hypos testing

2003 Pearson Prentice Hall 2003 Pearson Prentice Hall

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

8 8 -- 11


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 pExplain p--Value ConceptValue Concept

8 8 -- 22

3.3. Explain pExplain p--Value ConceptValue 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


Statistical MethodsStatistical Methods

StatisticalMethods

8 8 -- 33

DescriptiveStatistics

InferentialStatistics

EstimationHypothesis

Testing


Hypothesis Testing ConceptsHypothesis Testing Concepts

8 8 -- 44


Hypothesis TestingHypothesis Testing

8 8 -- 55



PopulationPopulation

8 8 -- 66




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

8 8 -- 77





8 8 -- 88

MeanMeanX X = 20= 20

Random Random samplesample





Reject hypothesis! Not close.

8 8 -- 99

MeanMeanX X = 20= 20

Random Random samplesample


Whats a Hypothesis?Whats a Hypothesis?

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

Parameter Is Parameter Is

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

8 8 -- 1010

Parameter Is Parameter Is PopulationPopulation Mean, Mean, Proportion, VarianceProportion, Variance

Must Be StatedMust Be StatedBeforeBefore AnalysisAnalysis

1984-1994 T/Maker Co.


Null HypothesisNull Hypothesis

1.1. What Is TestedWhat Is Tested

2.2. Has Serious Outcome If Incorrect Has Serious Outcome If Incorrect Decision MadeDecision Made

8 8 -- 1111

3.3. Designated HDesignated H00 (Pronounced H(Pronounced H--noughtnought))

4.4. Specified as HSpecified as H00: : Some Numeric Value Some Numeric Value Specified with = Sign Specified with = Sign , or , or Example, HExample, H00: : 33


Alternative HypothesisAlternative Hypothesis

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 H

8 8 -- 1212

3.3. Designated HDesignated Haa4.4. Specified HSpecified Haa: : < Some Value< Some Value

Example, HExample, Haa: : < 3< 3 will lead towill lead to twotwo--sided testssided tests will lead to one will lead to one--sided testssided tests


Identifying HypothesesIdentifying HypothesesStepsSteps

1.1. Example Problem: Test That the Example Problem: Test That the Population Mean Is Not 3Population Mean Is Not 3

2.2. StepsSteps

8 8 -- 1313

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 , , SignSign

State the Null Hypothesis (State the Null Hypothesis ( = 3)= 3)


State the question statistically: State the question statistically: = 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 Hypotheses?What Are the Hypotheses?

8 8 -- 1414

State the opposite statistically: State the opposite statistically: 12 12 Select the alternative hypothesis: Select the alternative hypothesis: HHaa: : 1212State the null hypothesis: State the null hypothesis: HH00: : = 12= 12


State the question statistically: State the question statistically: 1212

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


8 8 -- 1515

State the opposite statistically: State the opposite statistically: = 12= 12Select the alternative hypothesis: Select the alternative hypothesis: HHaa: : 1212State the null hypothesis: State the null hypothesis: HH00: : = 12= 12



Is the average cost per hat less than or Is the average cost per hat less than or equal to P20?equal to P20?


8 8 -- 1616

State the opposite statistically: State the opposite statistically: 2020Select the alternative hypothesis: Select the alternative hypothesis: HHaa: : 2020State the null hypothesis: State the null hypothesis: HH00: : 2020



Is the average amount spent in the Is the average amount spent in the bookstore greater than P25?bookstore greater than P25?


8 8 -- 1717

State the opposite statistically: State the opposite statistically: 25 25 Select the alternative hypothesis: Select the alternative hypothesis: HHaa: : 2525State the null hypothesis: State the null hypothesis: HH00: : 2525


Basic IdeaBasic Idea

8 8 -- 1818



Sampling DistributionSampling Distribution

8 8 -- 1919

Sample Mean = 50HH00HH00




It is unlikely It is unlikely that we would that we would get a sample get a sample

8 8 -- 2020

Sample Mean = 50

mean of this mean of this value ...value ...

20202020HH00HH00





8 8 -- 2121

Sample Mean = 50


... if in fact this were... if in fact this werethe population meanthe population mean

20202020HH00HH00





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

8 8 -- 2222

Sample Mean = 50


... if in fact this were... if in fact this werethe population meanthe population mean

hypothesis hypothesis that that = 50.= 50.

