| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Sample size determination |
| Difficulty | Standard +0.3 This is a straightforward application of variance properties for sample means and confidence interval width formula. Part (a) requires knowing Var(X̄) = σ²/n and that Var(X-Y) = Var(X) + Var(Y) for independent variables. Part (b) involves setting up the inequality 2×2.576×√(37.6/n) ≤ 5 and solving for n. While it requires multiple steps, all techniques are standard S3 material with no novel insight needed. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Var}(\bar{X}_A - \bar{X}_B) = \dfrac{18.8}{n} + \dfrac{18.8}{n} = \dfrac{37.6}{n}\) | M1 A1 | M1 for sum of two variance terms each \(\frac{18.8}{n}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Width of CI \(= 2 \times 2.576 \times \sqrt{\dfrac{37.6}{n}} \leq 5\) | M1 A1 | M1 for structure; A1 for \(z = 2.576\) |
| \(\sqrt{\dfrac{37.6}{n}} \leq \dfrac{5}{2 \times 2.576}\) | M1 | Rearranging |
| \(\dfrac{37.6}{n} \leq \left(\dfrac{5}{5.152}\right)^2 = 0.9415...\) | ||
| \(n \geq \dfrac{37.6}{0.9415} = 39.9...\) | A1 | Correct inequality |
| Minimum \(n = 40\) | A1 | Must be integer, rounded up |
# Question 6:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Var}(\bar{X}_A - \bar{X}_B) = \dfrac{18.8}{n} + \dfrac{18.8}{n} = \dfrac{37.6}{n}$ | M1 A1 | M1 for sum of two variance terms each $\frac{18.8}{n}$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Width of CI $= 2 \times 2.576 \times \sqrt{\dfrac{37.6}{n}} \leq 5$ | M1 A1 | M1 for structure; A1 for $z = 2.576$ |
| $\sqrt{\dfrac{37.6}{n}} \leq \dfrac{5}{2 \times 2.576}$ | M1 | Rearranging |
| $\dfrac{37.6}{n} \leq \left(\dfrac{5}{5.152}\right)^2 = 0.9415...$ | | |
| $n \geq \dfrac{37.6}{0.9415} = 39.9...$ | A1 | Correct inequality |
| Minimum $n = 40$ | A1 | Must be integer, rounded up |6 Population $A$ has a normal distribution with unknown mean $\mu _ { A }$ and a variance of 18.8.\\
Population $B$ has a normal distribution with unknown mean $\mu _ { B }$ but with the same variance as Population $A$.
The random variables $\bar { X } _ { A }$ and $\bar { X } _ { B }$ denote the means of independent samples, each of size $n$, from populations $A$ and $B$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find an expression, in terms of $n$, for $\operatorname { Var } \left( \bar { X } _ { A } - \bar { X } _ { B } \right)$.
\item Given that the width of a $99 \%$ confidence interval for $\mu _ { A } - \mu _ { B }$ is to be at most 5 , calculate the minimum value for $n$.\\[0pt]
[5 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2014 Q6 [5]}}