The comparison of two populations

The Comparison of Two
Populations

Slide 1

Shakeel Nouman
M.Phil Statistics

The Comparison of Two Populations By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

8

Slide 2

The Comparison of Two Populations

• Using Statistics
• Paired-Observation Comparisons
• A Test for the Difference between Two
•
•

•

Population Means Using Independent
Random Samples
A Large-Sample Test for the Difference
between Two Population Proportions
The F Distribution and a Test for the
Equality of Two Population Variances
Summary and Review of Terms


8-1 Using Statistics
•

Slide 3

Inferences about differences between parameters
of two populations
 Paired-Observations
 Observe the same group of persons or things
• At two different times: “before” and “after”
• Under two different sets of circumstances or “treatments”

 Independent Samples
» Observe different groups of persons or things
• At different times or under different sets of circumstances


8-2 Paired-Observation
Comparisons
•

•

Slide 4

Population parameters may differ at two different
times or under two different sets of
circumstances or treatments because:
The circumstances differ between times or
treatments
The people or things in the different groups are
themselves different
By looking at paired-observations, we are able to
minimize the “between group” , extraneous
variation.


Paired-Observation ComparisonsSlide 5
of Means
Test statistic for the paired - observations t test:
D -  D0
t
sD
n
w here D is the sample average differencebetw een each
pair of observations, s D is the sample standard deviation
of these difference and the sample size, n, is the number
s,
of pairs of observations. The symbol  D0 is the population
mean differenceunder the null hypothesis. When thenull
hypothesis is true and the population mean differenceis  D0 ,
the statistic has a t distribution w ith (n - 1) degrees of freedom.

Example 8-1

Slide 6

A random sample of 16 viewers of Home Shopping Network was selected for an experiment. All viewers in the
sample had recorded the amount of money they spent shopping during the holiday season of the previous year.
The next year, these people were given access to the cable network and were asked to keep a record of their total
purchases during the holiday season. Home Shopping Network managers want to test the null hypothesis that
their service does not increase shopping volume, versus the alternative hypothesis that it does.
Shopper
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

Previous
334
150
520
95
212
30
1055
300
85
129
40
440
610
208
880
25

Current
405
125
540
100
200
30
1200
265
90
206
18
489
590
310
995
75

Diff
71
-25
20
5
-12
0
145
-35
5
77
-22
49
-20
102
115
50

H0:  0
D
H1:  > 0
D

df = (n-1) = (16-1) = 15
D - D
0
Test Statistic:
t 
sD
n
Critical Value: t0.05 = 1.753

Do not reject H0 if : t 
1.753
Reject H0 if: t > 1.753


Example 8-1: Solution
D - D
32.81 - 0
0
t

 2.354
sD
55.75

t = 2.354 > 1.753, so H0 is rejected and we conclude that
there is evidence that shopping volume by network
viewers has increased, with a p-value between 0.01 an
0.025. The Template output gives a more exact p-value
of 0.0163. See the next slide for the output.

16

n

Slide 7

t Distribution: df=15
0.4

f(t)

0.3

0.2
Nonrejection
Region

0.1

Rejection
Region

0.0
-5

0

1.753
= t0.05

5

2.131
= t0.025

t

2.602
= t0.01

2.354=
test
statistic


Example 8-1: Template for
Testing Paired Differences

Slide 8


Example 8-2

Slide 9

It has recently been asserted that returns on stocks may change once a story about a company appears in The Wall
Street Journal column “Heard on the Street.” An investments analyst collects a random sample of 50 stocks that
were recommended as winners by the editor of “Heard on the Street,” and proceeds to conduct a two-tailed test of
whether or not the annualized return on stocks recommended in the column differs between the month before and
the month after the recommendation. For each stock the analysts computes the return before and the return after
the event, and computes the difference in the two return figures. He then computes the average and standard
deviation of the differences.

H0: D  0
H1: D > 0

D - D
0.1 - 0
0
z 

 14 .14
sD
0.05

n = 50
D = 0.1%
sD = 0.05%
Test Statistic:

n

z 

D - D
0
sD
n

50

p - value: p ( z > 14.14 )  0
This test result is highly significant,
and H 0 may be rejected at any reasonable
level of significance.


