1 1 Slide © 2003 South-Western /Thomson Learning™ Slides Prepared by JOHN S. LOUCKS St....

1 1 Slide

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© 2003 South-Western /Thomson Learning™© 2003 South-Western /Thomson Learning™

Slides Prepared bySlides Prepared byJOHN S. LOUCKSJOHN S. LOUCKS

St. Edward’s UniversitySt. Edward’s University

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Chapter 11Chapter 11 Inferences about Population Variances Inferences about Population Variances

Inference about a Population VarianceInference about a Population Variance Inferences about the Variances of Two Inferences about the Variances of Two

PopulationsPopulations

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Inferences about a Population VarianceInferences about a Population Variance

Chi-Square DistributionChi-Square Distribution Interval Estimation of Interval Estimation of 22

Hypothesis TestingHypothesis Testing

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Chi-Square DistributionChi-Square Distribution

The The chi-square distributionchi-square distribution is the sum of squared is the sum of squared standardized normal random variables such asstandardized normal random variables such as

((zz11))22+(+(zz22))22+(+(zz33))22 and so on. and so on. The chi-square distribution is based on sampling The chi-square distribution is based on sampling

from a normal population.from a normal population. The sampling distribution of (The sampling distribution of (nn - 1) - 1)ss22//22 has a has a

chi-square distribution whenever a simple chi-square distribution whenever a simple random sample of size random sample of size nn is selected from a is selected from a normal population.normal population.

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

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Interval Estimation of Interval Estimation of 22

Interval Estimate of a Population VarianceInterval Estimate of a Population Variance

where the where the values are based on a chi-square values are based on a chi-square distribution with distribution with nn - 1 degrees of freedom and - 1 degrees of freedom and where 1 - where 1 - is the confidence coefficient. is the confidence coefficient.

( ) ( )

/ ( / )

n s n s

1 12

22

22

1 22

( ) ( )

/ ( / )

n s n s

1 12

22

22

1 22

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Interval Estimation of Interval Estimation of

Interval Estimate of a Population Standard Interval Estimate of a Population Standard DeviationDeviation

Taking the square root of the upper and Taking the square root of the upper and lower limits of the variance interval provides lower limits of the variance interval provides the confidence interval for the population the confidence interval for the population standard deviation.standard deviation.

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Chi-Square Distribution With Tail Areas of .025Chi-Square Distribution With Tail Areas of .025

95% of thepossible 2 values 95% of thepossible 2 values

22

00

.025.025.025.025

.9752.9752 .025

2.0252


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Buyer’s Digest rates thermostats manufacturedBuyer’s Digest rates thermostats manufactured

for home temperature control. In a recent test, 10for home temperature control. In a recent test, 10

thermostats manufactured by ThermoRite were thermostats manufactured by ThermoRite were selected selected

and placed in a test room that was maintained at aand placed in a test room that was maintained at a

temperature of 68temperature of 68ooF. The temperature readings of the F. The temperature readings of the

ten thermostats are listed below. ten thermostats are listed below.

We will use the 10 readings to develop a 95%We will use the 10 readings to develop a 95%

confidence interval estimate of the population variance.confidence interval estimate of the population variance.

Therm.Therm. 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 9 109 10

Temp.Temp. 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.267.9 67.2

Example: Buyer’s Digest (A)Example: Buyer’s Digest (A)

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nn - 1 = 10 - 1 = 9 degrees of freedom and - 1 = 10 - 1 = 9 degrees of freedom and = .05= .05

22

00

.025.025.025.025

22 2.975 .0252

( 1)n s

2

2 2.975 .0252

( 1)n s

2.9752.975 2

.0252

.025

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22

00

.025.025

2.702.70

22.0252

( 1)2.70

n s

2

2.0252

( 1)2.70

n s


Area inArea inUpper TailUpper Tail

= .975= .975

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2 701

19 022

2.( )

. n s

2 70

119 02

2

2.( )

. n s

22

00

Area in UpperArea in UpperTail = .025Tail = .025Area in UpperArea in UpperTail = .025Tail = .025

.025.025

2.702.70 19.0219.02

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Sample variance Sample variance ss22 provides a point estimate of provides a point estimate of 22..

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

.33 .33 << 2 2 << 2.33 2.33

sx xni2

2

16 39

70

( ) .

.sx xni2

2

16 39

70

( ) .

.

( )..

( )..

10 1 7019 02

10 1 702 70

2 ( ).

.( ).

