Active management of labor (AML) is a group of interventions designed to help re

Question

Active management of labor (AML) is a group of interventions designed to help reduce the length of labor and the rate of cesarean deliveries. Physicians from the Department of Obstetrics and Gynecology at the University of New Mexico Health Sciences Center were interested in determining whether AML would also translate into reduced cost of delivery. According to their study, 200 AML deliveries had mean cost of $2480 with standard deviation of $766. At the time of the study, the average cost of having a baby in a U.S. hospital was $2528. (a) At 5% significance level, do the data provide sufficient evidence to conclude that, on average, AML reduces the average cost of delivery in a U.S. hospital? (b) Calculate the p-value of the test.

Explanation / Answer

Given that,
population mean(u)=2480
standard deviation, =766
sample mean, x =2528
number (n)=200
null, Ho: =2480
alternate, H1: >2480
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 2528-2480/(766/sqrt(200)
zo = 0.88619
| zo | = 0.88619
critical value
the value of |z | at los 5% is 1.645
we got |zo| =0.88619 & | z | = 1.645
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : right tail - ha : ( p > 0.88619 ) = 0.18776
hence value of p0.05 < 0.18776, here we do not reject Ho

ANSWERS
---------------
null, Ho: =2480
alternate, H1: >2480
test statistic: 0.88619
critical value: 1.645
decision: do not reject Ho
p-value: 0.18776
we don't have enough to support the claim

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