Confidence Intervals for Paired
Observations
A (1 -  ) 100% confidence interval for the mean difference 

D

Slide 10

:

s

D  t D
2 n

where t is the value of the t distributi on with (n - 1) degrees of freedom that cuts off an
2
area of



to its right, When the sample size is large, we may use z instead.
.
2
2


Observations – Example 8-2

Slide 11

95% confidence interval for the data in Example 8 - 2 :
s
D  z D  0.11.96 0.05  0.1 (1.96)(.0071)
n
50
2
 0.1 0.014  [0.086,0.114]
Note that this confidence interval does not include the value 0.


Observations – Example 8-2 Using
the Template


8-3 A Test for the Difference between Two
Population Means Using Independent Random
Samples

•

Slide 13

When paired data cannot be obtained, use
independent random samples drawn at different
times or under different circumstances.
Large sample test if:
» Both n1 30 and n2 30 (Central Limit Theorem), or
» Both populations are normal and s1 and s2 are both
known

Small sample test if:
» Both populations are normal and s1 and s2 are
unknown


Comparisons of Two Population
Means: Testing Situations
•

•

•

Slide 14

I: Difference between two population means is 0
 1= 2
» H0: 1 -2 = 0
» H1: 1 -2  0

II: Difference between two population means is less than
0
 1 2
» H0: 1 -2  0
» H1: 1 -2 > 0

III: Difference between two population means is less than
D
 1  2+D
» H0: 1 -2  D
» H1: 1 -2 > D

Means: Test Statistic

Slide 15

Large-sample test statistic for the difference between two
population means:
z

( x - x ) - ( -  )
1

2

s

1

2
1

n

+

2

s

0

2
2

n

The term (1- 2)0 is the difference between 1 an 2 under the
null hypothesis. Is is equal to zero in situations I and II, and it is
equal to the prespecified value D in situation III. The term in the
denominator is the standard deviation of the difference between
the two sample means (it relies on the assumption that the two
samples are independent).
1

2


Two-Tailed Test for Equality of
Two Population Means: Example
8-3

Slide 16

Is there evidence to conclude that the average monthly charge in the entire population of American Express Gold
Card members is different from the average monthly charge in the entire population of Preferred Visa
cardholders?

Population1 : Preferred Visa
H

n = 1200

0

: - 0
1
2

H : - 0
1
1
2

1

x = 452
1

s = 212
1

Population 2 : Gold Card

( x - x ) - ( -  )
2
1
2 0  ( 452 - 523) - 0
z  1
2
2
2
2
s
s
212
185
1 + 2
+
1200
800
n
n
1
2
- 71



80.2346



- 71

 -7.926

8.96

n = 800
2

x = 523

p - value : p(z < -7.926)  0

2

s = 185
2

H

0

is rejected at any common level of significan ce


Example 8-3: Carrying Out the
Test
Standard Normal Distribution
0.4

f(z)

0.3

0.2

0.1

0.0

-z0.01=-2.576

Rejection
Region
Test Statistic=-7.926

0

z
z0.01=2.576

Nonrejection Rejection
Region
Region

Slide 17

Since the vlue of the
test sttistic is fr
below the lower criticl
point, the null
hypothesis y be
rejected, nd we y
conclude tht there is 
sttisticlly significnt
difference between the
verge onthly chrges
of Gold Crd nd
Preferred Vis
crdholders.


Example 8-3: Using the
Template

Slide 18


Two-Tailed Test for Difference
Between Two Population Means:
Example 8-4

Slide 19

Is there evidence to substntite Durcells cli tht their btteries lst, on verge, t
lest 45 inutes longer thn Energizer btteries of the se size?

Population1 : Duracell

H :  -   45
0 1
2
H :  -  > 45
1 1
2

n = 100
1

x = 308
1

s = 84
1

Population 2 : Energizer

( x - x ) - ( -  )
2
1
2 0  (308 - 254) - 45
z 1
2
2
2
2
s
s
84
67
1 + 2
+
100 100
n
n
1
2


9
115.45



9

 0.838

10.75

n = 100
2

x = 254
2

s = 67
2

p - value : p(z > 0.838) = 0.201
H may not be rejected at any common
0
level of significan ce


Two-Tailed Test for Difference Slide 20
Between Two Population Means:
Example 8-4 – Using the Template
Is there evidence to substantiate Duracell’s claim that their batteries last, on average, at least 45 minutes longer
than Energizer batteries of the same size?


