| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Two-sample z-test (known variances) |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question on the difference of sample means with known variance. Part (a) requires standard results about linear combinations of normal distributions (routine for S3). Part (b) is a direct application of part (a) with clear numerical values and standard test procedure. The setup is given explicitly, requiring no novel insight—just careful arithmetic and standard hypothesis test mechanics. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{D}\) has a normal distribution | B1 | Normal |
| with mean \(= \mathbf{0}\) | B1 | CAO |
| and variance \(= \frac{\sigma^2}{n} + 1.5^2 \times \frac{\sigma^2}{n}\) | M1 | Must have (+ sign) & (1.5 or 1.5²) but allow no (±n) |
| \(= \frac{3.25\sigma^2}{n}\) | A1 | OE single expression |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \mu_{XL} = 1.5\mu_L\) | B1 | B1 both; allow any valid notation |
| \(H_1: \mu_{XL} \neq 1.5\mu_L\) | ||
| \(5\% \Rightarrow z = \mathbf{(±)1.96}\) | B1 | AWRT (1.95996) |
| \(z = \frac{2261 - 1.5 \times 1509}{\sqrt{\frac{3.25 \times 4.5^2}{50}}} = \frac{±2.5}{\sqrt{1.31625}}\) | M1 | Numerator; allow (2261 – 1509) |
| M1 | Denominator; allow \(\sqrt{2 \times 4.5^2/50}\) OE | |
| \(= \mathbf{(±)2.18}\) | A1 | AWRT (2.17907) |
| Evidence, at 5% level, that claim is not supported | Adep1 | Dep on z-value and CV; Must have consistent signs |
| 6 | ||
| Total | 10 |
## Part (a)
$\bar{D}$ has a **normal** distribution | B1 | Normal
with mean $= \mathbf{0}$ | B1 | CAO
and variance $= \frac{\sigma^2}{n} + 1.5^2 \times \frac{\sigma^2}{n}$ | M1 | Must have (+ sign) & (1.5 or 1.5²) but allow no (±n)
$= \frac{3.25\sigma^2}{n}$ | A1 | OE single expression
| | **4** |
## Part (b)
$H_0: \mu_{XL} = 1.5\mu_L$ | B1 | B1 both; allow any valid notation
$H_1: \mu_{XL} \neq 1.5\mu_L$ | |
$5\% \Rightarrow z = \mathbf{(±)1.96}$ | B1 | AWRT (1.95996)
$z = \frac{2261 - 1.5 \times 1509}{\sqrt{\frac{3.25 \times 4.5^2}{50}}} = \frac{±2.5}{\sqrt{1.31625}}$ | M1 | Numerator; allow (2261 – 1509)
| M1 | Denominator; allow $\sqrt{2 \times 4.5^2/50}$ OE
$= \mathbf{(±)2.18}$ | A1 | AWRT (2.17907)
Evidence, at 5% level, that **claim is not supported** | Adep1 | Dep on z-value and CV; Must have consistent signs
| | **6** |
**Total** | | **10** |
---\begin{enumerate}[label=(\alph*)]
\item The random variable $X$, which has distribution $\mathrm{N}(\mu_X, \sigma^2)$, is independent of the random variable $Y$, which has distribution $\mathrm{N}(\mu_Y, \sigma^2)$.
In order to test $\mathrm{H_0}: \mu_X = 1.5\mu_Y$, samples of size $n$ are taken on each of $X$ and $Y$ and the random variable $D$ is defined as
$$D = \overline{X} - 1.5\overline{Y}$$
State the distribution of $D$ assuming that $\mathrm{H_0}$ is true. [4 marks]
\item A machine that fills bags with rice delivers weights that are normally distributed with a standard deviation of 4.5 grams.
The machine fills two sizes of bags: large and extra-large.
The mean weight of rice in a random sample of 50 large bags is 1509 grams.
The mean weight of rice in an independent random sample of 50 extra-large bags is 2261 grams.
Test, at the 5\% level of significance, the claim that, on average, the rice in an extra-large bag is $1\frac{1}{3}$ times as heavy as that in a large bag. [6 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2016 Q5 [10]}}