### Statistics for Managers Using Microsoft Excel, 4/e

```Two-Sample Tests
Chap 9-1
Two Sample Tests
Two Sample Tests
Population
Means,
Independent
Samples
Means,
Related
Samples
Population
Proportions
Population
Variances
Examples:
Group 1 vs.
independent
Group 2
Same group
before vs. after
treatment
Proportion 1 vs.
Proportion 2
Variance 1 vs.
Variance 2
Chap 9-2
Difference Between Two Means
Population means,
independent
samples
*
σ1 and σ2 known
σ1 and σ2 unknown
Goal: Test hypotheses or form
a confidence interval for the
difference between two
population means, μ1 – μ2
The point estimate for the
difference is
X1 – X2
Chap 9-3
Independent Samples
Population means,
independent
samples
*
σ1 and σ2 known
σ1 and σ2 unknown
 Different data sources
 Unrelated
 Independent
 Sample selected from one
population has no effect on
the sample selected from
the other population
 Use the difference between 2
sample means
 Use Z test or pooled variance
t test
Chap 9-4
Difference Between Two Means
Population means,
independent
samples
*
σ1 and σ2 known
Use a Z test statistic
σ1 and σ2 unknown
Use S to estimate unknown
σ , use a t test statistic and
pooled standard deviation
Chap 9-5
σ1 and σ2 Known
Population means,
independent
samples
σ1 and σ2 known
σ1 and σ2 unknown
Assumptions:
*
 Samples are randomly and
independently drawn
 population distributions are
normal or both sample sizes
are  30
 Population standard
deviations are known
Chap 9-6
σ1 and σ2 Known
(continued)
Population means,
independent
samples
σ1 and σ2 known
When σ1 and σ2 are known and
both populations are normal or
both sample sizes are at least 30,
the test statistic is a Z-value…
*
…and the standard error of
X1 – X2 is
σ1 and σ2 unknown
σ X1  X2 
2
1
2
σ
σ2

n1
n2
Chap 9-7
σ1 and σ2 Known
(continued)
Population means,
independent
samples
σ1 and σ2 known
σ1 and σ2 unknown
The test statistic for
μ1 – μ2 is:
*

X
Z
1

 X 2   μ1  μ2 
2
1
2
σ
σ2

n1
n2
Chap 9-8
Hypothesis Tests for
Two Population Means
Two Population Means, Independent Samples
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: μ1  μ2
H1: μ1 < μ2
H0: μ1 ≤ μ2
H1: μ1 > μ2
H0: μ1 = μ2
H1: μ1 ≠ μ2
i.e.,
i.e.,
i.e.,
H0: μ1 – μ2  0
H1: μ1 – μ2 < 0
H0: μ1 – μ2 ≤ 0
H1: μ1 – μ2 > 0
H0: μ1 – μ2 = 0
H1: μ1 – μ2 ≠ 0
Chap 9-9
Hypothesis tests for μ1 – μ2
Two Population Means, Independent Samples
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: μ1 – μ2  0
H1: μ1 – μ2 < 0
H0: μ1 – μ2 ≤ 0
H1: μ1 – μ2 > 0
H0: μ1 – μ2 = 0
H1: μ1 – μ2 ≠ 0
a
a
-za
Reject H0 if Z < -Za
za
Reject H0 if Z > Za
a/2
-za/2
a/2
za/2
Reject H0 if Z < -Za/2
or Z > Za/2
Chap 9-10
Confidence Interval,
σ1 and σ2 Known
Population means,
independent
samples
σ1 and σ2 known
σ1 and σ2 unknown
The confidence interval for
μ1 – μ2 is:
*


2
1
2
σ
σ2
X1  X 2  Z

n1
n2
Chap 9-11
σ1 and σ2 Unknown
Assumptions:
Population means,
independent
samples
 Samples are randomly and
independently drawn
σ1 and σ2 known
σ1 and σ2 unknown
*
 Populations are normally
distributed or both sample
sizes are at least 30
 Population variances are
unknown but assumed equal
Chap 9-12
σ1 and σ2 Unknown
(continued)
Forming interval
estimates:
Population means,
independent
samples
σ1 and σ2 known
σ1 and σ2 unknown
*
 The population variances
are assumed equal, so use
the two sample standard
deviations and pool them to
estimate σ
 the test statistic is a t value
with (n1 + n2 – 2) degrees
of freedom
Chap 9-13
σ1 and σ2 Unknown
(continued)
Population means,
independent
samples
The pooled standard
deviation is
σ1 and σ2 known
σ1 and σ2 unknown
*
Sp 
n1  1S12  n2  1S2 2
(n1  1)  (n2  1)
Chap 9-14
σ1 and σ2 Unknown
(continued)
The test statistic for
μ1 – μ2 is:
Population means,
independent
samples

X  X   μ  μ 
t
1
σ1 and σ2 known
σ1 and σ2 unknown
2
1
2
1 1 
S   
 n1 n2 
2
p
*
Where t has (n1 + n2 – 2) d.f.,
and
2
2




n

1
S

n

1
S
1
2
2
S2  1
p
(n1  1)  (n2  1)
Chap 9-15
Confidence Interval,
σ1 and σ2 Unknown
Population means,
independent
samples
The confidence interval for
μ1 – μ2 is:
X
σ1 and σ2 known
σ1 and σ2 unknown
1

 X2  t n1 n2 -2
*
1 1 
S   
 n1 n2 
2
p
Where
2
p
S
2
2

n1  1S1  n2  1S2

(n1  1)  (n2  1)
Chap 9-16
Pooled Sp 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 (a = 0.05)?
