S4 June 2017 Q3
3. The lengths, \(X\) mm, of the wings of adult blackbirds follow a normal distribution. A random sample of 5 adult blackbirds is taken and the lengths of the wings are measured. The results are summarised below
\[\sum x = 655 \quad \text{and} \quad \sum x^2 = 85\,845\](a) Test, at the 10% level of significance, whether or not the mean length of an adult blackbird’s wing is less than 135 mm. State your hypotheses clearly. (7)
(b) Find the 90% confidence interval for the variance of the lengths of adult blackbirds’ wings. Show your working clearly. (4)
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : \mu = 135 \quad \mathrm{H}_1 : \mu \lt 135\) | B1 |
| \(\bar{x} = 131 \qquad s^2 = 10\) | B1 B1 |
| \(t = \dfrac{131 - 135}{\sqrt{10/5}} = -2.828\ldots\) | M1A1 |
| critical value \(t_4(10\%) = -1.533\) | B1 |
| sufficient evidence that the mean length of wing is less than 135 mm. | A1 ft |
| (7) |
Notes
B1 Both hypotheses
B1 131
B1 10 or awrt 3.16
M1 Allow \(\pm\dfrac{\text{"their 131"} - 135}{\sqrt{\text{"their 10"}/5}}\)
A1 awrt \(-2.83\) or \(-2\sqrt{2}\)
B1 \(\pm 1.533\) sign must match \(t\)-value or be \(\pm\)
A1ft ft \(t\)-value if awarded 1st and 4th B marks. The words ‘mean length’ and ‘135’ must be included in the context
| Scheme | Marks |
|---|---|
| 90% CI is given by \(\dfrac{4 \times 10}{9.488} \lt \sigma^2 \lt \dfrac{4 \times 10}{0.711}\) | M1 B1B1 |
| \((4.22,\ 56.3)\) | A1 |
| (4) | |
| (11 marks) |
Notes
M1 \(\dfrac{4 \times \text{"their 10"}}{\chi^2\,\textit{value}}\)
B1 awrt 9.49
B1 awrt 0.711
A1 awrt 4.22/4.21 and awrt 56.3