Confidence Intervals for the
Difference between Two
Population Means

Slide 21

A large-sample (1-)100% confidence interval for the difference
between two population means, 1- 2 , using independent
random samples:
(x - x )  z
1
2

2

2
2
s
1 + 2
n
n
1
2

s

A 95% confidence interval using the data in example 8-3:
(x - x )  z
1
2

2

2
2
s
2122 1852
1 + 2  (523 - 452)  1.96
+
 [53.44,88.56]
1200 800
n
n
1
2

s


8-4 A Test for the Difference between Two
Population Means: Assuming Equal Population
Variances

Slide 22

• If we might assume that the population variances s12 and s22
are equal (even though unknown), then the two sample
variances, s12 and s22, provide two separate estimators of
the common population variance. Combining the two
separate estimates into a pooled estimate should give us a
better estimate than either sample variance by itself.

* * ** * *** * * * *
*
Sample 1
x1
From sample 1 we get the estimate s12 with
(n1-1) degrees of freedom.

Deviation from the
mean. One for each
sample data point.

}

}

Deviation from the
mean. One for each
sample data point.

* ** * * * * * *
*
** *
Sample 2
x2
From sample 2 we get the estimate s22 with
(n2-1) degrees of freedom.

From both samples together we get a pooled estimate, sp2 , with (n1-1) + (n2-1) = (n1+ n2 -2)
total degrees of freedom.

Pooled Estimate of the
Population Variance

Slide 23

A pooled estimate of the common population variance, based on a sample
variance s12 from a sample of size n1 and a sample variance s22 from a sample
of size n2 is given by:

(n1 - 1) s12 + (n2 - 1) s22
s2 
p
n1 + n2 - 2

The degrees of freedom associated with this estimator is:
df = (n1+ n2-2)
The pooled estimate of the variance is a weighted average of the two
individual sample variances, with weights proportional to the sizes of the two
samples. That is, larger weight is given to the variance from the larger
sample.

Using the Pooled Estimate of the
Population Variance
The estimate of the standard deviation of (x1 - x 2 ) is given by:

Slide 24

1 
2 1
sp 
+

n1
n2 


Test statistic for the difference between two population means, assuming equal
population variances:
(x1 - x 2 ) - (  1 -  2 ) 0
t=
1
2 1
sp  +

n1 n2 


where (  1 -  2 ) 0 is the difference between the two population means under the null
hypothesis (zero or some other number D).

The number of degrees of freedom of the test statistic is df = ( n1 + n2 - 2 ) (the
2
number of degrees of freedom associated with s p , the pooled estimate of the
population variance.

Example 8-5

Slide 25

Do the data provide sufficient evidence to conclude that average percentage increase in the CPI differs when oil
sells at these two different prices?

Population 1: Oil price = $27.50
n1 = 14
x1 = 0.317%
s1 = 0.12%

Population 2: Oil price = $20.00
n2 = 9
x 2 = 0.21%
s 2 = 0.11%
df = (n + n - 2 )  (14 + 9 - 2 )  21
1
2

H 0 : 1 -  2  0
H1:  1 -  2  0

( x1 - x 2 ) - (  1 -  2 ) 0
t 
2
2
 ( n1 - 1) s1 + ( n2 - 1) s2   1 1 

 + 
n1 + n2 - 2

  n1 n2 
0.107
0.107


 2.154
0.00247 0.0497
Critical point: t

= 2.080

0.025
H 0 may be rejected at the 5% level of significance


Template

Slide 26

Do the data provide sufficient evidence to conclude that average percentage increase in the CPI differs when oil
sells at these two different prices?

P-value =
0.0430, so
reject H0 at
the 5%
significance
level.


Example 8-6
The manufacturers of compact disk players want
to test whether a small price reduction is enough
to increase sales of their product. Is there
evidence that the small price reduction is enough
to increase sales of compact disk players?

H : - 0
0
2
1
H : - >0
1
2
1
t

Population 1: Before Reduction
n 1 = 15
x 1 = $6598



s1 = $844

Population 2: After Reduction
n 2 = 12

Slide 27



( x - x ) - ( -  )
2
1
2
1 0
 ( n - 1) s 2 + ( n - 1) s 2  1 1
 1
1
2
2 
+

 n n
n +n -2
1
2

 1 2






( 6870 - 6598) - 0

 (14)844 2 + (11)669 2  1 1 

 + 

 15 12 
15 + 12 - 2


272



89375.25

272
 0.91
298.96

x 2 = $6870
s 2 = $669

Critical point : t

= 1.316
0.10

df = (n + n - 2 )  (15 + 12 - 2 )  25
1
2

H may not be rejected even at the 10% level of significan ce
0


Template

Slide 28

P-value =
0.1858, so
do not
reject H0 at
the 5%
significance
level.


