8 - 1 © 2000 Prentice-Hall, Inc. Statistics for Business and Economics Inferences Based on a Single...

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Statistics for Business Statistics for Business and Economicsand Economics

Inferences Based on a Single Sample: Inferences Based on a Single Sample: Tests of HypothesisTests of Hypothesis

Chapter 8Chapter 8

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Learning ObjectivesLearning Objectives

1.1. Distinguish Types of Hypotheses Distinguish Types of Hypotheses

2.2. Describe Hypothesis Testing ProcessDescribe Hypothesis Testing Process

3.3. Explain p-Value ConceptExplain p-Value Concept

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

5.5. Explain Power of a TestExplain Power of a Test

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Statistical MethodsStatistical Methods

StatisticalMethods

DescriptiveStatistics

InferentialStatistics

EstimationHypothesis

Testing

StatisticalMethods

DescriptiveStatistics

InferentialStatistics

EstimationHypothesis

Testing

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Hypothesis Testing Hypothesis Testing ConceptsConcepts

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Hypothesis TestingHypothesis Testing

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PopulationPopulation

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I believe the population mean age is 50 (hypothesis).

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MeanMean X X = 20= 20

Random Random samplesample

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MeanMean X X = 20= 20

Reject hypothesis! Not close.

Random Random samplesample

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What’s a What’s a Hypothesis?Hypothesis?

1.1. A Belief about a A Belief about a Population ParameterPopulation Parameter Parameter Is Parameter Is

PopulationPopulation Mean, Mean, Proportion, VarianceProportion, Variance

Must Be StatedMust Be StatedBeforeBefore Analysis Analysis

I believe the mean GPA I believe the mean GPA of this class is 3.5!of this class is 3.5!

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Null HypothesisNull Hypothesis

1.1. What Is TestedWhat Is Tested

2.2. Has Serious Outcome If Incorrect Decision Has Serious Outcome If Incorrect Decision MadeMade

3.3. Always Has Equality Sign: Always Has Equality Sign: , , , or , or 4.4. Designated HDesignated H00 (Pronounced H-oh) (Pronounced H-oh)

5.5. Specified as HSpecified as H00: : Some Numeric Value Some Numeric Value Specified with = Sign Even if Specified with = Sign Even if , or , or Example, HExample, H00: : 3 3

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Alternative Alternative HypothesisHypothesis

1.1. Opposite of Null HypothesisOpposite of Null Hypothesis

2.2. Always Has Inequality Sign:Always Has Inequality Sign: ,,, or , or

3.3. Designated HDesignated Haa

4.4. Specified HSpecified Haa: : < Some Value < Some Value Example, HExample, Haa: : < 3 < 3

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Identifying Identifying HypothesesHypotheses

StepsSteps1.1. Example Problem: Test That the Example Problem: Test That the

Population Mean Is Not 3Population Mean Is Not 3

2.2. StepsSteps State the Question Statistically (State the Question Statistically ( 3) 3) State the Opposite Statistically (State the Opposite Statistically ( = 3) = 3)

Must Be Mutually Exclusive & ExhaustiveMust Be Mutually Exclusive & Exhaustive Select the Alternative Hypothesis (Select the Alternative Hypothesis ( 3) 3)

Has the Has the , , <<, or , or > > SignSign State the Null Hypothesis (State the Null Hypothesis ( = 3) = 3)

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State the question statistically: State the question statistically: = 12 = 12

State the opposite statistically: State the opposite statistically: 12 12

Select the alternative hypothesis: Select the alternative hypothesis: HHaa: :

State the null hypothesis: State the null hypothesis: HH00: : = 12 = 12

Is the population average amount of TV Is the population average amount of TV viewing 12 hours?viewing 12 hours?

What Are the What Are the Hypotheses?Hypotheses?

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State the question statistically: State the question statistically: 12 12

State the opposite statistically: State the opposite statistically: = 12 = 12

Select the alternative hypothesis: Select the alternative hypothesis: HHaa: :

State the null hypothesis: State the null hypothesis: HH00: : = 12 = 12

Is the population average amount of TV Is the population average amount of TV viewing viewing differentdifferent from 12 hours? from 12 hours?

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Select the alternative hypothesis: Select the alternative hypothesis: HHaa: : 20 20

State the null hypothesis: State the null hypothesis: HH00: : 20 20

Is the average cost per hat less than or Is the average cost per hat less than or equal to $20?equal to $20?

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Select the alternative hypothesis: Select the alternative hypothesis: HHaa: : 25 25

State the null hypothesis: State the null hypothesis: HH00: : 25 25

Is the average amount spent in the Is the average amount spent in the bookstore greater than $25?bookstore greater than $25?

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Basic IdeaBasic Idea

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Sample Mean = 50 Sample Mean = 50

HH00HH00

Sampling DistributionSampling Distribution

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It is unlikely It is unlikely that we would that we would get a sample get a sample mean of this mean of this value ...value ...

20202020HH00HH00

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... if in fact this were... if in fact this were the population mean the population mean

20202020HH00HH00

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... if in fact this were... if in fact this were the population mean the population mean

... therefore, ... therefore, we reject the we reject the hypothesis hypothesis

that that = 50.= 50.

