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(i)
Let $\mu$ be the population mean time taken to install an electronic component.

$H_0 : \mu = 38$

$H_1 : \mu \textless 38$

(ii)
Under $H_0, \bar{T} \sim \mathrm{N}(38, \frac{25}{n})$

Test Statistic, $Z = \frac{\bar{T}-38}{\frac{5}{\sqrt{n}}} \sim N(0, 1)$

At 5% level of significance, we reject $H_0$ if $z \le -1.6449$

$\Rightarrow \frac{\bar{t}-38}{\frac{5}{\sqrt{n}}} \le -1.6449$

$\therefore, \{ \bar{t} \in \mathbb{R} | 0 \textless \bar{t} \le 36.8 \}$

(iii)
At 5% level of significance, we do not reject $H_0$ if $z > -1.6449$

$\Rightarrow \frac{37.1-38}{\frac{5}{\sqrt{n}}} \textless -1.6449$

$n \textless 83.5$

$\therefore, \{n \in \mathbb{R}| 0 \textless n \le 83 \}$

### KS Comments:

Some students forget to define $\mu$ appropriately. (ii), a few students actually do not realised that the rejection rule is when $p \le \alpha$ instead of $latex p < \alpha$, and did not consider the correct inequality. (iii), Students overlook that n refers to sample size and sample size must be an integer. The answers are required to be given in sets here. Some students still have trouble writing sets.

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