Statistics for Business and
Economics
Chapter 7
Inferences Based on Two Samples: Confidence
Intervals & Tests of Hypotheses
Learning Objectives
1. Distinguish Independent and Related Populations
2. Solve Inference Problems for Two Populations
• Mean
• Proportion
• Variance
3. Determine Sample Size
Thinking Challenge
How would you try to answer these questions?
• Who gets higher grades:
males or females?
• Which program is faster to
learn: Word or Excel?
Target Parameters
Difference between Means
1 – 2
Difference between
Proportions
p1 – p2
Ratio of Variances
( 1 )2
2
( 2 )
Possible Estimator
•
( X 1  X 2 ) for ( 1   2 )
( p1  p2 ) for ( p1  p2 )
2
S1
2
S2
1
for
2
2
2
Test Statistics
• What are the possible test statistics?
• Do we know the sampling distribution?
• Not any two populations we can make inferences
• In some cases, we can.
– Two independent populations
– Two related population, but paired samples
• In many other cases, we cannot…
– Two related population, but not paired samples
Independent & Related Populations
Independent
1. Different data sources
• Unrelated
• Independent
2. Use difference
between the two
sample means
•
X1 – X2
Related
1. Same data source
• Paired or matched
• Repeated measures
(before/after)
2. Use difference between
each pair of observations
•
di = x1i – x2i
Two Independent Populations Examples
1. An economist wishes to determine whether there is a
difference in mean family income for households in
two socioeconomic groups.
2. An admissions officer of a small liberal arts college
wants to compare the mean SAT scores of applicants
educated in rural high schools and in urban high
schools.
Two Related Populations Examples
1. Nike wants to see if there is a difference in durability
of two sole materials. One type is placed on one
shoe, the other type on the other shoe of the same
pair.
2. An analyst for Educational Testing Service wants to
compare the mean GMAT scores of students before
and after taking a GMAT review course.
Thinking Challenge
Are they independent or related?
1.
2.
3.
The life expectancy of light bulbs made in two different
factories
Difference in hardness between two metals: one contains an
alloy, one doesn’t
Tread life of two different motorcycle tires: one on the front,
the other on the back
Two Population Inference
Two
Populations
Mean
Paired
Proportion
Variance
Z
F
Indep.
Z
t
t
(Large
sample)
(Small
sample)
(Paired
sample)
Comparing Two Means
Two Population Inference
Two
Populations
Mean
Paired
Proportion
Variance
Z
F
Indep.
Z
t
t
(Large
sample)
(Small
sample)
(Paired
sample)
Comparing
Two Independent Means
Two Population Inference
Two
Populations
Mean
Paired
Proportion
Variance
Z
F
Indep.
Z
t
t
(Large
sample)
(Small
sample)
(Paired
sample)
Sampling Distribution
1
Population
1
2
2
1
Select simple random
sample, n1. Compute
X1
Compute X1 – X2
for every pair of
samples
Astronomical number
of X1 – X2 values
Population
2
Select simple random
sample, n2. Compute
X2
Sampling
Distribution
1 - 2
One Population Case
Base on CLT,
X1
σ12
N( 1 ,
)
n1
X2
σ22
N(  2 ,
)
n2
Two Population Case (independent)
•
So far we do not know the sampling distribution
of X 1  X 2
•
If these two populations are independent,
E ( X )  E ( X 1  X 2 )  1   2
V ( X )  V ( X1  X 2 )  V ( X1 )  V ( X 2 ) 
 12
n1

 22
n2
Since X 1 and X 2 are independent, and normally distributed,
X 1  X 2 should also be normally distributed.
Therefore, we have X 1  X 2
N ( 1  2 ,
 12
n1

 22
n2
)
Large-Sample Inference for
Two Independent Means
Two Population Inference
Two
Populations
Mean
Paired
Proportion
Variance
Z
F
Indep.
Z
t
t
(Large
sample)
(Small
sample)
(Paired
sample)
Conditions Required for Valid LargeSample Inferences about μ1 – μ2
Assumptions
• Independent, random samples
• Can be approximated by the normal distribution when n1 
30 and n2  30
Large-Sample Confidence Interval for
μ1 – μ2 (Independent Samples)
Confidence Interval
X
1
 X 2   Z 2

