S4 June 2013 Q2
2. Every 6 months some engineers are tested to see if their times, in minutes, to assemble a particular component have changed. The times taken to assemble the component are normally distributed. A random sample of 8 engineers was chosen and their times to assemble the component were recorded in January and in July. The data are given in the table below.
| Engineer | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) |
|---|---|---|---|---|---|---|---|---|
| January | 17 | 19 | 22 | 26 | 15 | 28 | 18 | 21 |
| July | 19 | 18 | 25 | 24 | 17 | 25 | 16 | 19 |
| Scheme | Marks |
|---|---|
| \(d\) = Jan - July: −2, 1, −3, 2, −2, 3, 2, 2 | M1 |
| \(\bar{d} = 0.375,\quad \sum d^2 = 39 \Rightarrow s^2 = 5.4107\ldots\) or \(s = 2.326\ldots\) | M1, M1 |
| \(t_7(0.025) = 2.365\) | B1 |
| Confidence Interval: \(0.375 \pm 2.365 \times \dfrac{2.326\ldots}{\sqrt{8}}\) | M1 |
| \(= \underline{\mathbf{(-1.57,\ 2.32)}}\) (o.e.) | A1,A1 |
| (7) |
Notes
1st M1 for attempting differences
2nd M1 for attempting \(\bar{d}\)
3rd M1 for attempting \(s_d^{\,2}\), correct expression with their \(\sum d^2\) and \(\bar{d}\) or correct calculation (to 2 sf or better)
4th M1 for use of a correct CI formula, using a value for \(t\) and ft their values.
1st A1 for lower limit of -1.57 or -2.32
2nd A1 for corresponding upper limit
S.C. Allow A1A1 for (0, 2.32)
(corrected from the printed mark scheme: the first line is printed as “d = Jan - June”; the second set of times is July) CHECK
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : \mu_D = 0 \qquad \mathrm{H}_1 : \mu_D \neq 0\) | B1 |
| Comment that 0 is in the interval | M1 |
| Not sig, no evidence of a change in mean time to assemble component | A1ft |
| (3) | |
| (10 marks) |
Notes
B1 for both hypotheses using \(\mu_D\)
M1 for a comment about 0 being in (or out) of their interval
A1 contextual conclusion – must include assemble components
S.C. If they have used difference in means test in part (a) to get the confidence interval then award the B1 for \(\mathrm{H}_0 : \mu_x - \mu_y = 0 \quad \mathrm{H}_1 : \mu_x - \mu_y \neq 0\) or the correct hypotheses.