20202020HH00HH00

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Level of SignificanceLevel of Significance

1.1. ProbabilityProbability

2.2. Defines Unlikely Values of Sample Defines Unlikely Values of Sample Statistic if Null Hypothesis Is TrueStatistic if Null Hypothesis Is True Called Rejection Region of Sampling Called Rejection Region of Sampling

DistributionDistribution

3.3. Designated Designated (alpha)(alpha) Typical Values Are .01, .05, .10Typical Values Are .01, .05, .10

4.4. Selected by Researcher at StartSelected by Researcher at Start

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Rejection Region Rejection Region (One-Tail Test) (One-Tail Test)

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HoValueCritical

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValueCritical

Sample Statistic

RejectionRegion

NonrejectionRegion

1 - 1 -

Level of ConfidenceLevel of Confidence

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HoValueCritical

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValueCritical

Sample Statistic

RejectionRegion

NonrejectionRegion

1 - 1 -

Observed sample statisticObserved sample statistic

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HoValueCritical

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValueCritical

Sample Statistic

RejectionRegion

NonrejectionRegion

1 - 1 -

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Rejection Regions Rejection Regions (Two-Tailed Test) (Two-Tailed Test)

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HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

NonrejectionRegion

1 - 1 -

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HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

NonrejectionRegion

1 - 1 -

Observed sample statisticObserved sample statistic

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HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

NonrejectionRegion

1 - 1 -

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HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

NonrejectionRegion

1 - 1 -

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Decision Making RisksDecision Making Risks

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Errors in Errors in Making DecisionMaking Decision

1.1. Type I ErrorType I Error Reject True Null HypothesisReject True Null Hypothesis Has Serious ConsequencesHas Serious Consequences Probability of Type I Error Is Probability of Type I Error Is (Alpha)(Alpha)Called Level of SignificanceCalled Level of Significance

2.2. Type II ErrorType II Error Do Not Reject False Null HypothesisDo Not Reject False Null Hypothesis Probability of Type II Error Is Probability of Type II Error Is (Beta)(Beta)

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Jury Trial H0 Test

Actual Situation Actual Situation

Verdict Innocent Guilty Decision H0 True H0

Innocent Correct ErrorDo NotReject

1 - Type IIError

Guilty Error Correct RejectH0

Type IError ()

Power(1 - )

Jury Trial H0 Test

Innocent Correct ErrorDo NotReject

1 - Type IIError

Type IError ()

Power(1 - )

Decision ResultsDecision Results

HH00: Innocent: Innocent

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Jury Trial H0 Test

Innocent Correct Error AcceptH0

1 - Type IIError

Type IError ()

Power(1 - )

Jury Trial H0 Test

Innocent Correct Error AcceptH0

1 - Type IIError

Type IError ()

Power(1 - )

Decision ResultsDecision Results

HH00: Innocent: Innocent

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& & Have an Have an Inverse RelationshipInverse Relationship

You can’t reduce both errors simultaneously!

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Factors Affecting Factors Affecting

1.1. True Value of Population ParameterTrue Value of Population Parameter Increases When Difference With Hypothesized Increases When Difference With Hypothesized

Parameter DecreasesParameter Decreases

2.2. Significance Level, Significance Level, Increases When Increases When DecreasesDecreases

3.3. Population Standard Deviation, Population Standard Deviation, Increases When Increases When Increases Increases

4.4. Sample Size, Sample Size, nn Increases When Increases When nn Decreases Decreases

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Hypothesis Testing Hypothesis Testing StepsSteps

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HH00 Testing Steps Testing Steps

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State HState H00

State HState Haa

Choose Choose

Choose Choose nn

Choose testChoose test

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Set up critical valuesSet up critical values

Collect dataCollect data

Compute test statisticCompute test statistic

Make statistical decisionMake statistical decision

Express decisionExpress decision

State HState H00

State HState Haa

Choose Choose

Choose Choose nn

Choose testChoose test

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One Population One Population TestsTests

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)

Mean Proportion Variance

2 Test(1 & 2tail)

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)

2 Test(1 & 2tail)

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Two-Tailed Z Test Two-Tailed Z Test of Mean (of Mean ( Known) Known)

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OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)

2 Test(1 & 2tail)

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)

2 Test(1 & 2tail)

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Two-Tailed Z Test Two-Tailed Z Test for Mean (for Mean ( Known) Known)

1.1. AssumptionsAssumptions Population Is Normally DistributedPopulation Is Normally Distributed If Not Normal, Can Be Approximated by If Not Normal, Can Be Approximated by

Normal Distribution (Normal Distribution (nn 30) 30)

2.2. Alternative Hypothesis Has Alternative Hypothesis Has Sign Sign

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Two-Tailed Z Test Two-Tailed Z Test for Mean (for Mean ( Known) Known)

2.2. Alternative Hypothesis Has Alternative Hypothesis Has Sign Sign

3.3. Z-Test StatisticZ-Test Statistic

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Two-Tailed Z TestTwo-Tailed Z Test Example Example

Does an average box of Does an average box of cereal contain cereal contain 368368 grams grams of cereal? A random of cereal? A random sample of sample of 2525 boxes boxes showedshowedX = 372.5X = 372.5. The . The company has specified company has specified to be to be 2525 grams. Test at grams. Test at the the .05.05 level. level. 368 gm.368 gm.

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Two-Tailed Z Test Two-Tailed Z Test SolutionSolution

HH00: :

HHaa: :

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

Test Statistic: Test Statistic:

Decision:Decision:

Conclusion:Conclusion:

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HH00: : = 368 = 368

HHaa: : 368 368

Decision:Decision:

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HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 2525

Decision:Decision:

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HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 2525

Decision:Decision:

Z0 1.96-1.96

Reject H 0 Reject H 0

Z0 1.96-1.96

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HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 2525

Decision:Decision:

Z0 1.96-1.96

372 5 3681525

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HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 2525

Decision:Decision:

Z0 1.96-1.96

372 5 3681525

Do not reject at Do not reject at = .05 = .05

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HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 2525

Decision:Decision:

Z0 1.96-1.96

372 5 3681525

No evidence No evidence average is not 368average is not 368

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Two-Tailed Z Test Two-Tailed Z Test Thinking ChallengeThinking Challenge

You’re a Q/C inspector. You want to You’re a Q/C inspector. You want to find out if a new machine is making find out if a new machine is making electrical cords to customer electrical cords to customer specification: specification: averageaverage breaking breaking strength of strength of 7070 lb. with lb. with = 3.5 = 3.5 lb. lb. You take a sample of You take a sample of 3636 cords & cords & compute a sample mean of compute a sample mean of 69.769.7 lb. lb. At the At the .05.05 level, is there evidence level, is there evidence that the machine is that the machine is notnot meeting the meeting the average breaking strength?average breaking strength?