2
1
n1


2
2
n2
Hypotheses for Means of Two
Independent Populations
Research Questions
Hypothesis
No Difference
Any Difference
Pop 1  Pop 2 Pop 1  Pop 2
Pop 1 < Pop 2 Pop 1 > Pop 2
H0
1  2  0 1  2  0
1  2  0
Ha
1  2  0 1  2  0
1  2  0
Large-Sample Test for μ1 – μ2
(Independent Samples)
Two Independent Sample Z-Test Statistic
z
( x1  x2 )  ( 1  2 )
 12
n1

 22
n2
Hypothesized
difference
Large-Sample Confidence Interval
Example
You’re a financial analyst for Charles Schwab.
You want to estimate the difference in dividend
yield between stocks listed on NYSE and
NASDAQ. You collect the following data:
NYSE
NASDAQ
Number
121
125
Mean
3.27
2.53
Std Dev
1.30
1.16
What is the 95% confidence interval
for the difference between the mean
dividend yields?
© 1984-1994 T/Maker Co.
Large-Sample Confidence Interval
Solution
X
1
 X 2   Z 2
 12
n1

 22
n2
(1.3)2 (1.16) 2

(3.27  2.53)  1.96
125
121
.43  1  2  1.05
Hypotheses for Means of Two
Independent Populations
Research Questions
Hypothesis
No Difference
Any Difference
Pop 1  Pop 2 Pop 1  Pop 2
Pop 1 < Pop 2 Pop 1 > Pop 2
H0
1  2  0 1  2  0
1  2  0
Ha
1  2  0 1  2  0
1  2  0
Large-Sample Test for μ1 – μ2
(Independent Samples)
Two Independent Sample Z-Test Statistic
z
( x1  x2 )  ( 1  2 )
 12
n1

 22
n2
Hypothesized
difference
Large-Sample Test
Example
You’re a financial analyst for Charles Schwab.
You want to find out if there is a difference in
dividend yield between stocks listed on NYSE
and NASDAQ. You collect the following data:
NYSE
NASDAQ
Number
121
125
Mean
3.27
2.53
Std Dev
1.30
1.16
Is there a difference in average
yield ( = .05)?
© 1984-1994 T/Maker Co.
Large-Sample Test
Solution
•
•
•
•
•
H0: 1 - 2 = 0 (1 = 2)
Ha: 1 - 2  0 (1  2)
  .05
n1= 121 , n2 = 125
Critical Value(s):
Reject H0
Reject H0
.025
-1.96 0 1.96
.025
z
Large-Sample Test
Solution
• Test Statistic:
(3.27  2.53)  0
z
 4.69
1.698 1.353

121
125
• Decision:
Reject at  = .05
• Conclusion:
There is evidence of a difference in means
Large-Sample Test
Thinking Challenge
You’re an economist for the Department of Education.
You want to find out if there is a difference in
spending per pupil between urban and rural high
schools. You collect the following:
Urban
Rural
Number
35
35
Mean
$ 6,012
$ 5,832
Std Dev
$ 602
$ 497
Is there any difference in population
means ( = .10)?
Large-Sample Test
Solution*
•
•
•
•
•
H0: 1 - 2 = 0 (1 = 2)
Ha: 1 - 2  0 (1  2)
  .10
n1 = 35 , n2 = 35
Critical Value(s):
Reject H 0
.05
-1.645
Reject H0
.05
0 1.645
z
Large-Sample Test
Solution*
• Test Statistic:
z
(6012  5832)  0
2
602 497

