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?

Question

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?

Answer 1

You can use a one-sample t-test since this sample size is fairly small.

Calculating:
t = (275.66 - 250)/(78.11/√25) = ?

I'll let you finish the calculation.

Use a t-table at .05 level of significance for a one-tailed test (the alternate hypothesis will show a specific direction) at 24 degrees of freedom (df = n - 1 = 25 - 1 = 24).

If the t-test statistic calculated above exceeds the critical or cutoff value from the t-table, reject the null. If the t-test statistic calculated above does not exceed the critical or cutoff value from the t-table, do not reject the null.

I'll let you take it from here.