State your hypotheses and decision rule. ?

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

(a) ANSWER: No! The "true average" out-of-pocket billing in excess of reimbursed amount to a significance value α = 0.05 does not exceed 250.00 dollars.

(b) ANSWER: Yes!  P-value > α, [0.0568 > 0.05] for statistical hypothesis testing is "a close decision" by (approx) 0.0068 proportion of the t-DISTRIBUTION.

Why??

SINGLE SAMPLE TEST, ONE-TAILED, 7 - Step Procedure for t Distributions, SMALL SAMPLE SIZE

1. Parameter of interest: "μ" = population mean out-of-pocket billing above reimbursed amount.

2. Null hypothesis Ho: μ = 250 dollars (out-of-pocket patient billing above reimbursed amount)

3. Alternative hypothesis Ha: μ > 250 dollars ("one-tailed" because "average patient should pay no more than $250 out-of-pocket")

4. Test statistic formula: t = (x-bar - μ)/(s/SQRT(n))

x-bar = estimate of the Population Mean (statistical mean of the sample) [275.66]

n = number of individuals in the sample [25]

s = sample standard deviation [78.11]

μ = Population Mean [250.00] (used for Test statistic)

5. Computation of Test statistic formula t = (approx) 1.64

6. Determination of the P-value based on n -1 = 24 df (degrees of freedom). Table "look-up" value shows area under the 24 df curve to the right of t = 1.64 is (approximately) 0.0568

Excel Formula =TDIST(1.64, 24, 1) which is [0.05675728] the value in the "right-tail" is "P-value". Note: Excel Formula requires "positive t" and t-DISTRIBUTION is symmetrical so that "right-tail" equals "left-tail".

SMALL SAMPLE SIZE requires "t-DISTRIBUTION".

7. Conclusion: with significance value α = 0.05 the above shows P-value > α, [0.0568 > 0.05]. Null hypothesis Ho: μ = 250.00 dollars should not be rejected. The "true average" out-of-pocket billing in excess of reimbursed amount to a significance value α = 0.05 does not exceed 250.00 dollars.