Given : In a restaurant, the proportion of people who order coffee with their dinner is p.
Sample size : n= 144
x= 120
[tex]\hat{p}=\dfrac{x}{n}=\dfrac{120}{144}=0.83333333\approx0.8333[/tex]
The null and the alternative hypotheses if you want to test if p is greater than or equal to 0.85 will be :-
Null hypothesis : [tex]H_0: p\geq0.85[/tex] [ it takes equality (=, ≤, ≥) ]
Alternative hypothesis : [tex]H_1: p<0.85[/tex] [its exactly opposite of null hypothesis]
∵Alternative hypothesis is left tailed, so the test is a left tailed test.
Test statistic : [tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
[tex]z=\dfrac{0.83-0.85}{\sqrt{\dfrac{0.85(1-0.85)}{144}}}\\\\=-0.561232257678\approx-0.56[/tex]
Using z-vale table ,
Critical value for 0.05 significance ( left-tailed test)=-1.645
Since the calculated value of test statistic is greater than the critical value , so we failed to reject the null hypothesis.
Conclusion : We have enough evidence to support the claim that p is greater than or equal to 0.85.