20202020HH00HH00


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

8 8 -- 2323

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


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

8 8 -- 2424



RejectionRegion

Sampling DistributionSampling DistributionLevel of ConfidenceLevel of Confidence

8 8 -- 2525

HoValueCritical

Value

Sample Statistic

NonrejectionRegion

1 1 --



RejectionRegion


8 8 -- 2626

HoValueCritical

Value

Sample Statistic

NonrejectionRegion

1 1 --

Observed sample statisticObserved sample statistic



RejectionRegion


8 8 -- 2727

HoValueCritical

Value

Sample Statistic

NonrejectionRegion

1 1 --


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

8 8 -- 2828



RejectionRegion

RejectionRegion


8 8 -- 2929

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

NonrejectionRegion

1 1 --



RejectionRegion

RejectionRegion


8 8 -- 3030

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

NonrejectionRegion

1 1 --

Observed sample statisticObserved sample statistic



RejectionRegion

RejectionRegion


8 8 -- 3131

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

NonrejectionRegion

1 1 --



RejectionRegion

RejectionRegion


8 8 -- 3232

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

NonrejectionRegion

1 1 --


Hypothesis Testing StepsHypothesis Testing Steps

8 8 -- 3333


HH00 Testing StepsTesting Steps

8 8 -- 3434



State HState H00

State HState Haa

Choose Choose

8 8 -- 3535

Choose Choose

Choose Choose nn

Choose testChoose test



Set up critical valuesSet up critical values

Collect dataCollect data

Compute test statisticCompute test statistic

State HState H00

State HState Haa

Choose Choose

8 8 -- 3636

Compute test statisticCompute test statistic

Make statistical decisionMake statistical decision

Express decisionExpress decision

Choose Choose

Choose Choose nn

Choose testChoose test


One Population TestsOne Population Tests

OnePopulation

8 8 -- 3737

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)

Mean Proportion Variance

2 Test(1 & 2tail)


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

8 8 -- 3838

of Mean (of Mean ( Known)Known)



OnePopulation

8 8 -- 3939

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)


TwoTwo--Tailed Z Test 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)

8 8 -- 4040

Normal Distribution (Normal Distribution (nn 30)30)2.2. Alternative Hypothesis Has Alternative Hypothesis Has SignSign


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



8 8 -- 4141

Normal Distribution (Normal Distribution (nn 30)30)2.2. Alternative Hypothesis Has Alternative Hypothesis Has SignSign3.3. ZZ--Test StatisticTest Statistic

ZX X

n

x

x


TwoTwo--Tailed Z TestTailed Z TestExample 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

8 8 -- 4242

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.


TwoTwo--Tailed Z Test Tailed Z Test SolutionSolution

HH00: :

HHaa: :

nn

Test Statistic: Test Statistic:

8 8 -- 4343

nn Critical Value(s):Critical Value(s): Decision:Decision:

Conclusion:Conclusion:



HH00: : = 368= 368HHaa: : 368368 nn


8 8 -- 4444

nn Critical Value(s):Critical Value(s): Decision:Decision:




HH00: : = 368= 368HHaa: : 368368 .05.05nn 2525


8 8 -- 4545

nn 2525Critical Value(s):Critical Value(s): Decision:Decision:




HH00: : = 368= 368HHaa: : 368368 .05.05nn 2525


8 8 -- 4646



Z0 1.96-1.96

.025

Reject H0 Reject H0

.025



HH00: : = 368= 368HHaa: : 368368 .05.05nn 2525


ZX

n

372 5 368

1525

150.

.

8 8 -- 4747



Z0 1.96-1.96

.025

Reject H0 Reject H0

.025

n 25



HH00: : = 368= 368HHaa: : 368368 .05.05nn 2525


ZX

n

372 5 368

1525

150.

.

8 8 -- 4848



Z0 1.96-1.96

.025

Reject H0 Reject H0

.025

n 25

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



HH00: : = 368= 368HHaa: : 368368 .05.05nn 2525


ZX

n

372 5 368

1525

150.

.

8 8 -- 4949



Z0 1.96-1.96

.025

Reject H0 Reject H0

.025

n 25


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


TwoTwo--Tailed Z Test Tailed Z Test Thinking ChallengeThinking Challenge

Youre a Q/C inspector. You want to Youre 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

8 8 -- 5050

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?


TwoTwo--Tailed Z Test Tailed Z Test Solution*Solution*

HH00: :

HHaa: :

= = nn = =


8 8 -- 5151

nn = = Critical Value(s):Critical Value(s):

Decision:Decision:




HH00: : = 70= 70HHaa: : 7070 = = nn ==


8 8 -- 5252

nn ==Critical Value(s):Critical Value(s):

Decision:Decision:




HH00: : = 70= 70HHaa: : 7070 = = .05.05nn = = 3636


8 8 -- 5353

nn = = 3636Critical Value(s):Critical Value(s):

Decision:Decision:




HH00: : = 70= 70HHaa: : 7070 = = .05.05nn = = 3636


8 8 -- 5454


Decision:Decision:


Z0 1.96-1.96

.025

Reject H0 Reject H0

.025



HH00: : = 70= 70HHaa: : 7070 = = .05.05nn = = 3636


ZX

n

69 7 70

3 536

51.

