John Loucks St . Edward’s University

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John LoucksSt. Edward’sUniversity

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Chapter 11 Inferences About Population Variances

Inference about a Population Variance Inferences about Two Populations Variances



Inferences About a Population Variance

If the sample variance is excessive, overfilling and underfilling may be occurring even though the mean is correct.

The mean filling weight is important, but also is the

variance of the filling weights.

Consider the production process of filling containers with a liquid detergent product.

A variance can provide important decision-making

information.

By selecting a sample of containers, we can compute

a sample variance for the amount of detergent placed in a container.



Inferences About a Population Variance

Chi-Square Distribution Interval Estimation of 2

Hypothesis Testing



Chi-Square Distribution

We can use the chi-square distribution to develop interval estimates and conduct hypothesis tests about a population variance.

The sampling distribution of (n - 1)s2/2 has a chi-

square distribution whenever a simple random sample of size n is selected from a normal population.

The chi-square distribution is based on sampling from a normal population.

The chi-square 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

0

With 2 degrees of freedom

2

2

( 1)n s





2 2 2.975 .025

Chi-Square Distribution

For example, there is a .95 probability of obtaining a 2 (chi-square) value such that

We will use the notation to denote the value for the chi-square distribution that provides an area of a to the right of the stated value.

2a

2a



95% of thepossible 2 values

2

0

.025

2.025

.025

2.975

Interval Estimation of 2

22 2.975 .0252

( 1)n s




( ) ( )

/ ( / )

n s n s

1 12

22

22

1 22

a a

( ) ( )

/ ( / )

n s n s

1 12

22

22

1 22

a a

2 2 2(1 / 2) / 2a a

22 2(1 / 2) / 22

( 1)n sa 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


( ) ( )

/ ( / )

n s n s

1 12

22

22

1 22

a a

( ) ( )

/ ( / )

n s n s

1 12

22

22

1 22

a a

where the values are based on a chi-squaredistribution with n - 1 degrees of freedom andwhere 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 confidenceinterval for the population standard deviation.

2 2

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, 10thermostats manufactured by ThermoRite

wereselected and placed in a test room that wasmaintained at a temperature of 68oF. Thetemperature readings of the ten thermostats

areshown on the next slide.


Example: Buyer’s Digest (A)




We will use the 10 readings below to develop a95% confidence interval estimate of the populationvariance.

Example: Buyer’s Digest (A)

Temperature 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.2Thermostat 1 2 3 4 5 6 7 8 9 10



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

5 0.554 0.831 1.145 1.610 9.236 11.070 12.832 15.0866 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.8127 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.4758 1.647 2.180 2.733 3.490 13.362 15.507 17.535 20.0909 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-Square Distribution Table

Our value

2.975

For n - 1 = 10 - 1 = 9 d.f. and a = .05




2

0

.025

2

2.0252

( 1)2.700 n s

Area inUpper Tail

= .975

2.700

For n - 1 = 10 - 1 = 9 d.f. and a = .05




5 0.554 0.831 1.145 1.610 9.236 11.070 12.832 15.0866 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.8127 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.4758 1.647 2.180 2.733 3.490 13.362 15.507 17.535 20.0909 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-Square Distribution Table

For n - 1 = 10 - 1 = 9 d.f. and a = .05

Our value 2

.025



2

0

.025

2.700


n - 1 = 10 - 1 = 9 degrees of freedom and a = .05

2

2( 1)2.700 19.023n s

19.023

Area in UpperTail = .025



Sample variance s2 provides a point estimate of 2.

s x xni2

2

16 39

70

( ) . .s x xni2

2

16 39

70

( ) . .

( )..

( )..

10 1 7019 02

10 1 702 70

2 ( ).

.( ).

.10 1 70

19 0210 1 70

2 702


.33 < 2 < 2.33

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



Left-Tailed Test

Hypothesis TestingAbout a Population Variance

22

021 ( )n s

2

2

021 ( )n s

where is the hypothesized valuefor the population variance

20

• Test Statistic

• Hypotheses 2 20 0: H

2 20: aH



Left-Tailed Test (continued)


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-squaredistribution with n - 1 d.f.

2(1 )a



Right-Tailed Test


H02

02: H0

202:

Ha : 202Ha : 202

22

021 ( )n s

2

2

021 ( )n s


20

• Test Statistic

• Hypotheses



Right-Tailed Test (continued)


Reject H0 if 2 2a

Reject H0 if p-value < a

2awhere is based on a chi-square

distribution with n - 1 d.f.

p-Value approach:




Two-Tailed Test


22

021 ( )n s

2

2

021 ( )n s


20

• Test Statistic

• Hypotheses

Ha : 202Ha : 202

H02

02: H0

202:



Two-Tailed Test (continued)




2 2 2 2(1 / 2) / 2 or a a Reject H0 if

where are based on achi-square distribution with n - 1 d.f.