.10 1 70

19 0210 1 70

2 702


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Using Excel to Construct an Interval Using Excel to Construct an Interval EstimateEstimate

of a Population Varianceof a Population Variance Formula WorksheetFormula Worksheet

A B C

1 Temp. Sample Size =COUNT(A2:A11)2 67.4 Variance =VAR(A2:A11)3 67.84 68.2 Confid. Coefficient 0.955 69.3 Level of Signif. (alpha) =1-C46 69.5 Chi-Sq. Value (low. tail) =CHIINV(1-C5/2,C1-1)7 67.0 Chi-Sq. Value (up. tail) =CHIINV(C5/2,C1-1)8 68.1 9 68.6 Point Estimate =C210 67.9 Lower Limit =((C1-1)*C2)/C711 67.2 Upper Limit =((C1-1)*C2)/C6

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Value WorksheetValue Worksheet

Using Excel to Construct an Interval Using Excel to Construct an Interval EstimateEstimate

of a Population Varianceof a Population Variance

A B C

1 Temp. Sample Size 102 67.4 Variance 0.7003 67.84 68.2 Confid. Coefficient 0.955 69.3 Level of Signif. (alpha) 0.056 69.5 Chi-Sq. Value (low. tail) 2.7007 67.0 Chi-Sq. Value (up. tail) 19.0238 68.1 9 68.6 Point Estimate 0.70010 67.9 Lower Limit 0.33111 67.2 Upper Limit 2.333

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Right-Tailed TestRight-Tailed Test

• HypothesesHypotheses

where is the hypothesized value where is the hypothesized value for the for the population variance population variance

• Test StatisticTest Statistic

Hypothesis Testing about a Population Hypothesis Testing about a Population VarianceVariance

H02

02: H0

202:

Ha : 202Ha : 202

22

02

1 ( )n s

22

02

1 ( )n s

20 20

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Right-Tailed Test (continued)Right-Tailed Test (continued)

• Rejection RuleRejection Rule

Using test statistic:Using test statistic:

Using Using pp-value:-value:

where is based on a chi-square where is based on a chi-square distribution distribution with with nn - 1 d.f. - 1 d.f.

Reject Reject HH00 if if 2 2 2 2

Reject Reject HH00 if if pp-value < -value <

22

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Left-Tailed TestLeft-Tailed Test


where is the hypothesized value where is the hypothesized value for the for the population variance population variance



2 20 0: H 2 20 0: H

2 20: aH 2 20: aH

22

02

1 ( )n s

22

02

1 ( )n s

20 20

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Left-Tailed Test (continued)Left-Tailed Test (continued)




where is based on a chi-square where is based on a chi-square distribution distribution with with nn - 1 d.f. - 1 d.f.


Reject Reject HH00 if if 2 2(1 ) 2 2(1 )


2(1 ) 2(1 )

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Two-Tailed TestTwo-Tailed Test




22

02

1 ( )n s

22

02

1 ( )n s

Ha : 202Ha : 202

H02

02: H0

202:

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2 2/ 2or 2 2/ 2or 2 2

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

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

Two-Tailed Test (continued)Two-Tailed Test (continued)



Reject Reject HH00 if if



where are based where are based on a on a

chi-square distribution with chi-square distribution with nn - 1 - 1 d.f.d.f.

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Example: Buyer’s Digest (B)Example: Buyer’s Digest (B)

Buyer’s Digest is rating ThermoRite thermostats Buyer’s Digest is rating ThermoRite thermostats made for home temperature control. Buyer’s Digest made for home temperature control. Buyer’s Digest gives an “acceptable” rating to a thermostat with a gives an “acceptable” rating to a thermostat with a temperature variance of 0.5 or less. In a recent test, 10 temperature variance of 0.5 or less. In a recent test, 10 ThermoRite thermostats were selected and placed in a ThermoRite thermostats were selected and placed in a test room that was maintained at a temperature of 68test room that was maintained at a temperature of 68ooF. F. The temperature readings of the thermostats are listed The temperature readings of the thermostats are listed below. below.

Using the 10 readings, we will conduct a hypothesis Using the 10 readings, we will conduct a hypothesis test (with test (with = .05) to determine whether the ThermoRite = .05) to determine whether the ThermoRite thermostat’s temperature variance is “acceptable”.thermostat’s temperature variance is “acceptable”.

