Chapter 10
Hypothesis
Testing with
Two Samples
© 2002 Thomson / South-Western
Slide 10-1
Learning Objectives
• Test hypotheses about the difference in two
population means using data from large
independent samples.
• Test hypotheses about the difference in two
population means using data from small
independent samples when the populations
are normally distributed.
© 2002 Thomson / South-Western
Slide 10-2
Learning Objectives, continued
• Test hypotheses about the mean
difference in two related populations when
the populations are normally distributed.
• Test hypotheses about the differences in
two population proportions.
• Test hypotheses about two population
variances when the populations are
normally distributed.
© 2002 Thomson / South-Western
Slide 10-3
Hypothesis Testing about the
Difference in Two Sample Means
X
Population 1
x
X  n
1
1
1
X X
1
X X
1
X
2

2
x
n
2
X
2
Population 2
© 2002 Thomson / South-Western
Slide 10-4
2
Hypothesis Testing about the
Difference in Two Sample Means
X
1

X2

 
1
X
2
X
© 2002 Thomson / South-Western
1

X2
1

X2


n
2
1

n
1
X
2
2
2
1

X
2
Slide 10-5
Z Formula for the Difference
in Two Sample Means
for n1  30, n2  30, and Independent Samples
Z
 X  X      
© 2002 Thomson / South-Western
1
2
1
 
n n
2
2
1
2
1
2
2
Slide 10-6
Example: Hypothesis Testing for Differences
Between Means (Part 1)
Computer Analysts
24.10
25.00
24.25
23.75
22.70
21.75
24.25
21.30
22.00
22.00
22.55
18.00
23.50
23.25
23.50
22.80
24.00
22.10
24.25
22.70
21.50
23.85
23.50
23.80
24.20
22.75
25.60
22.90
23.80
24.10
23.20
23.55
© 2002 Thomson / South-Western
n  32
.
X  2314
.
S  1373
.
S  1885
1
Registered Nurses
20.75
23.30
22.75
23.80
24.00
23.00
1
22.00
21.75
21.25
2
21.85
21.50
20.00
24.16
20.40
21.75
21.10
23.75
23.25
19.50
20.50
22.60
22.50
21.75
21.70
25.00
20.80
20.75
2
22.70
20.25
22.50
2
23.25
22.45
2
21.90
19.10
1
1
n  34
X  21.99
S  1.403
S  1.968
2
2
Slide 10-7
Example: Hypothesis Testing for
Differences Between Means (Part 2)
Ho: 1   2  0
Ha: 1   2  0
Rejection
Region
Rejection
Region

2

.01
2
.01
Nonrejection Region
X
1

X2
X X
1
2
Critical Values
© 2002 Thomson / South-Western
Slide 10-8
Example: Hypothesis Testing for
Differences Between Means (Part 3)
Rejection
Region
Rejection
Region

.01
2
If Z < - 2.33 or Z > 2.33, reject Ho.
If - 2.33  Z  2.33, do not reject Ho.

.01
2
Nonrejection Region
Z
c
 2.33
0
Z
c
 2.33
Critical Values
© 2002 Thomson / South-Western
Slide 10-9
Example: Hypothesis Testing for
Differences between Means (Part 4)
Rejection
Region
Rejection
Region
If Z < - 2.33 or Z > 2.33, reject Ho.
If - 2.33  Z  2.33, do not reject Ho.
 X  X      
Z
1

.01
2

.01
2
 2.33
c
0
Critical Values
Z
 2.33
c
1
2
2
1
2
1
2
S S
n n
Nonrejection Region
Z
2

2
 2314
.  2199
.    0
1885
.
1968
.

32
34
 3.36
Since Z = 3.36 > 2.33, reject Ho.
© 2002 Thomson / South-Western
Slide 10-10
Demonstration Problem 10.1 (Part 1)
Rejection
Region
Ho: 1   2  0
Ha: 1   2  0
 .001
Nonrejection Region
Z
c
 3.08
0
Critical Value
© 2002 Thomson / South-Western
Slide 10-11
Demonstration Problem 10.1 (Part 2)
Women
Men
X  $3,343
S  $1, 226
n  87
X  $5,568
S  $1,716
n  76
1
Rejection
Region
1
Nonrejection Region
c
 3.08
2
1
 .001
Z
2
 X      