Chap 9-17
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 9-18
Solution
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 = ± 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 a = 0.05
1 
 1
Conclusion:
1.5021  

 21 25 
There is evidence of a
difference in means.
Chap 9-19
Related Samples
Tests Means of 2 Related Populations
Related
samples



Paired or matched samples
Repeated measures (before/after)
Use difference between paired values:
D = X1 - X2
 Eliminates Variation Among Subjects
 Assumptions:
 Both Populations Are Normally Distributed
 Or, if Not Normal, use large samples
Chap 9-20
Mean Difference, σD Known
The ith paired difference is Di , where
Related
samples
Di = X1i - X2i
n
The point estimate for
the population mean
paired difference is D :
D
D
i 1
i
n
Suppose the population
standard deviation of the
difference scores, σD, is known
n is the number of pairs in the paired sample
Chap 9-21
Mean Difference, σD Known
(continued)
Paired
samples
The test statistic for the mean
difference is a Z value:
D  μD
Z
σD
n
Where
μD = hypothesized mean difference
σD = population standard dev. of differences
n = the sample size (number of pairs)
Chap 9-22
Confidence Interval, σD Known
Paired
samples
The confidence interval for D is
σD
DZ
n
Where
n = the sample size
(number of pairs in the paired sample)
Chap 9-23
Mean Difference, σD Unknown
Related
samples
If σD is unknown, we can estimate the
unknown population standard deviation
with a sample standard deviation:
The sample standard
deviation is
n
SD 
2
(D

D
)
 i
i1
n 1
Chap 9-24
Mean Difference, σD Unknown
(continued)
Paired
samples
The test statistic for D is now a
t statistic, with n-1 d.f.:
D  μD
t
SD
n
n
Where t has n - 1 d.f.
and SD is:
SD 
 (D
i1
i
 D)
2
n 1
Chap 9-25
Confidence Interval, σD Unknown
Paired
samples
The confidence interval for D is
SD
D  t n1
n
n
where
SD 
 (D  D)
i1
2
i
n 1
Chap 9-26
Hypothesis Testing for
Mean Difference, σD Unknown
Paired Samples
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: μD  0
H1: μD < 0
H0: μD ≤ 0
H1: μD > 0
H0: μD = 0
H1: μD ≠ 0
a
a
-ta
Reject H0 if t < -ta
ta
Reject H0 if t > ta
Where t has n - 1 d.f.
a/2
-ta/2
a/2
ta/2
Reject H0 if t < -ta/2
or t > ta/2
Chap 9-27
Paired Samples Example
 Assume you send your salespeople to a “customer
service” training workshop. Is the training effective?
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 
 (D  D)
2
i
n 1
 5.67
Chap 9-28
Paired Samples: 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
Critical Value = ± 4.604
d.f. = n - 1 = 4
Reject
Reject
a/2
a/2
- 4.604
4.604
- 1.66
Decision: Do not reject H0
(t stat is not in the reject region)
Test Statistic:
D  μD  4.2  0
t

 1.66
SD / n 5.67/ 5
Conclusion: There is not a
significant change in the
number of complaints.
Chap 9-29
Two Population Proportions
Population
proportions
Goal: test a hypothesis or form a
confidence interval for the difference
between two population proportions,
p 1 – p2
Assumptions:
n1p1  5 , n1(1-p1)  5
n2p2  5 , n2(1-p2)  5
The point estimate for
the difference is
ps1  ps2
Chap 9-30
Two Population Proportions
Population
proportions
Since we begin by assuming the null
hypothesis is true, we assume p1 = p2
and pool the two ps estimates
The pooled estimate for the
overall proportion is:
X1  X2
p
n1  n2
where X1 and X2 are the numbers from
samples 1 and 2 with the characteristic of
interest
Chap 9-31
Two Population Proportions
(continued)
The test statistic for
p1 – p2 is a Z statistic:
Population
proportions

p
Z
where
p
s1

 p s2   p1  p 2 
1 1
p (1  p)   
 n1 n2 
X1  X2
X
X
, ps1  1 , ps2  2
n1  n2
n1
n2
Chap 9-32
Confidence Interval for
Two Population Proportions
Population
proportions
p
s1
The confidence interval for
p1 – p2 is:

 p s2  Z
ps1 (1 p s1 )
n1

p s2 (1 p s2 )
n2
Chap 9-33
Hypothesis Tests for
Two Population Proportions
Population proportions
