Let [tex]\mu[/tex] be the population mean.
By considering the given information , we have
[tex]H_0:\mu\leq22\\\\H_a:\mu>22500[/tex], since the alternative hypothesis is right tailed , so the test is right tail test.
Given : Sample size : n=64, which is a large sample (n>30) so we use z-test.
Sample mean : [tex]\overline{x}=26.935[/tex]
Then, Standard deviation : [tex]\sigma=14 [/tex]
Test statistic for population mean :-
[tex]z=\dfrac{\overlien{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
i.e. [tex]z=\dfrac{26.935-22}{\dfrac{14}{\sqrt{64}}}=2.82[/tex]
P-value = [tex]P(z>2.82)=1-P(z<2.82)=1- 0.9975988= $0.0024012[/tex]
Since the p-value is less than the significance level , so we reject the null hypothesis that mean we can accept the alternative hypothesis.
Thus , we conclude that the true mean concentration can exceeds 22 blops .
