Answer:
The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.
Step-by-step explanation:
The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. Test if the mean cost has increased.
At the null hypothesis, we test if the mean cost is still the same, that is:

At the alternative hypothesis, we test if the mean cost has increased, that is:

The test statistic is:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question

In which X is the sample mean,
is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.
13252 is tested at the null hypothesis:
This means that 
The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500.
This means that 
Value of the test statistic:



P-value of the test and decision:
The p-value of the test is found using a t-score calculator, with a right-tailed test, with 20-1 = 19 degrees of freedom and t = 2.95. Thus, the p-value of the test is 0.0041.
The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.