S4 June 2011 Q7
7. A machine produces components whose lengths are normally distributed with mean 102.3 mm and standard deviation 2.8 mm. After the machine had been serviced, a random sample of 20 components were tested to see if the mean and standard deviation had changed. The lengths, \(x\) mm, of each of these 20 components are summarised as
\[\sum x = 2072 \qquad \sum x^2 = 214\,856\](a) Stating your hypotheses clearly, test, at the 5% level of significance, whether or not there is evidence of a change in standard deviation. (7)
(b) Stating your hypotheses clearly, test, at the 5% level of significance, whether or not the mean length of the components has changed from the original value of 102.3 mm using
(i) a normal distribution,
(ii) a \(t\) distribution. (9)
(c) Comment on the mean length of components produced after the service in the light of the tests from part (a) and part (b). Give a reason for your answer. (2)
| Scheme | Marks |
|---|---|
| \(s_x^{\,2} = \dfrac{214856 - 20 \times \left(\dfrac{2072}{20}\right)^2}{19} = 10.357\ldots\) awrt 10.4 | B1 |
| \(\mathrm{H}_0 : \sigma = 2.8\) (or \(\sigma^2 = \ldots\)) \(\mathrm{H}_1 : \sigma \neq 2.8\) (or \(\sigma^2 \neq \ldots\)) | B1 |
| \(\dfrac{(n-1)s^2}{\sigma^2} \sim \chi^2_{19}\) test statistic = 25.102… awrt 25.1 | M1A1 |
| \(\chi^2_{19}(0.025) = 32.852, \qquad \chi^2_{19}(0.975) = 8.907\) | B1B1 |
| Not significant so no evidence of a change in standard deviation | A1 |
| (7) |
Notes
M1 for use of the correct test statistic
| Scheme | Marks |
|---|---|
| (i) \(\mathrm{H}_0 : \mu = 102.3 \qquad \mathrm{H}_1 : \mu \neq 102.3\) | B1 |
| \(z = \dfrac{\frac{2072}{20} - 102.3}{\frac{2.8}{\sqrt{20}}} = 2.0763\ldots\) awrt 2.08 | M1A1 |
| Critical value is \(z = 1.96\) or awrt \(0.019 \lt 0.025\) | B1 |
| So a significant result, there is evidence of a change in mean length | A1ft |
| (ii) \(t = \dfrac{\frac{2072}{20} - 102.3}{\sqrt{\frac{10.357\ldots}{20}}} = 1.8064\ldots\) awrt 1.81 | M1A1 |
| Critical value of \(t_{19} = 2.093\) | B1 |
| Not significant, there is insufficient evidence of a change in mean length | A1 |
| (9) |
Notes
1st and 2nd M1 for use of correct test statistics
| Scheme | Marks |
|---|---|
| (a) suggests that \(\sigma\) is unchanged so can use \(\sigma = 2.8\) so normal test can be used | B1ft |
| So using (i) conclude that there is evidence of an increase in mean length | B1ft |
| (2) | |
| (18 marks) |
Notes
1st B1 for reason for selecting (i) or (ii) based on their conclusion from test in (a).
2nd B1 For a final conclusion about mean lengths based on their (a) and (b)
NB if both conclusions are the same it needs to be clear they have chosen (i)