..

8 8 -- 5555


Decision:Decision:


Z0 1.96-1.96

.025

Reject H0 Reject H0

.025

n 36



HH00: : = 70= 70HHaa: : 7070 = = .05.05nn = = 3636


ZX

n

69 7 70

3 536

51.

..

8 8 -- 5656


Decision:Decision:


Z0 1.96-1.96

.025

Reject H0 Reject H0

.025

n 36




HH00: : = 70= 70HHaa: : 7070 = = .05.05nn = = 3636


ZX

n

69 7 70

3 536

51.

..

8 8 -- 5757


Decision:Decision:


Z0 1.96-1.96

.025

Reject H0 Reject H0

.025

n 36


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


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

8 8 -- 5858

of Mean (of Mean ( Known)Known)


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



8 8 -- 5959

Normal Distribution (Normal Distribution (nn 30)30)2.2. Alternative Hypothesis Has < or > SignAlternative Hypothesis Has < or > Sign


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



8 8 -- 6060

Normal Distribution (Normal Distribution (nn 30)30)2.2. Alternative Hypothesis Has Alternative Hypothesis Has or > Signor > Sign3.3. ZZ--test Statistictest Statistic

ZX X

n

x

x


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

8 8 -- 6161


Reject H0


HH00::==0 H0 Haa: :


Reject H0Reject H0


HH00::==0 H0 Haa: : > 00

8 8 -- 6363

Z0

Z0

Must be Must be significantlysignificantlybelow below

Small values satisfy Small values satisfy HH0 0 . Dont reject!. Dont reject!


OneOne--Tailed Z Test Tailed Z Test Finding Critical ZFinding Critical Z

8 8 -- 6464


= 1


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

8 8 -- 6565

Z0

= .025= .025


= 1


.500 .500 -- .025.025


8 8 -- 6666

Z0

-- .025.025.475.475

= .025= .025


Z .05 .07 = 1


.500 .500 -- .025.025 .06

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


8 8 -- 6767

1.6 .4505 .4515 .4525

1.7 .4599 .4608 .4616

1.8 .4678 .4686 .4693

.4744 .4756

Z0

-- .025.025.475.475

1.9 .4750.4750

= .025= .025


Z .05 .07 = 1


.500 .500 -- .025.025 .06.06

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


8 8 -- 6868

1.6 .4505 .4515 .4525

1.7 .4599 .4608 .4616

1.8 .4678 .4686 .4693

.4744 .4756

Z0 1.96

-- .025.025.475.475

1.91.9 .4750

= .025= .025


OneOne--Tailed Z TestTailed Z TestExample Example

Does an average box of Does an average box of cereal contain cereal contain more thanmore than368368 grams of cereal? A grams of cereal? A random sample of random sample of 2525

8 8 -- 6969

random sample of random sample of 2525boxes showedboxes showedX = 372.5X = 372.5. . The company has The company has specified specified to be to be 2525grams. Test at the grams. Test at the .05.05level.level.

368 gm.368 gm.


OneOne--Tailed Z Test Tailed Z Test SolutionSolution

HH00: :

HHaa: :

= = n n = =


8 8 -- 7070

n n = =

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




HH00: : = 368= 368HHaa: : > 368> 368 = = n n = =


8 8 -- 7171

n n = =





HH00: : = 368= 368HHaa: : > 368> 368 = = .05.05n n = = 2525


8 8 -- 7272

n n = = 2525





HH00: : = 368= 368HHaa: : > 368> 368 = = .05.05n n = = 2525


8 8 -- 7373

n n = = 2525



Z0 1.645

.05

Reject



HH00: : = 368= 368HHaa: : > 368> 368 = = .05.05n n = = 2525


ZX

n

372 5 368

1525

150.

.

8 8 -- 7474

n n = = 2525



Z0 1.645

.05

Reject

n 25



HH00: : = 368= 368HHaa: : > 368> 368 = = .05.05n n = = 2525


ZX

n

372 5 368

1525

150.

.

8 8 -- 7575

n n = = 2525



Z0 1.645

.05

Reject

n 25




HH00: : = 368= 368HHaa: : > 368> 368 = = .05.05n n = = 2525


ZX

n

372 5 368

1525

150.

.

