An inspector of flow metering devices used to administer fluid intravenously wil

Question

An inspector of flow metering devices used to administer fluid intravenously will perform a hypothesis test to determine whether the mean flow rate is different from the flow rate setting of 200 milliliters per hour. Based on prior information, the standard deviation of the flow rate is assumed to be known and equal to 2 milters per hour. You intend to perform an appropriate hypothesis test using a level of significance of 0.05 on n = 18 devices. A) What is the probability that you reject the null hypothesis if the true mean flow rate is 200mhr B) Suppose you collect the data and observe a sample average of (Xbar) = 198.1 mL/hr. State and interpret the conclusion of this test. C) From part B, what is the smallest level of significance that would result in rejection of the null hypothesis.

Explanation / Answer

a.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 200
standard Deviation ( sd )= 12/ Sqrt ( 18 ) =2.8284
sample size (n) = 18
probability that
P(X < 200) = (200-200)/12/ Sqrt ( 18 )
= 0/2.8284= 0
= P ( Z <0) From Standard NOrmal Table
= 0.5 reject null hypothesis when true mean = 200ml/hr
b.
Given that,
population mean(u)=200
standard deviation, =12
sample mean, x =198.1
number (n)=18
null, Ho: =200
alternate, H1: !=200
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 198.1-200/(12/sqrt(18)
zo = -0.67175
| zo | = 0.67175
critical value
the value of |z | at los 5% is 1.96
we got |zo| =0.67175 & | z | = 1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -0.67175 ) = 0.50174
hence value of p0.05 < 0.50174, here we do not reject Ho
ANSWERS
---------------
null, Ho: =200
alternate, H1: !=200
test statistic: -0.67175
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.50174
we donot have enough evidence to support the claim

c.
Given that,
population mean(u)=200
standard deviation, =12
sample mean, x =198.1
number (n)=18
null, Ho: =200
alternate, H1: !=200
level of significance, = 0.51
from standard normal table, two tailed z /2 =0.659
since our test is two-tailed
reject Ho, if zo < -0.659 OR if zo > 0.659
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 198.1-200/(12/sqrt(18)
zo = -0.67175
| zo | = 0.67175
critical value
the value of |z | at los 51% is 0.659
we got |zo| =0.67175 & | z | = 0.659
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -0.67175 ) = 0.50174
hence value of p0.51 > 0.50174, here we reject Ho
ANSWERS
---------------
null, Ho: =200
alternate, H1: !=200
test statistic: -0.67175
critical value: -0.659 , 0.659
decision: reject Ho
p-value: 0.50174

level of significance, = 0.51 reject the null hypothesis

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.