S4 June 2010 Q2
2. As part of an investigation, a random sample of 10 people had their heart rate, in beats per minute, measured whilst standing up and whilst lying down. The results are summarised below.
| Person | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Heart rate lying down | 66 | 70 | 59 | 65 | 72 | 66 | 62 | 69 | 56 | 68 |
| Heart rate standing up | 75 | 76 | 63 | 67 | 80 | 75 | 65 | 74 | 63 | 75 |
(a) State one assumption that needs to be made in order to carry out a paired \(t\)-test. (1)
(b) Test, at the 5% level of significance, whether or not there is any evidence that standing up increases people’s mean heart rate by more than 5 beats per minute. State your hypotheses clearly. (8)
| Scheme | Marks |
|---|---|
| The differences in the mean heart rates are normally distributed. | B1 |
| (1) |
Notes
must have “The differences in (mean heart rate) are normally distributed)
| Scheme | Marks |
|---|---|
| \(D\) = standing up – lying down \(\mathrm{H}_0 : \mu_D = 5 \quad \mathrm{H}_1 : \mu_D \gt 5\) | B1 |
| \(d\): 9, 6, 4, 2, 8, 9, 3, 5, 7, 7 | M1 |
| \(\bar{d} = 6;\quad s_d = \sqrt{\dfrac{414 - 10 \times 36}{9}} = 2.45\) | M1;M1 |
| \(t_9 = \dfrac{6 - 5}{2.45/\sqrt{10}} = 1.29\) | M1A1 |
| \(t_9(5\%) = 1.833\) | B1 |
| insignificant. There is no evidence to suggest that heart rate rises by more than 5 beats when standing up. | A1 ft |
| (8) | |
| (9 marks) |
Notes
B1 both correct allow \(\mu_D - 5 \gt 0\) (\(\mu_D = -5 \quad \mathrm{H}_1 : \mu_D \lt -5\))
M1 finding differences
M1 finding \(\bar{d}\)
M1 \(\sqrt{\dfrac{\sum d^2 - 10 \times (\bar{d})^2}{9}}\) o.e
M1 \(\pm\left(\dfrac{6 - 5}{s_d/\sqrt{10}}\right)\) need to see full expression with numbers in
A1 awrt \(\pm 1.29\).
B1 \(\pm 1.833\) only
A1 ft their CV and \(t\). Need context. Heart rate and 5 beats