8 - 1 © 2003 pearson prentice hall chi-square ( 2 ) test of variance

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© 2003 Pearson Prentice Hall© 2003 Pearson Prentice Hall

Chi-Square (Chi-Square (22) Test ) Test of Varianceof Variance

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Chi-Square (Chi-Square (22) Test) Testfor Variancefor Variance

1.1. Tests One Population Variance or Tests One Population Variance or Standard DeviationStandard Deviation

2.2. Assumes Population Is Approximately Assumes Population Is Approximately Normally DistributedNormally Distributed

3.3. Null Hypothesis Is HNull Hypothesis Is H00: : 22 = = 0022

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Chi-Square (Chi-Square (22) Test) Testfor Variancefor Variance

1.1. Tests One Population Variance or Tests One Population Variance or Standard DeviationStandard Deviation

2.2. Assumes Population Is Approximately Assumes Population Is Approximately Normally DistributedNormally Distributed

3.3. Null Hypothesis Is HNull Hypothesis Is H00: : 22 = = 0022

4.4. Test StatisticTest Statistic

Hypothesized Pop. VarianceHypothesized Pop. Variance

Sample VarianceSample Variance

22

22

22

1)1)

(n(n SS

00

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Chi-Square (Chi-Square (22) ) DistributionDistribution

Select simple randomsample, size n.

Compute s2

Compute 2 =(n-1)s 2/2

Astronomical numberof 2 values

PopulationSampling Distributionsfor Different SampleSizes

21 2 30

Select simple randomsample, size n.

Compute s2

Compute 2 =(n-1)s 2/2

Astronomical numberof 2 values

PopulationSampling Distributionsfor Different SampleSizes

21 2 30

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Finding Critical Finding Critical Value ExampleValue Example

What is the critical What is the critical 22 value given: value given:HHaa: : 22 > 0.7 > 0.7

nn = 3 = 3 =.05? =.05?

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Upper Tail AreaDF .995 … .95 … .051 ... … 0.004 … 3.8412 0.010 … 0.103 … 5.991



20 20

22 Table Table (Portion)(Portion)


nn = 3 = 3 =.05? =.05?

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20

Reject

20

Reject

= .05= .05



nn = 3 = 3 =.05? =.05?

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20

Reject

20

Reject

= .05= .05



nn = 3 = 3 =.05? =.05?

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20

Reject

20

Reject

= .05= .05



nn = 3 = 3 =.05? =.05?

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20

Reject

20

Reject

= .05= .05



nn = 3 = 3 =.05? =.05?

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20

Reject

20

Reject

= .05= .05


dfdf = = nn - 1 = 2 - 1 = 2


nn = 3 = 3 =.05? =.05?

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20

Reject

20

Reject

= .05= .05


dfdf = = nn - 1 = 2 - 1 = 2


nn = 3 = 3 =.05? =.05?

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20

Reject

20

Reject

= .05= .05


dfdf = = nn - 1 = 2 - 1 = 2


nn = 3 = 3 =.05? =.05?

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20

Reject

20

Reject

= .05= .05


dfdf = = nn - 1 = 2 - 1 = 2


nn = 3 = 3 =.05? =.05?

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20 5.991

Reject

20 5.991

Reject

= .05= .05


dfdf = = nn - 1 = 2 - 1 = 2


nn = 3 = 3 =.05? =.05?

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What Do You Do If the Rejection Region Is on the Left?

What Do You Do If the Rejection Region Is on the Left?

What is the critical What is the critical 22 value given: value given:HHaa: : 22 << 0.7 0.7

nn = 3 = 3 =.05? =.05?

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20 20



nn = 3 = 3 =.05? =.05?

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20 20


nn = 3 = 3 =.05? =.05?




= .05= .05


RejectReject

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20 20


nn = 3 = 3 =.05? =.05?




= .05= .05


RejectReject Upper Tail Area Upper Tail Area for Lower Critical for Lower Critical Value = 1-.05 = .95Value = 1-.05 = .95

8 - 8 - 2020


20 20


nn = 3 = 3 =.05? =.05?




= .05= .05



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20 20


nn = 3 = 3 =.05? =.05?




= .05= .05



dfdf = = nn - 1 = 2 - 1 = 2

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20 20


nn = 3 = 3 =.05? =.05?




= .05= .05



dfdf = = nn - 1 = 2 - 1 = 2

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20 .103 20 .103


nn = 3 = 3 =.05? =.05?




= .05= .05


dfdf = = nn - 1 = 2 - 1 = 2

Upper Tail Area Upper Tail Area for Lower Critical for Lower Critical Value = 1-.05 = .95Value = 1-.05 = .95

RejectReject

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Chi-Square (Chi-Square (22) Test ) Test Example Example

Is the variation in boxes Is the variation in boxes of cereal, measured by of cereal, measured by the the variancevariance, equal to , equal to 1515 grams? A random grams? A random sample of sample of 2525 boxes had boxes had a standard deviation ofa standard deviation of 17.717.7 grams. Test at the grams. Test at the .05.05 level. level.

