Answer:
P-value for this hypothesis test is 0.00175.
Step-by-step explanation:
We are given that the alumni association conducted a survey to see if alumni were in favor of firing the coach.
A simple random sample of 100 alumni from the population of all living alumni was taken. Sixty-four of the alumni in the sample were in favor of firing the coach.
Let p = proportion of all living alumni who favored firing the coach
SO, Null Hypothesis, [tex]H_0[/tex] : p = 0.50 {means that the majority of alumni are not in favor of firing the coach}
Alternate Hypothesis, [tex]H_A[/tex] : p > 0.50 {means that the majority of alumni are in favor of firing the coach}
The test statistics that will be used here is One-sample z proportion statistics;
T.S. = [tex]\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of the alumni in the sample who were in favor of firing the coach = [tex]\frac{64}{100}[/tex] = 0.64
n = sample of alumni = 100
So, test statistics = [tex]\frac{0.64-0.50}{{\sqrt{\frac{0.64(1-0.64)}{100} } } } }[/tex]
= 2.92
Now, P-value of the hypothesis test is given by ;
P-value = P(Z > 2.92) = 1 - P(Z [tex]\leq[/tex] 2.92)
= 1 - 0.99825 = 0.00175
Therefore, the P-value for this hypothesis test is 0.00175.