Let [tex]\mu[/tex] represents the population mean .
By considering the given information, we have
a)
[tex]\text{Null hypothesis }H_0: \mu\leq135\\\\\text{Alternative hypothesis }H_a: \mu>135[/tex]
Since the alternative hypothesis is right-tailed , so the test is right-tailed test.
Given : n= 42 > 30 , so we use z-test.
[tex]\overline{x}=140[/tex] ; [tex]\sigma=24[/tex]
Test statistic : [tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
i.e. [tex]z=\dfrac{140-135}{\dfrac{24}{\sqrt{42}}}\approx1.35[/tex]
P-value (right tailed test)=[tex]P(Z>1.35)=1-P(z<1.35)=1-0.911492=0.088508[/tex]
Since , the p-value (0.088508) is less than the significance level,thus we reject the null hypothesis .
Conclusion : We have sufficient evidence to support the claim that Because of an improved production process, the company believes that there has been an increase in the life expectancy of its batteries.
