Answer:
Given Data :
Hypothesized Population Mean $(\mu) =1000$
Sample Standard Deviation $(s)=80$
Sample Size $ (n)=40$
Sample Mean $(x\bar)=990$
Significance Level $(\alpha)=0.05$
The null and alternative hypothesis :
$$H_0:\mu=1000$$
$$H_a:\mu!=1000$$
The test statistic :
$$t = \frac{\overline{x} – \mu}{\frac{s}{\sqrt{n}}} = \frac{990 – 1000}{\frac{80}{\sqrt{40}}} = -0.791$$
The p-value:
$$\text{The p-value} = 2 \times P(t < -0.791) = 0.4340 $$
The decision:
The p-value $(0.4340) > \alpha = 0.05 $. therefore, we fail to reject the null hypothesis.
The conclusion:
At the 0.05 level of significance, there is not enough evidence to reject the manufacturer’s claim that the mean lifetime of its lithium battery is 1000 hours.
