# here - Pearson ```Basic Business Statistics
12th Edition
Chapter 10
Two-Sample Tests
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
Chap 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
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
Chap 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
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
Proportion 1 vs.
Proportion 2
Variance 1 vs.
Variance 2
Chap 10-3
Difference Between Two Means
DCOVA
Population means,
independent
samples
*
σ1 and σ2 unknown,
assumed equal
σ1 and σ2 unknown,
not assumed equal
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
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
Chap 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.
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
Chap 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
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Chap 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
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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
Chap 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
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
 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
Chap 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
2
2




n

1
S

n

1
S
1
2
2
S2  1
p
*
(n 1  1)  (n 2  1)
• The test statistic is:
t STAT

X  X   μ

1
2
1
 μ2 
1 1 
S   
 n1 n 2 
2
p
σ1 and σ2 unknown,
not assumed equal
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
• Where tSTAT has d.f. = (n1 + n2 – 2)
Chap 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
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
*
The confidence interval for
μ1 – μ2 is:
X
1

 X 2  ta/2
1
1 
S   
 n1 n 2 
2
p
Where tα/2 has d.f. = n1 + n2 – 2
Chap 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)?
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Chap 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:
t STAT

X

1

 X 2  μ1  μ 2 
 1
1 
S   
 n1 n 2 
2
p

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)  (n 2  1)
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(21 - 1)  ( 25  1)
 2.040
 1.5021
Chap 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
Reject H0
.025
-2.0154
Reject H0
.025
0 2.0154
t
2.040
Test Statistic:
Decision:
3.27  2.53
t STAT 
 2.040 Reject H0 at a = 0.05
1 
 1
Conclusion:
1.5021   
 21 25 
There is evidence of a
difference in means.
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Chap 10-13
Excel Pooled-Variance t test
Comparing NYSE &amp; NASDAQ
DCOVA
Pooled-Variance t Test for the Difference Between Two Means
(assumes equal population variances)
Data
Hypothesized Difference
Level of Significance
Population 1 Sample
Sample Size
Sample Mean
Sample Standard Deviation
Population 2 Sample
Sample Size
Sample Mean
Sample Standard Deviation
Intermediate Calculations
Population 1 Sample Degrees of Freedom
Population 2 Sample Degrees of Freedom
Total Degrees of Freedom
Pooled Variance
Standard Error
Difference in Sample Means
t Test Statistic
0
0.05
21 =COUNT(DATA!\$A:\$A)
3.27 =AVERAGE(DATA!\$A:\$A)
1.3 =STDEV(DATA!\$A:\$A)
25 =COUNT(DATA!\$B:\$B)
2.53 =AVERAGE(DATA!\$B:\$B)
1.16 =STDEV(DATA!\$B:\$B)
Decision:
Reject H0 at a = 0.05
Conclusion:
There is evidence of a
difference in means.
20 =B7 - 1
24 =B11 - 1
44 =B16 + B17
1.502 =((B16 * B9^2) + (B17 * B13^2)) / B18
0.363 =SQRT(B19 * (1/B7 + 1/B11))
0.74 =B8 - B12
2.040 =(B21 - B4) / B20
Two-Tail Test
Lower Critical Value
Upper Critical Value
p-value
Reject the null hypothesis
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-2.015 =-TINV(B5, B18)
2.015 =TINV(B5, B18)
0.047 =TDIST(ABS(B22),B18,2)
=IF(B27&lt;B5,&quot;Reject the null hypothesis&quot;,
&quot;Do not reject the null hypothesis&quot;)
Chap 10-14
Minitab Pooled-Variance t test
Comparing NYSE &amp; NASDAQ
DCOVA
Two-Sample T-Test and CI
Sample
1
2
N Mean StDev SE Mean
21 3.27 1.30 0.28
25 2.53 1.16 0.23
Difference = mu (1) - mu (2)
Estimate for difference: 0.740
95% CI for difference: (0.009, 1.471)
T-Test of difference = 0 (vs not =): T-Value = 2.04 P-Value = 0.047 DF = 44
Both use Pooled StDev = 1.2256
Decision:
Reject H0 at a = 0.05
Conclusion:
There is evidence of a
difference in means.
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Chap 10-15
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  X   t
1
2
a/2
1
1 

S     0.74  2.0154  0.3628  (0.009, 1.471)
 n1 n 2 
2
p
Since 0 is less than the entire interval, we can be 95%
confident that &micro;NYSE &gt; &micro;NASDAQ
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Chap 10-16
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
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
*
 Population variances are
unknown and cannot be
assumed to be equal
Chap 10-17
Hypothesis tests for &micro;1 - &micro;2 with σ1 and
σ2 unknown and not assumed equal
(continued)
DCOVA
Population means,
independent
samples
σ1 and σ2 unknown,
assumed equal
The test statistic is:
t STAT

