Question

Data collected from a local high school found that 18% of the students do not have internet access at home, which puts these students at a disadvantage academically. One teacher felt this estimate was too low and decided to test if the true percentage of students without home internet access was greater than the data suggested. To do this, she sampled 200 students. In her sample, 25% of the students did not have internet access at home. If the significance level for the hypothesis test is 1%, can the teacher conclude that more than 18% of all such students do not have internet access?

When answering the questions below

• make sure a decimal representation begins with 0, like this: 0.286
• if a test has two Critical Values, type it like this: -2.34 & 2.34
• round z values to 2 decimal places, and t values to 3 decimal places
• for the result, type either Reject, or Fail to Reject

Please type in the Critical Value(s) , the Test Statistic , and the result of the test

Answer 1

= 0.25, n = 200 and p = 0.18, = 0.01

The Hypothesis:

H0: p = 0.18

Ha: p > 0.18

This is a Right tailed Test.

The Critical Value:   The critical value (Right tail) at = 0.01, Zcritical = +2.326

The Decision Rule is that if z test is > 2.326, Then Reject H0

____________________

The Test Statistic:

The Decision:    Since Z observed (2.58) is > Zcritical (2.326), We Reject H0

The Conclusion:   There is sufficient evidence at the 99% significance level to conclude that the true percentage of students without home internet access is greater than 18%.