# here - Pearson ```Statistics for Managers using
Microsoft Excel
6th Global Edition
Chapter 10
Two-Sample Tests
10-1
Learning Objectives
In this chapter, you learn:

How to use hypothesis testing for comparing the
difference between

The means of two independent populations

The means of two related populations

The proportions of two independent populations

The variances of two independent populations by
testing the ratio of the two variances
10-2
Two-Sample Tests
DCOVA
Two-Sample Tests
Population
Means,
Independent
Samples
Population
Means,
Related
Samples
Population
Proportions
Population
Variances
Examples:
Group 1 vs.
Group 2
Same group
before vs. after
treatment
Proportion 1 vs.
Proportion 2
Variance 1 vs.
Variance 2
10-3
Difference Between Two Means
DCOVA
Population means,
independent
samples
*
σ1 and σ2 unknown,
assumed equal
σ1 and σ2 unknown,
not assumed equal
Goal: Test hypothesis or form
a confidence interval for the
difference between two
population means, μ1 – μ2
The point estimate for the
difference is
X1 – X2
10-4
Difference Between Two Means:
Independent Samples DCOVA

Population means,
independent
samples
*
Different data sources


Unrelated
Independent

Sample selected from one
population has no effect on the
sample selected from the other
population
σ1 and σ2 unknown,
assumed equal
Use Sp to estimate unknown
σ. Use a Pooled-Variance t
test.
σ1 and σ2 unknown,
not assumed equal
Use S1 and S2 to estimate
unknown σ1 and σ2. Use a
Separate-variance t test
10-5
Hypothesis Tests for
Two Population Means DCOVA
Two Population Means, Independent Samples
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: μ1  μ2
H1: μ1 &lt; μ2
H0: μ1 ≤ μ2
H1: μ1 &gt; μ2
H0: μ1 = μ2
H1: μ1 ≠ μ2
i.e.,
i.e.,
i.e.,
H0: μ1 – μ2  0
H1: μ1 – μ2 &lt; 0
H0: μ1 – μ2 ≤ 0
H1: μ1 – μ2 &gt; 0
H0: μ1 – μ2 = 0
H1: μ1 – μ2 ≠ 0
10-6
Hypothesis tests for μ1 – μ2
DCOVA
Two Population Means, Independent Samples
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: μ1 – μ2  0
H1: μ1 – μ2 &lt; 0
H0: μ1 – μ2 ≤ 0
H1: μ1 – μ2 &gt; 0
H0: μ1 – μ2 = 0
H1: μ1 – μ2 ≠ 0
a
a
-ta
Reject H0 if tSTAT &lt; -ta
ta
Reject H0 if tSTAT &gt; ta
a/2
-ta/2
a/2
ta/2
Reject H0 if tSTAT &lt; -ta/2
or tSTAT &gt; ta/2
10-7
Hypothesis tests for &micro;1 - &micro;2 with σ1
and σ2 unknown and assumed equal
DCOVA
Assumptions:
Population means,
independent
samples
σ1 and σ2 unknown,
assumed equal
σ1 and σ2 unknown,
not assumed equal
 Samples are randomly and
independently drawn
*
 Populations are normally
distributed or both sample
sizes are at least 30
 Population variances are
unknown but assumed equal
10-8
Hypothesis tests for &micro;1 - &micro;2 with σ1
and σ2 unknown and assumed equal
(continued)
DCOVA
• The pooled variance is:
Population means,
independent
samples
σ1 and σ2 unknown,
assumed equal
Sp 
2
*
n 1  1S 12  n 2  1S 2 2
(n 1  1)  ( n 2  1)
• The test statistic is:
t STAT

X1  X 2   μ1  μ 2 

2
Sp
σ1 and σ2 unknown,
not assumed equal
1
1


n
n2
 1




• Where tSTAT has d.f. = (n1 + n2 – 2)
10-9
Confidence interval for &micro;1 - &micro;2 with σ1
and σ2 unknown and assumed equal
DCOVA
Population means,
independent
samples
σ1 and σ2 unknown,
assumed equal
σ1 and σ2 unknown,
not assumed equal
*
The confidence interval for
μ1 – μ2 is:
 X 1  X 2   t α /2
2 
Sp

