All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

Let \mu be the population mean time taken to install an electronic component.

H_0 : \mu = 38

H_1 : \mu \textless 38

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 \}

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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