8 8 -- 7676

n n = = 2525



Z0 1.645

.05

Reject

n 25


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


OneOne--Tailed Z Test Tailed Z Test Thinking ChallengeThinking Challenge

Youre an analyst for Ford. You Youre 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

8 8 -- 7777

least 32 mpg. Similar models least 32 mpg. Similar models have a standard deviation of have a standard deviation of 3.83.8mpg. You take a sample of mpg. You take a sample of 6060Escorts & compute a sample Escorts & compute a sample mean of mean of 30.730.7 mpg. At the mpg. At the .01.01level, is there evidence that the level, is there evidence that the miles per gallon is miles per gallon is at leastat least 3232??


OneOne--Tailed Z Test Tailed Z Test Solution*Solution*

HH00: :

HHaa: :

= = nn ==

Test Statistic: Test Statistic: Test Statistic: Test Statistic:

8 8 -- 7878

nn ==



Decision:Decision:




HH00: : = 32= 32HHaa: : < 32< 32 = = nn ==


8 8 -- 7979

nn ==



Decision:Decision:




HH00: : = 32= 32HHaa: : < 32< 32 == .01.01nn = = 6060


8 8 -- 8080

nn = = 6060



Decision:Decision:




HH00: : = 32= 32HHaa: : < 32< 32 = .01= .01nn = 60= 60


8 8 -- 8181

nn = 60= 60



Decision:Decision:


Z0-2.33

.01

Reject



HH00: : = 32= 32HHaa: : < 32< 32 = .01= .01nn = 60= 60


ZX

n

30 7 32

3 860

2 65.

..

8 8 -- 8282

nn = 60= 60



Decision:Decision:


Z0-2.33

.01

Reject

n 60



HH00: : = 32= 32HHaa: : < 32< 32 = .01= .01nn = 60= 60


ZX

n

30 7 32

3 860

2 65.

..

8 8 -- 8383

nn = 60= 60



Decision:Decision:


Z0-2.33

.01

Reject

n 60

Reject at Reject at = .01= .01



HH00: : = 32= 32HHaa: : < 32< 32 = .01= .01nn = 60= 60


ZX

n

30 7 32

3 860

2 65.

..

8 8 -- 8484

nn = 60= 60



Decision:Decision:


Z0-2.33

.01

Reject

n 60


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


Decision Making RisksDecision Making Risks

8 8 -- 8585


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)

8 8 -- 8686

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)


Jury Trial H0 Test

Actual Situation Actual Situation

Verdict Innocent Guilty Decision H0 True H0

Decision ResultsDecision Results

HH00: Innocent: Innocent

8 8 -- 8787

Verdict Innocent Guilty Decision H0 True H0False

Innocent Correct ErrorDo NotReject

H0

1 - Type IIError

()

Guilty Error Correct RejectH0

Type IError () Power(1 - )


Jury Trial H0 Test

Actual Situation Actual Situation

Verdict Innocent Guilty Decision H0 True H0

Decision ResultsDecision Results

HH00: Innocent: Innocent

8 8 -- 8888

Verdict Innocent Guilty Decision H0 True H0False

Innocent Correct Error AcceptH0

1 - Type IIError

()

Guilty Error Correct RejectH0

Type IError () Power(1 - )


& & Have an Have an Inverse RelationshipInverse Relationship

You cant reduce both errors simultaneously!

8 8 -- 8989


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,

8 8 -- 9090

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 DecreasesDecreases


Exercise 8.15Exercise 8.15

1000 subjects1000 subjects

500 told truth, 500 lied500 told truth, 500 lied

Lie detector says Lie detector says

8 8 -- 9191

185 truth tellers were liars185 truth tellers were liars

120 liars were truth tellers120 liars were truth tellers

Ho: truth tellerHo: truth teller

a)a) What is a typeWhat is a type--I error? TypeI error? Type--II error?II error?

b)b) What is Pr(typeWhat is Pr(type--I error)? Pr(typeI error)? Pr(type--II error)?II error)?


Observed Significance Observed Significance Levels: pLevels: p--ValuesValues

8 8 -- 9292

Levels: pLevels: p--ValuesValues


pp--ValueValue

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

8 8 -- 9393

2.2. Called Observed Level of SignificanceCalled Observed Level of Significance Smallest Value of Smallest Value of HH00 Can Be RejectedCan Be Rejected

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


TwoTwo--Tailed Z Test Tailed Z Test pp--Value Example 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

8 8 -- 9494

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 pp--Value.Value. 368 gm.368 gm.


TwoTwo--Tailed Z Test Tailed Z Test pp--Value SolutionValue Solution

8 8 -- 9595



ZX

n

372 5 368

1525

150.

.