Example 8-6: Continued

t Distribution: df = 25
0.4

f(t)

0.3
0.2
0.1

0.0
-5

-4

-3

-2

-1

0

1

Nonrejection
Region

2

3

4

t0.10=1.316
Rejection
Region

5

t

Slide 29

Since the test statistic is less
than t0.10, the null hypothesis
cannot be rejected at any
reasonable level of
significance. We conclude
that the price reduction does
not significantly affect sales.

Test Statistic=0.91


Confidence Intervals Using the
Pooled Variance

Slide 30

A (1-) 100% confidence interval for the difference between two
population means, 1- 2 , using independent random samples and
assuming equal population variances:
( x1 - x2 )  t


2 1

sp 

 n1

+



n2 
1

2

A 95% confidence interval using the data in Example 8-6:
( x1 - x 2 )  t



2
sp

 1 + 1


 n1 n2 

 ( 6870 - 6598 )  2 .06 ( 595835)( 0.15)  [ -343.85,887 .85]

2

Confidence Intervals Using the Pooled
Variance and the TemplateExample 8-6

Confidence Interval

8-5 A Large-Sample Test for the
Difference between Two Population
• Hypothesized difference is zero
Proportions
I: Difference between two population proportions is 0

Slide 32

• p1= p2
» H0: p1 -p2 = 0
» H1: p1 -p2  0
II: Difference between two population proportions is less than 0
• p1p2
» H0: p1 -p2  0
» H1: p1 -p2 > 0
• Hypothesized difference is other than zero:
III: Difference between two population proportions is less than
D
• p1  2+D
p
» H0:p-p2 D
The Comparison of H : p -p By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
» Two1Populations2 > D
1

Proportions When the Hypothesized
Difference Is Zero: Test Statistic

Slide 33

When the popultion proportions re hypothesized to be
p

equl, then  pooled estitor of the proportion ( ) y
be used in clculting the difference between
A lrge-sple test sttistic for thetest sttistic. two
popultion proportions, when the hypothesized difference is zero:

z
where

( p1 - p2 ) - 0
 
1 1
p 1- p  + 
( )

 n1 n2 

is the
x1 is the sple proportion in sple 1 nd 1
x
p1 

p1 

sple
n1
n1
p

proportion in sple 2. The sybol
stnds for the cobined

sple proportion in both sples, considered s  single sple.
Tht is:
x +x
p
ˆ
n +n
1

1

1

2


Proportions When the Hypothesized
Difference Is Zero: Example 8-8

Slide 34

Carry out a two-tailed test of the equality of banks’ share of the car loan market in 1980 and 1995.

n1 = 100

H 0 : p1 - p 2  0
H1: p1 - p 2  0

x1 = 53

z 

Population 1: 1980

p1 = 0.53


Population 2: 1995
n 2 = 100
x 2 = 43
p 2 = 0.43

x1 + x 2
53 + 43
p


 0.48
n1 + n 2 100 + 100

( p1 - p 2 ) - 0

p (1 


 1 1
p )

+

 n1 n2 

0.10



0.004992
Critical point: z

0.10



0.53 - 0.43

 1 + 1

 100 100

(.48)(.52) 

 1.415

0.07065
= 1.645

0.05
H 0 may not be rejected even at a 10%
level of significance.


Test

Slide 35

0.4

f(z)

0.3

0.2

0.1

0.0

z
-z0.05=-1.645

Rejection
Region

0

z0.05=1.645

Nonrejection
Region

Rejection
Region

Since the value of the test
statistic is within the
nonrejection region, even at a
10% level of significance, we
may conclude that there is no
statistically significant
difference between banks’
shares of car loans in 1980
and 1995.

Test Statistic=1.415


Template

Slide 36

P-value =
0.157, so do
not reject
H0 at the
5%
significance
level.


