BUSINESS STATISTICS
Chapter 10
INFERENCE: TWO POPULATIONS
Chap 10-1
Chapter Contents
This chapter includes two main parts:
1. Estimation for the difference of two population
parameters
2. Hypothesis Testing for the difference of two population
parameters
Chap 10-2
Part 1: Goals
After completing this part, you should be able to:
 Form confidence intervals for the mean difference from dependent
samples
 Form confidence intervals for the difference between two
independent population means (standard deviations known or
unknown)
 Compute confidence interval limits for the difference between two
independent population proportions
 Create confidence intervals for a population variance
 Find chi-square values from the chi-square distribution table
 Determine the required sample size to estimate a mean or
proportion within a specified margin of error
Chap 10-3
Estimation: Two populations
Chapter Topics
Population
Means,
Dependent
Samples
Population
Means,
Independent
Samples
Population
Proportions
Population
Variance
Proportion 1 vs.
Proportion 2
Variance of a
normal distribution
Examples:
Same group
before vs. after
treatment
Group 1 vs.
independent
Group 2
Chap 10-4
Dependent Samples
Tests Means of 2 Related Populations
Dependent
samples



Paired or matched samples
Repeated measures (before/after)
Use difference between paired values:
di = xi - yi
 Eliminates Variation Among Subjects
 Assumptions:
 Both Populations Are Normally Distributed
Chap 10-5
Mean Difference
The ith paired difference is di , where
Dependent
samples
di = x i - yi
The point estimate for
the population mean
paired difference is d :
The sample
standard
deviation is:
n
d
d
i 1
i
n
n
Sd 
2
(d

d
)
 i
i1
n 1
n is the number of matched pairs in the sample
Chap 10-6
Confidence Interval for
Mean Difference
Dependent
samples
The confidence interval for difference
between population means, μd , is
d  t n1,α/2
Sd
Sd
 μd  d  t n1,α/2
n
n
Where
n = the sample size
(number of matched pairs in the paired sample)
Chap 10-7
Confidence Interval for
Mean Difference
(continued)
Dependent
samples
 The margin of error is
ME  t n1,α/2
sd
n
 tn-1,/2 is the value from the Student’s t
distribution with (n – 1) degrees of freedom
for which
α
P(t n1  t n1,α/2 ) 
2
Chap 10-8
Paired Samples Example
 Six people sign up for a weight loss program. You
collect the following data:
Person
1
2
3
4
5
6
Weight:
Before (x)
After (y)
136
205
157
138
175
166
125
195
150
140
165
160
Difference, di
11
10
7
-2
10
6
42
di

d = n
= 7.0
Sd 
2
(d

d
)
 i
n 1
 4.82
Chap 10-9
Paired Samples Example
(continued)
 For a 95% confidence level, the appropriate t value is
tn-1,/2 = t5,.025 = 2.571
 The 95% confidence interval for the difference between
means, μd , is
Sd
Sd
d  t n 1,α/2
 μ d  d  t n 1,α/2
n
n
4.82
4.82
7  (2.571)
 μ d  7  (2.571)
6
6
1.94  μ d  12.06
Since this interval contains zero, we cannot be 95% confident, given this
limited data, that the weight loss program helps people lose weight
Chap 10-10
Difference Between Two Means
Population means,
independent
samples
Goal: Form a confidence interval
for the difference between two
population means, μx – μy
 Different data sources
 Unrelated
 Independent
 Sample selected from one population has no effect on the
sample selected from the other population
 The point estimate is the difference between the two
sample means:
x–y
Chap 10-11
Difference Between Two Means
(continued)
Population means,
independent
samples
σx2 and σy2 known
Confidence interval uses z/2
σx2 and σy2 unknown
σx2 and σy2
assumed equal
σx2 and σy2
assumed unequal
Confidence interval uses a value
from the Student’s t distribution
Chap 10-12
σx2 and σy2 Known
Population means,
independent
samples
σx2 and σy2 known
σx2 and σy2 unknown
Assumptions:
*
 Samples are randomly and
independently drawn
 both population distributions
are normal
 Population variances are
known
Chap 10-13
σx2 and σy2 Known
(continued)
When σx and σy are known and
both populations are normal, the
variance of X – Y is
Population means,
independent
samples
2
σx2 and σy2 known
σx2 and σy2 unknown
*
σ 2X Y
2
σy
σx


nx
ny
…and the random variable
Z
(x  y)  (μX  μY )
2
σ 2x σ y

nX nY
has a standard normal distribution
Chap 10-14
Confidence Interval,
σx2 and σy2 Known
Population means,
independent
samples
σx2 and σy2 known
σx2 and σy2 unknown
(x  y)  z α/2
*
The confidence interval for
μx – μy is:
σ 2X σ 2Y
σ 2X σ 2Y

