In 2006, 75.9% of first-year students said they used the Internet for research or homework. Administrators found that 168 of an SRS of 200 first-year students used the Internet for research or homework. Is the proportion of first-year students who used the Internet for research or homework larger than the 2006 national value of 75.9%?

Question

Also, carry out the hypothesis test described above and compute the P-value, and how does your value compare with the value given above?

Answer 1

You can try a proportional one-sample z-test for this one since this problem is using proportions.

Here's a few hints to get you started:

Null hypothesis:
Ho: p = .759 -->meaning: population proportion is equal to .759 (converting the 75.9% to a decimal).

Alternative hypothesis:
Ha: p > .759 -->meaning: population proportion is greater than .759 (this is a one-tailed test).

Using a formula for a proportional one-sample z-test with your data included, we have:

z = .84 - .759 -->test value (168/200 is .84) minus population value (.759)
divided by
√[(.759)(.241)/200] --> .241 represents 1-.759 and 200 is sample size.

Finish the calculation. Remember if the null is not rejected, then there is no difference. If you need to find the p-value for the test statistic, check a z-table. The p-value is the actual level of the test statistic.

I hope this will help get you started.

Answer 2

p=0.0031