S4 June 2014 Q4
4. A random sample of 8 people were given a new drug designed to help people sleep.
In a two-week period the drug was given for one week and a placebo (a tablet that contained no drug) was given for one week.
In the first week 4 people, selected at random, were given the drug and the other 4 people were given the placebo. Those who were given the drug in the first week were given the placebo in the second week. Those who were given the placebo in the first week were given the drug in the second week.
The mean numbers of hours of sleep per night for each of the people are shown in the table.
| Person | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) |
| Hours of sleep with drug | 10.8 | 7.2 | 8.7 | 6.8 | 9.4 | 10.9 | 11.1 | 7.6 |
| Hours of sleep with placebo | 10.0 | 6.5 | 9.0 | 5.6 | 8.7 | 8.0 | 9.8 | 6.8 |
| Scheme | Marks |
|---|---|
| The differences in the mean number of hours sleep are normally distributed | B1 |
| (1) |
Notes
B1 for a comment that mentions “differences” and “normal” distribution
| Scheme | Marks |
|---|---|
| Differences are 0.8, 0.7, −0.3, 1.2, 0.7, 2.9, 1.3, 0.8 | M1 |
| \(\bar{d} = \frac{8.1}{8} = 1.0125\) | M1 |
| \(s_d = \sqrt{\frac{13.89 - 8 \times 1.0125^2}{7}} = 0.901\ldots\) both \(\bar{d}\) and s | M1 |
| \(\mathrm{H}_0 : \mu_D = 1/6 \quad \mathrm{H}_1 : \mu_D \gt 1/6\) | B1 |
| \(t = \dfrac{1.0125 - 1/6}{0.901/\sqrt{8}} = \text{awrt } 2.65\) or \(\dfrac{c - \frac{1}{6}}{0.901/\sqrt{8}} = 2.998\ \therefore\ \text{CR } c \gt \text{awrt } 1.12\) | M1A1 |
| \(t_7(1\%) = 2.998\) (or prob. = awrt 0.0164) | B1 |
| There is insufficient evidence to suggest the drug increases the mean number of hours slept by more than 10 minutes. | A1ft |
| (8) | |
| (9 marks) |
Notes
1st M1 for attempting the \(d\)s
2nd M1 for attempting \(\bar{d}\)
1st M1 for \(s_d\) or \({s_d}^2\)
1st B1 for both hypotheses correct in terms of \(\mu\) or \(\mu_d\). (allow a defined symbol) Do not allow 10 instead of 1/6 (awrt 0.167) unless working in minutes throughout
3rd M1 for attempting the correct test statistic \(\dfrac{\bar{d} - \frac{1}{6}}{s_d/\sqrt{8}}\) or \(p = \text{awrt } 0.016\) or \(\dfrac{c - \frac{1}{6}}{0.901/\sqrt{8}} = t\) value
2nd A1 awrt 2.65 /2.655 or awrt 1.12 or awrt 0.016
2nd B1 2.998 or 0.0164
3rd A1ft for a correct comment in context based on their test statistic and their cv. Do not allow contradictions.
(Corrected from the printed mark scheme: the critical value is printed as 2.988 in the CR working and in the 2nd B1 note; \(t_7(1\%) = 2.998\).)