 μX  μY  (x  y)  z α/2

nx ny
nx ny
Chap 10-15
σx2 and σy2 Unknown,
Assumed Equal
Assumptions:
Population means,
independent
samples
 Samples are randomly and
independently drawn
σx2 and σy2 known
 Populations are normally
distributed
σx2 and σy2 unknown
σx2 and σy2
assumed equal
*
 Population variances are
unknown but assumed equal
σx2 and σy2
assumed unequal
Chap 10-16
σx2 and σy2 Unknown,
Assumed Equal
(continued)
Forming interval
estimates:
Population means,
independent
samples
 The population variances
are assumed equal, so use
the two sample standard
deviations and pool them to
estimate σ
σx2 and σy2 known
σx2 and σy2 unknown
σx2 and σy2
assumed equal
σx2 and σy2
assumed unequal
*
 use a t value with
(nx + ny – 2) degrees of
freedom
Chap 10-17
σx2 and σy2 Unknown,
Assumed Equal
(continued)
Population means,
independent
samples
The pooled variance is
σx2 and σy2 known
σx2 and σy2 unknown
σx2 and σy2
assumed equal
*
sp2 
(n x  1)s 2x  (n y  1)s 2y
nx  ny  2
σx2 and σy2
assumed unequal
Chap 10-18
Confidence Interval,
σx2 and σy2 Unknown, Equal
σx2 and σy2 unknown
σx2 and σy2
assumed equal
*
The confidence interval for
μ1 – μ2 is:
sp2
sp2
σx2 and σy2
assumed unequal
(x  y)  t nx ny 2,α/2
Where
sp2 
sp2
nx

ny
 μX  μY  (x  y)  t nx ny 2,α/2
nx

sp2
ny
(n x  1)s 2x  (n y  1)s 2y
nx  ny  2
Chap 10-19
Pooled Variance Example
You are testing two computer processors for speed.
Form a confidence interval for the difference in CPU
speed. You collect the following speed data (in Mhz):
CPUx
Number Tested
17
Sample mean
3004
Sample std dev
74
CPUy
14
2538
56
Assume both populations are
normal with equal variances,
and use 95% confidence
Chap 10-20
Calculating the Pooled Variance
The pooled variance is:
2
2




n

1
S

n

1
S

17  1742  14  1562
x
x
y
y
2
S 

p
(n x  1)  (ny  1)
(17 - 1)  (14  1)
 4427.03
The t value for a 95% confidence interval is:
tnx ny 2 , α/2  t 29 , 0.025  2.045
Chap 10-21
Calculating the Confidence Limits
 The 95% confidence interval is
(x  y)  t nx ny 2,α/2
(3004  2538)  (2.054)
sp2
nx

sp2
ny
 μX  μY  (x  y)  t nx ny 2,α/2
sp2
nx

sp2
ny
4427.03 4427.03
4427.03 4427.03

 μX  μY  (3004  2538)  (2.054)

17
14
17
14
416.69  μX  μY  515.31
We are 95% confident that the mean difference in
CPU speed is between 416.69 and 515.31 Mhz.
Chap 10-22
σx2 and σy2 Unknown,
Assumed Unequal
Assumptions:
Population means,
independent
samples
 Samples are randomly and
independently drawn
σx2 and σy2 known
 Populations are normally
distributed
σx2 and σy2 unknown
 Population variances are
unknown and assumed
unequal
σx2 and σy2
assumed equal
σx2 and σy2
assumed unequal
*
Chap 10-23
σx2 and σy2 Unknown,
Assumed Unequal
(continued)
Forming interval estimates:
Population means,
independent
samples
 The population variances are
assumed unequal, so a pooled
variance is not appropriate
σx2 and σy2 known
 use a t value with  degrees
of freedom, where
σx2 and σy2 unknown
2
σx2 and σy2
assumed equal
σx2 and σy2
assumed unequal
*
 s2x
s2y 
( )  ( )
n y 
 n x
v
2
2
2
2