Comparisons of Two Population Proportions
When the Hypothesized Difference Is Not Zero:
Example 8-9

Slide 37

Carry out a one-tailed test to determine whether the population proportion of traveler’s check buyers who buy at
least $2500 in checks when sweepstakes prizes are offered as at least 10% higher than the proportion of such
buyers when no sweepstakes are on.

n1 = 300

H 0 : p1 - p 2  0.10
H 1 : p1 - p 2 > 0.10

x1 = 120

z

Population 1: With Sweepstakes

p1 = 0.40


Population 2: No Sweepstakes
n 2 = 700
x 2 = 140
p 2 = 0.20




( p1 - p 2 ) - D





 p (1 - p )
1
1

 n1 +


p (1 - p ) 


2
2 

n2



( 0.40 - 0.20) - 0.10

 ( 0.40)( 0.60) ( 0.20)(.80) 
+


700
 300


Critical point: z



0.10

 3.118

0.03207

= 3.09

0.001
H 0 may be rejected at any common level of significance.


Test
0.4

f(z)

0.3

0.2

0.1

0.0
0

Nonrejection
Region

z
z0.001=3.09

Rejection
Region
Test Statistic=3.118

Slide 38

statistic is above the critical
point, even for a level of
significance as small as 0.001,
the null hypothesis may be
rejected, and we may conclude
that
the
proportion
of
customers buying at least
$2500 of travelers checks is at
least 10% higher when
sweepstakes are on.


Template

Slide 39

P-value =
0.0009, so
reject H0 at
the 5%
significance
level.


Confidence Intervals for the
Difference between Two
Population Proportions

Slide 40

A (1- 100% large-sample confidence interval for the difference
)
between two population proportions:
( p1 - p 2 )  z







 p (1 - p )
1
 1
 n1 +


p (1 - p ) 


2
2 

n2



2

A 95% confidence interval using the data in example 8-9:




 p1 (1 - p1 ) p 2 (1 - p 2 ) 

  ( 0.4 - 0.2)  1.96 ( 0.4 )( 0.6) + ( 0.2)( 0.8)
( p1 - p 2 )  z


+

n2
300
700
 
 n1

2
 0.2  (1.96)( 0.0321)  0.2  0.063  [ 0.137 ,0.263]

Confidence Intervals for the Difference
between Two Population Proportions –
Using the Template – Using the Data
from Example 8-9


8-6 The F Distribution and a Test for
Equality of Two Population
Variances

Slide 42

The F distribution is the distribution of the ratio of two chisquare random variables that are independent of each other, each
of which is divided by its own degrees of freedom.
An F random variable with k1 and k2 degrees of freedom:
c 12
k1
F( k ,k )  2
c2
k2
1

2


The F Distribution

F Distributions with different Degrees of Freedom

f(F)

• The F random variable cannot
be negative, so it is bound by
zero on the left.
• The F distribution is skewed to
the right.
• The F distribution is identified
the number of degrees of
freedom in the numerator, k1,
and the number of degrees of
freedom in the denominator,
k2 .

Slide 43

F(25,30)

1.0

F(10,15)
0.5

F(5,6)

0.0
0

1

2

3

4

5

F


Using the Table of the F
Distribution
Critical Points of the F Distribution Cutting Off a
Right-Tail Area of 0.05
k1

1

2

3

4

5

6

Slide 44

F Distribution with 7 and 11 De gre e s of Fre ed om
7

8

9
0.7

k2
161.4
18.51
10.13
7.71
6.61
5.99
5.59
5.32
5.12
4.96
4.84
4.75
4.67
4.60
4.54

199.5
19.00
9.55
6.94
5.79
5.14
4.74
4.46
4.26
4.10
3.98
3.89
3.81
3.74
3.68

215.7
19.16
9.28
6.59
5.41
4.76
4.35
4.07
3.86
3.71
3.59
3.49
3.41
3.34
3.29

224.6
19.25
9.12
6.39
5.19
4.53
4.12
3.84
3.63
3.48
3.36
3.26
3.18
3.11
3.06

230.2
19.30
9.01
6.26
5.05
4.39
3.97
3.69
3.48
3.33
3.20
3.11
3.03
2.96
2.90

234.0
19.33
8.94
6.16
4.95
4.28
3.87
3.58
3.37
3.22
3.09
3.00
2.92
2.85
2.79

236.8
19.35
8.89
6.09
4.88
4.21
3.79
3.50
3.29
3.14
3.01
2.91
2.83
2.76
3.01
2.71

238.9
19.37
8.85
6.04
4.82
4.15
3.73
3.44
3.23
3.07
2.95
2.85
2.77
2.70
2.64

240.5
19.38
8.81
6.00
4.77
4.10
3.68
3.39
3.18
3.02
2.90
2.80
2.71
2.65
2.59

0.6
0.5

f(F)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

0.4
0.3
0.2
0.1

F

0.0
0

1

2

3

4

5

F0.05=3.01

The left-hand critical point to go along with F(k1,k2) is given by:

1
F( k 2 ,k 1)

Where F(k1,k2) is the right-hand critical point for an F random variable with the
reverse number of degrees of freedom.