X
Z
1
2
1
2
2
1
2
1
2
S S
n n
0
Critical Value

If Z < - 3.08, reject Ho.
If Z   3.08, do not reject Ho.
2
2
 3343  5568   0 
2
1226  1716
87
2
 9.40
76
Since Z = -9.40 < -3.08, reject Ho.
© 2002 Thomson / South-Western
Slide 10-12
The t Test for Differences
in Population Means
• Each of the two populations is normally
distributed.
• The two samples are independent.
• At least one of the samples is small,
n < 30.
• The values of the population variances are
unknown.
• The variances of the two populations are
equal, 12 = 22
© 2002 Thomson / South-Western
Slide 10-13
t Formula to Test the Difference in
Means Assuming 12 = 22
t
 X  X      
1
2
S n  1  S n
n n 2
2
1
2
1
2
1
© 2002 Thomson / South-Western
2
2
1

1
2
1

1
n n
1
2
Slide 10-14
Hernandez Manufacturing Company
(Part 1)
Ho: 1   2  0
Ha: 1   2  0

Rejection
Region

.025
2
.05
.025
2
2
df  n1  n2  2  15  12  2  25
t

0.25, 25
 2.060
If t < - 2.060 or t > 2.060, reject Ho.
If - 2.060  t  2.060, do not reject Ho.
© 2002 Thomson / South-Western
Rejection
Region

.025
2
Nonrejection Region
t
.025, 25
 2.060
0
t
.025, 25
 2.060
Critical Values
Slide 10-15
Hernandez Manufacturing Company
(Part 2)
Training Method A
Training Method B
56
51
45
59
57
53
47
52
43
52
56
65
42
53
52
53
55
53
50
42
48
54
64
57
47
44
44
n  15
X  47.73
S  19.495
1
1
2
1
© 2002 Thomson / South-Western
n  12
X  56.5
S  18.273
2
2
2
2
Slide 10-16
Hernandez Manufacturing Company
(Part 3)
t
 X  X      
1
2
S n  1  S n
2
1
2
1
2
n1  n2  2

2
1
2

1
1

1
n n
 47.73  56.50  0
19.49514  18.27311
15  12  2
1
2
1
1

15 12
 5.20
If t < - 2.060 or t > 2.060, reject Ho.
If - 2.060  t  2.060, do not reject Ho.
© 2002 Thomson / South-Western
Since t = -5.20 < -2.060, reject Ho.
Slide 10-17
Dependent Samples
• Before and After
Measurements
on the same
individual
• Studies of twins
• Studies of
spouses
© 2002 Thomson / South-Western
Individual
Before
After
d
1
32
39
-7
2
11
15
-4
3
21
35
-14
4
17
13
4
5
30
41
-11
6
38
39
-1
Slide 10-18
Formulas for Dependent Samples
t
d D
S
d

d 
n
d
n
df  n  1
n  number of pairs
d = sample difference in pairs
S
d

 d  d 
2
n 1
 d


2

d
2
n 1
n
D = mean population difference
S
d
= standard deviation of sample difference
d = mean sample difference
© 2002 Thomson / South-Western
Slide 10-19
Sampling Distribution of Differences
in Sample Proportions
For large samples
1.
2.
3.
4.
n  p  5,
n  q  5,
n  p  5, and
n  q  5 where q
1
1
1
1
2
2
2
2
= 1 - p
the difference in sample proportions is normally distributed with
 p
1



p 2
p 1 p 2 
P P
1
and
2
P Q
n
1
1
1
Q
P

n
© 2002 Thomson / South-Western
2
2
2
Slide 10-20
Z Formula for the Difference
in Two Population Proportions
p  p    P  P 

Z
1
1
2
P Q
n
1
1
1

2
P Q
n
2
2
2
p  proportion from sample 1
p  proportion from sample 2
n  size of sample 1
n  size of sample 2
P  proportion from population 1
P  proportion from population 2
Q 1- P
Q  1- P
1
2
1
2
1
2
1
1
2
2
© 2002 Thomson / South-Western
Slide 10-21
Z Formula to Test the Difference
in Population Proportions
p  p    P  P 

Z
1
1
2
2
 1
1
 P  Q   
 n1 n2 
X
X
P
n n
p  n p
n

n n
1
2
1
1
2
1
2
1
2
2
Q  1 P
© 2002 Thomson / South-Western
Slide 10-22
Testing the Difference in Population
Proportions: Demonstration Problem 10.4
Ho: P1  P2  0
Ha: P1  P2  0
Rejection
Region
Rejection
Region