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Two-Tailed Z Test Two-Tailed Z Test Solution*Solution*

HH00: :

HHaa: :

nn = =

Decision:Decision:

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HH00: : = 70 = 70

HHaa: : 70 70

nn = =

Decision:Decision:

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HH00: : = 70 = 70

HHaa: : 70 70

= = .05.05

nn = = 3636

Decision:Decision:

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HH00: : = 70 = 70

HHaa: : 70 70

= = .05.05

nn = = 3636

Decision:Decision:

Z0 1.96-1.96

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HH00: : = 70 = 70

HHaa: : 70 70

= = .05.05

nn = = 3636

Decision:Decision:

Z0 1.96-1.96

69 7 703 536

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HH00: : = 70 = 70

HHaa: : 70 70

= = .05.05

nn = = 3636

Decision:Decision:

Z0 1.96-1.96

69 7 703 536

8 - 8 - 6363

HH00: : = 70 = 70

HHaa: : 70 70

= = .05.05

nn = = 3636

Decision:Decision:

Z0 1.96-1.96

69 7 703 536

No evidence No evidence average is not 70average is not 70

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One-Tailed Z Test One-Tailed Z Test of Mean (of Mean ( Known) Known)

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One-Tailed Z Test One-Tailed Z Test for Mean (for Mean ( Known) Known)

2.2. Alternative Hypothesis Has < or > SignAlternative Hypothesis Has < or > Sign

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One-Tailed Z Test One-Tailed Z Test for Mean (for Mean ( Known) Known)

2.2. Alternative Hypothesis Has Alternative Hypothesis Has or > Signor > Sign

3.3. Z-test StatisticZ-test Statistic

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One-Tailed Z Test One-Tailed Z Test for Mean Hypothesesfor Mean Hypotheses

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Reject H 0

HH00::==0 H0 Haa: : << 0 0

Must be Must be significantlysignificantly below below

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Reject H 0

HH00::==0 H0 Haa: : << 0 0 HH00::==0 H0 Haa: : >> 0 0

Must be Must be significantlysignificantly below below

Small values satisfy Small values satisfy HH0 0 . Don’t reject!. Don’t reject!

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One-Tailed Z Test One-Tailed Z Test Finding Critical ZFinding Critical Z

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What Is Z given What Is Z given = .025? = .025?

= .025= .025

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.500 .500 -- .025.025

.475.475

= .025= .025

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Z .05 .07

1.6 .4505 .4515 .4525

1.7 .4599 .4608 .4616

1.8 .4678 .4686 .4693

.4744 .4756

.500 .500 -- .025.025

.475.475

1.9 .4750.4750

Standardized Normal Standardized Normal Probability Table (Portion)Probability Table (Portion)

= .025= .025

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Z .05 .07

1.6 .4505 .4515 .4525

1.7 .4599 .4608 .4616

1.8 .4678 .4686 .4693

.4744 .4756

1.96 Z0

.500 .500 -- .025.025

.475.475.06.06

1.91.9 .4750

Standardized Normal Standardized Normal Probability Table (Portion)Probability Table (Portion)

= .025= .025

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One-Tailed Z TestOne-Tailed Z Test Example Example

Does an average box of Does an average box of cereal contain cereal contain more thanmore than 368368 grams of cereal? A grams of cereal? A random sample of random sample of 2525 boxes showedboxes showedX = 372.5X = 372.5. . The company has The company has specified specified to be to be 2525 grams. Test at the grams. Test at the .05.05 level.level.

368 gm.368 gm.

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One-Tailed Z Test One-Tailed Z Test SolutionSolution

HH00: :

HHaa: :

n n = =

Decision:Decision:

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HH00: : = 368 = 368

HHaa: : > 368 > 368

n n = =

Decision:Decision:

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HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 2525

Decision:Decision:

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HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 2525

Decision:Decision:

Z0 1.645

Reject

Z0 1.645

Reject

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HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 2525

Decision:Decision:

Z0 1.645

Reject

Z0 1.645

Reject

372 5 3681525

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HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 2525

Decision:Decision:

Z0 1.645

Reject

Z0 1.645

Reject

372 5 3681525

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HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 2525

Decision:Decision:

Z0 1.645

Reject

Z0 1.645

Reject

372 5 3681525

No evidence average No evidence average is more than 368is more than 368

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One-Tailed Z Test One-Tailed Z Test Thinking ChallengeThinking Challenge

You’re an analyst for Ford. You You’re an analyst for Ford. You want to find out if the average want to find out if the average miles per gallon of Escorts is at miles per gallon of Escorts is at least 32 mpg. Similar models least 32 mpg. Similar models have a standard deviation of have a standard deviation of 3.83.8 mpg. You take a sample of mpg. You take a sample of 6060 Escorts & compute a sample Escorts & compute a sample mean of mean of 30.730.7 mpg. At the mpg. At the .01.01 level, is there evidence that the level, is there evidence that the miles per gallon is miles per gallon is at leastat least 3232??

