October 2021 Paper 2 Q11
11 In 2010 the heights of adult women in the UK were found to have mean \(\mu = 161.6\) cm and variance \(\sigma^2 = 1.96\,\text{cm}^2\).
It is believed that the mean height of adult women in 2020 in the UK is greater than in 2010.
In 2020 a researcher collected a random sample of the heights of 200 adult women in the UK.
The researcher calculated the sample mean height and carried out a hypothesis test at the 5% level to investigate whether there was any evidence to suggest that the mean height of adult women in the UK had increased.
The researcher assumed that the variance was unaltered.
- State suitable hypotheses for the test, defining any variables you use.
- Explain whether the researcher conducted a 1-tail or a 2-tail test.
The researcher found that the sample mean was 161.9 cm and made the following statements.
- The sample mean is in the critical region.
- The null hypothesis is accepted.
- This proves that the mean height of adult women in the UK is unaltered at 161.6 cm.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H}_0 : \mu = 161.6\) \(\mathrm{H}_1 : \mu > 161.6\) | B1 | 1.1 |
| \(\mu\) is the population mean height of adult females (in 2020) | B1 | 2.5 |
| 1-tailed test because it is suspected that the mean height has increased oe | B1 | 2.4 |
| [3] |
Notes
B1: both hypotheses; allow other parameters (but not \(X\) or \(\bar{X}\)) if defined as mean
| Scheme | Marks | AO |
|---|---|---|
| use of \(\bar{X} \sim \mathrm{N}\left(161.6, \dfrac{1.96}{200}\right)\) soi | M1 | 3.3 |
| awrt \(\bar{X} > 161.8\) BC | A1 | 1.1 |
| [2] |
Notes
M1: may see \(1.645 = \dfrac{\bar{X} - 161.6}{\sqrt{\frac{1.96}{200}}}\)
\(\bar{X} = 161.76\ldots\) implies M1 only
A1: allow if inequality not strict
NB 161.76283…
allow eg sample mean > 161.8
| Scheme | Marks | AO |
|---|---|---|
| the 1st statement is correct because \(\bar{X} > 161.8\) or \(\bar{X} > 161.8\) so the sample mean is in the critical region | M1 | 2.3 |
| the 2nd statement is incorrect because the sample mean is in the critical region oe or because the result is significant oe or ‘…so/hence the null hypothesis is rejected.’ following from the first statement | A1 | 2.4 |
| the 3rd statement is incorrect; because it is only possible to infer, not prove, using a hypothesis test oe | B1 | 2.2b |
| [3] |
Notes
M1: FT their 161.8 (must be greater than 161.6) or allow if their \(P(\bar{X} > 161.9) [= 0.00122] < 0.05\) or their \(z > 1.645\) is considered
critical region must be an upper tail; their probability must be less than 0.5
A1: FT their calculated critical region or their calculated probability,
B1: ignore comments about rejecting the null hypothesis oe