s
 sx 
  /(n x  1)   y  /(n y  1)
n 
 nx 
 y
Chap 10-24
Confidence Interval,
σx2 and σy2 Unknown, Unequal
σx2 and σy2 unknown
σx2 and σy2
assumed equal
σx2 and σy2
assumed unequal
(x  y)  t ,α/2
*
The confidence interval for
μ1 – μ2 is:
2
2
s2x s y
s2x s y

 μX  μY  (x  y)  t ,α/2

nx ny
nx ny
Where
v
 s2x
s2y 
( )  ( )
n y 
 n x
2
2
 s2 
 s2x 
  /(n x  1)   y  /(n y  1)
n 
 nx 
 y
2
Chap 10-25
Two Population Proportions
Population
proportions
Goal: Form a confidence interval for
the difference between two
population proportions, Px – Py
Assumptions:
Both sample sizes are large (generally at
least 40 observations in each sample)
The point estimate for
the difference is
pˆ x  pˆ y
Chap 10-26
Two Population Proportions
(continued)
Population
proportions
 The random variable
Z
(pˆ x  pˆ y )  (p x  p y )
pˆ x (1 pˆ x ) pˆ y (1 pˆ y )

nx
ny
is approximately normally distributed
Chap 10-27
Confidence Interval for
Two Population Proportions
Population
proportions
The confidence limits for
Px – Py are:
(pˆ x  pˆ y )  Z / 2
pˆ x (1 pˆ x ) pˆ y (1 pˆ y )

nx
ny
Chap 10-28
Example:
Two Population Proportions
Form a 90% confidence interval for the
difference between the proportion of
men and the proportion of women who
have college degrees.
 In a random sample, 26 of 50 men and
28 of 40 women had an earned college
degree
Chap 10-29
Example:
Two Population Proportions
(continued)
Men:
ˆp x  26  0.52
50
Women:
ˆp y  28  0.70
40
pˆ x (1 pˆ x ) pˆ y (1 pˆ y )
0.52(0.48) 0.70(0.30)



 0.1012
nx
ny
50
40
For 90% confidence, Z/2 = 1.645
Chap 10-30
Example:
Two Population Proportions
(continued)
The confidence limits are:
(pˆ x  pˆ y )  Z α/2
pˆ x (1  pˆ x ) pˆ y (1  pˆ y )

nx
ny
 (.52  .70)  1.645 (0.1012)
so the confidence interval is
-0.3465 < Px – Py < -0.0135
Since this interval does not contain zero we are 90% confident that the
two proportions are not equal
Chap 10-31
Confidence Intervals for the
Population Variance
Population
Variance
 Goal: Form a confidence interval
for the population variance, σ2
 The confidence interval is based on
the sample variance, s2
 Assumed: the population is
normally distributed
Chap 10-32
Confidence Intervals for the
Population Variance
(continued)
Population
Variance
The random variable

2
n1
(n  1)s

2
σ
2
follows a chi-square distribution
with (n – 1) degrees of freedom
2

The chi-square value n1,  denotes the number for which
P( χn21  χn21, α )  α
Chap 10-33
Confidence Intervals for the
Population Variance
(continued)
Population
Variance
The (1 - )% confidence interval
for the population variance is
(n  1)s
(n  1)s
2
σ  2
2
χn1, α/2
χn1, 1 - α/2
2
2
Chap 10-34
Example
You are testing the speed of a computer processor. You
collect the following data (in Mhz):
CPUx
Sample size
17
Sample mean
3004
Sample std dev
74
Assume the population is normal.
Determine the 95% confidence interval for σx2
Chap 10-35
Finding the Chi-square Values
 n = 17 so the chi-square distribution has (n – 1) = 16
degrees of freedom
  = 0.05, so use the the chi-square values with area
0.025 in each tail:
2
χ n21, α/2  χ16
, 0.025  28.85
2
χ n21, 1 - α/2  χ16
, 0.975  6.91
probability
α/2 = .025
probability
α/2 = .025
216 = 6.91
216 = 28.85
216
Chap 10-36
Calculating the Confidence Limits
 The 95% confidence interval is
2
(n  1)s 2
(n