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One-Tailed Z Test One-Tailed Z Test Solution*Solution*

HH00: :

HHaa: :

nn = =

Decision:Decision:

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HH00: : = 32 = 32

HHaa: : < 32 < 32

nn = =

Decision:Decision:

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HH00: : = 32 = 32

HHaa: : < 32 < 32

== .01 .01

nn = = 6060

Decision:Decision:

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HH00: : = 32 = 32

HHaa: : < 32 < 32

= .01= .01

nn = 60= 60

Decision:Decision:

Z0-2.33

Reject

Z0-2.33

Reject

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HH00: : = 32 = 32

HHaa: : < 32 < 32

= .01= .01

nn = 60= 60

Decision:Decision:

Z0-2.33

Reject

Z0-2.33

Reject

30 7 323 860

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HH00: : = 32 = 32

HHaa: : < 32 < 32

= .01= .01

nn = 60= 60

Decision:Decision:

Z0-2.33

Reject

Z0-2.33

Reject

30 7 323 860

Reject at Reject at = .01 = .01

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HH00: : = 32 = 32

HHaa: : < 32 < 32

= .01= .01

nn = 60= 60

Decision:Decision:

Z0-2.33

Reject

Z0-2.33

Reject

30 7 323 860

There is evidence There is evidence average is less than 32average is less than 32

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Observed Significance Observed Significance Levels: p-ValuesLevels: p-Values

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p-Valuep-Value

1.1. Probability of Obtaining a Test Statistic Probability of Obtaining a Test Statistic More Extreme (More Extreme (or or than Actual than Actual Sample Value Given HSample Value Given H00 Is True Is True

2.2. Called Observed Level of SignificanceCalled Observed Level of Significance Smallest Value of Smallest Value of H H00 Can Be Rejected Can Be Rejected

3.3. Used to Make Rejection DecisionUsed to Make Rejection Decision If p-Value If p-Value , Do Not Reject H, Do Not Reject H00

If p-Value < If p-Value < , Reject H, Reject H00

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Two-Tailed Z Test Two-Tailed Z Test p-Value Example p-Value Example

Does an average box of Does an average box of cereal contain cereal contain 368368 grams grams of cereal? A random of cereal? A random sample of sample of 2525 boxes boxes showedshowedX = 372.5X = 372.5. The . The company has specified company has specified to be to be 2525 grams. Find the grams. Find the p-Value.p-Value. 368 gm.368 gm.

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Two-Tailed Z Test Two-Tailed Z Test p-Value Solutionp-Value Solution

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Z0 1.50-1.50 Z0 1.50-1.50

Z value of sample Z value of sample statistic (observed)statistic (observed)

372 5 3681525

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Z0 1.50-1.50 Z0 1.50-1.50

p-value is P(Z p-value is P(Z -1.50 or Z -1.50 or Z 1.50) 1.50)

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Z0 1.50-1.50

1/2 p-Value1/2 p-Value

Z0 1.50-1.50

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Z0 1.50-1.50

From Z table: From Z table: lookup 1.50lookup 1.50

.4332.4332

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Z0 1.50-1.50

.4332.4332

.5000.5000-- .4332.4332

.0668.0668

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Z0 1.50-1.50

1/2 p-Value.0668

Z0 1.50-1.50

1/2 p-Value.0668

p-value is P(Z p-value is P(Z -1.50 or Z -1.50 or Z 1.50) 1.50) = = .1336.1336

Z value of sample Z value of sample statisticstatistic

.4332.4332

.5000.5000-- .4332.4332

.0668.0668

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0 1.50-1.50 Z

RejectReject

0 1.50-1.50 Z

RejectReject

1/2 p-Value = .06681/2 p-Value = .06681/2 p-Value = .06681/2 p-Value = .0668

1/2 1/2 = .025 = .0251/2 1/2 = .025 = .025

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0 1.50-1.50 Z

RejectReject

0 1.50-1.50 Z

RejectReject

(p-Value = .1336) (p-Value = .1336) ( ( = .05). = .05). Do not reject.Do not reject.

1/2 p-Value = .06681/2 p-Value = .06681/2 p-Value = .06681/2 p-Value = .0668

1/2 1/2 = .025 = .0251/2 1/2 = .025 = .025

Test statistic is in ‘Do not reject’ regionTest statistic is in ‘Do not reject’ region

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One-Tailed Z Test One-Tailed Z Test p-Value Example p-Value Example

Does an average box of Does an average box of cereal contain cereal contain more thanmore than 368368 grams of cereal? A grams of cereal? A random sample of random sample of 2525 boxes showedboxes showedX = 372.5X = 372.5. . The company has The company has specified specified to be to be 2525 grams. Find the p-Value.grams. Find the p-Value. 368 gm.368 gm.

8 - 8 - 104104

One-Tailed Z Test One-Tailed Z Test p-Value Solutionp-Value Solution

8 - 8 - 105105

Z0 1.50 Z0 1.50

372 5 3681525

8 - 8 - 106106

Z0 1.50

p-Value

Z0 1.50

p-ValueUse Use alternative alternative hypothesis hypothesis to find to find directiondirection

p-Value is P(Z p-Value is P(Z 1.50) 1.50)

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Z0 1.50

p-Value

Z0 1.50

.4332.4332

8 - 8 - 108108

Z0 1.50

p-Value

Z0 1.50

.4332.4332

.5000.5000-- .4332.4332

.0668.0668

8 - 8 - 109109

Z0 1.50

p-Value.0668

Z0 1.50

p-Value.0668

.4332.4332

Use Use alternative alternative hypothesis hypothesis to find to find directiondirection