1 



n
n
2 
 1
1
Where tα/2 has d.f. = n1 + n2 – 2
10-10
Pooled-Variance t Test Example
DCOVA
You are a financial analyst for a brokerage firm. Is there
a difference in dividend yield between stocks listed on the
NYSE &amp; NASDAQ? You collect the following data:
NYSE NASDAQ
Number
21
25
Sample mean
3.27
2.53
Sample std dev 1.30
1.16
Assuming both populations are
approximately normal with
equal variances, is
there a difference in mean
yield (a = 0.05)?
10-11
Pooled-Variance t Test Example:
Calculating the Test Statistic
(continued)
H0: μ1 - μ2 = 0 i.e. (μ1 = μ2)
H1: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)
DCOVA
The test statistic is:

X 1  X 2   μ 1  μ 2 
t

2
Sp
S 
2
p
 1
1


n
n2
 1
n 1  1S 12  n 2
 1S 2
(n 1  1)  ( n 2  1)




3.27
 2.53   0
 2.040
1 
 1
1 . 5021 


25 
 21
2

21  11.30
2
 25  11.16
(21 - 1)  ( 25  1)
2
 1.5021
10-12
Pooled-Variance t Test Example:
Hypothesis Test Solution
DCOVA
H0: μ1 - μ2 = 0 i.e. (μ1 = μ2)
H1: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)
a = 0.05
df = 21 + 25 - 2 = 44
Critical Values: t = &plusmn; 2.0154
t
1 
 1
1 . 5021 


 21 25 
.025
-2.0154
Reject H0
.025
0 2.0154
t
2.040
Test Statistic:
3.27  2.53
Reject H0
 2.040
Decision:
Reject H0 at a = 0.05
Conclusion:
There is evidence of a
difference in means.
10-13
Pooled-Variance t Test Example:
Confidence Interval for &micro;1 - &micro;2
DCOVA
Since we rejected H0 can we be 95% confident that &micro;NYSE
&gt; &micro;NASDAQ?
95% Confidence Interval for &micro;NYSE - &micro;NASDAQ
X
1

 X 2  t a /2
 1
1 

  0 . 74  2 . 0154  0 . 3628  ( 0 . 009 , 1 . 471 )
S 


 n1 n 2 
2
p
Since 0 is less than the entire interval, we can be 95%
confident that &micro;NYSE &gt; &micro;NASDAQ
10-14
Hypothesis tests for &micro;1 - &micro;2 with σ1
and σ2 unknown, not assumed equal
DCOVA
Assumptions:
Population means,
independent
samples
 Samples are randomly and
independently drawn
 Populations are normally
distributed or both sample
sizes are at least 30
σ1 and σ2 unknown,
assumed equal
σ1 and σ2 unknown,
not assumed equal
*
 Population variances are
unknown and cannot be
assumed to be equal
10-15
Related Populations
The Paired Difference Test
DCOVA
Tests Means of 2 Related Populations
Related
samples



Paired or matched samples
Repeated measures (before/after)
Use difference between paired values:
Di = X1i - X2i


Eliminates Variation Among Subjects
Assumptions:
 Both Populations Are Normally Distributed
 Or, if not Normal, use large samples
10-16
Related Populations
The Paired Difference Test
DCOVA
(continued)
The ith paired difference is Di , where
Related
samples
Di = X1i - X2i
The point estimate for the
paired difference
D
population mean μD is D :
n
D
i
i 1
n
n
The sample standard
deviation is SD
SD 
 (Di  D)
2
i 1
n 1
n is the number of pairs in the paired sample
10-17
The Paired Difference Test:
Finding tSTAT
DCOVA

Paired
samples
The test statistic for μD is:
t STAT 
D μD
SD
n

Where tSTAT has n - 1 d.f.
10-18
The Paired Difference Test:
Possible Hypotheses DCOVA
Paired Samples
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: μD  0
H1: μD &lt; 0
H0: μD ≤ 0
H1: μD &gt; 0
H0: μD = 0
H1: μD ≠ 0
a
a
-ta
Reject H0 if tSTAT &lt; -ta
ta
Reject H0 if tSTAT &gt; ta
Where tSTAT has n - 1 d.f.
a/2
-ta/2
a/2
ta/2
Reject H0 if tSTAT &lt; -ta/2
or tSTAT &gt; ta/2
10-19
The Paired Difference
Confidence Interval
Paired
samples
DCOVA
The confidence interval for μD is
D  ta / 2
SD
n
n
where
SD 
 (D
i
 D)
2
i 1
n 1
10-20
Paired Difference Test:
Example
DCOVA
Assume you send your salespeople to a “customer
service” training workshop. Has the training made a
difference in the number of complaints? You collect
the following data:

Number of Complaints:
(2) - (1)
Salesperson Before (1) After (2)
Difference, Di
C.B.
T.F.
M.H.
R.K.
M.O.
6
20
3
0
4
4
6
2
0
0
- 2
-14
- 1
0
- 4
-21
D =
 Di
n
= -4.2
SD 
 (Di  D)
2
n 1
 5.67
10-21
Paired Difference Test:
DCOVA
Solution
 Has the training made a difference in the number of
complaints (at the 0.01 level)?
H0: μD = 0
H1: μD  0
a = .01
D = - 4.2
t0.005 = &plusmn; 4.604
t STAT 

SD / n
5.67/ 5
a/2
a/2
- 4.604
4.604
- 1.66
(tstat is not in the reject region)
Test Statistic:
 4.2  0
Reject
Decision: Do not reject H0
d.f. = n - 1 = 4
D  μD
Reject
 1.66
Conclusion: There is not a
significant change in the
number of complaints.
10-22
Two Population Proportions
DCOVA
Goal: test a hypothesis or form a
confidence interval for the difference
between two population proportions,
π1 – π2
Population
proportions
Assumptions:
n1 π1  5 , n1(1- π1)  5
n2 π2  5 , n2(1- π2)  5
The point estimate for
the difference is
p1  p 2
10-23
Two Population Proportions
DCOVA
Population
proportions
In the null hypothesis we assume the
null hypothesis is true, so we assume π1
= π2 and pool the two sample estimates
The pooled estimate for the
overall proportion is:
p 
X1  X 2
n1  n 2
where X1 and X2 are the number of items of
interest in samples 1 and 2
10-24
Two Population Proportions
(continued)
DCOVA
The test statistic for
π1 – π2 is a Z statistic:
Population
proportions
Z STAT 
where
p 
 p1  p 2    π1  π 2 
 1
1 

p (1  p ) 

 n1 n 2 
X1  X 2
n1  n 2
, p1 
X1
n1
, p2 
X2
n2
10-25
Hypothesis Tests for
Two Population Proportions
DCOVA
Population proportions
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: π1  π2
H1: π1 &lt; π2
H0: π1 ≤ π2
H1: π1 &gt; π2
H0: π1 = π2
H1: π1 ≠ π2
i.e.,
i.e.,
i.e.,
H0: π1 – π2  0
H1: π1 – π2 &lt; 0
H0: π1 – π2 ≤ 0
H1: π1 – π2 &gt; 0
H0: π1 – π2 = 0
H1: π1 – π2 ≠ 0
10-26
Hypothesis Tests for
Two Population Proportions
(continued)
Population proportions
DCOVA
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: π1 – π2  0
H1: π1 – π2 &lt; 0
H0: π1 – π2 ≤ 0
H1: π1 – π2 &gt; 0
H0: π1 – π2 = 0
H1: π1 – π2 ≠ 0
a
a
-za
Reject H0 if ZSTAT &lt; -Za
za
Reject H0 if ZSTAT &gt; Za
a/2
-za/2
a/2
za/2
Reject H0 if ZSTAT &lt; -Za/2
or ZSTAT &gt; Za/2
10-27
Hypothesis Test Example:
Two population Proportions
DCOVA
Is there a significant difference between the
proportion of men and the proportion of
women who will vote Yes on Proposition A?

