Question

In constructing 95% confidence interval estimate for the difference between the means of two populations, where the unknown population variances are assumed not to be equal, summary statistics computed from two independent samples are: ?1=45,?̅1=756,?1=18,?2=40,?̅2=762,?2=15

(using 2-sample T menu)

a. Calculate the 95% confidence interval for the true difference of two means.

b. Base on the interval in the previous question, can one conclude there is a difference in means of two populations?

Justify your answer.

Answer 1

a.
TRADITIONAL METHOD
given that,
mean(x)=756
standard deviation , s.d1=18
number(n1)=45
y(mean)=762
standard deviation, s.d2 =15
number(n2)=40
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((324/45)+(225/40))
= 3.581
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 39 d.f is 2.023
margin of error = 2.023 * 3.581
= 7.245
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (756-762) ± 7.245 ]
= [-13.245 , 1.245]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=756
standard deviation , s.d1=18
sample size, n1=45
y(mean)=762
standard deviation, s.d2 =15
sample size,n2 =40
CI = x1 - x2 ± t 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)
ta/2 = t-table value
CI = confidence interval
CI = [( 756-762) ± t a/2 * sqrt((324/45)+(225/40)]
= [ (-6) ± t a/2 * 3.581]
= [-13.245 , 1.245]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-13.245 , 1.245] contains the true population proportion
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 true population proportion
b.
Given that,
mean(x)=756
standard deviation , s.d1=18
number(n1)=45
y(mean)=762
standard deviation, s.d2 =15
number(n2)=40
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.023
since our test is two-tailed
reject Ho, if to < -2.023 OR if to > 2.023
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =756-762/sqrt((324/45)+(225/40))
to =-1.6754
| to | =1.6754
critical value
the value of |t α| with min (n1-1, n2-1) i.e 39 d.f is 2.023
we got |to| = 1.67542 & | t α | = 2.023
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.6754 ) = 0.102
hence value of p0.05 < 0.102,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -1.6754
critical value: -2.023 , 2.023
decision: do not reject Ho
p-value: 0.102
we do not have enough evidence to support the claim that difference in means of two populations.