Ho: μ=$250
Ha: μ>$250
Calculate a Z-score.
Z = (x-μ)/SD
Z = ($275.66 - $250)/$78.11
Since you are only interested in finding if the amount exceeds the expected mean, you have a one-tailed test.
In a table in the back of your statistics text labeled something like "area under the normal distribution," find the calculated Z-score. If the proportion distant from the mean (in the smaller portion) is <.05, the difference is significant. How much the proportion differs from .05 will determine how close you think the decision is.
I hope this helps. Thanks for asking.
An auditor reviewed 25 oral surgery insurance claims from a particular surgical office, determining that the mean out-of-pocker patient billing above the reimbursed amount was $275.66 with a standard deviation of $78.11. (a)at the 5% level of significance, soes this sample prove a violation of the guideline that the average patient should pay no more than $250 ou-of-pocket? State your hypotheses and decision rule. (b) is this a close decision.?
3 answers
Thank you so very much for your help.
You are welcome.