S4 June 2013 Q6
6. The carbon content, measured in suitable units, of steel is normally distributed. Two independent random samples of steel were taken from a refining plant at different times and their carbon content recorded. The results are given below.
Sample \(A\): 1.5 0.9 1.3 1.2
Sample \(B\): 0.4 0.6 0.8 0.3 0.5 0.4
(a) Stating your hypotheses clearly, carry out a suitable test, at the 10% level of significance, to show that both samples can be assumed to have come from populations with a common variance \(\sigma^2\). (7)
(b) Showing your working clearly, find the 99% confidence interval for \(\sigma^2\) based on both samples. (6)
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : \sigma_A^2 = \sigma_B^2 \qquad \mathrm{H}_1 : \sigma_A^2 \neq \sigma_B^2\) | B1 |
| \(s_A^{\,2} = (0.25)^2 = 0.0625 \qquad s_B^{\,2} = (0.178885\ldots)^2 = 0.032\) | B1B1 |
| \(F = \dfrac{0.0625}{0.032} = 1.953\ldots\) | M1A1 |
| Critical Value: \(F_{3,5} = 5.41\) | B1 |
| not sig, samples come from populations with common variance | A1cso |
| (7) |
| Scheme | Marks |
|---|---|
| \(s_p^{\,2} = \dfrac{3 \times 0.25^2 + 5 \times 0.032}{8} = 0.04343\ldots = (0.2084\ldots)^2\) | M1A1 |
| Use \(\dfrac{8s_p^{\,2}}{\sigma^2} \sim \chi_8^{\,2}\) | M1 |
| \(1.344 \lt \dfrac{8 \times 0.0434\ldots}{\sigma^2} \lt 21.955\) | B1,B1 |
| 99% confidence interval is (0.0158, 0.259) | A1 |
| (6) | |
| (13 marks) |
Notes
(corrected from the printed mark scheme: printed as “= (0.0284…)2”; \(\sqrt{0.04343\ldots} = 0.2084\ldots\)) CHECK