Ch8 Inference concerning variance
Dr. Deshi Ye
yedeshi@zju.edu.cn
Outline
The estimation of variance
Hypothesis concerning one variance
Hypothesis concerning two variances
One Population Tests
One
Population
Mean
Proportion
Variance
Z Test
t Test
Z Test
c2 Test
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)
8.1 The Estimation of Variances
Standard deviation S
Variance 
n
Let
S2 
2
(
X

X
)
 i
i 1
n 1
be the sample variance
based on any population having variance

2
Unbiased estimation of a population variance
The sample variance
n
S2 
2
(
X

X
)
 i
i 1
n 1
is an unbiased estimator of

2
Remark: the sample standard deviation S is not
an unbiased estimator of 
However, for large samples the bias is small, and it is common
practice to estimate
with S

Confidence interval
By Theorem 6.4
(n  1) S 2
2
is a random variable having the chi-square distribution
with n-1 degrees of freedom.
c
2
1 / 2

(n  1) S 2

2
100(1-a)% confidence interval for
(n  1) S 2
c / 2
2
 
2

(n  1) S 2
c12 / 2
 c / 2
2
8.2 Hypothesis concerning one variance
Consider the problem of testing the null
hypothesis that a population variance
equals a specified constant against a
suitable one-sided or two-sided alternative.
Null Hypothesis
c2 
(n  1) S 2

2
0
H0 :  2   02
is a random sample from a normal population with
2
the variance  0 is a random variable having the chisquare distribution with n-1 degree of freedom.
Criterion Region for testing (Normal population)
2 2
0
Alternative
hypothesis
Reject null
hypothesis if
 2   02
c 2  c12
 
c 2  c2
2
2
0
 2   02
c 2  c12 / 2 or c 2  c2 / 2
EX. Testing hypothesis concerning a standard deviation
The lapping process which is used to grind certain silicon
wafers to the proper thickness is acceptable only if , the
population standard deviation of the thickness of dice cut
from the wafers, is at most 0.5 mil. Use the 0.05 level of
significance to test the null hypothesis   0.5 against the
alternative hypothesis   0.5 ,if the thickness of 15 dice
cut from such wafers have a standard deviation of 0.64 mil.
Solution
1. Null hypothesis:
  0.5
Alternative hypothesis   0.5
2. Level of significance: 0.05
2
c
 23.685
3. Criterion: Reject the null hypothesis if
4. Calculation:
c 
2
(n  1) S 2
 02
(15  1)(0.64) 2

 22.94
2
0.5
5. The null hypothesis cannot be rejected at level 0.05.
6. P-value: 1-0.9387=0.0613 > level of significance
8.3 Hypothesis concerning two variances
If independent random samples of size n1
and n2
are taken from normal
population having the same variance, it
follows from Theorem 6.5 that
S12
F 2
S2
is a random variable having the F
distribution with n1  1 and n2  1 degrees of
freedom
Testing two variances
Null hypothesis
 12   22
Criterion Region for testing (Normal population)
2 2
0
Alternative
hypothesis
Test
statistic
Reject null hypothesis if
 12   22
S 22
F 2
S1
F  F (n2  1, n1  1)
 
S12
F 2
S2
F  F (n1  1, n2  1)
2
1
 
2
1
2
2
2
2
S M2
F 2
Sm
F  F (nM  1, nm  1)
EX.
It is desired to determine whether there is less variability
in the silver plating done by Company 1 than in that done
by Company 2. If independent random samples of size
12 of the two companies’ work yield s1  0.035 and
against
s2  0.062, test the null hypothesis  12   22
the alternative hypothesis  12   22 at the 0.05 level of
significance.
Solution
1. Null hypothesis:
 12   22
Alternative hypothesis  12   22
2. Level of significance: 0.05
S22
3. Criterion: Reject the null hypothesis if F  2.82, F  2
S1
2
4. Calculation:
(0.062)
F
 3.14
2
(0.035)
5. The null hypothesis must be rejected at level 0.05.
6. P-value: 1-0.965=0.035 < level of significance