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Chapter 11 Inferences About Population Variances

 Inference about a Population Variance

 Inferences about Two Population Variances

Inferences About a Population Variance

 Chi-Squared Distribution

 Interval Estimation

 Hypothesis Testing

Chi-Squared Distribution

 We can use the chi-squared distribution to construct interval estimates and do hypothesis tests about a population variance.

 The sampling distribution of (n - 1)s2/ 2 has a chi- squared distribution whenever a simple random sample of size n is selected from a normal population.

 The chi-squared distribution is the sum of squared

standardized normal random variables such as

(Z1) 2 + (Z2)

2 + (Z3) 2 and so on.

Examples of Sampling Distribution of (n - 1)s2/ 2

With 2 degrees of freedom

( 1)n s





With 5 degrees of freedom

With 10 degrees of freedom

2 2 2

.975 .025    

Chi-Squared Distribution

 For example, there is a 0.95 probability of obtaining a 2 (chi-squared) value such that

 We shall use the notation to denote the value for the chi-squared distribution that gives an area of a to the right of the stated value.

a 

95% of the possible 2 values

2

0.025

.025 

0.025

.975 

Interval Estimation of 2

2 2 2 .975 .0252

( 1)n s  



  

Interval Estimation of 2

( ) ( )

/ ( / )

n s n s  



1 1 2

2 2

1 2 2

 

a a

( ) ( )

/ ( / )

n s n s  



1 1 2

2 2

1 2 2

 

a a

2 2 2

(1 / 2) / 2a a   

  

2 2 2

(1 / 2) / 22

( 1)n s a a

  



  

 Substituting (n – 1)s2/2 for the 2 we get

 Performing algebraic manipulation we get

 There is a (1 – a) probability of obtaining a 2 value

such that

 Interval Estimate of a Population Variance

Interval Estimation of 2

( ) ( )

/ ( / )

n s n s  



1 1 2

2 2

1 2 2

 

a a

( ) ( )

/ ( / )

n s n s  



1 1 2

2 2

1 2 2

 

a a

where the  values are based on a chi-squared

distribution with n - 1 degrees of freedom and

where 1 - a is the confidence coefficient.

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.

2 2

/ 2 (1 / 2)

( 1) ( 1)n s n s

a a

  



   

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 20oC.

The temperature readings of the ten thermostats are

shown on the next slide.

Interval Estimation of 2

 Example: Buyer’s Digest (A)

Interval Estimation of 2

We will use the 10 readings below to

Construct a 95% confidence interval

estimate of the population variance.

 Example: Buyer’s Digest (A)

Temperature 19.7 19.9 20.1 20.7 20.8 19.4 20.1 20.3 19.9 19.6

Thermostat 1 2 3 4 5 6 7 8 9 10

Degrees

of Freedom .99 .975 .95 .90 .10 .05 .025 .01

5 0.554 0.831 1.145 1.610 9.236 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

Area in Upper Tail

Interval Estimation of 2

Selected Values from the Chi-Squared Distribution Table

Our value 2 .975

For n - 1 = 10 - 1 = 9 d.f. and a = 0.05

Interval Estimation of 2

2

0.025

 

  

2 2 .0252

( 1) 2.700

n s

Area in Upper Tail

= 0.975

2.700

For n - 1 = 10 - 1 = 9 d.f. and a = 0.05

Degrees

of Freedom .99 .975 .95 .90 .10 .05 .025 .01

5 0.554 0.831 1.145 1.610 9.236 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

Area in Upper Tail

Interval Estimation of 2

Selected Values from the Chi-Squared Distribution Table

For n - 1 = 10 - 1 = 9 d.f. and a = 0.05

Our value  2 .025

2

0.025

2.700

Interval Estimation of 2

n - 1 = 10 - 1 = 9 degrees of freedom and a = 0.05



  

( 1) 2.700 19.023

n s

19.023

Area in Upper Tail = 0.025

 Sample variance s2 provides a point estimate of  2. 2

2 ( ) 1.94

0.216 1 9

i x x

s n

   





2(10 1)0.216 (10 1)0.216

19.02 2.70 

   

Interval Estimation of 2

0.102 < 2 < 0.72

 A 95% confidence interval for the population variance is given by:

 Left-Tailed Test

Hypothesis Testing About a Population Variance

 

2 2

0 2

1 

( )n s 



2 2

0 2

1 

( )n s

where is the hypothesized value for the population variance

2 0

•Test Statistic

•Hypotheses 2 2

0 0: H  

  2 2

1 0: H

 Left-Tailed Test (continued)

Hypothesis Testing About a Population Variance

Reject H0 if p-value < ap-Value approach:

Critical value approach:

•Rejection Rule Reject H0 if

2 2 ( 1 )a  

where is based on a chi-squared distribution with n - 1 d.f.

