An auditor for a government agency is assigned the task of evaluating reimbursem

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 had a 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

PART A.
Given that,
population mean(u)=100
sample mean, x =93.7
standard deviation, s =34.55
number (n)=75
null, Ho: =100
alternate, H1: <100
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.666
since our test is left-tailed
reject Ho, if to < -1.666
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =93.7-100/(34.55/sqrt(75))
to =-1.5791
| to | =1.5791
critical value
the value of |t | with n-1 = 74 d.f is 1.666
we got |to| =1.5791 & | t | =1.666
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :left tail - Ha : ( p < -1.5791 ) = 0.05928
hence value of p0.05 < 0.05928,here we do not reject Ho
ANSWERS
---------------
null, Ho: =100
alternate, H1: <100
test statistic: -1.5791
critical value: -1.666
decision: do not reject Ho
p-value: 0.05928

no evidence that population mean is less than 100

PART B.
Given that,
possibile chances (x)=12
sample size(n)=75
success rate ( p )= x/n = 0.16
success probability,( po )=0.1
failure probability,( qo) = 0.9
null, Ho:p=0.1  
alternate, H1: p>0.1
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.64
since our test is right-tailed
reject Ho, if zo > 1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.16-0.1/(sqrt(0.09)/75)
zo =1.7321
| zo | =1.7321
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =1.732 & | z | =1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 1.73205 ) = 0.04163
hence value of p0.05 > 0.04163,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.1
alternate, H1: p>0.1
test statistic: 1.7321
critical value: 1.64
decision: reject Ho
p-value: 0.04163

the proportion of incorrect reimbursements in the population was greater than 0.10

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