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.
Null hypothesis:
Ho: p = .80 -->meaning: population proportion is equal to .80
Alternate hypothesis:
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.
To determine if there is a reason to believe that the individual auditor's overall "pay more" percentage might be different from 80%, we can conduct a hypothesis test.
First, let's state the null and alternative hypotheses:
- Null Hypothesis (H0): The individual auditor's overall "pay more" percentage is 80%.
- Alternative hypothesis (H1): The individual auditor's overall "pay more" percentage is different from 80%.
Next, we need to calculate the test statistic and the corresponding p-value. Since we are comparing a sample proportion to a known population proportion, we can use the Z-test.
Here are the steps to calculate the test statistic and p-value:
1. Calculate the standard error: SE = √(p * (1-p) / n), where p is the hypothesized proportion (80%) and n is the sample size (400).
SE = √(0.8 * 0.2 / 400) = 0.02
2. Calculate the test statistic (Z-score): Z = (sample proportion - hypothesized proportion) / SE.
Z = (0.77 - 0.80) / 0.02 = -1.5
3. Determine the p-value associated with the test statistic. Since this is a two-tailed test (we're looking for deviations in either direction), we need to find the probability of observing a Z-score as extreme as -1.5 or more extreme in either direction. We can use a standard normal distribution table or a statistical software to find this probability.
Looking up the Z-score of -1.5 in the standard normal distribution table, we find that the cumulative probability is 0.0668. However, since this is a two-tailed test, we need to double this probability.
p-value = 2 * 0.0668 = 0.1336
4. Interpret the p-value: The p-value represents the probability of observing a result as extreme as, or more extreme than, the one observed, assuming that the null hypothesis is true. In this case, the p-value is 0.1336, which is greater than the specified significance level of 0.10.
Since the p-value is greater than the significance level, we fail to reject the null hypothesis. There is not enough evidence to support the claim that the individual auditor's overall "pay more" percentage is different from 80% at the 0.10 level of significance.
Therefore, based on the given data, we do not have a reason to believe that her overall "pay more" percentage might be some value other than 80%.