Let [tex]\mu[/tex] the population mean.
The hypothesis for the given situation :-
[tex]H_0:\mu=24\\\\H_a: \mu<24[/tex], since the alternative hypothesis is left-tailed , so the test a left-tailed test.
The sample size : [tex]n=50[/tex], since n>30 so its a large sample . We use z-test.
Sample mean : [tex]\overline{x}=23.6[/tex]
Standard deviation : [tex]\sigma= 0.70[/tex]
Significance level : [tex]\alpha=0.01[/tex]
The test statistic for population mean :-
[tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
i.e. [tex]z=\dfrac{23.6-24}{\dfrac{0.70}{\sqrt{50}}}=-4.04[/tex]
The p-value : [tex]P(z<-4.04)=0.0000267[/tex]
Since, the p-value is less than the significance level , so we reject the null hypothesis.
Hence, we have evidence to accept the agency's claim that the brewery is cheating its customers.