35
35
2
 1.36
• Decision:
Do not reject at  = .10
• Conclusion:
There is no evidence of a difference in means
Small-Sample Inference
for Two Independent Means
Two Population Inference
Two
Populations
Mean
Paired
Proportion
Variance
Z
F
Indep.
Z
t
t
(Large
sample)
(Small
sample)
(Paired
sample)
Conditions Required for Valid SmallSample Inferences about μ1 – μ2
Assumptions
•
•
•
Independent, random samples
Populations are approximately normally distributed
Population variances are equal
Small-Sample Confidence Interval for
μ1 – μ2 (Independent Samples)
Confidence Interval
X
SP
1
2
 X 2   t
SP
2
2
1 1 
  
 n1 n2 
n1  1  S


  n2  1  S 2
n1  n2  2
2
1
df  n1  n2  2
2
Small-Sample Confidence Interval
Example
You’re a financial analyst for Charles Schwab.
You want to estimate the difference in dividend
yield between stocks listed on the NYSE and
NASDAQ? You collect the following data:
NYSE
NASDAQ
Number
11
15
Mean
3.27
2.53
Std Dev
1.30
1.16
Assuming normal populations, what
is the 95% confidence interval
for the difference between the
mean dividend yields?
© 1984-1994 T/Maker Co.
Small-Sample Confidence Interval
Solution
df = n1 + n2 – 2 = 11 + 15 – 2 = 24
SP
t.025 = 2.064
2
2
n

1

S

n

1

S




2
1
2
2
 1
n1  n2  2
11  1  1.30   15  1  1.16 


2
11  15  2
2
 1.489
1 1
 3.27  2.53  2.064 1.489    
 11 15 
.26  1  2  1.74
Small-Sample Test for μ1 – μ2
(Independent Samples)
Two Independent Sample t–Test Statistic

X
t
1
 X 2   1   2 
SP
SP
2
2
1 1
   
 n1 n2 

n1  1  S1

 n2  1  S 2
n1  n2  2
2
df  n1  n2  2
Hypothesized
difference
2
Small-Sample Test
Example
You’re a financial analyst for Charles Schwab. Is
there a difference in dividend yield between stocks
listed on the NYSE and NASDAQ? You collect
the following data:
NYSE
NASDAQ
Number
11
15
Mean
3.27
2.53
Std Dev
1.30
1.16
Assuming normal populations,
is there a difference in average
yield ( = .05)?
© 1984-1994 T/Maker Co.
Small-Sample Test
Solution
•
•
•
•
•
H0: 1 - 2 = 0 (1 = 2) Test Statistic:
Ha: 1 - 2  0 (1  2)
  .05
df  11 + 15 - 2 = 24
Critical Value(s):
Decision:
Reject H 0
Reject H 0
.025
.025
-2.064
0 2.064
t
Conclusion:
Small-Sample Test
Solution
 n1  1  S1   n2  1  S2
2
SP 
2
2
n1  n2  2
11  1  1.30   15  1  1.16 


2
11  15  2
X

t
1
 X 2    1  2 
1 1
SP    
 n1 n2 
2
2
 1.489
3.27  2.53   0 


 1.53
1 1
1.489    
 11 15 
Small-Sample Test
Solution
• Test Statistic:
t
3.27  2.53
1 1
1.489    
 11 15 
 1.53
• Decision:
Do not reject at  = .05
• Conclusion:
There is no evidence of a difference in means
Small-Sample Test
Thinking Challenge
You’re a research analyst for General Motors. Assuming
equal variances, is there a difference in the average
miles per gallon (mpg) of two car models ( = .05)?
You collect the following:
Sedan
Van
Number
15
11
Mean
22.00
20.27
Std Dev
4.77
3.64
Small-Sample Test
Solution*
•
•
•
•
•
H0: 1 - 2 = 0 (1 = 2) Test Statistic:
Ha: 1 - 2  0 (1  2)
  .05
df  15 + 11 - 2 = 24
Critical Value(s):
Decision:
Reject H 0
Reject H 0
.025
.025
-2.064
0 2.064
t
Conclusion:
Small-Sample Test
Solution*
n1  1  S1   n2  1  S 2