2 ( 1 )a 

 Right-Tailed Test

Hypothesis Testing About a Population Variance

H0 2

0 2

:  H0 2

0 2

:  

2 2

1 0 : H  

 

2 2

0 2

1 

( )n s 



2 2

0 2

1 

( )n s

where is the hypothesized value for the population variance

2 0

•Test Statistic

•Hypotheses

 Right-Tailed Test (continued)

Hypothesis Testing About a Population Variance

Reject H0 if 2 2

a  

Reject H0 if p-value < a

2 awhere is based on a chi-squared

distribution with n - 1 d.f.

p-Value approach:

Critical value approach:

•Rejection Rule

 Two-Tailed Test

Hypothesis Testing About a Population Variance

 

2 2

0 2

1 

( )n s 



2 2

0 2

1 

( )n s

where is the hypothesized value for the population variance

2 0

•Test Statistic

•Hypotheses

2 2

1 0 : H  

H0 2

0 2

:  H0 2

0 2

:  

 Two-Tailed Test (continued)

Hypothesis Testing About a Population Variance

Reject H0 if p-value < a

p-Value approach:

Critical value approach:

•Rejection Rule

2 2 2 2 ( 1 /2 ) /2 or a a    Reject H0 if

where are based on a chi-squared distribution with n - 1 d.f.

2 2 ( 1 /2 ) /2 and a a 

Recall that Buyer’s Digest is rating

ThermoRite thermostats. Buyer’s Digest

gives an “acceptable” rating to a thermo-

stat with a temperature variance of 0.15

or less.

Hypothesis Testing About a Population Variance

 Example: Buyer’s Digest (B)

We will do a hypothesis test (with

a = 0.10) to determine whether the ThermoRite

thermostat’s temperature variance is “acceptable”.

Hypothesis Testing About a Population Variance

Using the 10 readings, we will

conduct a hypothesis test (with a = 0.10)

to determine whether the ThermoRite

thermostat’s temperature variance is

“acceptable”.

 Example: Buyer’s Digest (B)

Temperature 19.7 19.9 20.1 20.7 20.8 19.4 20.1 20.3 19.9 19.6

Thermostat 1 2 3 4 5 6 7 8 9 10

 Hypotheses

  2

0 : 0.15H

  2

1 : 0.15H

Hypothesis Testing About a Population Variance

Reject H0 if  2 > 14.684

 Rejection Rule

Degrees

of Freedom .99 .975 .95 .90 .10 .05 .025 .01

5 0.554 0.831 1.145 1.610 9.236 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

Area in Upper Tail

Selected Values from the Chi-Squared Distribution Table

For n - 1 = 10 - 1 = 9 d.f. and a = 0.10

Hypothesis Testing About a Population Variance

Our value  2 .10

2

0 14.684

Area in Upper Tail = 0.10

Hypothesis Testing About a Population Variance

 Rejection Region

 

  

2 2 2

( 1) 9

0.15

n s s

Reject H0

 Test Statistic

   2 9(0.216)

12.96 0.15

Hypothesis Testing About a Population Variance

Because 2 = 12.96 is less than 14.684, we cannot

reject H0. The sample variance s 2 = 0.216 is insufficient

evidence to conclude that the temperature variance

for ThermoRite thermostats is not acceptable.

 Conclusion

The sample variance s2 = 0.216

 Using the p-Value

• The sample variance of s2 = 0.216 is insufficient evidence to conclude that the temperature variance is not acceptable (>0.15).

• Because the p–value > a = 0.10, we cannot reject the null hypothesis.

• The rejection region for the ThermoRite thermostat example is in the upper tail; so the

appropriate p-value is less than 0.90 (2 = 4.168)

and greater than 0.10 (2 = 14.684).

