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A manufacturer claims that the average lifetime of its LED bulbs is $\mathbf{1000}$ hours.
A consumer agency randomly tests $\mathbf{50}$ bulbs and finds their mean lifetime $=\mathbf{980}$ hours with a population standard deviation $= \mathbf{60}$ hours.

At $\mathbf{5 \%}$ significance level, test whether the manufacturer's claim is valid.
 

\[
\begin{array}{|c|c|c|}
\hline
\textbf{Z-value} & \textbf{Cumulative area} & \textbf{Tail area (two-tail)} \\
\hline
1.645 & 0.950 & 0.10 \\
\hline
1.96 & 0.975 & 0.05 \\
\hline
2.33 & 0.990 & 0.02 \\
\hline
\end{array}
\]
 

  1. $\mathrm{Z}=-1.67$, Do not reject $H_0$
     
  2. $\mathrm{Z}=-2.36$, Reject $H_0$
     
  3. $\mathrm{Z}=-3.24$, Reject $H_0$
     
  4. $\mathrm{Z}=-1.12$, Do not reject $H_0$

 

1 Answer

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$$
\begin{aligned}
& \mu=1000 \\
& \bar{x}=980 \\
& \sigma=60 \\
& n=50 \\
& \alpha=0.05
\end{aligned}
$$


Formula:

$$
Z=\frac{\bar{X}-\mu}{\sigma / \sqrt{n}}
$$


Substitute:

$$
Z=\frac{980-1000}{60 / \sqrt{50}}=\frac{-20}{60 / 7.071}=\frac{-20}{8.485}=-2.36
$$


Critical value (two-tailed, $\alpha=0.05$ ): $\pm 1.96$


Decision: $$
|\mathrm{Z}|=2.36>1.96 \rightarrow \textbf { Reject } \mathbf{H_0}
$$

Answer:
Position:
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