(i) Let
μ {\mu} μ denote the population mean time in minutes for an employee to install an electronic component, and
H 0 {\mathrm{H}_0} H 0 and
H 1 {\mathrm{H}_1} H 1 be the null and alternative hypothesis respectively.
H 0 : μ = 38 {\textrm{H}_0: \mu = 38} H 0 : μ = 38 H 1 : μ < 38 ■ {\textrm{H}_1: \mu < 38 \; \blacksquare} H 1 : μ < 38 ■
(ii) Under
H 0 , {\mathrm{H}_0,} H 0 , Z = T ‾ − μ σ n ∼ N ( 0 , 1 ) Z= \frac{\overline{T} - \mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1) Z = n σ T − μ ∼ N ( 0 , 1 ) For the test to reject the null hypothesis,
t ‾ − 38 5 50 < − 1.6449 t ‾ < 36.8 (3 sf) \begin{align*}
\frac{\overline{t}-38}{\frac{5}{\sqrt{50}}} &< -1.6449 \\
\overline{t} &< 36.8 \textrm{ (3 sf)}
\end{align*} 50 5 t − 38 t < − 1.6449 < 36.8 (3 sf) Set of values of
t ‾ : {\overline{t}:} t : { t ‾ ∈ R + : t ‾ < 36.8 } ■ \{ \overline{t} \in \mathbb{R}^+: \overline{t} < 36.8 \} \; \blacksquare { t ∈ R + : t < 36.8 } ■
(iii) For the test to not reject the null hypothesis,
37.1 − 38 5 n > − 1.6449 − 0.9 > − 1.6449 ( 5 n ) − 0.9 n > − 1.6449 ( 5 ) n < 9.1381 n < 83.504 \begin{align*}
\frac{37.1-38}{\frac{5}{\sqrt{n}}} &> -1.6449 \\
-0.9 &> -1.6449 \left( \frac{5}{\sqrt{n}} \right) \\
-0.9 \sqrt{n} &> -1.6449 \left( 5 \right) \\
\sqrt{n} &< 9.1381 \\
n &< 83.504
\end{align*} n 5 37.1 − 38 − 0.9 − 0.9 n n n > − 1.6449 > − 1.6449 ( n 5 ) > − 1.6449 ( 5 ) < 9.1381 < 83.504 Set of values of
n : {n:} n : { n ∈ Z + : n ≤ 83 } ■ \{ n \in \mathbb{Z}^+: n \leq 83 \} \; \blacksquare { n ∈ Z + : n ≤ 83 } ■