Final answer:
At the 0.05 level of significance, the test statistic calculated from the sample proportion is greater than the critical value. Therefore, there is sufficient evidence to reject the college's claim that no more than 25% of students commute more than fifteen miles to school.
Step-by-step explanation:
To determine whether we can reject the college's claim at the 0.05 level of significance, we can perform a hypothesis test. The null hypothesis (H0) is that the proportion p of students who commute more than fifteen miles to school is no more than 25% (p ≤ 0.25). The alternative hypothesis (H1) is that the proportion is greater than 25% (p > 0.25).
We can use the sample proportion to calculate the test statistic using the following formula:
Test statistic (z) = (p - P0) / sqrt(P0(1 - P0) / n)
Where p is the sample proportion, P0 is the hypothesized proportion, and n is the sample size. Substituting our values we get:
p = 81 / 255 = 0.3176
P0 = 0.25
n = 255
Test statistic (z) = (0.3176 - 0.25) / sqrt(0.25 * 0.75 / 255) ≈ 2.92
Next, we compare this test statistic to the critical value from the standard normal distribution for a one-tailed test at the 0.05 significance level, which is approximately 1.645. Since 2.92 > 1.645, we reject the null hypothesis.
Therefore, at the 0.05 level of significance, there is sufficient evidence to reject the claim made by the local college that no more than 25% of students commute more than fifteen miles to school.