A manufacturer claims that the mean lifetime of its lithium battery is 1000 hours. A homeowner selects 40 batteries and finds the mean lifetime to be 990 hours with a standard deviation of 80 hours. Test the manufacturer’s claim. Use α=0.05.

Answer:

Given Data :

Hypothesized Population Mean $(\mu )=1000$

Sample Standard Deviation $(s)=80$

Sample Size $(n)=40$

Sample Mean $(x\overline{)}=990$

Significance Level $(\alpha )=0.05$

The null and alternative hypothesis :

$$H_0:\mu =1000$$

$$H_a:\mu !=1000$$

The test statistic :

$$t=\frac{\bar{x}–\mu }{\frac{s}{\sqrt{n}}}=\frac{990–1000}{\frac{80}{\sqrt{40}}}=-0.791$$

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.