2
SP
2
2
n1  n2  2
15  1   4.77   11  1   3.64 


2
15  11  2
X

t
1
 X 2    1  2 
1 1
SP    
 n1 n2 
2
2
 18.793
22.00  20.27    0 


 1.00
1 1
18.793    
 15 11 
Small-Sample Test
Solution*
• Test Statistic:
t
22.00  20.27
1 1
18.793    
 15 11 
 1.00
• Decision:
Do not reject at  = .05
• Conclusion:
There is no evidence of a difference in means
Paired Difference
Experiments
Small-Sample
Two Population Inference
Two
Populations
Mean
Paired
Proportion
Variance
Z
F
Indep.
Z
t
t
(Large
sample)
(Small
sample)
(Paired
sample)
Paired-Difference Experiments
1. Compares means of two related populations
• Paired or matched
• Repeated measures (before/after)
2. Eliminates variation among subjects
Conditions Required for Valid SmallSample Paired-Difference Inferences
Assumptions
• Random sample of differences
• Both population are approximately normally distributed
Paired-Difference Experiment
Data Collection Table
Observation
1
2
Group 1
x11
x12
Group 2
x21
x22
Difference
d1 = x11 – x21
d2 = x12 – x22

i

x1i

x2i


n

x1n

x2n
di = x1i – x2i

dn = x1n – x2n
成對樣本
• 譬如說, 為了檢視投影片教學對於統計課是否有幫
助, 我們可以隨機選取 n 個學生, 並記錄其投影片
教學前與投影片教學後的成績
• 如果以X1i 與X2i 分別代表第i 個同學在投影片教學
前後的成績, 則我們知道X1i 與X1j 相互獨立, 但是
X1i 與X2i 則非獨立
• 諸如此類的樣本, 我們稱之為成對樣本
Paired-Difference Experiment Small-Sample
Confidence Interval
d  t
2
Sample Mean
sd
nd
df = nd – 1
Sample Standard Deviation
n
d 
d i
i 1
nd
n
Sd 
 (di - d)2
i 1
nd  1
Paired-Difference Experiment
Confidence Interval Example
You work in Human Resources. You want to see if
there is a difference in test scores after a training
program. You collect the following test score data:
Name
Before (1) After (2)
Sam
85
94
Tamika
94
87
Brian
78
79
Mike
87
88
Find a 90% confidence interval for the
mean difference in test scores.
Computation Table
Observation
Before
After
Difference
Sam
85
94
-9
Tamika
94
87
7
Brian
78
79
-1
Mike
87
88
-1
Total
-4
d = –1
Sd = 6.53
Paired-Difference Experiment
Confidence Interval Solution
df = nd – 1 = 4 – 1 = 3
d  t 2
Sd
nd
6.53
1  2.353
4
8.68  d  6.68
t.05 = 2.353
Hypotheses for Paired-Difference
Experiment
Research Questions
Pop 1  Pop 2 Pop 1  Pop 2
Pop 1 < Pop 2 Pop 1 > Pop 2
Hypothesis
No Difference
Any Difference
H0
d  0
d  0
d  0
Ha
d  0
d  0
d  0
Note: di = x1i – x2i for ith observation
Paired-Difference Experiment SmallSample Test Statistic
d  D0
t 
Sd
df = n – 1
nd
Sample Mean
Sample Standard Deviation
n
d 
 di
i 1
nd
n
Sd 
 (di - d)2
i 1
nd  1
Paired-Difference Experiment SmallSample Test Example
You work in Human Resources. You want to see if a
training program is effective. You collect the
following test score data:
Name
Before
After
Sam
85
94
Tamika
94
87
Brian
78
79
Mike
87
88
At the .10 level of significance, was the
training effective?
Null Hypothesis Solution
1. Was the training effective?
2. Effective means ‘Before’ < ‘After’.
3. Statistically, this means B < A.
4. Rearranging terms gives B – A < 0.
5. Defining d = B – A and substituting into (4)
gives d  .
6. The alternative hypothesis is Ha: d  0.
Computation Table
Observation
Before
After
Difference
Sam
85
94
-9
Tamika
94
87
7
Brian
78
79
-1
Mike
87
88
-1
Total
-4
d = –1
Sd = 6.53
Paired-Difference Experiment SmallSample Test Solution
•
•
•
•
•
H0: d = 0 (d = B - A)
Ha: d < 0
 = .10
df = 4 - 1 = 3
Critical Value(s):
Test Statistic:
Decision:
Reject H0
.10
-1.638
Conclusion:
0
t
Paired-Difference Experiment SmallSample Test Solution
• Test Statistic:
d  D0 1  0
t