.005
2

.01

.005
2
2
Z .005  2.575
If Z < - 2.575 or Z > 2.575, reject Ho.
If - 2.575  Z  2.575, do not reject Ho.
© 2002 Thomson / South-Western

.005
2
Nonrejection Region
Z
c
 2.575
0
Z
c
 2.575
Critical Values
Slide 10-23
Demonstration Problem 10.4, continued
n  95
X  39
39
p  95 .41
n  100
X  24
24
p  100 .24
1
2
1
p  p    P  P 

Z
1

X X
n n
1
1
2
2
24  39
100  95
.323


1
2

P
2
P  Q  1  1 
n n
2
1
1
2
2
.24 .41  0
.323.677
1
1
 
100 95 
.17
.067
 2.54

Since - 2.575  Z = - 2.54  2.575, do not reject Ho.
© 2002 Thomson / South-Western
Slide 10-24
Hypothesis Testing about the Difference
in Two Population Variances
F Test for Two Population Variances
S
F
S
© 2002 Thomson / South-Western
1
2
2
  n 1
  n 1
dfnumerator 
dfdeno min ator
2
1
2
1
2
Slide 10-25
Example: An F Distribution
for 1 = 10 and 2 = 8
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0.00
1.00
© 2002 Thomson / South-Western
2.00
3.00
4.00
5.00
6.00
Slide 10-26
A Portion of the F Distribution Table
for  = 0.025
F
.025,9 ,11
Numerator Degrees of Freedom
1
2
3
4
Denominator
5
Degrees of Freedom 6
7
8
9
10
11
12
1
647.79
38.51
17.44
12.22
10.01
8.81
8.07
7.57
7.21
6.94
6.72
6.55
2
799.48
39.00
16.04
10.65
8.43
7.26
6.54
6.06
5.71
5.46
5.26
5.10
© 2002 Thomson / South-Western
3
864.15
39.17
15.44
9.98
7.76
6.60
5.89
5.42
5.08
4.83
4.63
4.47
4
899.60
39.25
15.10
9.60
7.39
6.23
5.52
5.05
4.72
4.47
4.28
4.12
5
921.83
39.30
14.88
9.36
7.15
5.99
5.29
4.82
4.48
4.24
4.04
3.89
6
937.11
39.33
14.73
9.20
6.98
5.82
5.12
4.65
4.32
4.07
3.88
3.73
7
948.20
39.36
14.62
9.07
6.85
5.70
4.99
4.53
4.20
3.95
3.76
3.61
8
956.64
39.37
14.54
8.98
6.76
5.60
4.90
4.43
4.10
3.85
3.66
3.51
9
963.28
39.39
14.47
8.90
6.68
5.52
4.82
4.36
4.03
3.78
3.59
3.44
Slide 10-27
Hypothesis Test for Equality of Two
Population Variances:
Sheet Metal Example (Part 1)
Ho: 1   2
2
  0.05
2
n  10
n  12
Ha: 1   2
2
dfdeno min ator
2
 359
.
S
S
1
F
1

359
.
 0.28
1
2
2
1
1
2
2
© 2002 Thomson / South-Western
F.975,11,9 =
.025,9,11
2
  n 1
   n 1
dfnumerator 
.025,9,11
1
2
F
F
If F < 0.28 or F > 3.59, reject Ho.
If 0.28  F  359
. , do reject Ho.
Slide 10-28
Sheet Metal Example (Part 2)
Rejection Regions
If F < 0.28 or F > 3.59, reject Ho.
If 0.28  F  359
. , do reject Ho.
Nonrejection
Region
F
.975,11,9
 0.28
F
.025,9 ,11
 359
.
Critical Values
© 2002 Thomson / South-Western
Slide 10-29
Sheet Metal Example (Part 3)
Machine 1
Machine 2
22.3
21.8
22.2
22.0
22.2
22.0
21.8
21.9
21.6
22.1
22.0
22.1
22.3
22.4
21.8
21.7
21.9
21.6
22.5
21.9
21.9
22.1
n  10
.
S  01138
S
F
S
1
2
1
2
1
2
2
01138
.

 5.63
0.0202
n
S
2
2
2
 12
 0.0202
Since F = 5.63 > Fc = 3.59, reject Ho.
© 2002 Thomson / South-Western
Slide 10-30