Module 27: Two Sample t-tests With
Unequal Variances
This module shows how to test the hypothesis that two
population means are equal when there is evidence that
the requirement that the two populations have the same
variance is not met.
REVIEWED 19 July 05 /MODULE 27
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The General Situation
Earlier we tested hypotheses about two
population means, based on data from two
independent samples under the assumption that
the two populations had the same variance.
For this situation, we calculated the pooled
estimate of the common variance with:
s
2
p

n1  1s  n2  1s

n1  1  n2  1
2
1
2
2
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We then tested the null hypothesis:
H0 : µ1= µ2 vs. H1: µ1 ≠ µ2
with the test statistic:
x1  x2
t
sp
1
1

n1
n2
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What if σ12 ≠ σ22
If the two population variances are not the same,
then they are estimated separately and:
 2  2
Var ( x1  x2 )   1  2
 n1
n2

is estimated by:




s2 s 2
 1  2 
 n1

n
2


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When we have two independent random samples
from two different populations and there is evidence
that the two population variances differ, i.e.
12 ≠ 22, then we can base a test about the
difference between the population means 1 and 2
on the following:
t
x1  x2
2
1
2
2
 t( f )
s
s

n1 n2
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To deal with 12 ≠ 22, we are adjusting the
degrees of freedom in order to obtain a better
approximation than would otherwise be the case.
This adjustment requires that we calculate:
2
s
s 
  
n1 n2 

f 
2
2
2
2
2
 s1 
 s2 
 
 
 n1    n2 
n1  1
n2  1
2
1
2
2
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Example
Independent random samples were taken from
populations of histidine levels for males and
females. The measurements were approximately
normally distributed .
The statistics from the two samples were:
n
s
s2
Males
5
300.8
15,291.7
Females
10
153.2
2,484.62
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Test the hypothesis H0: M = F vs. H1: M  F
1. The hypothesis:
H0: M = F vs H1: M  F
2. The assumptions:
Independent samples, normal
distributions
3. The  - level:
 = 0.05
4. The test statistic:
t
x1  x2
2
1
2
2
s
s

n1 n2
t f 
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5. The critical region:
Reject if t is not between
 t0.975(f), where
2
15, 291.7 2, 484.6 


5
10 
f 
2
2
2
15, 291.7 
 2, 484.6 


 10 
5

5 1
10  1
so that f = 6.99-2 = 4.99. We will use f = 5, and
t0.975(5) = 2.57058.
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6. The result:
300.8  153.2
t
15, 291.7 2, 486.6

5
10
147.6
t
 2.57
57.50
7. The conclusion: The value calculated for t above is
very close to the cut point so that P  0.05
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99% Confidence Interval

s12 s22
s12 s22 
C  x1  x2  t1 / 2 ( f )
  1  2  x1  x2  t1 / 2 ( f )
   0.99
n1 n2
n1 n2 

for n1 = 5,
n2 =10,
for which
x1  300.8 ,
s12  15, 291.70
x2  153.2 ,
s22  2484.62
s12
s22

 3, 306.8
n1
n2
f = 5,
t0.995(5) = 4.0321
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We have


C 300.8  153.2  4.0321 3,306.8  1   2  300.8  153.2  4.0321 3,306.8  0.99
C147.6  231.87  1  2  147.6  231.87  0.99
C 84.27  1  2  379.47  0.99
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Example: Am J Public Health, 1996;Oct:1436
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Example: Am J Public Health, 1999;Dec:1692
Table 1 in the article by Sargent JD et al provides data
on blood lead levels for samples from two communities.
n
x
SD
Providence
(1)
136
6.7
2.06
Test the hypothesis that
 12  6.0
Test the hypothesis that
1  2
Test the hypothesis that
 12   22
Worcester
(2)
153
5.4
1.10
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