<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%.