1)s
2

σ
 2
2
χn1, α/2
χn1, 1 - α/2
2
(17  1)(74)2
(17

1)(74)
 σ2 
28.85
6.91
3037  σ 2  12683
Converting to standard deviation, we are 95%
confident that the population standard deviation of
CPU speed is between 55.1 and 112.6 Mhz
Chap 10-37
Sample Size Determination
Determining
Sample Size
For the
Mean
For the
Proportion
Chap 10-38
Margin of Error
 The required sample size can be found to reach
a desired margin of error (ME) with a specified
level of confidence (1 - )
 The margin of error is also called sampling error
 the amount of imprecision in the estimate of the
population parameter
 the amount added and subtracted to the point
estimate to form the confidence interval
Chap 10-39
Sample Size Determination
Determining
Sample Size
For the
Mean
x  z α/2
σ
n
Margin of Error
(sampling error)
ME  z α/2
σ
n
Chap 10-40
Sample Size Determination
(continued)
Determining
Sample Size
For the
Mean
ME  z α/2
σ
n
Now solve
for n to get
z σ
n
2
ME
2
α/2
2
Chap 10-41
Sample Size Determination
(continued)
 To determine the required sample size for the
mean, you must know:
 The desired level of confidence (1 - ), which
determines the z/2 value
 The acceptable margin of error (sampling error), ME
 The standard deviation, σ
Chap 10-42
Required Sample Size Example
If  = 45, what sample size is needed to
estimate the mean within ± 5 with 90%
confidence?
z σ
(1.645) (45)
n

 219.19
2
2
ME
5
2
α/2
2
2
2
So the required sample size is n = 220
(Always round up)
Chap 10-43
Sample Size Determination
Determining
Sample Size
For the
Proportion
pˆ  z α/2
pˆ (1 pˆ )
n
ME  z α/2
pˆ (1  pˆ )
n
Margin of Error
(sampling error)
Chap 10-44
Sample Size Determination
(continued)
Determining
Sample Size
For the
Proportion
ME  z α/2
pˆ (1  pˆ )
n
pˆ (1 pˆ ) cannot
be larger than
0.25, when p̂ =
0.5
Substitute
0.25 for pˆ (1 pˆ )
and solve for
n to get
0.25 z
n
2
ME
2
α/2
Chap 10-45
Sample Size Determination
(continued)
 The sample and population proportions, p̂ and P, are
generally not known (since no sample has been taken
yet)
 P(1 – P) = 0.25 generates the largest possible margin
of error (so guarantees that the resulting sample size
will meet the desired level of confidence)
 To determine the required sample size for the
proportion, you must know:
 The desired level of confidence (1 - ), which determines the
critical z/2 value
 The acceptable sampling error (margin of error), ME
 Estimate P(1 – P) = 0.25
Chap 10-46
Required Sample Size Example
How large a sample would be necessary
to estimate the true proportion defective in
a large population within ±3%, with 95%
confidence?
Chap 10-47
Required Sample Size Example
(continued)
Solution:
For 95% confidence, use z0.025 = 1.96
ME = 0.03
Estimate P(1 – P) = 0.25
0.25 z
n
2
ME
2
α/2
2
(0.25)(1.9 6)

 1067.11
2
(0.03)
So use n = 1068
Chap 10-48
Part 1: Summary
 Compared two dependent samples (paired samples)
 Formed confidence intervals for the paired difference
 Compared two independent samples
 Formed confidence intervals for the difference between two
means, population variance known, using z
 Formed confidence intervals for the differences between two
means, population variance unknown, using t
 Formed confidence intervals for the differences between two
population proportions
 Formed confidence intervals for the population variance
using the chi-square distribution
 Determined required sample size to meet confidence
and margin of error requirements
Chap 10-49
Part 2
Hypothesis Testing: Two
populations
Chap 10-50
Part 2: Goals
After completing this part, you should be able to:
 Test hypotheses for the difference between two population means
 Two means, matched pairs
 Independent populations, population variances known
 Independent populations, population variances unknown but
equal
 Complete a hypothesis test for the difference between two
proportions (large samples)
 Use the chi-square distribution for tests of the variance of a normal
distribution
 Use the F table to find critical F values
 Complete an F test for the equality of two variances
Chap 10-51
Two Sample Tests
Two Sample Tests
Population
Means,
Matched
Pairs
Population
Means,
Independent
Samples
Population
Proportions
Population
Variances
Examples:
Same group
before vs. after
treatment
Group 1 vs.
independent
Group 2
Proportion 1 vs.
Proportion 2
Variance 1 vs.
Variance 2
(Note similarities to part 1)
Chap 10-52
Matched Pairs
Tests Means of 2 Related Populations
Matched
Pairs



