4. We are given two independent samples from normally distributed populations. T

Question

4. We are given two independent samples from normally distributed populations. The data is summarizeod in the table below SampleI Sample 2 121.5 12.5 96.3 16.5 15 (a) Construct a 98% confidence interval for the difference in means 2-A- Use two procedures. assuming i) equal and ii) unequal variances (that is, the pooled procedure and Welch's procedure How do the confidence intervals differ? (b) Do a hypothesis test of Ho . 2-1 against Ha : ,12 > 4-Again, use two procedures, assuming i) equal and ii) unequal variances. Report a P-value.

Explanation / Answer

a.i)
TRADITIONAL METHOD
given that,
mean(x)=96.3
standard deviation , s.d1=16.5
number(n1)=15
y(mean)=121.5
standard deviation, s.d2 =12.5
number(n2)=8
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (14*272.25 + 7*156.25) / (23- 2 )
s^2 = 233.583
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 233.583 * (1/15+1/8) )
=6.691
III.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.02
from standard normal table, two tailed and value of |t | with (n1+n2-2) i.e 21 d.f is 2.518
margin of error = 2.518 * 6.691
= 16.848
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (96.3-121.5) ± 16.848 ]
= [-42.048 , -8.352]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=96.3
standard deviation , s.d1=16.5
sample size, n1=15
y(mean)=121.5
standard deviation, s.d2 =12.5
sample size,n2 =8
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 96.3-121.5) ± t a/2 * sqrt( 233.583 * (1/15+1/8) ]
= [ (-25.2) ± 16.848 ]
= [-42.048 , -8.352]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 98% sure that the interval [-42.048 , -8.352]contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population proportion
ii)unequal
TRADITIONAL METHOD
given that,
mean(x)=96.3
standard deviation , s.d1=16.5
number(n1)=15
y(mean)=121.5
standard deviation, s.d2 =12.5
number(n2)=8
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((272.25/15)+(156.25/8))
= 6.139
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 7 d.f is 2.998
margin of error = 2.998 * 6.139
= 18.403
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (96.3-121.5) ± 18.403 ]
= [-43.603 , -6.797]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=96.3
standard deviation , s.d1=16.5
sample size, n1=15
y(mean)=121.5
standard deviation, s.d2 =12.5
sample size,n2 =8
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 = [( 96.3-121.5) ± t a/2 * sqrt((272.25/15)+(156.25/8)]
= [ (-25.2) ± t a/2 * 6.139]
= [-43.603 , -6.797]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 98% sure that the interval [-43.603 , -6.797] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population proportion
i) equal
confidence interval = [ (96.3-121.5) ± 16.848 ]
= [-42.048 , -8.352]
ii)unequal
confidence interval = [ (96.3-121.5) ± 18.403 ]
= [-43.603 , -6.797]
there is small difference between both equal and unequal variances in t test
b.
i) equal
Given that,
mean(x)=96.3
standard deviation , s.d1=16.5
number(n1)=15
y(mean)=121.5
standard deviation, s.d2 =12.5
number(n2)=8
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, = 0.02
from standard normal table,left tailed t /2 =2.19
since our test is left-tailed
reject Ho, if to < -2.19
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (14*272.25 + 7*156.25) / (23- 2 )
s^2 = 233.5833
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=96.3-121.5/sqrt((233.5833( 1 /15+ 1/8 ))
to=-25.2/6.691
to=-3.7662
| to | =3.7662
critical value
the value of |t | with (n1+n2-2) i.e 21 d.f is 2.19
we got |to| = 3.7662 & | t | = 2.19
make decision
hence value of | to | > | t | and here we reject Ho
p-value: left tail - ha : ( p < -3.7662 ) = 0.00057
hence value of p0.02 > 0.00057,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -3.7662
critical value: -2.19
decision: reject Ho
p-value: 0.00057
ii) unequal
Given that,
mean(x)=96.3
standard deviation , s.d1=16.5
number(n1)=15
y(mean)=121.5
standard deviation, s.d2 =12.5
number(n2)=8
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, = 0.02
from standard normal table,left tailed t /2 =2.52
since our test is left-tailed
reject Ho, if to < -2.52
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =96.3-121.5/sqrt((272.25/15)+(156.25/8))
to =-4.105
| to | =4.105
critical value
the value of |t | with min (n1-1, n2-1) i.e 7 d.f is 2.52
we got |to| = 4.10523 & | t | = 2.52
make decision
hence value of | to | > | t | and here we reject Ho
p-value:left tail - Ha : ( p < -4.1052 ) = 0.00227
hence value of p0.02 > 0.00227,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -4.105
critical value: -2.52
decision: reject Ho
p-value: 0.00227

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