| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2015 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Test using proportion |
| Difficulty | Moderate -0.3 This is a standard S3 hypothesis testing question covering routine procedures: (a) normal approximation test for proportion with continuity correction, (b) binomial exact test, (c) sample size calculation. All parts follow textbook methods with no novel problem-solving required, though the multi-part structure and need for careful application of different techniques makes it slightly easier than average overall. |
| Spec | 5.05a 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 |
| \(H_0: p = 0.6\), \(H_1: p \neq 0.6\) | B1 | Both hypotheses correct |
| \(X \sim B(250, 0.6)\) under \(H_0\) | ||
| Normal approximation: \(X \sim N(150, 60)\) | M1 | Mean = \(250 \times 0.6 = 150\), Variance = \(250 \times 0.6 \times 0.4 = 60\) |
| Continuity correction: \(P(X \geq 164.5)\) or \(P(X \leq 164.5)\) | M1 | Correct continuity correction applied |
| \(z = \frac{164.5 - 150}{\sqrt{60}}\) | M1 | Standardising with their mean and variance |
| \(z = \frac{14.5}{7.746} = 1.872\) | A1 | Correct \(z\) value |
| Critical value \(z = 1.96\) (two-tailed, 5%) | B1 | Correct critical value stated |
| \(1.872 < 1.96\), do not reject \(H_0\) | A1 | Correct conclusion: insufficient evidence that UK proportion differs from USA |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: p = 0.25\), \(H_1: p < 0.25\) | B1 | Both hypotheses correct |
| \(X \sim B(40, 0.25)\) under \(H_0\) | M1 | Correct distribution identified |
| \(P(X \leq 5) = 0.0713\) | M1 | Correct probability calculated |
| \(0.0713 > 0.05\) | A1 | Comparison with 5% significance level |
| Do not reject \(H_0\); insufficient evidence that fewer than 25% own a 4G phone | A1 | Correct contextualised conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z = 2.326\) for 98% confidence | B1 | Correct \(z\) value for 98% CI |
| \(n \geq \left(\frac{z}{0.05}\right)^2 p(1-p)\) | M1 | Correct formula structure |
| Using \(p = 0.3\) (maximum variance with \(p \leq 0.3\)) | M1 | Using \(p = 0.3\) as it gives largest \(p(1-p)\) |
| \(n \geq \frac{(2.326)^2 \times 0.3 \times 0.7}{(0.05)^2}\) | A1 | Correct substitution |
| \(n \geq 457\) | A1 | Correct minimum sample size (rounding up) |
# Question 4:
## Part (a) [7 marks]
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p = 0.6$, $H_1: p \neq 0.6$ | B1 | Both hypotheses correct |
| $X \sim B(250, 0.6)$ under $H_0$ | | |
| Normal approximation: $X \sim N(150, 60)$ | M1 | Mean = $250 \times 0.6 = 150$, Variance = $250 \times 0.6 \times 0.4 = 60$ |
| Continuity correction: $P(X \geq 164.5)$ or $P(X \leq 164.5)$ | M1 | Correct continuity correction applied |
| $z = \frac{164.5 - 150}{\sqrt{60}}$ | M1 | Standardising with their mean and variance |
| $z = \frac{14.5}{7.746} = 1.872$ | A1 | Correct $z$ value |
| Critical value $z = 1.96$ (two-tailed, 5%) | B1 | Correct critical value stated |
| $1.872 < 1.96$, do not reject $H_0$ | A1 | Correct conclusion: insufficient evidence that UK proportion differs from USA |
## Part (b) [5 marks]
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p = 0.25$, $H_1: p < 0.25$ | B1 | Both hypotheses correct |
| $X \sim B(40, 0.25)$ under $H_0$ | M1 | Correct distribution identified |
| $P(X \leq 5) = 0.0713$ | M1 | Correct probability calculated |
| $0.0713 > 0.05$ | A1 | Comparison with 5% significance level |
| Do not reject $H_0$; insufficient evidence that fewer than 25% own a 4G phone | A1 | Correct contextualised conclusion |
## Part (c) [5 marks]
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = 2.326$ for 98% confidence | B1 | Correct $z$ value for 98% CI |
| $n \geq \left(\frac{z}{0.05}\right)^2 p(1-p)$ | M1 | Correct formula structure |
| Using $p = 0.3$ (maximum variance with $p \leq 0.3$) | M1 | Using $p = 0.3$ as it gives largest $p(1-p)$ |
| $n \geq \frac{(2.326)^2 \times 0.3 \times 0.7}{(0.05)^2}$ | A1 | Correct substitution |
| $n \geq 457$ | A1 | Correct minimum sample size (rounding up) |4
\begin{enumerate}[label=(\alph*)]
\item A large survey in the USA establishes that 60 per cent of its residents own a smartphone.
A survey of 250 UK residents reveals that 164 of them own a smartphone.\\
Assuming that these 250 UK residents may be regarded as a random sample, investigate the claim that the percentage of UK residents owning a smartphone is the same as that in the USA. Use the 5\% level of significance.
\item A random sample of 40 residents in a market town reveals that 5 of them own a 4 G mobile phone.
Use an exact test to investigate, at the $5 \%$ level of significance, the belief that fewer than 25 per cent of the town's residents own a 4 G mobile phone.
\item A marketing company needs to estimate the proportion of residents in a large city who own a 4 G mobile phone. It wishes to estimate this proportion to within 0.05 with a confidence of 98\%.
Given that the proportion is known to be at most 30 per cent, estimate the sample size necessary in order to meet the company's need.\\[0pt]
[5 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2015 Q4 [17]}}