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2200

Chi-Square (Chi-Square (22) Test ) Test SolutionSolution

HH00: :

HHaa: :

= =

df = df =

Critical Value(s):Critical Value(s):

Test Statistic: Test Statistic:

Decision:Decision:

Conclusion:Conclusion:

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2200


HH00: : 22 = 15 = 15

HHaa: : 22 15 15

= =

df = df =



Decision:Decision:


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2200


HH00: : 22 = 15 = 15

HHaa: : 22 15 15

= = .05.05

df = df = 25 - 1 = 2425 - 1 = 24



Decision:Decision:


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2200


HH00: : 22 = 15 = 15

HHaa: : 22 15 15

= = .05.05

df = df = 25 - 1 = 2425 - 1 = 24



Decision:Decision:


/2 = .025/2 = .025

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2200 39.36439.36412.40112.401


HH00: : 22 = 15 = 15

HHaa: : 22 15 15

= = .05.05

df = df = 25 - 1 = 2425 - 1 = 24



Decision:Decision:


/2 = .025/2 = .025

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2200 39.36439.36412.40112.401


HH00: : 22 = 15 = 15

HHaa: : 22 15 15

= = .05.05

df = df = 25 - 1 = 2425 - 1 = 24



Decision:Decision:


/2 = .025/2 = .025

2222

22

22

22

1)1) (25 -(25 -1)1) 1717 77

1515

3333 4242

(n(n SS

00

..

..

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2200 39.36439.36412.40112.401


HH00: : 22 = 15 = 15

HHaa: : 22 15 15

= = .05.05

df = df = 25 - 1 = 2425 - 1 = 24



Decision:Decision:


Do Not Reject at Do Not Reject at = .05 = .05 /2 = .025/2 = .025

2222

22

22

22

1)1) (25 -(25 -1)1) 1717 77

1515

3333 4242

(n(n SS

00

..

..

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2200 39.36439.36412.40112.401


HH00: : 22 = 15 = 15

HHaa: : 22 15 15

= = .05.05

df = df = 25 - 1 = 2425 - 1 = 24



Decision:Decision:


Do Not Reject at Do Not Reject at = .05 = .05

There Is No Evidence There Is No Evidence 22 Is Not 15 Is Not 15

/2 = .025/2 = .025

2222

22

22

22

1)1) (25 -(25 -1)1) 1717 77

1515

3333 4242

(n(n SS

00

..

..

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Calculating Type II Calculating Type II Error ProbabilitiesError Probabilities

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Power of TestPower of Test

1.1. Probability of Rejecting False HProbability of Rejecting False H00

Correct DecisionCorrect Decision

2.2. Designated 1 - Designated 1 -

3.3. Used in Determining Test AdequacyUsed in Determining Test Adequacy

4.4. Affected byAffected by True Value of Population ParameterTrue Value of Population Parameter Significance Level Significance Level Standard Deviation & Sample Size Standard Deviation & Sample Size nn

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XX00 = 368= 368

RejectRejectDo NotDo NotRejectReject

Finding PowerFinding PowerStep 1Step 1

Hypothesis:Hypothesis:HH00: : 00 368 368

HH11: : 00 < 368 < 368 = .05= .05

n =n =15/15/2525

DrawDraw

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XX11 = 360= 360

XX00 = 368= 368


Finding PowerFinding PowerSteps 2 & 3Steps 2 & 3


HH11: : 00 < 368 < 368

‘‘True’ Situation:True’ Situation: 11 = 360 = 360

= .05= .05

n =n =15/15/2525

DrawDraw

DrawDraw

SpecifySpecify

1-1-

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XX11 = 360= 360 363.065363.065

XX00 = 368= 368




HH11: : 00 < 368 < 368


065.363

25

1564.13680

n

ZX L

065.363

25

1564.13680

n

ZX L

= .05= .05

n =n =15/15/2525

DrawDraw

DrawDraw

SpecifySpecify

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XX11 = 360= 360 363.065363.065

XX00 = 368= 368




HH11: : 00 < 368 < 368


= .05= .05

n =n =15/15/2525

= .154= .154

1-1- =.846 =.846

DrawDraw

DrawDraw

SpecifySpecify

Z TableZ Table

065.363

25

1564.13680

n

ZX L

065.363

25

1564.13680

n

ZX L

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Power CurvesPower Curves

PowerPower PowerPower

PowerPower

Possible True Values for Possible True Values for 11 Possible True Values for Possible True Values for 11

Possible True Values for Possible True Values for 11

HH00: : 00 HH00: : 00

HH00: : = =00

= 368 in = 368 in

ExampleExample

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ConclusionConclusion

1.1. Distinguished Types of Hypotheses Distinguished Types of Hypotheses

2.2. Described Hypothesis Testing ProcessDescribed Hypothesis Testing Process

3.3. Explained p-Value ConceptExplained p-Value Concept

4.4. Solved Hypothesis Testing Problems Solved Hypothesis Testing Problems Based on a Single SampleBased on a Single Sample

5.5. Explained Power of a TestExplained Power of a Test

End of Chapter

Any blank slides that follow are blank intentionally.

8 - 1 © 2003 pearson prentice hall chi-square ( 2 ) test of variance

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