An auditor for a government agency is assigned the task of evaluating reimbursem
ID: 3433969 • Letter: A
Question
An auditor for a government agency is assigned the task of evaluating reimbursement for office visits to physicians paid by Medicare. The audit was conducted on a random sample of 75 of the reimbursements, with the following results:
- In 12 of the office visits, an incorrect amount of reimbursement was provided.
- The amount of reimbursement was Xbar = $93.70, s = $34.55.
a. At the 0.05 level of significance, is there evidence that the population mean reimbursement was less than $100?
b. At the 0.05 level of significance, is there evidence that the proportion of incorrect reimbursements in the population was greater than 0.10?
Explanation / Answer
(a) Let mu be the population mean
The test hypothesis:
Ho: mu=100 (i.e. null hypothesis)
Ha: mu<100 (i.e. alternative hypothesis)
The test statistic is
Z=(xbar-mu)/(s/vn)
=(93.7-100)/(34.55/sqrt(75))
=-1.58
It is a left-tailed test.
Given a=0.05, the critical value is Z(0.05) =-1.645 (from standard normal table)
The rejection region is if Z<-1.645, we reject the null hypothesis.
Since Z=-1.58 is larger than -1.645, we do not reject the null hypothesis.
So we can not conclude that the population mean reimbursement was less than $100
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(b) Ho: p=0.1
Ha: p>0.1
The test statistic is
Z=(phat-p)/sqrt(p*(1-p)/n)
=(12/75-0.1)/sqrt(0.1*0.9/75)
=1.73
It is a right-tailed test.
Given a=0.05, the critical value is Z(0.05) = 1.645 (from standard normal table)
The rejection region is if Z>1.645, we reject the null hypothesis.
Since Z=1.73 is larger than 1.645, we reject the null hypothesis.
So we can conclude that the proportion of incorrect reimbursements in the population was greater than 0.10
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