Critical Points of the F Distribution:
F(6, 9),  = 0.10

F Distribution with 6 and 9 Degrees of Freedom
0.7

0.05

0.90

0.6

f(F)

0.5

Slide 45

The right-hand critical point
read directly from the table of
the F distribution is:

0.4
0.3

F(6,9) =3.37

0.05

0.2
0.1
0.0
0

1

F0.95=(1/4.10)=0.2439

2

3

4

F0.05=3.37

5

F

The corresponding left-hand
critical point is given by:
1
1

 0.2439
F( 9 , 6) 410
.


Test Statistic for the Equality of
Two Population Variances

Slide 46

Test statistic for the equality of the variances of two normally
distributed populations:
F( n -1,n -1)
1

2

s12
 2
s2

 I: Two-Tailed Test
• s1  s2
H 0 : s1  s2
H 1 : s1  s2
 II: One-Tailed Test
• s1s2
H 0 : s1  s2
H 1 : s1 > s2

Example 8-10

Slide 47

The economist wants to test whether or not the event (interceptions and prosecution of insider traders) has
decreased the variance of prices of stocks.

Population1 : Before
n = 25
1
s 2  9.3
1
Population 2 : After
n = 24
2
s 2  3.0
2

  0.05
F

(24,23)

 2.01

  0.01
F

(24,23)

H 0: s
H1: s

2
1
2
1

s
>s

2

2

21
2
2

s2
9.3
1
F
 F


 3.1
3.0
n1 - 1, n 2 - 1
24,23
s2
2

(

)

(

)

H 0 may be rejected at a 1% level of significance.

 2.70


Distribution with 24 and 23 Degrees of Freedom
0.7
0.6

f(F)

0.5
0.4
0.3
0.2
0.1

F

0.0
0

1

2

F0.01=2.7

3

4

5

Test Statistic=3.1

Slide 48

statistic is above the critical
point, even for a level of
significance as small as 0.01,
the null hypothesis may be
rejected, and we may conclude
that the variance of stock
prices is reduced after the
interception and prosecution
of inside traders.


Example 8-10: Solution Using
the Template

Slide 49

Observe that the pvalue for the test is
0.0042 which is less
than 0.01. Thus the null
hypothesis must be
rejected at this level of
significance of 0.01.


Example 8-11: Testing the
Equality of Variances for
Example 8-5

Slide 50

Population 1 Population 2
n = 14
1

n =9
2

2
2
s  0.12
1

2
2
s  0.11
2

  0.05
F

(13,8)

 3.28

  0.10
F

(13,8)

 2.50

2
2
H :s  s
0 1
2
2
2
H :s  s
1 1
2
s2
2
1  0.12  119
F
F

.
n1 - 1, n2 - 1)
13,8) s2 0.112
(
(
2
H may not be rejected at the 10% level of significance.
0


F Distribution with 13 and 8 Degrees of Freedom
0.7

0.10

0.80

0.6

f(F)

0.5
0.4
0.3

0.10

0.2
0.1
0.0
0

1

F0.90=(1/2.20)=0.4545

2

3

4

F0.10=3.28

5

F

Slide 51

statistic is between the critical
points, even for a 20% level of
significance, we can not reject
the null hypothesis. We
conclude the two population
variances are equal.

Test Statistic=1.19


Template to test for the Difference
between Two Population
Variances: Example 8-11
Observe that the pvalue for the test is
0.8304 which is larger
than 0.05. Thus the
null
hypothesis
cannot be rejected at
this
level
of
significance of 0.05.
That is, one can
assume
equal
variance.


The F Distribution Template to


The Template for Testing Equality
of Variances


Name
Religion
Domicile
Contact #
E.Mail
M.Phil (Statistics)

Shakeel Nouman
Christian
Punjab (Lahore)
0332-4462527. 0321-9898767
sn_gcu@yahoo.com
sn_gcu@hotmail.com
GC University, .
(Degree awarded by GC University)

M.Sc (Statistics)
Statitical Officer
(BS-17)
(Economics & Marketing
Division)

GC University, .
(Degree awarded by GC University)

Livestock Production Research Institute
Bahadurnagar (Okara), Livestock & Dairy Development
Department, Govt. of Punjab


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