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

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HypothesesHypotheses

Rejection RuleRejection Rule

Reject Reject HH00 if if 22 > 14.6837 > 14.6837

20 : 0.5H 20 : 0.5H

2: 0.5aH 2: 0.5aH

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Rejection RegionRejection Region

22

00

.10.10

2 22

2

( 1) 9.5

n s s

2 2

22

( 1) 9.5

n s s

14.683714.6837

Reject Reject HH00Reject Reject HH00

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Test StatisticTest Statistic

The sample variance The sample variance ss22 = 0.7 = 0.7

ConclusionConclusion

Because Because 22 = 12.6 is less than 14.6837, we = 12.6 is less than 14.6837, we cannot reject cannot reject HH00. The sample variance . The sample variance ss22 = .7 = .7 is insufficient evidence to conclude that the is insufficient evidence to conclude that the temperature variance for ThermoRite temperature variance for ThermoRite thermostats is unacceptable.thermostats is unacceptable.


2 9(.7)12.6

.5 2 9(.7)

12.6.5

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Using Excel to Conduct a Hypothesis TestUsing Excel to Conduct a Hypothesis Testabout a Population Varianceabout a Population Variance

Formula WorksheetFormula WorksheetA B C

1 Temp. Sample Size =COUNT(A2:A11)2 67.4 Variance =VAR(A2:A11)3 67.84 68.2 Hypothesized Value 0.55 69.36 69.5 Test Statistic =((C1-1)*C2)/C47 67.0 Degrees of Freedom =C1-18 68.1 9 68.6 p -Value (Lower Tail) =1-CHIDIST(C6,C7)10 67.9 p -Value (Upper Tail) =CHIDIST(C6,C7)11 67.2 p -Value (Two Tail) =2*MIN(C9,C10)

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A B C

1 Temp. Sample Size 102 67.4 Variance 0.73 67.84 68.2 Hypothesized Value 0.55 69.36 69.5 Test Statistic 12.67 67.0 Degrees of Freedom 98 68.1 9 68.6 p -Value (Lower Tail) 0.81844336710 67.9 p -Value (Upper Tail) 0.18155663311 67.2 p -Value (Two Tail) 0.363113265

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Using the Using the pp-Value-Value

• The rejection region for the ThermoRite The rejection region for the ThermoRite thermostat example is in the upper tail; thermostat example is in the upper tail; thus, the appropriate thus, the appropriate pp-value is .1816.-value is .1816.

• Because .1816 > Because .1816 > = .10, we cannot reject = .10, we cannot reject the null hypothesis.the null hypothesis.

• The sample variance of The sample variance of ss22 = .7 is insufficient = .7 is insufficient evidence to conclude that the temperature evidence to conclude that the temperature variance is unacceptable (>.5)variance is unacceptable (>.5)

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One-Tailed TestOne-Tailed Test


Denote the population providing theDenote the population providing the larger sample variance as population larger sample variance as population

1.1.


Hypothesis Testing about the Variances Hypothesis Testing about the Variances of Two Populationsof Two Populations

2 20 1 2: H 2 20 1 2: H

2 21 2: aH 2 21 2: aH

21

22

sFs

21

22

sFs

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One-Tailed Test (continued)One-Tailed Test (continued)



where the value of where the value of FF is based on an is based on an FF distribution with distribution with nn11 - 1 (numerator) and - 1 (numerator) and

nn2 2 - 1 (denominator) d.f.- 1 (denominator) d.f.



Reject Reject HH00 if if FF > > FF


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Two-Tailed TestTwo-Tailed Test


Denote the population providing theDenote the population providing the larger sample variance as population larger sample variance as population

1.1.



H0 12

22: H0 1

222:

Ha : 12

22Ha : 1

222

21

22

sFs

21

22

sFs

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Two-Tailed Test (continued)Two-Tailed Test (continued)



where the value of where the value of FF/2 /2 is based on an is based on an FF distribution with distribution with nn11 - 1 (numerator) and - 1 (numerator) and

nn2 2 - 1 (denominator) d.f.- 1 (denominator) d.f.



Reject Reject HH00 if if FF > > FF/2/2


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Buyer’s Digest has conducted the same test, as wasBuyer’s Digest has conducted the same test, as was

described earlier, on another 10 thermostats, this time described earlier, on another 10 thermostats, this time

manufactured by TempKing. The temperature readingsmanufactured by TempKing. The temperature readings

of the ten thermostats are listed below. of the ten thermostats are listed below.

We will conduct a hypothesis test with We will conduct a hypothesis test with = .10 to see = .10 to see

if the variances are equal for ThermoRite’s thermostatsif the variances are equal for ThermoRite’s thermostats

and TempKing’s thermostats.and TempKing’s thermostats.