2 2(1 / 2) / 2 and a a



Recall that Buyer’s Digest is rating ThermoRitethermostats. Buyer’s Digest gives an “acceptable”rating to a thermostat with a temperature varianceof 0.5 or less.


Example: Buyer’s Digest (B)

We will conduct a hypothesis test (with a = .10)to determine whether the ThermoRite thermostat’stemperature variance is “acceptable”.




Using the 10 readings, we will conduct ahypothesis test (with a = .10) to determine whetherthe ThermoRite thermostat’s temperature variance is“acceptable”.

Example: Buyer’s Digest (B)




Hypotheses2

0 : 0.5H 2: 0.5aH


Reject H0 if 2 > 14.684

Rejection Rule

Right-tailedtest




5 0.554 0.831 1.145 1.610 9.236 11.070 12.832 15.0866 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.8127 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.4758 1.647 2.180 2.733 3.490 13.362 15.507 17.535 20.0909 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 TailSelected Values from the Chi-Square Distribution Table

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


Our value 2

.10



2

0 14.684

Area in UpperTail = .10


Rejection Region

2 22

2( 1) 9

.5n s s

Reject H0



Test Statistic

2 9(.7) 12.6.5


Because 2 = 12.6 is less than 14.684, we cannotreject H0. The sample variance s2 = .7 is insufficientevidence to conclude that the temperature variancefor ThermoRite thermostats is unacceptable.

Conclusion

The sample variance s 2 = 0.7



Using the p-Value

• The sample variance of s 2 = .7 is insufficient evidence to conclude that the temperature variance is unacceptable (>.5).

• Because the p –value > a = .10, we cannot reject the null hypothesis.

• The rejection region for the ThermoRite thermostat example is in the upper tail; thus, the appropriate p-value is less than .90 (2 = 4.168) and greater than .10 (2 = 14.684).


The exact p-value is .18156.



Inferences About Two Population Variances

The two sample variances will be the basis for making

inferences about the two population variances.

We use data collected from two independent random sample, one from population 1 and another from population 2.

We may want to compare the variances in: product quality resulting from two different

production processes,

assembly times for two assembly methods.

temperatures for two heating devices, or



One-Tailed Test

• Test Statistic

• Hypotheses

Hypothesis Testing About theVariances of Two Populations

Denote the population providing thelarger sample variance as population 1.

2 20 1 2: H

2 21 2: aH

21 2

2

sF s



One-Tailed Test (continued)

Reject H0 if p-value < a

where the value of Fa is based on anF distribution with n1 - 1 (numerator)and n2 - 1 (denominator) d.f.

p-Value approach:



Reject H0 if F > Fa



Two-Tailed Test

• Test Statistic

• Hypotheses


H0 12

22: H0 1

222:

Ha : 12

22Ha : 1

222

Denote the population providing thelarger sample variance as population 1.

21 2

2

sF s



Two-Tailed Test (continued)




Reject H0 if F > Fa/2

where the value of Fa/2 is based on anF 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.


Example: Buyer’s Digest (C)

We will conduct a hypothesis test with a = .10 to seeif the variances are equal for ThermoRite’s thermostatsand TempKing’s thermostats.




Example: Buyer’s Digest (C)ThermoRite Sample

TempKing Sample





HypothesesH0 1

222: H0 1

222:

Ha : 12

22Ha : 1

222


Reject H0 if F > 3.18

The F distribution table (on next slide) shows that withwith a = .10, 9 d.f. (numerator), and 9 d.f. (denominator),F.05 = 3.18.

(Their variances are not equal)

(TempKing and ThermoRite thermostatshave the same temperature variance)

Rejection Rule



Denominator Area inDegrees Upper

of Freedom Tail 7 8 9 10 158 .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 FreedomSelected Values from the F Distribution Table




Test Statistic


We cannot reject H0. F = 2.53 < F.05 = 3.18.There is insufficient evidence to conclude thatthe population variances differ for the twothermostat brands.

Conclusion

21 2

2

sF s = 1.768/.700 = 2.53

TempKing’s sample variance is 1.768ThermoRite’s sample variance is .700



Determining and Using the p-Value


• Because a = .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 .20 and .10.

• Because F = 2.53 is between 2.44 and 3.18, the area in the upper tail of the distribution is between .10 and .05.

Area in Upper Tail .10 .05 .025 .01F Value (df1 = 9, df2 = 9) 2.44 3.18 4.03 5.35



End of Chapter 11

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John Loucks St . Edward’s University

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