An auditor reviewed 25 oral surgery insurance claims from a particular surgical office, determining that the mean out-of-pocket patient billing above the reimbursed amount was $275.66 with a standard deviation of $78.11.
(a) At the 5 percent level of significance, does this sample prove a violation of the guideline that the average patient should pay no more than $250 out-of-pocket? State your hypotheses and decision rule.
(b) Is this a close decision
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
Since only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
Ho: No difference
Ha: Difference (State in more specific terms.)
(Use .025 for two-tailed test.)
To determine whether there is a violation of the guideline that the average patient should pay no more than $250 out-of-pocket, we need to conduct a hypothesis test.
(a) Hypotheses:
Null hypothesis (H0): The mean out-of-pocket patient billing is equal to or less than $250.
Alternative hypothesis (H1): The mean out-of-pocket patient billing is greater than $250.
Decision Rule:
Since the alternative hypothesis is that the mean out-of-pocket patient billing is greater than $250, this is a one-tailed test.
At the 5 percent level of significance, we will reject the null hypothesis if the test statistic is greater than the critical value for a one-tailed test with the significance level of 5 percent.
To perform the hypothesis test, we will use the sample mean, standard deviation, and the number of observations.
Z = (sample mean - population mean) / (standard deviation / sqrt(sample size))
Z = ($275.66 - $250) / ($78.11 / sqrt(25)) = $25.66 / ($78.11 / 5) = 1.65
Using a Z-table or calculator, we can find that the critical value for a one-tailed test at the 5 percent level of significance is approximately 1.645.
Since the calculated test statistic (1.65) is greater than the critical value (1.645), we reject the null hypothesis.
Decision:
At the 5 percent level of significance, we can conclude that there is evidence to support a violation of the guideline that the average patient should pay no more than $250 out-of-pocket.
(b) Since the calculated test statistic is only slightly greater than the critical value, this can be considered a close decision. However, the decision is still made to reject the null hypothesis.