Therm.Therm. 11 22 33 44 55 66 77 88 99 10 10 Temp.Temp. 66.466.467.867.868.268.270.370.3 69.5 69.5 68.0 68.1 68.6 67.9 68.0 68.1 68.6 67.9

66.266.2

Example: Buyer’s Digest (C)Example: Buyer’s Digest (C)

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Hypothesis Testing about the Variances of Two Hypothesis Testing about the Variances of Two PopulationsPopulations


(TempKing and ThermoRite (TempKing and ThermoRite thermo-thermostats have same stats have same temperature variance)temperature variance)

(Their variances are not equal)(Their variances are not equal)


The The FF distribution table shows that with distribution table shows that with = .10,= .10,

9 d.f. (numerator), and 9 d.f. (denominator), 9 d.f. (numerator), and 9 d.f. (denominator),

FF.05.05 = 3.18. = 3.18.

Reject Reject HH00 if if FF > 3.18 > 3.18

H0 12

22: H0 1

222:

Ha : 12

22Ha : 1

222


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Hypothesis Testing about the Variances of Two Hypothesis Testing about the Variances of Two PopulationsPopulations


TempKing’s sample variance is 1.52.TempKing’s sample variance is 1.52.

ThermoRite’s sample variance is .70.ThermoRite’s sample variance is .70.

= 1.52/.70 = 2.17= 1.52/.70 = 2.17

• ConclusionConclusion

We We cannotcannot reject reject HH00. There is . There is insufficient insufficient

evidence to conclude that the populationevidence to conclude that the population

variances differ for the two thermostat variances differ for the two thermostat brands.brands.


21

22

sFs

21

22

sFs

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Excel’s F-Test Two-Sample for Variances ToolExcel’s F-Test Two-Sample for Variances Tool

Step 1: Step 1: Select the Select the ToolsTools pull-down menu pull-down menu

Step 2:Step 2: Choose the Choose the Data AnalysisData Analysis option option

Step 3:Step 3: When the Data Analysis dialog box When the Data Analysis dialog box appears:appears:

Choose Choose F-Test Two Sample for F-Test Two Sample for VariancesVariances

Click Click OKOK

… … continuecontinue

Using Excel to Conduct a Hypothesis TestUsing Excel to Conduct a Hypothesis Testabout the Variances of Two Populationsabout the Variances of Two Populations

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Excel’s F-Test Two-Sample for Variances ToolExcel’s F-Test Two-Sample for Variances Tool

Step 4: Step 4: When the F-Test Two-Sample for When the F-Test Two-Sample for Variances Variances dialog box appears: dialog box appears:

Enter A1:A11 in the Enter A1:A11 in the Variable 1 Variable 1 RangeRange box box

Enter B1:B11 in the Enter B1:B11 in the Variable 2 Variable 2 RangeRange box box

Select Select LabelsLabels

Enter .05 in the Enter .05 in the AlphaAlpha box box

Select Select Output RangeOutput Range

Enter C1 in the Enter C1 in the Output RangeOutput Range box box


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A B C D E

1Temp-King

Therm-oRite F-Test Two-Sample for Variances

2 66.4 67.43 67.8 67.8 Temp-King Therm-oRite4 68.2 68.2 Mean 68.1 68.15 70.3 69.3 Variance 1.5222 0.70006 69.5 69.5 Observations 10 107 68.0 67.0 df 9 98 68.1 68.1 F 2.17469 68.6 68.6 P(F<=f) one-tail 0.1314

10 67.9 67.9 F Critical one-tail 3.178911 66.2 67.2

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Determining and Using the Determining and Using the pp-Value-Value

• The output labeled P(F<=f) one-tail, 0.1314, The output labeled P(F<=f) one-tail, 0.1314, can be used to determine the can be used to determine the pp-value for the -value for the hypothesis test.hypothesis test.

• If the thermostat example had been a one-If the thermostat example had been a one-tailed hypothesis test, this would have been tailed hypothesis test, this would have been the the pp-value.-value.

• Because the thermostat example is a two-Because the thermostat example is a two-tailed test, we must multiply the 0.1314 value tailed test, we must multiply the 0.1314 value by 2 to obtain the correct by 2 to obtain the correct pp-value, 0.2628.-value, 0.2628.

• Because .2628 > Because .2628 > = .10, we cannot reject = .10, we cannot reject the null hypothesis.the null hypothesis.

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End of Chapter 11End of Chapter 11

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