 .306
Sd
6.53
4
nd
• Decision:
Do not reject at  = .10
• Conclusion:
There is no evidence training was effective
Paired-Difference Experiment SmallSample Test Thinking Challenge
You’re a marketing research
analyst. You want to
compare a client’s calculator
to a competitor’s. You
sample 8 retail stores. At
the .01 level of significance,
does your client’s calculator
sell for less than their
competitor’s?
(1)
(2)
Store Client Competitor
1
$ 10
$ 11
2
8
11
3
7
10
4
9
12
5
11
11
6
10
13
7
9
12
8
8
10
Paired-Difference Experiment SmallSample Test Solution*
•
•
•
•
•
H0:d = 0 (d = 1 - 2)
Ha:d < 0
 = .01
df = 8 - 1 = 7
Critical Value(s):
Test Statistic:
Decision:
Reject H0
Conclusion:
.01
-2.998 0
t
Paired-Difference Experiment SmallSample Test Solution*
• Test Statistic:
d  D0 2.25  0
t

 5.486
Sd
1.16
8
nd
• Decision:
Reject at  = .01
• Conclusion:
There is evidence client’s brand (1) sells for less
Comparing Two Population
Proportions
Two Population Inference
Two
Populations
Mean
Paired
Proportion
Variance
Z
F
Indep.
Z
t
t
(Large
sample)
(Small
sample)
(Paired
sample)
Conditions Required for Valid LargeSample Inference about p1 – p2
Assumptions
• Independent, random samples
• Normal approximation can be used if
n1 pˆ1  15, n1q1  15, n2 pˆ 2  15, and n2 q2  15
Two Population Case (independent)
^
^
• A natural candidate of estimator would be p1  p 2
but we do not know its sampling distribution
If two populations are independent
^
Based on CLT： p1
^
p1 q1 ^
N ( p1 ,
), p 2
n1
N ( p2 ,
p2 q2
)
n1
^
E ( p1  p 2 )  p1  p2
p1 q1 p 2 q 2
V ( p1  p 2 )  V ( p1 )  V ( p 2 ) 

( p1 , p2 are independent)
n1
n2
^
^
^
^
^
Therefore, p1  p 2
^
p1 q1 p 2 q 2
N ( p1  p2 ,

)
n1
n2
Large-Sample Confidence Interval
for p1 – p2
Confidence Interval
 pˆ1  pˆ 2   Z 2
pˆ1qˆ1 pˆ 2 qˆ2

n1
n2
Confidence Interval for p1 – p2 Example
As personnel director, you
want to test the perception of
fairness of two methods of
performance evaluation. 63 of
78 employees rated Method 1
as fair. 49 of 82 rated Method
2 as fair. Find a 99%
confidence interval for the
difference in perceptions.
Confidence Interval for p1 – p2 Solution
63
p?1   .808 q1  1  .808  .192
78
49
p?2   .598 q2  1  .598  .402
82
.808  .192 .598  .402

