S4 June 2014 (R) Q3
3. A farmer is investigating the milk yields of two breeds of cow. He takes a random sample of 9 cows of breed \(A\) and an independent random sample of 12 cows of breed \(B\). For a 5 day period he measures the amount of milk, \(x\) gallons, produced by each cow. The results are summarised in the table below.
| Breed | Sample size | Mean (\(\bar{x}\)) | Standard deviation (\(s_x\)) |
|---|---|---|---|
| \(A\) | 9 | 6.23 | 2.98 |
| \(B\) | 12 | 7.13 | 2.33 |
The amount of milk produced by each cow can be assumed to follow a normal distribution.
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : {\sigma_A}^2 = {\sigma_B}^2 \qquad \mathrm{H}_1 : {\sigma_A}^2 \ne {\sigma_B}^2\) | B1 |
| \(\left(F_{8,11} =\right) \dfrac{2.98^2}{2.33^2} = (1.6357\ldots)\) | M1 |
| \(F_{8,11}\) 10% (two-tail) cv = 2.95 (or prob. = awrt 0.22) | B1 |
| Not significant so can accept the assumption that variances are equal. | A1 |
| (4) |
Notes
1st B1 allow \(\sigma\) or \(\sigma^2\)
M1 for use of the correct test statistic
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : \mu_A = \mu_B \qquad \mathrm{H}_1 : \mu_A \ne \mu_B\) | B1 |
| \({s_p}^2 = \dfrac{8 \times 2.98^2 + 11 \times 2.33^2}{19},= 6.88216\ldots\) or \(s_p = 2.62338\ldots\) | M1, A1 |
| \(\left(t_{19} =\right)(\pm)\dfrac{7.13 - 6.23}{s_p\sqrt{\frac{1}{9} + \frac{1}{12}}} = (\pm)0.7780047\ldots\) = awrt 0.778 | M1 A1 |
| \(t_{19}(0.05)\) two-tail cv = 2.093 | B1 |
| [Not significant] Insufficient evidence of a difference in mean milk yields between the two breeds | A1 |
| (7) |
Notes
1st M1 for attempting \(s_p\) or \({s_p}^2\)
1st A1 for awrt 6.90 or 2.63
2nd M1 for use of a correct test statistic
2nd A1 for awrt 0.77 (accept \(\pm\))
2nd B1 for 2.093 (allow \(\pm\) 1.729 for one-tailed \(\mathrm{H}_1\))
| Scheme | Marks |
|---|---|
| Test in part(b) requires the variances to be equal. The test in part (a) showed that the variances could be assumed to be equal. | B1 |
| (1) | |
| (12 marks) |