.5000.5000-- .4332.4332

.0668.0668

p-Value is P(Z p-Value is P(Z 1.50) = .0668 1.50) = .0668

8 - 8 - 110110

0 1.50 Z

Reject

0 1.50 Z

Reject

p-Value = .0668p-Value = .0668

= .05= .05

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0 1.50 Z

Reject

0 1.50 Z

Reject

(p-Value = .0668) (p-Value = .0668) ( ( = .05). = .05). Do not reject.Do not reject.

p-Value = .0668p-Value = .0668

= .05= .05

Test statistic is in ‘Do not reject’ regionTest statistic is in ‘Do not reject’ region

8 - 8 - 112112

p-Value p-Value Thinking ChallengeThinking Challenge

You’re an analyst for Ford. You You’re an analyst for Ford. You want to find out if the average want to find out if the average miles per gallon of Escorts is miles per gallon of Escorts is at at least 32 least 32 mpg. Similar models mpg. Similar models have a standard deviation of have a standard deviation of 3.83.8 mpg. You take a sample of mpg. You take a sample of 6060 Escorts & compute a sample Escorts & compute a sample mean of mean of 30.730.7 mpg. What is the mpg. What is the value of the observed level of value of the observed level of significance (significance (p-Valuep-Value)?)?

8 - 8 - 113113

p-Value p-Value Solution*Solution*

Z0-2.65

p-Value.004

Z0-2.65

p-Value.004

Z value of Z value of sample statisticsample statistic

.4960.4960

Use Use alternative alternative hypothesis hypothesis to find to find directiondirection

.5000.5000-- .4960.4960

.0040.0040

p-Value is P(Z p-Value is P(Z -2.65) = .004. -2.65) = .004.p-Value < (p-Value < ( = .01). Reject H = .01). Reject H00..

8 - 8 - 114114

Two-Tailed t Test Two-Tailed t Test of Mean (of Mean ( Unknown) Unknown)

8 - 8 - 115115

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)

2 Test(1 & 2tail)

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)

2 Test(1 & 2tail)

8 - 8 - 116116

t Test for Mean t Test for Mean (( Unknown) Unknown)

1.1. AssumptionsAssumptions Population Is Normally DistributedPopulation Is Normally Distributed If Not Normal, Only Slightly Skewed & If Not Normal, Only Slightly Skewed &

Large Sample (Large Sample (nn 30) Taken 30) Taken

2.2. Parametric Test ProcedureParametric Test Procedure

8 - 8 - 117117

t Test for Mean t Test for Mean (( Unknown) Unknown)

1.1. AssumptionsAssumptions Population Is Normally DistributedPopulation Is Normally Distributed If Not Normal, Only Slightly Skewed & If Not Normal, Only Slightly Skewed &

Large Sample (Large Sample (nn 30) Taken 30) Taken

2.2. Parametric Test ProcedureParametric Test Procedure

3.3. t Test Statistict Test Statistic

8 - 8 - 118118

Two-Tailed t TestTwo-Tailed t Test Finding Critical t Finding Critical t

ValuesValues

8 - 8 - 119119

ValuesValuesGiven: n = 3; Given: n = 3; = .10 = .10

8 - 8 - 120120

ValuesValues

/2 = .05/2 = .05

Given: n = 3; Given: n = 3; = .10 = .10

8 - 8 - 121121

ValuesValues

/2 = .05/2 = .05

Given: n = 3; Given: n = 3; = .10 = .10

df = n - 1 = 2df = n - 1 = 2

8 - 8 - 122122

v t.10 t.05 t.025

1 3.078 6.314 12.706

2 1.886 2.920 4.303

3 1.638 2.353 3.182

v t.10 t.05 t.025

1 3.078 6.314 12.706

2 1.886 2.920 4.303

3 1.638 2.353 3.182t0 t0

ValuesValuesCritical Values of t Table Critical Values of t Table

(Portion)(Portion)

/2 = /2 = .05.05

/2 = .05/2 = .05

Given: n = 3; Given: n = 3; = .10 = .10

df = n - 1 = df = n - 1 = 22

8 - 8 - 123123

v t.10 t.05 t.025

1 3.078 6.314 12.706

2 1.886 2.920 4.303

3 1.638 2.353 3.182

v t.10 t.05 t.025

1 3.078 6.314 12.706

2 1.886 2.920 4.303

3 1.638 2.353 3.182t0 2.920-2.920 t0 2.920-2.920

ValuesValuesCritical Values of t Table Critical Values of t Table

(Portion)(Portion)

/2 = .05/2 = .05

Given: n = 3; Given: n = 3; = .10 = .10

df = n - 1 = 2df = n - 1 = 2

8 - 8 - 124124

Two-Tailed t TestTwo-Tailed t Test Example Example

Does an average box of Does an average box of cereal contain cereal contain 368368 grams of cereal? A grams of cereal? A random sample of random sample of 3636 boxes had a mean of boxes had a mean of 372.5372.5 & a standard & a standard deviation ofdeviation of 1212 grams. grams. Test at the Test at the .05.05 level. level. 368 gm.368 gm.

