# Sampling Dist. Ratio of 2 Variances

advertisement ```IT233: Applied Statistics
TIHE 2005
Example 2: A sample of size 16 is to be taken from a population having a
normal distribution with  2 = 6.25. Find a number C such that
P(S 2  C)  0.025
Solution:
 2 = 6.25
Given:
n = 16

d. f. ( ) = n 1 = 15
P(S 2  C)  0.025
 (n  1) S 2
 P

(n 1)C 
= 0.025
 2
 2 

15  C 

 P 2 
= 0.025
6.25 

 P  2  2.4C = 0.025

0.025

0
2.4 C
From Chi-Square Table we get

2.4 C = 27.488
C = 11.453
Example 3: For the Example 2, find a constants C1 and C2 such that
P(C1  S 2  C2 )  0.90
1
Solution:
P(C1  S 2  C2 )  0.90
0.05
15C2 
 15C
1  2 
 P
 = 0.90
 6.25

6.25


 P 2.4C   2  2.4C = 0.90
1
2

0.05
0.90

0 2.4 C1
2.4 C2
From Chi-Square Table we find C1 and C2 as follows:

2.4 C1 = 7.261
2.4 C2 = 24.996

C1 = 3.025
C2 = 10.415
S12
Sampling Distribution of the Ratio of Two Sample Variances - 2
S2
Theorem: If S 2 and S 2 are the variances of independent random samples of
1
2
size n1 and n2 taken from normal populations with variances  2 and  2 ,
1
respectively, then the statistic
S12 12  22 S12
F

S22  22 12 S22
has an F - distribution with 1  n1 1 and  2  n2  1 d. f.
Applications:

The sampling distribution of
S12
S22
To find the probabilities about
2
S12
S22
could be used:
2
 12

To find the confidence interval for

To test the hypothesis H 0 : 12   22
 22
Example 1: (Page 229, No.9) If S 2 and S 2 represent the variances of
1
2
independent random samples of size n1 = 25 and n2 = 31, taken from normal
populations with variances  2 = 10 and  2 = 15, respectively, find
1
2

P S
2
1

2
S 1.26 .
2
Solution:
We know,
F
 22 S12
12 S22
Fn  1, n  1
1
2
2
 S2

 2 S2




P  1  1.26   P  2 1  1.26 2 
 S2

 2 S2
12 
 2

 1 2
15 

 P  F  1.26  
10 

0.05
 P  F  1.89 
0
1.89
Looking F-table for 1 = 24 and  2 = 30 at  = 0.05 we found P  F  1.89 =
0.05

 S2


1
P
 1.26  = 0.05
2
S

 2

3
Example 2: In the above example, find a value C such that
 S2


1
P
 C  = 0.01
2
S

 2

Solution:
 S2


1
P
 C  = 0.01
 S2

 2

 2 S2
 22 

2
1
 P
C
= 0.01
2 
 2 S2

1 
 1 2
1 = 24,  2 = 30
0.01
 P  F  1.5C  = 0.01
0
1.5 C
Looking F-table for 1 = 24 and  2 = 30 at  = 0.01 we get

NOTE:
1.5 C = 2.47
C = 1.647
Since F-distribution is also used for finding confidence interval
and hypothesis testing, sometimes we need to find lower tail as
well as upper tail values of F.
0.95
0.05

f 0.95 1,  2


f0.05 1, 2

Upper-tail value: Its cuts off a
small area to the right.
Lower-tail value: Its cuts off a
large area to the right.
4
The F-table gives only upper-tail values. The lower-tail values are computed as


f1   1,  2 


f 0.95 1,  2 
Hence,


f 0.975 1,  2 
Similarly,
Example:
19, 20  


2
1
2
f 0.99 with  = 28 and  = 12
1
2.1555
f 0.99  28,12  



1
= 0.4639
(ii)

f 0.05  2 , 1
f 0.025  2 , 1
1
f 0.05  20,19 
0.95 

1
1
Solution:
f
f  2 , 1
f 0.95 with  = 19 and  = 20
Find (i)
(ii)
(i)

1
1
f 0.01 12, 28 
1
2.90
= 0.3448
5
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