## Question 6:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{h} = 65.4$ | B1 | 65.4 only |
| $s^2 = \frac{214676 - 50\times(\text{"65.4"})^2}{49}$ | M1 | Correct method to find $s^2$ using their $\bar{h}$ |
| $= 16.693\ldots$ | A1 | awrt 16.7 |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu_{\text{do}} = \mu_{\text{do not}}$, $H_1: \mu_{\text{do}} < \mu_{\text{do not}}$ | B1 | Both hypotheses correct – must be clear which is exercise and which is not |
| $z = \pm\frac{\text{"65.4"} - 70.8}{\sqrt{\frac{\text{"16.693"}}{50} + \frac{29.6}{40}}}$ | M1M1 | For the denominator (ft their 16.693); correct ft their 65.4 and 16.693 |
| $= \pm5.21\ldots$ | A1 | awrt 5.21, allow $|z| = 5.21\ldots$ |
| CV 1.6449 | B1 | $|CV| = 1.6449$ or better |
| Amala's belief is supported | A1ft | ft their $z$ value and CV if hypotheses are correct way round. May be in words with heart and exercise e.g. resting heart rate is lower in men who exercise regularly |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| CLT enables you to assume that (the sampling distribution of the sample mean of) resting heart rate is normally distributed for both groups | B1 | For the idea both groups normally distributed. Allow sample sizes big enough for CLT to hold |

### Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Each population/sample is independent **or** each male is independent of the other males | B1 | For identifying the need for the groups or males to be independent |
| Assume the $\sigma_{\text{do}}^2 = s_{\text{do}}^2$ and $\sigma_{\text{do not}}^2 = s_{\text{do not}}^2$ | B1 | Realising the $\sigma^2 = s^2$ |