<h2>
Answer with explanation:</h2>
By considering the given information, we have
, since the alternative hypothesis is right tail, so the test is a right tail test.
Given : Sample size : n= 200
Number of students have PC's in their home = 65
Then,
Test statistic :
Using the standard normal distribution table for z, we have
P-value for right tail test :
Since the p-value is less that the significance level (), so we reject the null hypothesis.
Hence, we have evidence to reject the null hypothesis i.e. we may support the alternative hypothesis that the proportion exceeds 25%.