## Question 9:

### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| e.g. randomly select $N$ different businesses and then randomly select $P$ computers from each business; may be implied by correct description | B1 | $N \times P = 120$ where $N$ and $P$ are integers greater than 1; e.g. 20 and 6 or 15 and 8 |

### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 5$ oe; $H_1: \mu \neq 5$ oe | B1 | allow any parameter apart from $\bar{x}$ for population mean as long as clearly defined as (population) mean |
| $H_1$ takes this form as Claud is testing whether the mean length of time is different to 5 oe | B1 | |
| $\mu$ is the **population** mean time for which computers are kept before being replaced | B1 | |
| use of $N\!\left(5, \frac{2.7054^2}{120}\right)$ to find $P(\bar{X} < 4.8855)$ or invNorm$\!\left(p, 5, \frac{2.7054}{\sqrt{120}}\right)$ where $p = 0.025$ or $0.05$; may be implied by $0.3215$ or $4.51595\ldots$ or $4.59377\ldots$ | M1 | condone use of $2.6941$ (may be rounded) instead of $2.7054$ for M1; or $z = \frac{4.8855-5}{\frac{2.7054}{\sqrt{120}}}$; awrt $-0.46$ A1 (may be implied by $-0.466$ or $-0.465567\ldots$ if $2.6941$ used) |
| $P(\bar{X} < 4.8855) =$ awrt $0.32$ | A1 | or CR is $(\bar{X}) < 4.5 - 4.6$ |
| $0.32 > 0.025$ or $4.8855 > 4.5(2)$ | M1 | comparison of their probability with $0.025$ or comparison of $4.8855$ with their critical value from use of $0.025$, as long as previous M1 awarded; their $-0.4436 > -1.96$ oe; M1 dep on award of previous M1 |
| not significant or accept $H_0$ or do not reject $H_0$ or reject $H_1$ | A1FT | may be embedded in conclusion in context |
| insufficient evidence to **suggest** (at 5% level) that the (population) **mean** length of time (computers are kept) is not 5 years | A1FT | do not allow e.g. conclude/prove/indicate or other assertive statement instead of suggest; A0 if answer spoiled |