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

(i)
Let \mu be the population mean number of minutes that the bus is late.
H_0: \mu = 4.3
H_1: \mu \textless 4.3

(ii)
s^{2} = \frac{10}{9}(3.2) = \frac{32}{9}

Under H_0. \frac{\bar{T}-4.3}{\sqrt{\frac{32/9}{10}}}~\sim~ t(9)

Since H_0 is not rejected, t_{calc}~>~t_{crit}

\frac{\bar{t}-4.3}{\sqrt{\frac{32/9}{10}}} > -1.3830287

\bar{t} > 3.48

Required set =~\{ \bar{t} \in \mathbb{R} ~|~ \bar{t} > 3.48\}

(iii)
For H_0 to be rejected, t_{calc}~\le~t_{crit}

\frac{4.0-4.3}{\sqrt{\frac{k^{2}}{9}}} \le -1.3830287

k^{2} \le 0.423

Required set =~\{ k^{2} \in \mathbb{R} ~|~ 0 \textless k^{2} \le 0.423\}

Personal Comments:
To find the t_{crit} value in (ii), students are required to know how to use InvT(0.1,9) or read the T-table. The last part wants range of k^{2} and not k, do read carefully and note that variance cannot be negative so it is necessary to bound it by zero. Also the question asked for set of values, so students need to give answers in set.

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