posted by Julia Brown .
It has been reported that 80% of taxpayers who are audited by the IRS end up paying more money in taxes. Assume that auditors are randomly assigned to cases, and that one of the ways the IRS oversees its auditors is to monitor the percentage of cases that result in the taxpayer paying more taxes. If a sample of 400 cases handled by an individual auditor has 77.0% of those she audited paying more taxes, is there a reason to believe her overall "pay more" percentage might be some value other than 80%? Use the 0.10 level of significance in reaching a conclusion. Determine and interpret the p-value for the test.
Ho: p = .80 -->meaning: population proportion is equal to .80
Ha: p does not equal .80 -->meaning: population proportion does not equal .80
Using a formula for a binomial proportion one-sample z-test with your data included, we have:
z = .77 - .80 / √[(.80)(.20)/400]
Finish the calculation.
Use a z-table to find the p-value.
(The p-value is the actual level of the test statistic.) Determine whether or not to reject the null.
I hope this will help get you started.