The​ quality-control manager at a compact fluorescent light bulb​ (CFL) factory needs to determine whether the mean life of a large shipment of CFLs is equal to 7450 hours. The population standard deviation is 1 comma 350 hours. A random sample of 81 light bulbs indicates a sample mean life of 7 comma 240 hours. a. At the 0.05 level of​ significance, is there evidence that the mean life is different from 7450 hours question mark b. Compute the​ p-value and interpret its meaning. c. Construct a 95​% confidence interval estimate of the population mean life of the light bulbs. d. Compare the results of​ (a) and​ (c). What conclusions do you​ reach

Answer with explanation:-Let [tex]\mu[/tex] be the population mean .By observing the given information, we have :-[tex]H_0:\mu=7450\\\\H_a:\mu\neq7450[/tex]Since the alternative hypotheses is two tailed so the test is a two-tailed test.We assume that the  life of a large shipment of CFLs is normally distributed.(a) Given : Sample size :  n=81 , since n>30 so we use z-test.Sample mean : [tex]\overline{x}=7240[/tex]Standard deviation : [tex]\sigma=1350[/tex]Test statistic for population mean :-[tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex][tex]\Rightarrow\ z=\dfrac{7240-7450}{\dfrac{1350}{\sqrt{81}}}\\\\\Rightarrow\ z=-1.4[/tex]The critical value (two-tailed) corresponds to the given significance level :-[tex]z_{\alpha/2}=z_{0.025}=1.96[/tex]Since the observed value of z (-1.4) is less than the critical value (1.96) , so we do not reject the null hypothesis.Hence, we conclude that we have enough evidence to accept that the mean life is different from 7450 hours .(b) The p-value : [tex]2P(z>-1.4)=0.1615[/tex] , it means that the probability that the life of CFLs less than 7240 and greater than 7240 is 0.1615.(c) The confidence interval for population mean is given by :-[tex]\overline{x} \pm\ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=7240\pm(1.96)\dfrac{1350}{\sqrt{81}}\\\\=7240\pm294\\\\=(6946,7534)[/tex],