Paired or matched samples
Repeated measures (before/after)
Use difference between paired values:
di = xi - yi
 Assumptions:
 Both Populations Are Normally Distributed
Chap 10-53
Test Statistic: Matched Pairs
Matched
Pairs
The test statistic for the mean
difference is a t value, with
n – 1 degrees of freedom:
d  D0
t
sd
n
Where
D0 = hypothesized mean difference
sd = sample standard dev. of differences
n = the sample size (number of pairs)
Chap 10-54
Decision Rules: Matched Pairs
Paired Samples
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: μx – μy  0
H1: μx – μy < 0
H0: μx – μy ≤ 0
H1: μx – μy > 0
H0: μx – μy = 0
H1: μx – μy ≠ 0


-t
t
Reject H0 if t < -tn-1, 
Where
Reject H0 if t > tn-1, 
t
d  D0
sd
n
/2
-t/2
/2
t/2
Reject H0 if t < -tn-1 , /2
or t > tn-1 , /2
has n - 1 d.f.
Chap 10-55
Matched Pairs Example
 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:
di

d = n
Number of Complaints:
(2) - (1)
Salesperson
C.B.
T.F.
M.H.
R.K.
M.O.
Before (1)
After (2)
Difference, di
6
20
3
0
4
4
6
2
0
0
- 2
-14
- 1
0
- 4
-21
= - 4.2
Sd 
2
(d

d
)
 i
n 1
 5.67
Chap 10-56
Matched Pairs: Solution
 Has the training made a difference in the number of
complaints (at the  = 0.01 level)?
H0: μx – μy = 0
H1: μx – μy  0
 = .01
d = - 4.2
Critical Value = ± 4.604
d.f. = n - 1 = 4
Reject
Reject
/2
/2
- 4.604
4.604
- 1.66
Decision: Do not reject H0
(t stat is not in the reject region)
Test Statistic:
d  D0  4.2  0
t

 1.66
sd / n 5.67/ 5
Conclusion: There is not a
significant change in the
number of complaints.
Chap 10-57
Difference Between Two Means
Population means,
independent
samples
Goal: Form a confidence interval
for the difference between two
population means, μx – μy
 Different data sources
 Unrelated
 Independent
 Sample selected from one population has no effect on the
sample selected from the other population
Chap 10-58
Difference Between Two Means
(continued)
Population means,
independent
samples
σx2 and σy2 known
Test statistic is a z value
σx2 and σy2 unknown
σx2 and σy2
assumed equal
σx2 and σy2
assumed unequal
Test statistic is a a value from the
Student’s t distribution
Chap 10-59
σx2 and σy2 Known
Population means,
independent
samples
σx2 and σy2 known
σx2 and σy2 unknown
Assumptions:
*
 Samples are randomly and
independently drawn
 both population distributions
are normal
 Population variances are
known
Chap 10-60
σx2 and σy2 Known
(continued)
When σx2 and σy2 are known and
both populations are normal, the
variance of X – Y is
Population means,
independent
samples
2
σx2 and σy2 known
σx2 and σy2 unknown
*
σ 2X Y
2
σy
σx


nx
ny
…and the random variable
Z
(x  y)  (μX  μY )
2
σ 2x σ y

nX nY
has a standard normal distribution
Chap 10-61
Test Statistic,
σx2 and σy2 Known
Population means,
independent
samples
σx2 and σy2 known
σx2 and σy2 unknown
The test statistic for
μx – μy is:
*

