This module discusses confidence intervals and hypothesis tests for variances for the one sample situation.
Reviewed 19 July 05/ MODULE 25
25 - 1

Earlier we selected from the population of weights numerous samples of sizes n = 5, 10, and 20 where we assumed we knew that the population parameters were:

= 150 lbs,

2 = 100 lbs 2 ,

= 10 lbs.
25 - 2

For the population mean 
, point estimates, confidence intervals and hypothesis tests were based distributions.
For the population variance 
2 , point estimates, confidence intervals and hypothesis tests are based on the sample variance s 2 and the chi-squared distribution for
( n


1)
2 s
2


( )
2
 
2
25 - 3

For a 95% confidence interval, or

= 0.05, we use
C


( n

1) s
2

2
1
 
/ 2
 
2 
( n

1) s
2

2

/ 2

 
0.95
For hypothesis tests we calculate

2 
( n

1) s
2

2 and compare the results to the χ 2 tables.
25 - 4

Population of Weights Example x 2 = 166.41
s 2 = 166.41 is sample estimate of

2 = 100 s = 12.9 is sample estimate of

= 10
For a 95% confidence interval, we use
C


( n

1) s
2

2
0.975
 
2 
( n

1) s
2

2
0.025



0.95

2
0.975(4)

2
0.025(4)

11.143

0.484
df = n - 1 = 4
25 - 5

C


( n

1) s
2

2
0.975
 
2 
( n

1) s
2

2
0.025



0.95
C


11.143
 
2 
0.484
C


665.64
11.143
 
2 
665.64
0.484



0.95
C  59.74
 
2 
1, 375.29


0.95
Length

1, 315.55
lbs
2



0.95
25 - 6

From the Population of weights, for n = 5, we had x

2
146.4
x

3
153.2
s
2
= 5.4 s
2 
2
29.16
x

4
149.0
x

5
153.6
s
3
= 18.6 s
4
= 8.1 s
2 
3
345.96
s
2 
4
65.61
s
5
= 7.7 s
2

5
59.29
25 - 7

95% CI for

2 , n = 5, df = 4 s
2
2
C


4(29.16)
11.143
 
2 
4(29.16)
0.484



C  10.47
 
2 
240.99


0.95
Length = 230.52 lbs 2 s
3
2
C


4(345.96)
11.143
 
2 
4(345.96)
0.484



C  124.19
 
2 
2,859.17


0.95
Length = 2,734.98 lbs 2
25 - 8

s
4
2
C


4(65.61)
11.143
 
2 
4(65.61)
0.484



C 23.55
 
2 
542.23


0.95
Length = 518.68 lbs 2 s
5
2
C


4(59.29)
11.143
 
2 
4(59.29)
0.484



C 21.28
 
2 
490.00


0.95
Length = 468.72 lbs 2
25 - 9

x

1
151.6
x

2
151.3
x

3
150.4
x
4

151.4
x

5
150.1
s
1 s
2
= 10.2
= 8.4 s
2 
1
104.04
s
2

2
70.55
s
2

3
129.96
s
3
= 11.4 s
4
= 11.5 s
2 
4
132.25
s
5
= 8.4 s
2 
5
70.56
25 - 10

s
1
2 s
2
2 s
3
2
95% CIs for

2 , n = 20, df = 19
C


( n

1) s
2

2
0.975

32.852
 
2 

2
( n
0.025


1) s
2
8.907



0.95
C

19(104.04)

32.852
 
2 
19(104.04)
8.907



C 60.17
 
2 
221.93


0.95
Length = 161.76 lbs 2
C 


19(70.55)
32.852
 
2 
19(70.55)
8.907




C 

40.80
 
2 
149.44

 
0.95
C



19(129.96)
32.852
 
2 
19(129.96)
8.907




C 

75.16
 
2 
277.22

 
0.95
25 - 11

Example: For the first sample from the samples with n = 5, we had s 2 = 166.41.
Test whether or not

2 = 200.
1. The hypothesis:
2. The assumptions:
3. The α-level:
H
0
:

2 = 200, vs H
1
:

2 ≠ 200
Independent observations normal distribution
α = 0.05
25 - 12

4. The test statistic:

2 
( n

1) s
2

2
5. The critical region
:
Reject H
0
: σ 2 = 200 if the value calculated for χ 2 is not between
χ 2
0.025
χ 2
0.975
(4) =0 .484, and
(4) =11.143
6. The Result:

2 
( n


1)
2 s
2

4(166.41)
200

3.33
7. The conclusion: Accept H
0
:

2 = 200.
25 - 13

25 - 14

25 - 15

25 - 16

Table 3 indicates that the mean Global Stress Index for Lesbians is 16 with SD = 6.8. Suppose that previous work in this area had indicated that the SD for the population was about 
= 10. Hence, we would be interested in testing whether or not

2 = 100.
25 - 17

1. The hypothesis: H
0
:

2 = 100, vs H
1
:

2 ≠ 100
2. The assumptions: Independence, normal distribution
3. The α-level: α
= 0.05
4. The test statistic:

2 
( n

1) s
2

2
5. The critical region
:
Reject H
0
: σ 2 = 100 if the value calculated for χ 2 is not between
χ 2
0.025
χ 2
0.975
(549) = 615.82, and
(549) = 485.97
6. The Result: 
2 
( n


1)
2 s
2

549(46.24)
100

253.86
7. The conclusion: Reject H
0
:

2 = 100.
25 - 18