X

1

 X 2   μ1  μ 2 
S12 S 22

n1 n 2
tSTAT has d.f. ν =
2
σ1 and σ2 unknown,
not assumed equal
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*
 S1 2 S 2 2 


n  n 
1
2 
  2
2
 S1 2 
 S22 




n 
n 
 1   2 
n1  1
n2  1
Chap 10-18
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
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Chap 10-19
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
population mean μD is D : D 
n
D
i 1
i
n
n
The sample standard
deviation is SD
SD 
2
(D

D
)
 i
i 1
n1
n is the number of pairs in the paired sample
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Chap 10-20
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.
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Chap 10-21
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
ta
Reject H0 if tSTAT &lt; -ta
Reject H0 if tSTAT &gt; ta
Where tSTAT has n - 1 d.f.
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a/2
-ta/2
a/2
ta/2
Reject H0 if tSTAT &lt; -ta/2
or tSTAT &gt; ta/2
Chap 10-22
The Paired Difference
Confidence Interval
Paired
samples
DCOVA
The confidence interval for μD is
D  ta / 2
SD
n
n
where
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SD 
 (D  D)
i1
2
i
n 1
Chap 10-23
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
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4
6
2
0
0
- 2
-14
- 1
0
- 4
-21
D =
 Di
n
= -4.2
SD 
2
(D

D
)
 i
n 1
 5.67
Chap 10-24
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
d.f. = n - 1 = 4
Test Statistic:
t STAT
D  μ D  4.2  0


 1.66
SD / n 5.67/ 5
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Reject
Reject
a/2
a/2
- 4.604
4.604
- 1.66
Decision: Do not reject H0
(tstat is not in the reject region)
Conclusion: There is
insufficient evidence there
is significant change in the
number of complaints.
Chap 10-25
Paired t Test In Excel
DCOVA
Paired t Test
Data
Hypothesized Mean Diff.
Level of Significance
Intermediate Calculations
Sample Size
Dbar
Degrees of Freedom
SD
Standard Error
t-Test Statistic
Two-Tail Test
Lower Critical Value
Upper Critical Value
p-value
Do not reject the null
Hypothesis
0
0.05
Since -2.776 &lt; -1.66 &lt; 2.776
we do not reject the null hypothesis.
Or
5 =COUNT(I2:I6)
-4.2 =AVERAGE(I2:I6)
4 =B8 - 1
5.67 =STDEV(I2:I6)
2.54 =B11/SQRT(B8)
-1.66 =(B9 - B4)/B12
Since p-value = 0.173 &gt; 0.05 we
do not reject the null hypothesis.
Thus we conclude that there is
Insufficient evidence to conclude
there is a difference in the average
number of complaints.
-2.776 =-TINV(B5,B10)
2.776 =TINV(B5,B10)
0.173 =TDIST(ABS(B13),B10,2)
=IF(B18&lt;B5,&quot;Reject the null
hypothesis&quot;,
&quot;Do not reject the null hypothesis&quot;)
Data not shown is in column I
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Chap 10-26
Paired t Test In Minitab Yields
The Same Conclusions
Paired T-Test and CI: After, Before
DCOVA
Paired T for After - Before
After
Before
Difference
N
5
5
5
Mean
2.40
6.60
-4.20
StDev
2.61
7.80
5.67
SE Mean
1.17
3.49
2.54
95% CI for mean difference: (-11.25, 2.85)
T-Test of mean difference = 0 (vs not = 0): T-Value = -1.66 P-Value = 0.173
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Chap 10-27
Two Population Proportions
DCOVA
Goal: test a hypothesis or form a
confidence interval for the difference
between two population proportions,
π1 – π2
Population
proportions
The point estimate for
the difference is
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
p1  p2
Chap 10-28
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:
X1  X2
p
n1  n 2
where X1 and X2 are the number of items of
interest in samples 1 and 2
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Chap 10-29
Two Population Proportions
(continued)
The test statistic for
π1 – π2 is a Z statistic:
Population
proportions
Z STAT 
where
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DCOVA
 p1  p2    π1  π2 
 1
1
p (1  p )   
 n1 n2 
X1  X 2
X1
p
, p1 
n1  n2
n1
X2
, p2 
n2
Chap 10-30
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
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Chap 10-31
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
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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
Chap 10-32
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
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Chap 10-33
Hypothesis Test Example:
Two population Proportions
(continued)

The hypothesis test is:
DCOVA
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:
X 1  X 2 36  35 71
p


 0 .582
n1  n2
72  50 122
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Chap 10-34
Hypothesis Test Example:
Two population Proportions
DCOVA
(continued)
The test statistic for π1 – π2 is:
z STAT 

