Sampling distribution of s2
The chi-square distribution results when independent variables
with normal distributions are squared and summed.
2

(
x

x
)
i
s2 
n 1
Sampling distribution of 2
The chi-square distribution results when independent variables
with normal distributions are squared and summed.
 -stat  (n  1) 
2
s2
2
2  ( n  1)
0
n–1
Sampling distribution of 2
The chi-square distribution results when independent variables
with normal distributions are squared and summed.
 -stat  (n  1) 
2
s2
2
.025
2
 .025
Sampling distribution of 2
The chi-square distribution results when independent variables
with normal distributions are squared and summed.
 -stat  (n  1) 
2
.975
2
.975
s2
2
Interval Estimation of  2
To derive the interval estimate of  2, first substitute
(n  1)s2/ 2 for 2 into the following inequality
2
2
.975
  2  .025
2
.975

(n  1)s 2
2
2
 .025
.025
Now write the above as two inequalities
Interval Estimation of  2
Next, multiply the inequalities by  2
2
(
n

1)
s
22
2
2
2



.975



(
n

1)
s
.975
2
2

((nn  1)
1)ss22 22
2
(22n  1)
s  .025
 2

Divide both of the inequalities above by the respective chisquare critical value:

22
.975
.975


2
2
2
 
22
.975
.975
(n  1)s
2
.975
.975 2
(n  1)s
2
.025
2
 2 
22
22



(n  1)s
25  
2
..0
025
  22
22
.02
.025
5
.025
.025
22
(n  1)s 2
2
.975
Interval Estimation of  2
The 95% confidence interval for the population variance
(n  1)s 2
2
.025
 2 
(n  1)s 2
2
.975
Interval Estimation of  2
The 1 - a confidence interval for the population variance
(n  1)s 2
2
a.025
/2
 2 
(n  1)s 2
2
1.975
 a/2
Interval Estimation of  2
Example 1
Buyer’s Digest rates thermostats manufactured for home
temperature control. In a recent test, ten thermostats
manufactured by ThermoRite were selected at random and
placed in a test room that was maintained at a temperature of
68oF.
Use the ten readings in the table below to develop a 95%
confidence interval estimate of the population variance.
Thermostat
1
2
3
4
5
6
7
8
9
10
Temperature 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.2
Interval Estimation of  2
xi
x
xix
68.1
i 
( x(ixi 68x.)12)2
67.4
67.8
68.2
69.3
69.5
67.0
68.1
68.6
67.9
67.2
-0.7
-0.3
0.1
1.2
1.4
-1.1
0.0
0.5
-0.2
-0.9
0.49
0.09
0.01
1.44
1.96
1.21
0.00
0.25
0.04
0.81
x  68.1
sum = 6.3
s 2 = 0.7
2

(
x

x
)
i
s2 
n 1
Interval Estimation of  2
((10
n -1)(0.7)
1)s 2
a2 /2
2 
2
((10
n
1)
s
-1)(0.7)
 12a /2
Interval Estimation of  2
((10
n -1)(0.7)
1)s 2
2
a0.25
/2
2 
2
((10
n
1)
s
-1)(0.7)
2
0.975
1 a /2
1a  .95
.025
.025
0
2
.975
9
2
.025
2
Interval Estimation of  2
Selected Values from the Chi-Square Distribution Table
Degrees
of Freedom
Area in Upper Tail
.99
.975
.95
.90
.10
.025
.01
5
0.554 0.831 1.145 1.610
11.070 12.832
15.086
6
0.872 1.237 1.635 2.204 10.645 12.592 14.449
16.812
7
1.239 1.690 2.167 2.833 12.017 14.067 16.013
18.475
8
1.647 2.180 2.733 3.490 13.362 15.507 17.535
20.090
9
2.088 2.700 3.325 4.168 14.684 16.919 19.023
21.666
10
2.558 3.247 3.940 4.865 15.987 18.307 20.483
23.209
2
.975
9.236
.05

2
.025
Interval Estimation of  2
((10
n -1)(0.7)
1)s 2
2
19.023
a0.25
/2
2 
2
((10
n
1)
s
-1)(0.7)
2
2.700

