S4 June 2018 Q5
5. A machine makes posts. The length of a post is normally distributed with unknown mean \(\mu\) and standard deviation 4 cm.
A random sample of size \(n\) is taken to test, at the 5% significance level, the hypotheses
\[\mathrm{H}_0 : \mu = 150 \qquad\qquad \mathrm{H}_1 : \mu \gt 150\]The manufacturer requires the probability of a Type II error to be less than 0.1 when the actual value of \(\mu\) is 152
| Scheme | Marks |
|---|---|
| 0.05 or 5% | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| Let the CR be \(\bar{X} \gt k\) \(\mathrm{P}\left(\bar{X} \gt k \mid \bar{X} \text{ is } \mathrm{N}\left(150, \dfrac{16}{n}\right)\right) = 0.05\) | |
| \(\therefore \dfrac{\bar{k} - 150}{\frac{4}{\sqrt{n}}} = 1.6449\) | M1B1A1 |
| \(\bar{k} = 150 + 1.6449 \times \dfrac{4}{\sqrt{n}}\) | |
| \(\therefore \dfrac{\bar{k} - 152}{\frac{4}{\sqrt{n}}} = -1.2816\) | M1B1A1 |
| \(\bar{k} = 152 - 1.2816 \times \dfrac{4}{\sqrt{n}}\) | |
| \(150 + 1.6449 \times \dfrac{4}{\sqrt{n}} \lt 152 - 1.2816 \times \dfrac{4}{\sqrt{n}}\) or \(\dfrac{150 + \frac{6.5796}{\sqrt{n}} - 152}{\frac{4}{\sqrt{n}}} = -1.2816\) | M1dd |
| \(\left[\sqrt{n}\right] \gt 5.853\) | A1 |
| \(\left[n \gt\right] 34.25\) | M1 |
| \(n = 35\) | A1cso |
| (10) | |
| (11 marks) |
Notes
M1 \(\therefore \dfrac{\bar{k} - 150}{4/\sqrt{n}} = z\text{-value},\ |z| \gt 1.5\)
B1 awrt \(\pm 1.6449\)
A1 correct equation = awrt 1.65/1.64
M1 \(\therefore \dfrac{\bar{k} - 152}{4/\sqrt{n}} = z\text{-value},\ 1 \lt |z| \lt 1.5\)
B1 \(\pm\) awrt 1.2816
A1 correct equation = awrt \(-1.28\)
M1dd dependent on both previous M marks being awarded. forming an equation and solving leading to \(n = \ldots\) or \(\sqrt{n} = \ldots\)
A1 awrt 5.85
M1 for squaring
A1cso 35 only