x  y   D0
z
2
σy
σx

nx
ny
2
Chap 10-62
Hypothesis Tests for
Two Population Means
Two Population Means, Independent Samples
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: μx  μy
H1: μx < μy
H0: μx ≤ μy
H1: μx > μy
H0: μx = μy
H1: μx ≠ μy
i.e.,
i.e.,
i.e.,
H0: μx – μy  0
H1: μx – μy < 0
H0: μx – μy ≤ 0
H1: μx – μy > 0
H0: μx – μy = 0
H1: μx – μy ≠ 0
Chap 10-63
Decision Rules
Two Population Means, Independent
Samples, Variances Known
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: μx – μy  0
H1: μx – μy < 0
H0: μx – μy ≤ 0
H1: μx – μy > 0
H0: μx – μy = 0
H1: μx – μy ≠ 0


-z
Reject H0 if z < -z
z
Reject H0 if z > z
/2
-z/2
/2
z/2
Reject H0 if z < -z/2
or z > z/2
Chap 10-64
σx2 and σy2 Unknown,
Assumed Equal
Assumptions:
Population means,
independent
samples
 Samples are randomly and
independently drawn
σx2 and σy2 known
 Populations are normally
distributed
σx2 and σy2 unknown
σx2 and σy2
assumed equal
*
 Population variances are
unknown but assumed equal
σx2 and σy2
assumed unequal
Chap 10-65
σx2 and σy2 Unknown,
Assumed Equal
(continued)
Forming interval
estimates:
Population means,
independent
samples
 The population variances
are assumed equal, so use
the two sample standard
deviations and pool them to
estimate σ
σx2 and σy2 known
σx2 and σy2 unknown
σx2 and σy2
assumed equal
σx2 and σy2
assumed unequal
*
 use a t value with
(nx + ny – 2) degrees of
freedom
Chap 10-66
Test Statistic,
σx2 and σy2 Unknown, Equal
The test statistic for
μx – μy is:
σx2 and σy2 unknown
σx2 and σy2
assumed equal
σx2 and σy2
assumed unequal
*
t
 x  y    μx  μy 
1 1 
S   
n n 
y 
 x
2
p
Where t has (n1 + n2 – 2) d.f.,
and
sp2 
(n x  1)s 2x  (n y  1)s 2y
nx  ny  2
Chap 10-67
σx2 and σy2 Unknown,
Assumed Unequal
Assumptions:
Population means,
independent
samples
 Samples are randomly and
independently drawn
σx2 and σy2 known
 Populations are normally
distributed
σx2 and σy2 unknown
 Population variances are
unknown and assumed
unequal
σx2 and σy2
assumed equal
σx2 and σy2
assumed unequal
*
Chap 10-68
σx2 and σy2 Unknown,
Assumed Unequal
(continued)
Forming interval estimates:
Population means,
independent
samples
 The population variances are
assumed unequal, so a pooled
variance is not appropriate
σx2 and σy2 known
 use a t value with  degrees
of freedom, where
σx2 and σy2 unknown
2
σx2 and σy2
assumed equal
σx2 and σy2
assumed unequal
*
 s2x
s2y 
( )  ( )
n y 
 n x
v
2
2
2
2


s
 sx 
  /(n x  1)   y  /(n y  1)
n 
 nx 
 y
Chap 10-69
Test Statistic,
σx2 and σy2 Unknown, Unequal
σx2 and σy2 unknown
The test statistic for
μx – μy is:
σx2 and σy2
assumed equal
σx2 and σy2
assumed unequal
*
t
(x  y)  D0
Where t has  degrees of freedom:
σ
σ

nX nY
2
y
2
x
v
 s2x
s2y 
( )  ( )
n y 
 n x
2
2
 s2 
 s2x 
  /(n x  1)   y  /(n y  1)
n 
 nx 
 y
2
Chap 10-70
Decision Rules
Two Population Means, Independent
Samples, Variances Unknown
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: μx – μy  0
H1: μx – μy < 0
H0: μx – μy ≤ 0
H1: μx – μy > 0
H0: μx – μy = 0
H1: μx – μy ≠ 0


-t
Reject H0 if t < -tn-1, 
t
Reject H0 if t > tn-1, 
/2
-t/2
/2
t/2
Reject H0 if t < -tn-1 , /2
or t > tn-1 , /2
Where t has n - 1 d.f.
Chap 10-71
Pooled Variance t Test: Example
You are a financial analyst for a brokerage firm. Is there a
difference in dividend yield between stocks listed on the
NYSE & 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 average
yield ( = 0.05)?
Chap 10-72
Calculating the Test Statistic
The test statistic is:

X  X   μ  μ 
t

1
2
1
1 1
S   
 n1 n2 
2
p
2
3.27  2.53   0
1 
 1
1.5021 

 21 25 
2
2
2
2








n

1
S

n

1
S
21

1
1.30

25

1
1.16
1
2
2
S2  1

p
(n1  1)  (n2  1)
(21 - 1)  (25  1)
 2.040
 1.5021
Chap 10-73
Solution
H0: μ1 - μ2 = 0 i.e. (μ1 = μ2)
H1: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)
 = 0.05
df = 21 + 25 - 2 = 44
Critical Values: t = ± 2.0154
Reject H0
.025
-2.0154
Reject H0
.025
0 2.0154
t
2.040
Test Statistic:
Decision:
3.27  2.53
t
 2.040 Reject H0 at  = 0.05
1 
 1
Conclusion:
1.5021  

 21 25 
There is evidence of a
difference in means.
Chap 10-74
Two Population Proportions
Population
proportions
Goal: Test hypotheses for the
difference between two population
proportions, Px – Py
Assumptions:
Both sample sizes are large,
nP(1 – P) > 9
Chap 10-75
Two Population Proportions
(continued)
Population
proportions
 The random variable
Z
(pˆ x  pˆ y )  (p x  p y )
pˆ x (1 pˆ x ) pˆ y (1 pˆ y )

nx
ny
is approximately normally distributed
Chap 10-76
Test Statistic for
Two Population Proportions
The test statistic for
H0: Px – Py = 0
is a z value:
Population
proportions
z
 pˆ
x
 pˆ y 
pˆ 0 (1  pˆ 0 ) pˆ 0 (1  pˆ 0 )

nx
ny
Where
pˆ 0 
nxpˆ x  nypˆ y
nx  ny
Chap 10-77
Decision Rules: Proportions
Population proportions
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: px – py  0
H1: px – py < 0
H0: px – py ≤ 0
H1: px – py > 0
H0: px – py = 0
H1: px – py ≠ 0


-z
Reject H0 if z < -z
z
Reject H0 if z > z
/2
-z/2
/2
z/2
Reject H0 if z < -z/2
or z > z/2
Chap 10-78
Example:
Two Population Proportions
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 31 of
50 women indicated they would vote Yes
 Test at the .05 level of significance
Chap 10-79
Example:
Two Population Proportions
(continued)
 The hypothesis test is:
H0: PM – PW = 0 (the two proportions are equal)
H1: PM – PW ≠ 0 (there is a significant difference between
proportions)
 The sample proportions are:
 Men:
p̂M = 36/72 = .50
p̂ W = 31/50 = .62
 Women:
 The estimate for the common overall proportion is:
pˆ 0 
nxpˆ x  nypˆ y
nx  ny
72(36/72)  50(31/50) 67


 .549
72  50
122
Chap 10-80
Example:
Two Population Proportions
(continued)
The test statistic for PM – PW = 0 is:
z

 pˆ
 pˆ W 
pˆ 0 (1  pˆ 0 ) pˆ 0 (1  pˆ 0 )

n1
n2
Reject H0
Reject H0
.025
.025
M
-1.96
-1.31
1.96
 .50  .62 
 .549 (1  .549) .549 (1  .549)  Decision: Do not reject H0



72
50

 Conclusion: There is not
  1.31
Critical Values = ±1.96
For  = .05
significant evidence of a
difference between men
and women in proportions
who will vote yes.
Chap 10-81
Hypothesis Tests of
one Population Variance
Population
Variance
 Goal: Test hypotheses about the
population variance, σ2
 If the population is normally distributed,

2
n1
(n  1)s

σ2
2
follows a chi-square distribution with
(n – 1) degrees of freedom
Chap 10-82
Confidence Intervals for the
Population Variance
(continued)
Population
Variance
The test statistic for
hypothesis tests about one
population variance is
χ
2
n 1
(n  1)s

σ 02
2
Chap 10-83
Decision Rules: Variance
Population variance
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: σ2  σ02
H1: σ2 < σ02
H0: σ2 ≤ σ02
H1: σ2 > σ02
H0: σ2 = σ02
H1: σ2 ≠ σ02