1 a /2
0.975
0.331   2  2.333
We are 95% confident that the population
variance is in this interval
Hypothesis Testing – One Variance
Example 2
Recall that Buyer’s Digest is rating ThermoRite thermostats.
Buyer’s Digest gives an “acceptable” rating to a thermostat with
a temperature variance of 0.5 or less.
Conduct a hypothesis test--at the 10% significance level--to
determine whether the ThermoRite thermostat’s temperature
variance is “acceptable”.
Hypotheses:
H0 :  2  0.5
H a :  2  0.5
Recall that s2 = 0.7 and df = 9. With  0 = 0.5,
2
2
(
n

1)s
(9)
(
0.
7)
 2 -stat  12.6 2
0
0.5
Hypothesis Testing – One Variance
a = .10 (column) and df = 10 – 1 = 9 (row)
Selected Values from the Chi-Square Distribution Table
Degrees
of Freedom
Area in Upper Tail
.99
.975
.95
.90
.10
.025
.01
5
0.554 0.831 1.145 1.610
11.070 12.832
15.086
6
0.872 1.237 1.635 2.204 10.645 12.592 14.449
16.812
7
1.239 1.690 2.167 2.833 12.017 14.067 16.013
18.475
8
1.647 2.180 2.733 3.490 13.362 15.507 17.535
20.090
9
2.088 2.700 3.325 4.168 14.684 16.919 19.023
21.666
10
2.558 3.247 3.940 4.865 15.987 18.307 20.483
23.209
2
Our .10
value
9.236
.05
Hypothesis Testing – One Variance
H0 :  2  0.5
  n 1  9
Do not reject H0
Reject H0
.10
9
a = .10
12.6 14.684
2
 2 -stat
There is insufficient evidence to conclude that the temperature
variance for ThermoRite thermostats is unacceptable.
Sampling distribution of F
The F-distribution results from taking the ratio of variances of
normally distributed variables.
 12
1
2
2
if 12 = 22
Sampling distribution of F
The F-distribution results from taking the ratio of variances of
normally distributed variables.
Bigger
s12
F -stat  2 ≈1
s2
0
1
if 12 = 22
Sampling distribution of F
The F-distribution results from taking the ratio of variances of
normally distributed variables.
s12
F -stat  2 ≈1
s2
.025
F.025
Sampling distribution of F
The F-distribution results from taking the ratio of variances of
normally distributed variables.
s12
F -stat  2 ≈1
s2
.975
F.975
Hypothesis Testing – Two Variances
Example 3
Buyer’s Digest has conducted the same test, but on 10
other thermostats. This time it test thermostats manufactured
by TempKing. The temperature readings of the 10
thermostats are listed below.
We will conduct a hypothesis at a 10% level of significance
to see if the variances are equal for both thermostats.
ThermoRite Sample
Temperature 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.2
s2 = 0.7 and df = 9
TempKing Sample
Temperature 67.7 66.4 69.2 70.1 69.5 69.7 68.1 66.6 67.3 67.5
s2 = ? and df = 9
Hypothesis Testing – Two Variances
TempKing
xi
xi 68.21
x
xi 
67.7
66.4
69.2
70.1
69.5
69.7
68.1
66.6
67.3
67.5
-0.51
-1.81
0.99
1.89
1.29
1.49
-0.11
-1.61
-0.91
-0.71
x  68.21
2
( xi(
xi 68
 x.2)12)
0.2601
3.2761
0.9801
3.5721
1.6641
2.2201
0.0121
2.5921
0.8281
0.5041
sum = 15.909
2
ss21 = 1.768
1.768
 ( xi  x ) 2
s 
n 1
2
s22  0.7
Since this is larger
Than ThermoRite’s
Hypothesis Testing – Two Variances
H 0 :  12   22
H a :  12   22
Hypotheses:
a/2 = .05 (row)
& n2  1  9
n1 = 10 – 1 = 9 (column)
Selected Values from the F Distribution Table
Denominator
Area in
Degrees
Upper
of Freedom
Tail
9
Numerator Degrees of Freedom
7
8
9
10
15
.01
6.18
6.03
5.91
5.81
5.52
.10
2.51
2.47
2.44
2.42
2.34
.05
3.29
3.23
3.18
3.14
3.01
.025
4.20
4.10
4.03
3.96
3.77
.01
5.61
5.47
5.35
5.26
F.05
4.96
Hypothesis Testing – Two Variances
s12
1.768
F -stat  2.532
s2
0.70
Reject H0
Do not Reject H0
.05
.05
F.95
Reject H0
There is insufficient
evidence to conclude
that the population
variances differ for the
two thermostat brands.
≈1
2.53 3.18
F