An auditor reviewed 25 oral surgery insurance claims from a particular surgical

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?
(b) Is this a close decision?

Explanation / Answer

In order for this test to be valid the data must come from a normal population. If this is not the case then this test is not valid and other methods, such as a randomization test or permutation test should be used. Assuming the normality assumption is valid to test the null hypothesis H0: µ = ? or H0: µ = ? or H0: µ = ? Find the test statistic t = (xbar - ? ) / (sx / v (n)) where xbar is the sample average sx is the sample standard deviation, if you know the population standard deviation, s , then replace sx with s in the equation for the test statistic. n is the sample size and t follows the Student t distribution with n - 1 degrees of freedom. We use the Student t distribution to account for the uncertainty in the estimate of the variance. As the degrees of freedom approach infinity the Student t converges in probability to the Standard Normal. In most cases the values of the percentiles of the Student t are close enough to the Standard Normal when the degrees of freedom are greater than 30. This is the source of the empirical rule of thumb that samples of size > 30 have a mean that is normally distributed. Keep that in mind as well, for these hypothesis tests we are assuming the mean is normally distributed. This assumption is easy to verify if the data is normally distributed. The Central Limit Theorem accounts of all other means. The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis. H1: µ > ?; p-value is the area to the right of t H1: µ < ?; p-value is the area to the left of t H1: µ ? ?; p-value is the area in the tails greater than |t| If the p-value is less than or equal to the significance level a, i.e., p-value = a, then we reject the null hypothesis and conclude the alternate hypothesis is true. If the p-value is greater than the significance level, i.e., p-value > a, the significance level then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible. The hypothesis test in this question is: H0: µ = 250 vs. H1: µ > 250 The test statistic is: t = ( 275.66 - 250 ) / ( 78.11 / v ( 25 )) t = 1.642555 The p-value = P( t_ 24 > t ) = P( t_ 24 > 1.642555 ) = 0.05675728 Since the p-value is greater than the significance level of 0.05 we fail to reject the null hypothesis and conclude µ = 250 is plausible.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.