.808  .598   2.58
78
82
.029  p1  p2  .391
Hypotheses for
Two Proportions
Research Questions
Hypothesis
No Difference
Any Difference
Pop 1  Pop 2 Pop 1  Pop 2
Pop 1 < Pop 2 Pop 1 > Pop 2
H0
p1  p2  0
p1  p2  0
p1  p2  0
Ha
p1  p2  0
p1  p2  0
p1  p2  0
Large-Sample Test for p1 – p2
Z-Test Statistic for Two Proportions
Z
( pˆ1  pˆ 2 )  ( p1  p2 )
1 1
ˆ   
pq
 n1 n2 
where pˆ 
X1  X 2
n1  n2
Test for Two Proportions
Example
As personnel director, you
want to test the perception of
fairness of two methods of
performance evaluation. 63 of
78 employees rated Method 1
as fair. 49 of 82 rated Method
2 as fair. At the .01 level of
significance, is there a
difference in perceptions?
Test for Two Proportions Solution
•
•
•
•
•
H0: p1 - p2 = 0
Ha: p1 - p2  0
 =.01
n1 = 78 n2 = 82
Critical Value(s):
Reject H0
Test Statistic:
Decision:
Reject H0
.005
-2.58 0 2.58
.005
z
Conclusion:
Test for Two Proportions Solution
X 1 63
pˆ1 

 0.808
n1 78
X 2 49
p2 

 0.598
n2 82
X 1  X 2 63  49
pˆ 

 0.70 (H0 implies equal variance)
n1  n2
78  82
Z
( pˆ1  pˆ 2 )  ( p1  p2 )
1 1
pˆ  (1  pˆ )    
 n1 n2 
 2.90

(0.808  0.598)  (0)
1 
 1
(0.70)  (1  0.70)    
 78 82 
Test for Two Proportions Solution
• Test Statistic:
Z = +2.90
• Decision:
Reject at  = .01
• Conclusion:
There is evidence of a difference in proportions
Test for Two Proportions
Thinking Challenge
You’re an economist for the
Department of Labor. You’re
studying unemployment rates. In
MA, 74 of 1500 people surveyed
were unemployed. In CA, 129
of 1500 were unemployed. At
the .05 level of significance, does
MA have a lower unemployment
rate than CA?
MA
CA
Test for Two Proportions Solution*
•
•
•
•
•
H0: pMA – pCA = 0
Test Statistic:
Ha: pMA – pCA < 0
 =.05
nMA =1500 nCA = 1500
Critical Value(s):
Decision:
Reject H0
.05
-1.645 0
Conclusion:
Z
Test for Two Proportions Solution*
X MA
74
pˆ MA 

 0.0493
nMA 1500
X CA 129
pCA 

 0.0860
nCA 1500
X MA  X CA
74  129
pˆ 

 0.0677
nMA  nCA 1500  1500
(0.0493  0.0860 )  (0)
Z
1 
 1
(0.0677 )  (1  0.0677 )  


 1500 1500 
 4.00
Test for Two Proportions Solution*
• Test Statistic:
Z = –4.00
• Decision:
Reject at  = .05
• Conclusion:
There is evidence MA is less than CA
Determining Sample Size
Determining Sample Size
• Sample size for estimating μ1 – μ2
Z  


2
n1  n2
 2
2
1

2
2

( ME )2
• Sample size for estimating p1 – p2
Z  pq  p q 


2
n1  n2
 2
1 1
( ME )
2 2
2
ME = Margin
of Error
Sample Size Example
What sample size is needed to estimate μ1 – μ2 with
95% confidence and a margin of error of 5.8? Assume
prior experience tells us σ1 =12 and σ2 =18.
2
2
1.96
12

18
 

2
n1  n2 
(5.8)
2
 53.44  54
Sample Size Example
What sample size is needed to estimate p1 – p2 with 90%
confidence and a width of .05?
width .05
ME 

 .025
2
2
1.645 .5  .5  .5  .5 


 2164.82  2165
2
n1  n2
(.025)2
Comparing
Two Population Variances
Two Population Inference
Two
Populations
Mean
Paired
Proportion
Variance
Z
F
Indep.
Z
t
t
(Large
sample)
(Small
sample)
(Paired
sample)
F Distribution
1.if x1 ~ N ( 1 ,  )
2
1
x2 ~ N ( 2 ,  2 )
2
S
2
1
(x  x )


i
1
n1  1
S 
則
S 
2
1
2
2
2
1
2
2
2
, S2
2
(x  x )


~ Fv1 ,v2  Fn1 1,n2 1
i
n2  1
2
2
2.F 為一右偏分配，其型態決定於 1 & 2
3.查表  , 1 , 2 
4.Fn1 1,n2 1,1 
1
Fn2 1,n1 1,1
F-Test for Equal Variances
Critical Values
Reject H 0
/2
Reject H 0
/2
Do Not
Reject H 0
F
0
FL ( / 2;  1, 2 ) 
1
FU ( / 2;  2 , 1 )
FU ( / 2;  1, 2 )
Note!