8 - 8 - 125125

Two-Tailed t Test Two-Tailed t Test SolutionSolution

HH00: :

HHaa: :

df = df = Critical Value(s):Critical Value(s):

Decision:Decision:

8 - 8 - 126126

HH00: : = 368 = 368

HHaa: : 368 368

df = df = Critical Value(s):Critical Value(s):

Decision:Decision:

8 - 8 - 127127

HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05

df = df = 36 - 1 = 3536 - 1 = 35Critical Value(s):Critical Value(s):

Decision:Decision:

8 - 8 - 128128

HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05

Decision:Decision:

t0 2.0301-2.0301

8 - 8 - 129129

HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05

Decision:Decision:

t0 2.0301-2.0301

372 5 3681236

8 - 8 - 130130

HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05

Decision:Decision:

t0 2.0301-2.0301

372 5 3681236

8 - 8 - 131131

HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05

Decision:Decision:

t0 2.0301-2.0301

372 5 3681236

There is evidence pop. There is evidence pop. average is not 368average is not 368

8 - 8 - 132132

Two-Tailed t TestTwo-Tailed t TestThinking ChallengeThinking Challenge

You work for the FTC. A You work for the FTC. A manufacturer of detergent manufacturer of detergent claims that the mean weight claims that the mean weight of detergent is of detergent is 3.253.25 lb. You lb. You take a random sample of take a random sample of 6464 containers. You calculate the containers. You calculate the sample average to be sample average to be 3.2383.238 lb. with a standard deviation lb. with a standard deviation of of .117.117 lb. At the lb. At the .01.01 level, is level, is the manufacturer correct?the manufacturer correct?

3.25 lb.3.25 lb.

8 - 8 - 133133

Two-Tailed t Test Two-Tailed t Test Solution*Solution*

HH00: :

HHaa: :

Decision:Decision:

8 - 8 - 134134

HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

Decision:Decision:

8 - 8 - 135135

HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

.01.01

df df 64 - 1 = 6364 - 1 = 63

Decision:Decision:

8 - 8 - 136136

HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

.01.01

df df 64 - 1 = 6364 - 1 = 63

Decision:Decision:

t0 2.6561-2.6561

8 - 8 - 137137

HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

.01.01

df df 64 - 1 = 6364 - 1 = 63

Decision:Decision:

t0 2.6561-2.6561

3 238 3 2511764

8 - 8 - 138138

HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

.01.01

df df 64 - 1 = 6364 - 1 = 63

Decision:Decision:

t0 2.6561-2.6561

3 238 3 2511764

8 - 8 - 139139

HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

.01.01

df df 64 - 1 = 6364 - 1 = 63

Decision:Decision:

t0 2.6561-2.6561

3 238 3 2511764

There is no evidence There is no evidence average is not 3.25average is not 3.25

8 - 8 - 140140

One-Tailed t Test One-Tailed t Test of Mean (of Mean ( Unknown) Unknown)

8 - 8 - 141141

One-Tailed t TestOne-Tailed t TestExample Example

Is the average capacity of Is the average capacity of batteries batteries at least 140 at least 140 ampere-hours? A random ampere-hours? A random sample of sample of 2020 batteries had batteries had a mean of a mean of 138.47138.47 & a & a standard deviation of standard deviation of 2.662.66. . Assume a normal Assume a normal distribution. Test at the distribution. Test at the .05.05 level.level.

8 - 8 - 142142

One-Tailed t Test One-Tailed t Test SolutionSolution

HH00: :

HHaa: :

df =df =

Decision:Decision:

8 - 8 - 143143

HH00: : = 140 = 140

HHaa: : < 140 < 140

df = df =

Decision:Decision:

8 - 8 - 144144

HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

df = df = 20 - 1 = 1920 - 1 = 19

Decision:Decision:

8 - 8 - 145145

t0-1.7291

Reject

t0-1.7291

Reject

HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

df = df = 20 - 1 = 1920 - 1 = 19

Decision:Decision:

8 - 8 - 146146

t0-1.7291

Reject

t0-1.7291

Reject

HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

df = df = 20 - 1 = 1920 - 1 = 19

Decision:Decision:

138 47 1402 66

8 - 8 - 147147

t0-1.7291

Reject

t0-1.7291

Reject

HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

df = df = 20 - 1 = 1920 - 1 = 19

Decision:Decision:

138 47 1402 66

8 - 8 - 148148

t0-1.7291

Reject

t0-1.7291

Reject

HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

df = df = 20 - 1 = 1920 - 1 = 19

Decision:Decision:

138 47 1402 66

There is evidence pop. There is evidence pop. average is less than 140average is less than 140

8 - 8 - 149149

One-Tailed t TestOne-Tailed t Test Thinking Challenge Thinking Challenge

You’re a marketing analyst for You’re a marketing analyst for Wal-Mart. Wal-Mart had teddy Wal-Mart. Wal-Mart had teddy bears on sale last week. The bears on sale last week. The weekly sales ($ 00) of bears weekly sales ($ 00) of bears sold in sold in 1010 stores was: stores was: 8 11 0 8 11 0 4 7 8 10 5 8 34 7 8 10 5 8 3. . At the At the .05.05 level, is there level, is there evidence that the average bear evidence that the average bear sales per store is sales per store is moremore thanthan 5 5 ($ 00)?($ 00)?

8 - 8 - 150150

One-Tailed t Test One-Tailed t Test Solution*Solution*

HH00: :

HHaa: :

df = df =

Decision:Decision:

8 - 8 - 151151

HH00: : = 5 = 5

HHaa: : > 5 > 5

df =df =

Decision:Decision:

8 - 8 - 152152

HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

df = df = 10 - 1 = 910 - 1 = 9

Decision:Decision:

8 - 8 - 153153

t0 1.8331

Reject

t0 1.8331

Reject

HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

df = df = 10 - 1 = 910 - 1 = 9

Decision:Decision:

8 - 8 - 154154

t0 1.8331

Reject

t0 1.8331

Reject

HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

df = df = 10 - 1 = 910 - 1 = 9

Decision:Decision:

6 4 53 373

8 - 8 - 155155

t0 1.8331

Reject

t0 1.8331

Reject

HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

df = df = 10 - 1 = 910 - 1 = 9

Decision:Decision:

6 4 53 373

8 - 8 - 156156

t0 1.8331

Reject

t0 1.8331

Reject

HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

df = df = 10 - 1 = 910 - 1 = 9

Decision:Decision:

6 4 53 373

There is no evidence There is no evidence average is more than 5average is more than 5

8 - 8 - 157157

Z Test of ProportionZ Test of Proportion

8 - 8 - 158158

Data TypesData Types

Numerical Qualitative

Discrete Continuous

Numerical Qualitative

Discrete Continuous

8 - 8 - 159159

Qualitative DataQualitative Data

1.1. Qualitative Random Variables Yield Qualitative Random Variables Yield Responses That ClassifyResponses That Classify e.g., Gender (Male, Female)e.g., Gender (Male, Female)

2.2. Measurement Reflects # in CategoryMeasurement Reflects # in Category

3.3. Nominal or Ordinal ScaleNominal or Ordinal Scale

4.4. ExamplesExamples Do You Own Savings Bonds? Do You Own Savings Bonds? Do You Live On-Campus or Off-Campus?Do You Live On-Campus or Off-Campus?

8 - 8 - 160160

ProportionsProportions

1.1. Involve Qualitative VariablesInvolve Qualitative Variables

2.2. Fraction or % of Population in a CategoryFraction or % of Population in a Category

3.3. If Two Qualitative Outcomes, Binomial If Two Qualitative Outcomes, Binomial DistributionDistribution Possess or Don’t Possess CharacteristicPossess or Don’t Possess Characteristic

8 - 8 - 161161

ProportionsProportions

1.1. Involve Qualitative VariablesInvolve Qualitative Variables

2.2. Fraction or % of Population in a CategoryFraction or % of Population in a Category

3.3. If Two Qualitative Outcomes, Binomial If Two Qualitative Outcomes, Binomial DistributionDistribution Possess or Don’t Possess CharacteristicPossess or Don’t Possess Characteristic

4.4. Sample Proportion (Sample Proportion (pp))

number of successes

sample sizep

number of successes

sample size

8 - 8 - 162162

1.1. Approximated by Approximated by Normal Normal

DistributionDistribution

Excludes 0 or nExcludes 0 or n

2.2. MeanMean

3.3. Standard ErrorStandard Error

Sampling Sampling Distribution Distribution of Proportionof Proportion

P p P p

where where pp00 = Population Proportion = Population Proportionpp̂̂pp

.0.0 .2.2 .4.4 .6.6 .8.8 1.01.0

P(PP(P^̂ )) ˆ1ˆ3ˆ ppnpn ˆ1ˆ3ˆ ppnpn

8 - 8 - 163163

Z Z = 0= 0

zz= 1= 1

Standardizing Standardizing Sampling Distribution Sampling Distribution

of Proportionof Proportion

Sampling Sampling DistributionDistribution

Standardized Standardized Normal DistributionNormal Distribution

PP^̂PP

ZZpp pp pp

(( ))11

8 - 8 - 164164

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)

2 Test(1 & 2tail)

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)

2 Test(1 & 2tail)

8 - 8 - 165165

One-Sample Z Test One-Sample Z Test for Proportionfor Proportion

8 - 8 - 166166

1.1. AssumptionsAssumptions Two Categorical OutcomesTwo Categorical Outcomes Population Follows Binomial DistributionPopulation Follows Binomial Distribution Normal Approximation Can Be UsedNormal Approximation Can Be Used Does Not Contain 0 or nDoes Not Contain 0 or n ˆ1ˆ3ˆ ppnpn ˆ1ˆ3ˆ ppnpn

8 - 8 - 167167

1.1. AssumptionsAssumptions Two Categorical OutcomesTwo Categorical Outcomes Population Follows Binomial DistributionPopulation Follows Binomial Distribution Normal Approximation Can Be UsedNormal Approximation Can Be Used Does Not Contain 0 or nDoes Not Contain 0 or n

2.2. Z-test statistic for proportionZ-test statistic for proportion

p pp p

0 01Hypothesized Hypothesized population proportionpopulation proportion

ˆ1ˆ3ˆ ppnpn ˆ1ˆ3ˆ ppnpn

8 - 8 - 168168

One-Proportion Z Test One-Proportion Z Test

Example Example The present packaging The present packaging system produces system produces 10%10% defective cereal boxes. defective cereal boxes. Using a new system, a Using a new system, a random sample of random sample of 200200 boxes hadboxes had1111 defects. defects. Does the new system Does the new system produce produce fewerfewer defects? defects? Test at the Test at the .05.05 level. level.

8 - 8 - 169169

One-Proportion Z One-Proportion Z Test SolutionTest Solution

HH00: :

HHaa: :

nn = =

Decision:Decision:

8 - 8 - 170170

HH00: : pp = .10 = .10

HHaa: : pp < .10 < .10

nn = =

Decision:Decision:

8 - 8 - 171171

HH00: : pp = =.10.10

HHaa: : pp < .10 < .10

= = .05.05

nn = = 200200

Decision:Decision:

8 - 8 - 172172

HH00: : pp = .10 = .10

HHaa: : pp < .10 < .10

= = .05.05

nn = = 200200

Decision:Decision:

Z0-1.645

Reject

Z0-1.645

Reject

8 - 8 - 173173

HH00: : pp = .10 = .10

HHaa: : pp < .10 < .10

= = .05.05

nn = = 200200

Decision:Decision:

Z0-1.645

Reject

Z0-1.645

Reject

. ( . ).0

10 1 10200

212Zp p

. ( . ).0

10 1 10200

8 - 8 - 174174

HH00: : pp = .10 = .10

HHaa: : pp < .10 < .10

= = .05.05

nn = = 200200

Decision:Decision:

