Math 4030 – 10aInferences Concerning

Variances

Sample variance is defined as

2

2

1

1

1

n

ii

S X Xn

22 SE

2

1

2

2

1

2

1

,1

:Note

n

ii

n

ii

XXn

E

Xn

E

If S2 is the variance of a random sample of size n taken from a normally distributed population with variance σ2, then

2

22 1

2 2

1

n

ii

X Xn S

has chi-square distribution with parameter (df) ν = n – 1.

Chi-square distribution• Is a special case of Gamma distribution when

• Density function:

• Mean is .

, 22

12 2

2

1, for 0,

( ) 22

0, otherwise.

x

x e x

f x

Table 5 on Page 517:

Table values are cut-off scores. (Same as t-Table)

For sample of size n = 10,

025.0023.199

2

2

S

P975.07.29

2

2

S

P

05.0023.199

7.22

2

S

P

025.07.29

2

2

S

P

95% Confidence Interval for variance?

Confidence Interval for Population Variance (Sec. 9.2):

Objective: Estimate the population variance.Assumption: The population is normally distributed.Given: Sample variance s2 from a random

sample of size n. value.Step 1. From the chi-square table (Table 5)

with degree of freedom n – 1, find and .

Step 2. The (1-)100% confidence interval for the population variance is

22/1

22/

.)1()1(

22/1

22

22/

2

snsn

Example 1.A sample of size 20 (from a normally distributed population) results a sample variance of 0.00012. Construct the 90% confidence interval for estimating the population standard deviation .

.)1()1(

22/1

22

22/

2

snsn

Hypothesis Testing regarding Variance or standard deviation (Sec. 9.2)

• Null and alternative hypotheses regarding the population variance or standard deviation

• Level of significance, tail(s) of the test.• Under the normality assumption, use Chi-square

distribution to find the critical value or (for one-tail test), and (for two-tails test). And determine the critical region.

• Calculate the test statistic

• Make conclusion.

21

2

.)1(

20

22

Sn

22/1

22/

Example 2.Playing 10 rounds of golf on his home course, a golf professional averaged 71.3 with a standard deviation of 1.32.Test the claim that he is actually less consistent than = 1.20 which is indicated in his profile. (Use = 0.05)

Compare variances from two samples (Sec. 9.4)

If S21 and S2

2 are the variances of two independent random samples of sizes n1 and n2, respectively, taken from two normally distributed populations with the same σ2, then

2122

SF

S

Has F distribution with parameters

1 1 2 21, 1v n v n

• Parameters v1 and v2, called numerator and denominator degrees of freedom;

• Take only positive values;

• Skewed to the right;

Table 6 on Page 518-519

11

Limitation of the Tables: How can we find F1-?

By definition, F is the cut-off value such that

112

2

21 FS

SP

12

2

21 FS

SP

121

22 1

FS

SP

12211

,),(

1

FF

Case 1. 22

211 : H

21

22

S

SF compare with )1,1( 12 nnF

Case 2. 22

211 : H

22

21

S

SF compare with )1,1( 21 nnF

Case 3. 22

211 : H

2

2

m

M

S

SF compare with )1,1(2/ mM nnF

rianceSmaller va

ianceLarger varF

)1riancesmaller va withsample theof size

,1iancelarger var withsample theof size(2/

F

Hypothesis Testing to compare two Variances (Sec. 9.3)

• Null and alternative hypotheses regarding the ratio of two population variances.

• Level of significance, tail(s) of the test.• Under the normality assumption, use F

distribution to find the critical value(s). And determine the critical region.

• Calculate the test statistic

• Make conclusion.

.21

22

S

SF

Example 3. (One Side F-test)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 a sample of size 10 from Company 1’s work and a sample of 16 from Company 2’s work yield

Test the null hypothesis 1 = 2 against the alternative hypothesis 1 < 2 at = 0.05 level. (Assuming both populations are normally distributed.)

062.0 ,053.0 21 ss

Example 4. (Two Side F-test)It is desired to determine whether there is any difference in the variability in the silver plating done by Companies 1 and 2. If a sample of size 10 from Company 1’s work and a sample of 16 from Company 2’s work yield

Test the null hypothesis 1 = 2 against the alternative hypothesis 1 ≠ 2 at = 0.02 level. (Assuming both populations are normally distributed.)

062.0 ,053.0 21 ss