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, $H_0$ : p = 0.50 {means that the majority of alumni are not in favor of firing the coach}

Alternate Hypothesis, $H_A$ : 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. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of the alumni in the sample who were in favor of firing the coach = $\frac{64}{100}$ = 0.64

n = sample of alumni = 100

So, test statistics = $\frac{0.64-0.50}{{\sqrt{\frac{0.64(1-0.64)}{100} } } } }$

= 2.92

Now, P-value of the hypothesis test is given by ;

P-value = P(Z > 2.92) = 1 - P(Z $\leq$ 2.92)

= 1 - 0.99825 = 0.00175

Therefore, the P-value for this hypothesis test is 0.00175.

4 0

4 years ago

The value of a company's equipment is $25,000 and decreases at a rate of 15% per year. What is the value after 9 years?

Sindrei [870]

Answer:

6,812.26

Step-by-step explanation:

I found the answer key

7 0

3 years ago