# SOLUTION: 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

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 Click here to see ALL problems on Probability-and-statistics Question 174373: 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 by stanbon(60782)   (Show Source): You can put this solution on YOUR website!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. --- Ho: u = 250 Ha: u > 250 ---- Critical value for a one-tail test with alpha = 5%: z = 1.645 --- Test statistic: z(275.66) = (275.66-250)/[78.11/Sqrt(25)] = 1.6456 ------------------- Conclusion: Since the TS is not in the reject interval, Fail to reject Ho. ------------------- (b) Is this a close decision? Yes, the test statistic is very close to the critical value. p-value = P(z > 1.6456) = 0.05024, which is very close to the alpha value. There are only 5.024% of test results that could have provided stronger evidence for rejecting Ho. ============================ Cheers, Stan H.