3125$, use Normal approximation | M1 | Normal approximation stated or implied |
| $X \sim N(93.75, 70.3125)$ approximately | A1 | Both parameters correct |
| $P(X \geq 108) = P\left(Z \geq \frac{107.5 - 93.75}{\sqrt{70.3125}}\right)$ | M1 | Continuity correction, standardising |
| $= P(Z \geq 1.639...)$ | A1 | Correct $z$-value |
| $= 1 - \Phi(1.64) = 0.0505$ | | |
| Critical value at 2% is $z = 2.054$ | M1 | Comparison with correct critical value |
| $1.64 < 2.054$, insufficient evidence to reject $H_0$ | A1 | Correct conclusion in context |
| The consumer report's claim is not supported at the 2% significance level | | |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| The 375 visitors form a random sample (of all visitors who used the catering facilities) | B1 | Must mention random sample |
---1 A consumer report claimed that more than 25 per cent of visitors to a theme park were dissatisfied with the catering facilities provided.
In a survey, 375 visitors who had used the catering facilities were interviewed independently, and 108 of them stated that they were dissatisfied with the catering facilities provided.
\begin{enumerate}[label=(\alph*)]
\item Test, at the $2 \%$ level of significance, the consumer report's claim.
\item State an assumption about the 375 visitors that was necessary in order for the hypothesis test in part (a) to be valid.
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2011 Q1 [7]}}