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

(i)

$H_0: \mu = 12$

$H_1: \mu \ne 12$

Under $H_0, \bar{X} ~\sim~ \mathrm{N} (12, \frac{0.8^{2}}{20})$

For $H_0$ to be not rejected at 5% level of significance, then

$latex -1.95996 < \frac{m-12}{0.8/ \sqrt{20}} < 1.95996$ $latex \therefore, \{m \in \mathbb{R} |11.6 < m < 12.4\}$ (ii) $latex H_0: \mu = 12$ $latex H_1: \mu < 12$ Under $latex H_0, \bar{X} ~\sim~ \mathrm{N} (12, \frac{0.8^{2}}{40})$ From the graphing calculator, p-value $latex = 0.02405 < 0.05$, we reject $latex H_0$. Thus, there is sufficient evidence at 5% level of significant to conclude that the mean salt content has been reduced.