S4 June 2014 Q3
3. A large number of chicks were fed a special diet for 10 days. A random sample of 9 of these chicks is taken and the weight gained, \(x\) grams, by each chick is recorded. The results are summarised below.
\[\sum x = 181 \qquad \sum x^2 = 3913\]You may assume that the weights gained by the chicks are normally distributed.
Calculate a 95% confidence interval for
A chick which gains less than 16 g has to be given extra feed.
| Scheme | Marks |
|---|---|
| (i) \(\bar{x} = \frac{181}{9} = 20.111\ldots\) | B1 |
| \({s_x}^2 = \left(\dfrac{3913 - 9 \times \bar{x}^2}{8} =\right) 34.1111 \quad \left(s_x = 5.84\right)\) | B1 |
| \(t_8(0.025)\) cv \(= 2.306\) | B1 |
| 95% CI for \(\mu\) is \(= 20.111 \pm 2.306 \times \frac{5.84}{\sqrt{9}}\) | M1 |
| \(= (15.6, 24.6)\) awrt (15.6, 24.6) | A1, A1 |
| (ii) \({\chi_8}^2(0.025) = 2.18(0), \quad {\chi_8}^2(0.975) = 17.535\) | B1B1 |
| 95% CI for \(\sigma^2\) is given by \(2.180 \lt \dfrac{8{s_x}^2}{\sigma^2} \lt 17.535\) | M1 |
| So 95% CI for \(\sigma^2\) is \(=\) awrt (15.6, 125) | A1 |
| (10) |
Notes
(i) 1st M1 ‘their \(\bar{x}\)’ \(\pm\ t\ \textit{value} \times \dfrac{\text{‘their } s\text{’}}{\sqrt{9}}\)
1st A1 awrt 15.6
2nd A1 awrt 24.6
(ii) 2nd M1 \(\chi^2 \lt \dfrac{8s^2}{\sigma^2} \lt \chi^2\)
A1 awrt 15.6 and 125
| Scheme | Marks |
|---|---|
| Require \(\mathrm{P}(X \lt 16) = \mathrm{P}\left(Z \lt \dfrac{16 - \mu}{\sigma}\right)\) to be as small as possible OR \(\dfrac{16 - \mu}{\sigma}\) to be as large as possible but negative; imply lowest \(\boldsymbol{\sigma}\) and largest \(\boldsymbol{\mu}\). | M1 |
| \(\mathrm{P}\left(Z \lt \dfrac{16 - 24.6}{\sqrt{15.6}}\right)\); \(= 1 - 0.9854 =\) 0.0146 or 0.0147 | M1A1ft;A1 |
| (4) | |
| (14 marks) |
Notes
M1 Identify must use lowest \(\boldsymbol{\sigma}\) and largest \(\boldsymbol{\mu}\)
M1 standardising and finding correct area use either limit for \(\mu\) and \(\sigma\)
A1 ft their lowest \(\boldsymbol{\sigma}\) and largest \(\boldsymbol{\mu}\)
A1 awrt 0.0146 or 0.0147