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]
,
