Answer:
Explanation:
A hypothesis test is performed to determine if the proportion of schools needing such a waiver is above 25%
At the null hypothesis, we test that the proportion is of 25%, that is:

At the alternate hypothesis, we test that the proportion is above 25%, that is:

The test statistic is:

In which X is the sample mean,
is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
0.25 is tested at the null hypothesis:
This means that

Sample of 85 schools, 32 need a waiver:
This means that

Value of the test statistic:



P-value of the test and decision:
The p-value of the test is the probability of finding a sample proportion above 0.3765, which is 1 subtracted by the p-value of z = 2.69.
Looking at the z-table, z = 2.69 has a p-value of 0.9964
1 - 0.9964 = 0.0036
The p-value of the test is 0.0036 < 0.05, thus we reject the null hypothesis that the proportion is 25%, and accept the alternate hypothesis that the proportion of schools needing such a waiver is above 25%.