Hypothesis Testing About a Population Variance

A precise p-value can be found using Minitab, SPSS or Excel.

 One-Tailed Test

•Test Statistic

•Hypotheses

Hypothesis Testing About the Variances of Two Populations

Denote the population providing the

larger sample variance as population 1.

2 2 0 1 2: H  

  2 2

1 1 2: H

2 1

2 2

s F

s 

 One-Tailed Test (continued)

Reject H0 if p-value < a

where the value of Fa is based on an

F distribution with n1 - 1 (numerator)

and n2 - 1 (denominator) d.f.

p-Value approach:

Critical value approach:

•Rejection Rule

Hypothesis Testing About the Variances of Two Populations

Reject H0 if F > Fa

 Two-Tailed Test

•Test Statistic

•Hypotheses

Hypothesis Testing About the Variances of Two Populations

H0 1 2

2 2

:  H0 1 2

2 2

:  

2 2

1 1 2 : H  

Denote the population providing the

larger sample variance as population 1.

2 1

2 2

s F

s 

 Two-Tailed Test (continued)

Reject H0 if p-value < ap-Value approach:

Critical value approach:

•Rejection Rule

Hypothesis Testing About the Variances of Two Populations

Reject H0 if F > Fa/2

where the value of Fa/2 is based on an

F distribution with n1 - 1 (numerator)

and n2 - 1 (denominator) d.f.

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 on the next slide.

Hypothesis Testing About the Variances of Two Populations

 Example: Buyer’s Digest (C)

We will do a hypothesis test with a = 0.10 to see

if the variances are equal for ThermoRite’s thermostats

and TempKing’s thermostats.

Hypothesis Testing About the Variances of Two Populations

 Example: Buyer’s Digest (C)

ThermoRite Sample

TempKing Sample

Temperature 19.7 19.9 20.1 20.7 20.8 19.4 20.1 20.3 19.9 19.6

Thermostat 1 2 3 4 5 6 7 8 9 10

Temperature 19.8 19.1 20.1 21.2 20.8 20.9 20.1 19.2 19.6 19.7

Thermostat 1 2 3 4 5 6 7 8 9 10

 Hypotheses

H0 1 2

2 2

:  H0 1 2

2 2

:  

2 2

1 1 2 : H  

Hypothesis Testing About the Variances of Two Populations

Reject H0 if F > 3.18

The F distribution table (on next slide) shows that with

with a = 0.10, 9 d.f. (numerator), and 9 d.f. (denominator),

F.05 = 3.18.

(Their variances are not equal)

(TempKing and ThermoRite thermostats have the same temperature variance)

 Rejection Rule

Denominator Area in

Degrees Upper

of Freedom Tail 7 8 9 10 15

8 .10 2.62 2.59 2.56 2.54 2.46

.05 3.50 3.44 3.39 3.35 3.22

.025 4.53 4.43 4.36 4.30 4.10

.01 6.18 6.03 5.91 5.81 5.52

9 .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 4.96

Numerator Degrees of Freedom

Selected Values from the F Distribution Table

Hypothesis Testing About the Variances of Two Populations

 Test Statistic

Hypothesis Testing About the Variances of Two Populations

We cannot reject H0. F = 2.53 < F.05 = 3.18.

There is insufficient evidence to conclude that

the population variances differ for the two

thermostat brands.

 Conclusion

2 1

2 2

s F

s  = 0.546/0.216 = 2.53

TempKing’s sample variance is 0.546

ThermoRite’s sample variance is 0.216

 Determining and Using the p-Value

Hypothesis Testing About the Variances of Two Populations

• Because a = 0.10, we have p-value > a and therefore we cannot reject the null hypothesis.

• But this is a two-tailed test; after doubling the upper- tail area, the p-value is between 0.20 and 0.10. (A precise p-value can be found using SPSS, Minitab or Excel.)

• Because F = 2.53 is between 2.44 and 3.18, the area in the upper tail of the distribution is between 0.10 and 0.05.

Area in Upper Tail .10 .05 .025 .01

F Value (df1 = 9, df2 = 9) 2.44 3.18 4.03 5.35

End of Chapter 11