統計學
Spring 2004
授課教師：統計系余清祥
日期：2004年3月16日
第五週：比較變異數
Slide 1
Chapter 11
Inferences About Population Variances


Inference about a Population Variance
Inferences about the Variances of Two Populations
Slide 2
Inferences About a Population Variance



Chi-Square Distribution
Interval Estimation of 2
Hypothesis Testing
Slide 3
Chi-Square Distribution




The chi-square distribution is the sum of squared
standardized normal random variables such as
(z1)2+(z2)2+(z3)2 and so on.
The chi-square distribution is based on sampling from
a normal population.
The sampling distribution of (n - 1)s2/ 2 has a chisquare distribution whenever a simple random
sample of size n is selected from a normal population.
We can use the chi-square distribution to develop
interval estimates and conduct hypothesis tests about
a population variance.
Slide 4
Interval Estimation of 2

Interval Estimate of a Population Variance
( n  1) s 2
 2 / 2
 2 
( n  1) s 2
 2(1  / 2)
where the  values are based on a chi-square
distribution with n - 1 degrees of freedom and where
1 -  is the confidence coefficient.
Slide 5
Interval Estimation of 

Interval Estimate of a Population Standard Deviation
Taking the square root of the upper and lower
limits of the variance interval provides the
confidence interval for the population standard
deviation.
Slide 6
Interval Estimation of 2

Chi-Square Distribution With Tail Areas of .025
.025
.025
95% of the
possible 2 values
0
2
.975
2
.025
2
Slide 7
Example: Buyer’s Digest
Buyer’s Digest rates thermostats manufactured
for home temperature control. In a recent test, 10
thermostats manufactured by ThermoRite were selected
and placed in a test room that was maintained at a
temperature of 68oF. The temperature readings of the
ten thermostats are listed below.
We will use the 10 readings to develop a 95%
confidence interval estimate of the population variance.
Therm. 1
2
3 4
5
6
7
8
9 10
Temp. 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.2
Slide 8
Example: Buyer’s Digest

Interval Estimation of 2
n - 1 = 10 - 1 = 9 degrees of freedom and  = .05
2
.975

.025
0
2
.975
(n  1)s 2
2
2
 .025
.025
2
.025
2
Slide 9
Example: Buyer’s Digest

Interval Estimation of 2
n - 1 = 10 - 1 = 9 degrees of freedom and  = .05
2.70 
.025
(n  1)s 2
2
2
  .025
Area in
Upper Tail
= .975
2
0 2.70
Slide 10
Example: Buyer’s Digest

Interval Estimation of 2
n - 1 = 10 - 1 = 9 degrees of freedom and  = .05
2. 70 
( n  1) s 2
.025

2
 19 . 02
Area in Upper
Tail = .025
2
0 2.70
19.02
Slide 11
Example: Buyer’s Digest

Interval Estimation of  2
Sample variance s2 provides a point estimate of  2.
2
(
x

x
)
6. 3

i
s2 

 . 70
n 1
9
A 95% confidence interval for the population variance
is given by:
(10  1). 70
(10  1). 70
 2 
19. 02
2. 70
.33 < 2 < 2.33
Slide 12
Hypothesis Testing
About a Population Variance

Left-Tailed Test
• Hypotheses
H 0 :  2   20
H a :  2   20
• Test Statistic
2 
( n  1) s 2
 20
• Rejection Rule
Reject H0 if  2   2 (where 2 is based on
a chi-square distribution with n - 1 d.f.) or
Reject H0 if p-value < 
Slide 13
Hypothesis Testing
About a Population Variance

Right-Tailed Test
• Hypotheses
H 0 :  2   20
H a :  2   20
• Test Statistic
2 
( n  1) s 2
 20
• Rejection Rule
2
2
2
Reject H0 if    (1  ) (where  (1 ) is based on
a chi-square distribution with n - 1 d.f.) or
Reject H0 if p-value < 
Slide 14
Hypothesis Testing
About a Population Variance

Two-Tailed Test
• Hypotheses
H 0 :  2   20
H a :  2   20
• Test Statistic
2 
( n  1) s 2
 20
• Rejection Rule
Reject H0 if  2   2(1  / 2 ) or  2   2 / 2 (where
(12  /2) and 2 /2 are based on a chi-square distribution with n - 1 d.f.) or Reject H0 if p-value < 
Slide 15
Hypothesis Testing About the
Variances of Two Populations

One-Tailed Test
• Hypotheses
H 0 :  12   22
H a :  12   22
• Test Statistic
s12
F 2
s2
• Rejection Rule
Reject H0 if F > F where the value of F is based
on an F distribution with n1 - 1 (numerator) and
n2 - 1 (denominator) d.f.
Slide 16
Hypothesis Testing About the
Variances of Two Populations

Two-Tailed Test
• Hypotheses
H 0 :  12   22
Ha : 12   22
• Test Statistic
s12
F 2
s2
• Rejection Rule
Reject H0 if F > F/2 where the value of F/2 is
based on an F distribution with n1 - 1
(numerator) and n2 - 1 (denominator) d.f.
Slide 17
Example: Buyer’s Digest
Buyer’s Digest has conducted the same test, as was
described earlier, on another 10 thermostats, this time
manufactured by TempKing. The temperature readings
of the ten thermostats are listed below.
We will conduct a hypothesis test with  = .10 to see
if the variances are equal for ThermoRite’s thermostats
and TempKing’s thermostats.
Therm. 1
2
3
4
5
6
7
8
9 10
Temp. 66.4 67.8 68.2 70.3 69.5 68.0 68.1 68.6 67.9 66.2
Slide 18
Example: Buyer’s Digest

Hypothesis Testing About the Variances of Two
Populations
• Hypotheses
H 0 :  12   22 (ThermoRite and TempKing thermostats have same temperature variance)
H a :  12   22 (Their variances are not equal)
• Rejection Rule
The F distribution table shows that with  = .10,
9 d.f. (numerator), and 9 d.f. (denominator),
F.05 = 3.18.
Reject H0 if F > 3.18
Slide 19
Example: Buyer’s Digest

Hypothesis Testing About the Variances of Two
Populations
• Test Statistic
ThermoRite’s sample variance is .70.
TempKing’s sample variance is 1.52.
F = 1.52/.70 = 2.17
• Conclusion
We cannot reject H0. There is insufficient
evidence to conclude that the population
variances differ for the two thermostat brands.
Slide 20
End of Chapter 11
Slide 21