Batteries produced by a manufacturing company have had a life expectancy of 135 hours. Because of an improved production process, the company believes that there has been an increase in the life expectancy of its batteries. A sample of 42 batteries showed an average life of 140 hours. From past information it is known the standard deviation of the population is 24 hours. Test to determine whether there has been an increase in the life expectancy of batteries. What is the test statistic and what conclusion can be made at the.10 level of significance? a. z =.208; Do not reject the null hypothesis b. z =.088; Do not reject the null hypothesis c. z = 1.35; Reject the null hypothesis d. z = 2.33; Reject the null hypothesis

Answer:  c. z = 1.35; Reject the null hypothesisStep-by-step explanation:Let [tex]\mu[/tex] be the average life expectancy of its batteries.As per given , we haveNull hypothesis : [tex]H_0 : \mu =135[/tex]Alternative hypothesis : [tex]H_a : \mu >135[/tex]Since [tex]H_a[/tex] is right-tailed and population standard deviation is also known, so we perform right-tailed z-test.Test statistic : [tex]\dfrac{\overlien{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]where, n= sample size[tex]\overlien{x}[/tex]= sample mean[tex]\mu[/tex]= Population mean[tex]\sigma[/tex]=population standard deviationFor [tex]n=42,\ \overline{x}=140\ \&\ \sigma=24[/tex], we have[tex]z=\dfrac{140-135}{\dfrac{24}{\sqrt{42}}}=1.35015431217\approx1.35[/tex]Critical one-tailed test value for 0.10 significance level :[tex]z_{0.1}=1.28[/tex]  Decision : Since critical z value (1.28) < test statistic (1.35), so we reject the null hypothesis .[When critical value is less than the test statistic value , we reject the null hypothesis .]