 p1  p2    π1  π2 
 1
1 
p ( 1  p)   
 n1 n2 
 .50  .70   0
1 
 1
.582 ( 1  .582 )   
 72 50 
Critical Values = &plusmn;1.96
For a = .05
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
Reject H0
Reject H0
.025
.025
-1.96
-2.20
  2 .20
1.96
Decision: Reject H0
Conclusion: There is
evidence of a difference in
proportions who will vote
yes between men and
women.
Chap 10-35
Two Proportion Test In Excel
DCOVA
Z Test for Differences in Two Proportions
Data
Hypothesized Difference
Level of Significance
Group 1
Number of items of interest
Sample Size
Group 2
Number of items of interest
Sample Size
Intermediate Calculations
Group 1 Proportion
Group 2 Proportion
Difference in Two Proportions
Average Proportion
Z Test Statistic
Two-Tail Test
Lower Critical Value
Upper Critical Value
p-value
Reject the null hypothesis
0
0.05
36
72
35
50
Since -2.20 &lt; -1.96
Or
Since p-value = 0.028 &lt; 0.05
We reject the null hypothesis
0.5 =B7/B8
0.7 =B10/B11
-0.2 =B14 - B15
Decision:
0.582 =(B7 + B10)/(B8 + B11)
-2.20 =(B16-B4)/SQRT((B17*(1-B17))*(1/B8+1/B11))
-1.96 =NORMSINV(B5/2)
1.96 =NORMSINV(1 - B5/2)
0.028 =2*(1 - NORMSDIST(ABS(B18)))
=IF(B23 &lt; B5,&quot;Reject the null hypothesis&quot;,
&quot;Do not reject the null hypothesis&quot;)
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
Reject H0
Conclusion: There is
evidence of a difference in
proportions who will vote
yes between men and
women.
Chap 10-36
Two Proportion Test In Minitab
Shows The Same Conclusions
DCOVA
Test and CI for Two Proportions
Sample X
1
36
2
35
N
72
50
Sample p
0.500000
0.700000
Difference = p (1) - p (2)
Estimate for difference: -0.2
95% CI for difference: (-0.371676, -0.0283244)
Test for difference = 0 (vs not = 0): Z = -2.28 P-Value = 0.022
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
Chap 10-37
Confidence Interval for
Two Population Proportions
DCOVA
Population
proportions
The confidence interval for
π1 – π2 is:
 p1  p 2   Za/2
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
p1 (1  p1 ) p 2 (1  p 2 )

n1
n2
Chap 10-38
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
2
1
2
2
S
S
Where:
S12 = Variance of sample 1 (the larger sample variance)
n1 = sample size of sample 1
S 22 = 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
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
Chap 10-39
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
S

S

When FSTAT

In the F table,
2
1
2
2
df1 = n1 – 1 ; df2 = n2 – 1

numerator degrees of freedom determine the column

denominator degrees of freedom determine the row
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
Chap 10-40
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
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
a
F
0
Do not
reject H0
Fα
Reject H0
F
Reject H0 if FSTAT &gt; Fα
Chap 10-41
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?
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
Chap 10-42
F Test: Example Solution
DCOVA

Form the hypothesis test:
(there is no difference between variances)
H 0: σ12  σ 22
(there is a difference between variances)
H 1: σ12  σ 22

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
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
Chap 10-43
F Test: Example Solution
DCOVA
(continued)

The test statistic is:
FSTAT
H0: σ12 = σ22
H1: σ12 ≠ σ22
S12 1.30 2
 2 
 1.256
2
S 2 1.16
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 insufficient evidence of
a difference in variances at a = .05
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
Reject H0
F
F0.025=2.33
Chap 10-44
Two Variance F Test In Excel
DCOVA
F Test for Differences in Two Variables
Data
Level of Significance
Larger-Variance Sample
Sample Size
Sample Variance
Smaller-Variance Sample
Sample Size
Sample Variance
0.05
21
1.6900 =1.3^2
25
1.3456 =1.16^2
Intermediate Calculations
F Test Statistic
Population 1 Sample Degrees of Freedom
Population 2 Sample Degrees of Freedom
Conclusion:
There is insufficient evidence
of a difference in variances
at a = .05 because:
1.256 1.255945
20 =B6 - 1
24 =B9 - 1
F statistic =1.256 &lt; 2.327 Fα/2
or
p-value = 0.589 &gt; 0.05 = α.
Two-Tail Test
Upper Critical Value
p-value
Do not reject the null hypothesis
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
2.327 =FINV(B4/2,B14,B15)
0.589 =2*FDIST(B13,B14,B15)
=IF(B19&lt;B4,&quot;Reject the null hypothesis&quot;,
&quot;Do not reject the null hypothesis&quot;)
Chap 10-45
Two Variance F Test In Minitab
Yields The Same Conclusion
DCOVA
Test and CI for Two Variances
Null hypothesis
Sigma(1) / Sigma(2) = 1
Alternative hypothesis Sigma(1) / Sigma(2) not = 1
Significance level
Alpha = 0.05
Statistics
Sample
1
2
N
21
25
StDev
1.300
1.160
Variance
1.690
1.346
Ratio of standard deviations = 1.121
Ratio of variances = 1.256
95% Confidence Intervals
Distribution
of Data
Normal
CI for StDev
Ratio
(0.735, 1.739)
CI for
Variance
Ratio
(0.540, 3.024)
Tests
Test
Method
DF1 DF2 Statistic P-Value
F Test (normal) 20 24
1.26 0.589
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
Chap 10-46
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
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
Chap 10-47
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
Copyright &copy;2012 Pearson Education, Inc. publishing as Prentice Hall
Chap 10-48
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