Section 6.4 Inferences for Variances

If zzz ,..., 21 are independent standard normal random variables then 2 2 2

1 2 ...z z z has a chi-squared distribution with 2 , Xdf

Chi-square probability densities

Use of Chi-square distribution

Sample variance S2 is used to estimate the population variance 2

Assume that the population is normal.

has a chi-squared distribution with v=n-1 degrees of freedom.

Chi-square probabilities can be found in table B.5.

2

22 )1(

Sn

Then we get a 95% confidence interval:

2

2

2

2 22

( (0.025) (0.975)) 0.95

( 1)( (0.025) (0.975)) 0.95

( 1) ( 1)( ) 0.95(0.975) (0.025)

P Q Q

n sP Q Q

n s n sPQ Q

Confidence interval for 2

ExampleThe soft-drink company wants to control the variability in the amount of fill.

A sample of size 28 was drawn and the sample variance s2=0.0007.

Give a 95% confidence interval about the variance 2.

Solution

• v=n-1=27.•

• What is the CI for ?

20.025 14.573 20.975 43.194

2( 1) (28 1)(0.0007) 0.00044(0.975) 43.194n SQ

2( 1) (28 1)(0.0007) 0.00130(0.025) 14.573n SQ

00130.000044.0 2

To test #: 2 Ho vs 2: #Ha

22 2

1( 1) ~

# nn s

Reject Ho if

)975.0(#

)1(

)025.0(#

)1(

2

2

Qsn

or

Qsn

Equivalently, see if # is in the 95% confidence interval.

Example

The soft-drink company wants to control the variability in the amount of fill.

A sample of size 28 was drawn and the sample variance s2=0.0007.

Use the five-step significance testing format to assess the evidence that the variance is greater than 0.0005.

• 1 &2.

• 3. The test statistic is

• 4. The sample gives

• 5. The observed level of significance is P(a chi-square random variable with 27 d.f>37.8)0.05<P-value<0.1

20

2

: 0.0005

: 0.0005

H

Ha

22

2

( 1)n s

2 (28 1)*0.0007 37.80.0005

Inference for the Ratio of two variances

• For two independent samples from normal distributions, sometimes we want to compare two variances.

• A new distribution called F distribution can be of use here.

F distribution

F-distribution

• A F-distribution has two degrees of freedom:– Numerator degrees of freedom v1– Denominator degrees of freedom v2

– Tables B.6 give quantiles of F-distributions

• For F-distribution

So

• This is particularly useful since in table B.6 only quantiles for p larger than 0.5 are given.

1 2

2 1

,,

1( )(1 )

Q pQ p

3,55,3

1(0.1)(.99)

QQ

The ratio of two variances

• When s12 and s2

2 come from independent samples from normal distributions, the variable

where n1-1 and n2-1 are associated degrees of freedom for s1 and s2.

1 2

2 21 1

1, 12 22 2

has an distribution. n nsF Fs

1 21, 1

It is possible to pick appropriate F quantiles L and U such that the probability that the variable falls between L and U corresponds to a desired confidence level.

( ) 0.95( (0.025)n n n

P L F UP Q F Q

1 21, 1

2 21 12 22 2

2 2 21 1 1

2 2 22 2 2

(0.975)) 0.95

( (0.025) (0.975)) 0.95

( ) 0.95(0.975) (0.025)

n

sP Q Qs

s sPQ s Q s

Confidence interval for 12 /2

2

A confidence interval for 22

21 / if given by

2 21 12 22 2

1 1 to S SU S L S

1 2

1 2

1, 1

1, 1

(0.95)

(0.05)n n

n n

U Q

L Q

90%CI

The book’s F tables only give upper quantiles since we can find lower quantiles by interchanging 11 n and 12 n

)95.0(1)05.0(

05.0))95.0(

1(

95.0))95.0(

1(

95.0))95.0((

1,11,1

1,122

21

1,122

21

1,121

22

12

21

12

12

12

nnnn

nn

nn

nn

QQ

Qss

P

Qss

P

Qss

P

Example 14 (p.395) Hardness of carbon steel Heat Treated n = 10 22

1 52.7s Cold Treated n = 5 22

2 52.3s

1: 22

21

Ho

Let’s try to find a 90% CI for the ratio of the variances.

2 212 22

9,4

9,44,9

(7.52) 4.6(3.52)

(0.95) 6.00

1 1(0.05) (0.95)3.63

ssU Q

L QF

90% confidence interval 1 14.6 to 4.66 1/ 3.63

So a 90% CI for the ratio between the variances : 0.77 to 16.70Consequently a 90% CI for the ratio between the standard deviationsshould be .87 to 4.07.

is

At 10.0 level we do not reject 1: 22

21

Ho

A five step significance test of equality of variances:2

10 2

2

2

1

2

2

2 2 2 2 2

1 2 1 2 1

2 2 2

1 2 2

2

2

9,4

1. : 1.

2. : 1.

3. Test statistic is 1

7.524. The samples give 4.6

3.525. The p-value is 2P[an F random variable 4.6]=2*(0.05<p<0.1)

So p-value is betwee

H

Ha

s s s s sF

s

f

n 0.1 and 0.2.