An auditor reviewed 22 oral surgery insurance claims from a particular surgical

Question

An auditor reviewed 22 oral surgery insurance claims from a particular surgical office, determining that the mean out-of-pocket patient billing above the reimbursed amount was $281.54 with a standard deviation of $77.58.

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?

H0: $250 versus H1: > $250. Choose the right option.

Calculate the test statistic. (Round your answer to 2 decimal places.)

An auditor reviewed 22 oral surgery insurance claims from a particular surgical office, determining that the mean out-of-pocket patient billing above the reimbursed amount was $281.54 with a standard deviation of $77.58.

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?

Explanation / Answer

Given that,
population mean(u)=250
sample mean, x =281.54
standard deviation, s =77.58
number (n)=22
null, Ho: <250
alternate, H1: >250
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.721
since our test is right-tailed
reject Ho, if to > 1.721
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =281.54-250/(77.58/sqrt(22))
to =1.907
| to | =1.907
critical value
the value of |t | with n-1 = 21 d.f is 1.721
we got |to| =1.907 & | t | =1.721
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 1.9069 ) = 0.03516
hence value of p0.05 > 0.03516,here we reject Ho
ANSWERS
---------------
null, Ho: <=250
alternate, H1: >250
test statistic: 1.907
critical value: 1.721
decision: reject Ho
p-value: 0.03516

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