Z0-1.645

Reject

Z0-1.645

Reject Reject at Reject at = .05 = .05

. ( . ).0

10 1 10200

212Zp p

. ( . ).0

10 1 10200

8 - 8 - 175175

HH00: : pp = .10 = .10

HHaa: : pp < .10 < .10

= = .05.05

nn = = 200200

Decision:Decision:

Z0-1.645

Reject

Z0-1.645

Reject Reject at Reject at = .05 = .05

There is evidence new There is evidence new system < 10% defectivesystem < 10% defective

. ( . ).0

10 1 10200

212Zp p

. ( . ).0

10 1 10200

8 - 8 - 176176

One-Proportion Z One-Proportion Z Test Thinking Test Thinking

ChallengeChallengeYou’re an accounting You’re an accounting manager. A year-end audit manager. A year-end audit showed showed 4%4% of transactions of transactions had errors. You implement had errors. You implement new procedures. A random new procedures. A random sample of sample of 500500 transactions transactions had had 2525 errors. Has the errors. Has the proportionproportion of incorrect of incorrect transactions transactions changedchanged at at the the .05.05 levellevel? ?

8 - 8 - 177177

One-Proportion Z One-Proportion Z Test Solution*Test Solution*

HH00: :

HHaa: :

nn = =

Decision:Decision:

8 - 8 - 178178

HH00: : pp = .04 = .04

HHaa: : pp .04 .04

nn = =

Decision:Decision:

8 - 8 - 179179

HH00: : pp = .04 = .04

HHaa: : pp .04 .04

= = .05.05

nn = = 500500

Decision:Decision:

8 - 8 - 180180

HH00: : pp = .04 = .04

HHaa: : pp .04 .04

= = .05.05

nn = = 500500

Decision:Decision:

Z0 1.96-1.96

8 - 8 - 181181

HH00: : pp = .04 = .04

HHaa: : pp .04 .04

= = .05.05

nn = = 500500

Decision:Decision:

Z0 1.96-1.96

. ( . ).0

04 1 04500

114Zp p

. ( . ).0

04 1 04500

8 - 8 - 182182

HH00: : pp = .04 = .04

HHaa: : pp .04 .04

= = .05.05

nn = = 500500

Decision:Decision:

Z0 1.96-1.96

. ( . ).0

04 1 04500

114Zp p

. ( . ).0

04 1 04500

8 - 8 - 183183

HH00: : pp = .04 = .04

HHaa: : pp .04 .04

= = .05.05

nn = = 500500

Decision:Decision:

Z0 1.96-1.96

There is evidence There is evidence proportion is still 4% proportion is still 4%

. ( . ).0

04 1 04500

114Zp p

. ( . ).0

04 1 04500

8 - 8 - 184184

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

8 - 8 - 185185

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)

2 Test(1 & 2tail)

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)

2 Test(1 & 2tail)

8 - 8 - 186186

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

8 - 8 - 187187

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

(n(n SS

8 - 8 - 188188

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

8 - 8 - 189189

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?

8 - 8 - 190190

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

22 Table Table (Portion)(Portion)

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

8 - 8 - 191191

Reject

= .05= .05

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

8 - 8 - 192192

Reject

= .05= .05

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

8 - 8 - 193193

Reject

= .05= .05

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

8 - 8 - 194194

Reject

= .05= .05

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

8 - 8 - 195195

Reject

= .05= .05

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

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

8 - 8 - 196196

Reject

= .05= .05

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

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

8 - 8 - 197197

Reject

= .05= .05

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

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

8 - 8 - 198198

Reject

= .05= .05

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

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

8 - 8 - 199199

20 5.991

Reject

20 5.991

Reject

= .05= .05

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

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

8 - 8 - 200200

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?

8 - 8 - 201201

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

8 - 8 - 202202

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

= .05= .05

RejectReject

8 - 8 - 203203

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 - 204204

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

= .05= .05

8 - 8 - 205205

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

= .05= .05

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

8 - 8 - 206206

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

= .05= .05

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

8 - 8 - 207207

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

8 - 8 - 208208

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.

8 - 8 - 209209

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

HH00: :

HHaa: :

df = df =

Decision:Decision:

8 - 8 - 210210

HH00: : 22 = 15 = 15

HHaa: : 22 15 15

df = df =

Decision:Decision:

8 - 8 - 211211

HH00: : 22 = 15 = 15

HHaa: : 22 15 15

= = .05.05

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

Decision:Decision:

8 - 8 - 212212

HH00: : 22 = 15 = 15

HHaa: : 22 15 15

= = .05.05

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

Decision:Decision:

/2 = .025/2 = .025

8 - 8 - 213213

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

8 - 8 - 214214

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

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

3333 4242

(n(n SS

8 - 8 - 215215

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

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

3333 4242

(n(n SS

8 - 8 - 216216

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

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

3333 4242

(n(n SS

8 - 8 - 217217

Calculating Type II Calculating Type II Error ProbabilitiesError Probabilities

8 - 8 - 218218

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

8 - 8 - 219219

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

8 - 8 - 220220

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

SpecifySpecify

8 - 8 - 221221

XX11 = 360= 360 363.065363.065

XX00 = 368= 368

HH11: : 00 < 368 < 368

065.363

1564.13680

065.363

1564.13680

= .05= .05

n =n =15/15/2525

DrawDraw

SpecifySpecify

8 - 8 - 222222

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

SpecifySpecify

Z TableZ Table

065.363

1564.13680

065.363

1564.13680

8 - 8 - 223223

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

8 - 8 - 224224

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

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