In a random sample, 36 of 72 men and 35 of
50 women indicated they would vote Yes

Test at the .05 level of significance
10-28
Hypothesis Test Example:
Two population Proportions
(continued)

DCOVA
The hypothesis test is:
H0: π1 – π2 = 0 (the two proportions are equal)
H1: π1 – π2 ≠ 0 (there is a significant difference between proportions)

The sample proportions are:

Men:
p1 = 36/72 = 0.50

Women:
p2 = 35/50 = 0.70
 The pooled estimate for the overall proportion is:
p 
X1  X 2
n1  n 2

36  35
72  50

71
 .582
122
10-29
Hypothesis Test Example:
Two population Proportions
DCOVA
(continued)
The test statistic for π1 – π2 is:
z STAT 

 p1  p 2     1   2 
 1
1 

p (1  p ) 


 n1 n 2 
 .50
 .70    0 
1 
 1
.582 (1  .582) 


72
50


Critical Values = &plusmn;1.96
For a = .05
Reject H0
Reject H0
.025
.025
-1.96
-2.20
  2.20
1.96
Decision: Do not reject H0
Conclusion: There is not
significant evidence of a
difference in proportions
who will vote yes between
men and women.
10-30
Confidence Interval for
Two Population Proportions
DCOVA
Population
proportions
 p1  p 2 
The confidence interval for
π1 – π2 is:
 Z a /2
p 1 (1  p 1 )
n1

p 2 (1  p 2 )
n2
10-31
Testing for the Ratio Of Two
Population Variances DCOVA
Tests for Two
Population
Variances
F test statistic
*
Hypotheses
H0: σ12 = σ22
H1: σ12 ≠ σ22
H0: σ12 ≤ σ22
H1: σ12 &gt; σ22
FSTAT
S12 / S22
Where:
S12 = Variance of sample 1 (the larger sample variance)
n1 = sample size of sample 1
S22 = Variance of sample 2 (the smaller sample variance)
n2 = sample size of sample 2
n1 –1 = numerator degrees of freedom
n2 – 1 = denominator degrees of freedom
10-32
The F Distribution
DCOVA

The F critical value is found from the F table

There are two degrees of freedom required: numerator
and denominator

The larger sample variance is always the numerator
2


When F STAT 
S1
S
2
2
df1 = n1 – 1 ; df2 = n2 – 1
In the F table,

numerator degrees of freedom determine the column

denominator degrees of freedom determine the row
10-33
Finding the Rejection Region
DCOVA
H0: σ12 = σ22
H1: σ12 ≠ σ22
H0: σ12 ≤ σ22
H1: σ12 &gt; σ22
a/2
0
Do not
reject H0
Fα/2
Reject H0
Reject H0 if FSTAT &gt; Fα/2
a
F
0
Do not
reject H0
Fα
Reject H0
F
Reject H0 if FSTAT &gt; Fα
10-34
F Test: An Example
DCOVA
You are a financial analyst for a brokerage firm. You
want to compare dividend yields between stocks listed
on the NYSE &amp; NASDAQ. You collect the following data:
NYSE
NASDAQ
Number
21
25
Mean
3.27
2.53
Std dev
1.30
1.16
Is there a difference in the
variances between the NYSE
&amp; NASDAQ at the a = 0.05 level?
10-35
F Test: Example Solution
DCOVA

Form the hypothesis test:
H0: σ21 = σ22 (there is no difference between variances)
H1: σ21 ≠ σ22 (there is a difference between variances)

Find the F critical value for a = 0.05:

Numerator d.f. = n1 – 1 = 21 –1 =20

Denominator d.f. = n2 – 1 = 25 –1 = 24

Fα/2 = F.025, 20, 24 = 2.33
10-36
F Test: Example Solution
DCOVA
(continued)

The test statistic is:
2
F STAT 
S1
S
2
2

1 . 30
2
1 . 16
2
H0: σ12 = σ22
H1: σ12 ≠ σ22
 1 . 256
a/2 = .025
0
Do not
reject H0

FSTAT = 1.256 is not in the rejection
region, so we do not reject H0

Conclusion: There is not sufficient evidence
of a difference in variances at a = .05
Reject H0
F
F0.025=2.33
10-37
Chapter Summary

Compared two independent samples




Performed pooled-variance t test for the difference in
two means
Performed separate-variance t test for difference in
two means
Formed confidence intervals for the difference
between two means
Compared two related samples (paired
samples)

Performed paired t test for the mean difference

Formed confidence intervals for the mean difference
10-38
Chapter Summary
(continued)

Compared two population proportions



Formed confidence intervals for the difference
between two population proportions
Performed Z-test for two population proportions
Performed F test for the ratio of two
population variances