Question

Healthcare Associates operates 33 medical clinics in Iowa. As part of the company’s ongoing efforts to examine its customer service, Healthcare worked with an MBA class at the University of Iowa on a project in which a team of students observed patients at the clinics to estimate the difference in mean time spent per visit for male and female patients. Previous studies show that the standard deviation is 11 minutes for men and 16 minutes for women. The MBA students observe 100 male and 100 female patients and discover the mean time per visit for men is fifty minutes, and for women is fifty-five.

(a) Develop a 95% confidence interval estimate for the difference in population mean time per visit by gender.

(b) Test at 5% significance the hypothesis that mean visit time is exactly 50 minutes for women. Based on your conclusion, are you more likely to have made a type I or type II error? What is the probability of this error?

(c) What is the p-value for the test in part (b)?

Answer 1

a.
TRADITIONAL METHOD
given that,
mean(x)=50
standard deviation , σ1 =11
population size(n1)=100
y(mean)=55
standard deviation, σ2 =16
population size(n2)=100
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((121/100)+(256/100))
= 1.9416
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 1.9416
= 3.8056
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (50-55) ± 3.8056 ]
= [-8.8056 , -1.1944]
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DIRECT METHOD
given that,
mean(x)=50
standard deviation , σ1 =11
number(n1)=100
y(mean)=55
standard deviation, σ2 =16
number(n2)=100
CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 50-55) ±Z a/2 * Sqrt( 121/100+256/100)]
= [ (-5) ± Z a/2 * Sqrt( 3.77) ]
= [ (-5) ± 1.96 * Sqrt( 3.77) ]
= [-8.8056 , -1.1944]
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interpretations:
1. we are 95% sure that the interval [-8.8056 , -1.1944] contains the difference between
true population mean U1 - U2
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean U1 - U2
3. Since this Cl does contain a zero we can conclude at 0.05 true mean
difference is zero
b.
Given that,
population mean(u)=50
standard deviation, σ =16
sample mean, x =55
number (n)=100
null, Ho: μ=50
alternate, H1: μ!=50
level of significance, α = 0.05
from standard normal table, two tailed z α/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 = 55-50/(16/sqrt(100)
zo = 3.125
| zo | = 3.125
critical value
the value of |z α| at los 5% is 1.96
we got |zo| =3.125 & | z α | = 1.96
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 3.125 ) = 0.002
hence value of p0.05 > 0.002, here we reject Ho
ANSWERS
---------------
null, Ho: μ=50
alternate, H1: μ!=50
test statistic: 3.125
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.002
we have enough evidence to support the claim that mean visit time is exactly 50 minutes for women.
conclusion:
Based on your conclusion, are you more likely to have made a type I error is possible
because it is reject the null hypothesis.
c.
p-value: 0.002