χ n21,
χn21,1
Reject H0 if
χ
2
n 1
χ
2
n 1,1
Reject H0 if
χ n21  χ n21,
/2
/2
χ n21,1 / 2
χn21, / 2
Reject H0 if
χ n21  χ n21, / 2
or
χn21  χ n21,1 / 2
Chap 10-84
Hypothesis Tests for Two Variances
Tests for Two
Population
Variances
F test statistic
 Goal: Test hypotheses about two
population variances
H0: σx2  σy2
H1: σx2 < σy2
H0: σx2 ≤ σy2
H1: σx2 > σy2
H0: σx2 = σy2
H1: σx2 ≠ σy2
Lower-tail test
Upper-tail test
Two-tail test
The two populations are assumed to be
independent and normally distributed
Chap 10-85
Hypothesis Tests for Two Variances
(continued)
Tests for Two
Population
Variances
F test statistic
The random variable
2
x
2
y
s /σ
F
s /σ
2
x
2
y
Has an F distribution with (nx – 1)
numerator degrees of freedom and
(ny – 1) denominator degrees of
freedom
Denote an F value with 1 numerator and 2
denominator degrees of freedom by
Chap 10-86
Test Statistic
Tests for Two
Population
Variances
F test statistic
The critical value for a hypothesis test
about two population variances is
s
F
s
2
x
2
y
where F has (nx – 1) numerator
degrees of freedom and (ny – 1)
denominator degrees of freedom
Chap 10-87
Decision Rules: Two Variances
Use sx2 to denote the larger variance.
H0: σx2 = σy2
H1: σx2 ≠ σy2
H0: σx2 ≤ σy2
H1: σx2 > σy2
/2

0
Do not
reject H0
Reject H0
Fnx 1,ny 1,α
Reject H0 if F  Fnx 1,ny 1,α
F
0
Do not
reject H0
F
Reject H0
Fnx 1,ny 1,α / 2
rejection region for a twotail test is:

Reject H0 if F  Fnx 1,ny 1,α / 2
where sx2 is the larger of
the two sample variances
Chap 10-88
Example: F Test
You are a financial analyst for a brokerage firm. You
want to compare dividend yields between stocks listed
on the NYSE & 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
& NASDAQ at the  = 0.10 level?
Chap 10-89
F Test: Example Solution
 Form the hypothesis test:
H0: σx2 = σy2 (there is no difference between variances)
H1: σx2 ≠ σy2 (there is a difference between variances)
 Find the F critical values for  = .10/2:
Degrees of Freedom:
 Numerator
(NYSE has the larger
standard deviation):
 nx – 1 = 21 – 1 = 20 d.f.
Fnx 1, ny 1, α / 2
 F20 , 24 , 0.10/2  2.03
 Denominator:
 ny – 1 = 25 – 1 = 24 d.f.
Chap 10-90
F Test: Example Solution
(continued)
 The test statistic is:
H0: σx2 = σy2
H1: σx2 ≠ σy2
s 2x 1.30 2
F 2 
 1.256
2
s y 1.16
 F = 1.256 is not in the rejection
region, so we do not reject H0
/2 = .05
Do not
reject H0
Reject H0
F
F20 , 24 , 0.10/2  2.03
 Conclusion: There is not sufficient evidence
of a difference in variances at  = .10
Chap 10-91
Two-Sample Tests in EXCEL
For paired samples (t test):
 Tools | data analysis… | t-test: paired two sample for means
For independent samples:
 Independent sample Z test with variances known:
 Tools | data analysis | z-test: two sample for means
For variances…
 F test for two variances:
 Tools | data analysis | F-test: two sample for variances
Chap 10-92
Chapter Summary
 Compared two dependent samples (paired
samples)
 Performed paired sample t test for the mean
difference
 Compared two independent samples
 Performed z test for the differences in two means
 Performed pooled variance t test for the differences
in two means
 Compared two population proportions
 Performed z-test for two population proportions
Chap 10-93
Part 1: Summary
(continued)
 Used the chi-square test for a single population
variance
 Performed F tests for the difference between
two population variances
 Used the F table to find F critical values
Chap 10-94