A2 June 2023 Q3
3. Two machines, \(A\) and \(B\), are used to fill bottles of water. The amount of water dispensed by each machine is normally distributed.
Samples are taken from each machine and the amount of water, \(x\) ml, dispensed in each bottle is recorded. The table shows the summary statistics for Machine \(A\).
| Sample size | \(\sum x\) | \(\sum x^2\) | |
|---|---|---|---|
| Machine \(A\) | 9 | 2268 | 571 700 |
For Machine \(B\), a random sample of 11 bottles is taken. The sample variance of the amount of water dispensed in bottles is 12.7 ml2
You should state the hypotheses and the critical value used. (4)
| Scheme | Marks | AO |
|---|---|---|
| \(\left[\bar{x} = \frac{2268}{9} = 252\right] \quad s_A^2 = \dfrac{571700 - 9 \times 252^2}{8}\,[= 20.5]\) | M1 | 2.1 |
| \(\chi^2_{8,\,0.025} = 17.535 \qquad \chi^2_{8,\,0.975} = 2.180\) | B1 | 3.3 |
| \(\dfrac{8 \times 20.5}{17.535} \lt \sigma_A^2 \lt \dfrac{8 \times 20.5}{2.180}\) | M1 | 1.1b |
| \(9.3527\ldots \lt \sigma_A^2 \lt 75.2293\ldots\) | A1 | 1.1b |
| (4) |
Notes
M1: correct expression for \(s_A^2\)
B1: for selecting \(\chi^2\) with \(v = 8\). May indicated in notation or either of 17.535, 2.180
M1: setting up 95% CI with their sample variance, chi-squared values, and 8
A1: correct CI with awrt 9.35 and awrt 75.2
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H}_0 : \sigma_A^2 = \sigma_B^2,\ \ \mathrm{H}_1 : \sigma_A^2 \ne \sigma_B^2\) | B1 | 2.5 |
| \(\dfrac{s_A^2}{s_B^2} = \dfrac{\text{“}20.5\text{”}}{12.7} = 1.61\ldots\) | M1 | 3.3 |
| \(\mathrm{F}_{8,\,10}(0.05) = 3.07\) | B1 | 1.1b |
| There is insufficient evidence to suggest that the variances are different. | A1 | 2.2b |
| (4) | ||
| (8 marks) |
Notes
B1: both hypotheses correct using \(\sigma\) or \(\sigma^2\)
M1: using the \(F\)-distribution as the model eg \(\dfrac{s_A^2}{s_B^2}\)
B1: awrt 3.07
A1: Drawing a correct inference following through their CV and value for
Allow \(\sigma_B^2 = \sigma_A^2\)
Allow standard deviation instead of variance
NB: Allow candidates to use their \(\dfrac{s_B^2}{s_A^2}\) with \(\dfrac{1}{3.35}\) = awrt 0.299 for the final M1B1A1 in (b)