S
區間估計  推估 (變異數比) 

S
2
1
2
2
2
1
S
S
2
2

2
1

2
2
S 


S 
2
1
2
2
2
2
2
1
Fv1 ,v2
2
1
2
2
Sampling Distribution
1
Population
1
1
Select simple random
sample, size n1.
Compute S12
Astronomical number
of S12/S22 values
2
Population
2
2
Compute F = S12/S22
for every pair of n1
& n2 size samples
Select simple random
sample, size n2.
Compute S22
Sampling
Distributions for
Different Sample
Sizes
Conditions Required for a Valid F-Test
for Equal Variances
Assumptions
•
Both populations are normally distributed
—
•
Test is not robust to violations
Independent, random samples
F-Test for Equal Variances
Hypotheses
• Hypotheses
H0: 12 = 22
Ha: 12  22
OR H0: 12  22 (or  )
Ha: 12  22 (or >)
• Test Statistic
• F = s12 /s22
• Two sets of degrees of freedom
— 1 = n1 – 1;
2 = n2 – 1
• Follows F distribution
F-Test for Equal Variances
Critical Values
Reject H 0
/2
Reject H 0
/2
Do Not
Reject H 0
F
0
FL ( / 2;  1, 2 ) 
1
FU ( / 2;  2 , 1 )
FU ( / 2;  1, 2 )
Note!
F-Test for Equal Variances
Example
You’re a financial analyst for Charles Schwab.
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 variances
between the NYSE & NASDAQ
at the .05 level of significance?
© 1984-1994 T/Maker Co.
F-Test for Equal Variances
Solution
•
•
•
•
•
H0: 12 = 22
Ha: 12  22
  .05
1  20 2  24
Critical Value(s):
Reject H0
Decision:
Reject H0
.025
0 .415
Test Statistic:
.025
2.33
Conclusion:
F
F-Test for Equal Variances
Solution
• Test Statistic:
2
S1 1.30 2
F 2 
 1.25
2
S 2 1.16
• Decision:
Do not reject at  = .05
• Conclusion:
There is no evidence of a difference in variances
F-Test for Equal Variances
Solution
Reject H0
/2 = .025
Reject H0
Do Not
Reject H0
0
FL(.025;20,24)
/2 = .025
F
FU (.025;20,24 )  2 .33
1
1


 .415
FU (.025;24,20) 2.41
F-Test for Equal Variances
Thinking Challenge
You’re an analyst for the Light & Power Company.
You want to compare the electricity consumption of
single-family homes in two towns. You compute the
following from a sample of homes:
Town 1
Town 2
Number
25
21
Mean
$ 85
$ 68
Std Dev
$ 30
$ 18
At the .05 level of significance, is there evidence
of a difference in variances between the two towns?
F-Test for Equal Variances
Solution*
•
•
•
•
•
H0: 12 = 22
Ha: 12  22
  .05
1  24 2  20
Critical Value(s):
Reject H0
Reject H0
.025
.025
0 .429
2.41
Test Statistic:
Decision:
Conclusion:
F
Critical Values
Solution*
Reject H0
/2 = .025
Reject H0
Do Not
Reject H0
0
FL(.025;24,20)
/2 = .025
F
FU (.025;24,20 )  2 .41
1
1


 .429
FU (.025;20,24) 2.33
F-Test for Equal Variances
Solution*
• Test Statistic:
S12 30 2
F  2  2  2.778
S2
18
• Decision:
Reject at  = .05
• Conclusion:
There is evidence of a difference in variances
Conclusion
1. Distinguished Independent and Related Populations
2. Solved Inference Problems for Two Populations
• Mean
• Proportion
• Variance
3. Determined Sample Size