After a college football team once again lost a game to their archrival, 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 represent the proportion of all living alumni who favored firing the coach. Suppose the alumni association wished to see if the majority of alumni are in favor of firing the coach. To do this they test the hypotheses H0: p = 0.50 versus Ha: p > 0.50.
(A) What is the P-value for this hypothesis test?

Respuesta :

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.

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