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: p1  p2
H1: p1 < p2
H0: p1 ≤ p2
H1: p1 > p2
H0: p1 = p2
H1: p1 ≠ p2
i.e.,
i.e.,
i.e.,
H0: p1 – p2  0
H1: p1 – p2 < 0
H0: p1 – p2 ≤ 0
H1: p1 – p2 > 0
H0: p1 – p2 = 0
H1: p1 – p2 ≠ 0
Chap 9-34
Hypothesis Tests for
Two Population Proportions
(continued)
Population proportions
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: p1 – p2  0
H1: p1 – p2 < 0
H0: p1 – p2 ≤ 0
H1: p1 – p2 > 0
H0: p1 – p2 = 0
H1: p1 – p2 ≠ 0
a
a
-za
Reject H0 if Z < -Za
za
Reject H0 if Z > Za
a/2
-za/2
a/2
za/2
Reject H0 if Z < -Za/2
or Z > Za/2
Chap 9-35
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 9-36
Example:
Two population Proportions
(continued)
 The hypothesis test is:
H0: p1 – p2 = 0 (the two proportions are equal)
H1: p1 – p2 ≠ 0 (there is a significant difference between proportions)
 The sample proportions are:
 Men:
ps1 = 36/72 = .50
 Women:
ps2 = 31/50 = .62
 The pooled estimate for the overall proportion is:
X1  X2 36  31 67
p


 .549
n1  n2 72  50 122
Chap 9-37
Example:
Two population Proportions
(continued)
The test statistic for p1 – p2 is:

p
z

s1

 p s2   p1  p 2 
1 1
p (1 p)   
 n1 n2 
 .50  .62   0
1 
 1
.549 (1 .549) 


72
50


Reject H0
Reject H0
.025
.025
-1.96
-1.31
  1.31
Critical Values = ±1.96
For a = .05
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.
Chap 9-38
Hypothesis Tests for Variances
Tests for Two
Population
Variances
F test statistic
*
H0: σ12 = σ22
H1: σ12 ≠ σ22
Two-tail test
H0: σ12  σ22
H1: σ12 < σ22
Lower-tail test
H0: σ12 ≤ σ22
H1: σ12 > σ22
Upper-tail test
Chap 9-39
Hypothesis Tests for Variances
(continued)
Tests for Two
Population
Variances
F test statistic
The F test statistic is:
*
2
1
2
2
S
F
S
S12 = Variance of Sample 1
n1 - 1 = numerator degrees of freedom
S22 = Variance of Sample 2
n2 - 1 = denominator degrees of freedom
Chap 9-40
The F Distribution
 The F critical value is found from the F table
 The are two appropriate degrees of freedom:
numerator and denominator
S12
F 2
S2
where df1 = n1 – 1 ; df2 = n2 – 1
 In the F table,
 numerator degrees of freedom determine the column
 denominator degrees of freedom determine the row
Chap 9-41
Finding the Rejection Region
H0: σ12  σ22
H1: σ12 < σ22
a
H0: σ12 = σ22
H1: σ12 ≠ σ22
a/2
0
Reject
H0
FL
0
Reject H0 if F < FL
Reject
H0
H0: σ1 ≤ σ2
H1: σ12 > σ22
2
2
Do not
reject H0
FU
Reject H0
FL
Do not
reject H0
FU
rejection
region for a
two-tail test is:

a
0
a/2
F
Do not
reject H0
F
F
Reject H0
S12
F  2  FU
S2
S12
F  2  FL
S2
Reject H0 if F > FU
Chap 9-42
Finding the Rejection Region
(continued)
a/2
H0: σ12 = σ22
H1: σ12 ≠ σ22
a/2
0
Reject
H0
FL
Do not
reject H0
FU
Reject H0
F
To find the critical F values:
1. Find FU from the F table
for n1 – 1 numerator and
n2 – 1 denominator
degrees of freedom
1
2. Find FL using the formula: FL 
FU*
Where FU* is from the F table
with n2 – 1 numerator and n1 – 1
denominator degrees of freedom
(i.e., switch the d.f. from FU)
Chap 9-43
F Test: An Example
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 a = 0.05 level?
Chap 9-44
F Test: Example Solution
 Form the hypothesis test:
H0: σ21 – σ22 = 0 (there is no difference between variances)
H1: σ21 – σ22 ≠ 0 (there is a difference between variances)
 Find the F critical values for a = .05:
FU:
 Numerator:
 n1 – 1 = 21 – 1 = 20 d.f.
 Denominator:
 n2 – 1 = 25 – 1 = 24 d.f.
FU = F.025, 20, 24 = 2.33
FL:
 Numerator:
 n2 – 1 = 25 – 1 = 24 d.f.
 Denominator:
 n1 – 1 = 21 – 1 = 20 d.f.
FL = 1/F.025, 24, 20 = 1/2.41
= .41
Chap 9-45
F Test: Example Solution
(continued)
 The test statistic is:
H0: σ12 = σ22
H1: σ12 ≠ σ22
S12 1.302
F 2 
 1.256
2
S2 1.16
a/2 = .025
a/2 = .025
0
Reject H0
 F = 1.256 is not in the rejection
region, so we do not reject H0
Do not
reject H0
FL=0.41
Reject H0
F
FU=2.33
 Conclusion: There is not sufficient evidence
of a difference in variances at a = .05
Chap 9-46
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