8 8 -- 9696

Z0 1.50-1.50

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



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

8 8 -- 9797

Z0 1.50-1.50




1/2 p-Value1/2 p-Value


8 8 -- 9898

Z0 1.50-1.50




1/2 p-Value1/2 p-Value


8 8 -- 9999

Z0 1.50-1.50


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

.4332.4332



1/2 p-Value1/2 p-Value.5000.5000


8 8 -- 100100

Z0 1.50-1.50



.4332.4332

.5000.5000-- .4332.4332

.0668.0668



1/2 p-Value.0668

1/2 p-Value.0668

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

.5000.5000

8 8 -- 101101

Z0 1.50-1.50

.0668.0668

Z value of sample Z value of sample statisticstatistic


.4332.4332

.5000.5000-- .4332.4332

.0668.0668



RejectReject

1/2 p1/2 p--Value = .0668Value = .06681/2 p1/2 p--Value = .0668Value = .0668

8 8 -- 102102

0 1.50-1.50 Z

RejectReject

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



RejectReject

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

1/2 p1/2 p--Value = .0668Value = .06681/2 p1/2 p--Value = .0668Value = .0668

8 8 -- 103103

0 1.50-1.50 Z

RejectReject

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


OneOne--Tailed Z Test Tailed Z Test pp--Value Example Value Example

Does an average box of Does an average box of cereal contain cereal contain more thanmore than368368 grams of cereal? A grams of cereal? A random sample of random sample of 2525

8 8 -- 104104

random sample of random sample of 2525boxes showedboxes showedX = 372.5X = 372.5. . The company has The company has specified specified to be to be 2525grams. Find the pgrams. Find the p--Value.Value. 368 gm.368 gm.


OneOne--Tailed Z Test Tailed Z Test pp--Value SolutionValue Solution

8 8 -- 105105



ZX

n

372 5 368

1525

150.

.

8 8 -- 106106

Z0 1.50




p-ValueUse Use

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

8 8 -- 107107

Z0 1.50

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




p-ValueUse Use


8 8 -- 108108

Z0 1.50




.4332.4332



p-ValueUse Use


8 8 -- 109109

Z0 1.50




.4332.4332

.5000.5000-- .4332.4332

.0668.0668



p-Value.0668Use Use

pp--Value is P(Z Value is P(Z 1.50) = .06681.50) = .0668

8 8 -- 110110

Z0 1.50

.0668



.4332.4332

Use Use alternative alternative hypothesis hypothesis to find to find directiondirection

.5000.5000-- .4332.4332

.0668.0668



Reject

pp--Value = .0668Value = .0668

8 8 -- 111111

0 1.50 Z

Reject

= .05= .05



Reject

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

pp--Value = .0668Value = .0668

8 8 -- 112112

0 1.50 Z

Reject

= .05= .05

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


pp--Value Value Thinking ChallengeThinking Challenge

Youre an analyst for Ford. You Youre 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

8 8 -- 113113

least 32 least 32 mpg. Similar models mpg. Similar models have a standard deviation of have a standard deviation of 3.83.8mpg. You take a sample of mpg. You take a sample of 6060Escorts & 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 (pp--ValueValue)?)?


pp--Value Value Solution*Solution*

p-Value.004Use Use .5000.5000

pp--Value is P(Z Value is P(Z --2.65) = .004.2.65) = .004.pp--Value < (Value < ( = .01). Reject H= .01). Reject H00..

8 8 -- 114114

Z0-2.65

.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


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

8 8 -- 115115

of Mean (of Mean ( Unknown)Unknown)



OnePopulation

8 8 -- 116116

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)


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


If Not Normal, Only Slightly Skewed & If Not Normal, Only Slightly Skewed & Large Sample (Large Sample (nn 30) Taken30) Taken

8 8 -- 117117

Large Sample (Large Sample (nn 30) Taken30) Taken2.2. Parametric Test ProcedureParametric Test Procedure


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


If Not Normal, Only Slightly Skewed & If Not Normal, Only Slightly Skewed & Large Sample (Large Sample (nn 30) Taken30) Taken

8 8 -- 118118

Large Sample (Large Sample (nn 30) Taken30) Taken2.2. Parametric Test ProcedureParametric Test Procedure

3.3. t Test Statistict Test Statistic

tX

Sn


TwoTwo--Tailed t TestTailed t TestFinding Critical t ValuesFinding Critical t Values

8 8 -- 119119



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

8 8 -- 120120

t0



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

8 8 -- 121121

t0

/2 = .05/2 = .05

/2 = .05/2 = .05



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

df = n df = n -- 1 = 21 = 2

8 8 -- 122122

t0

/2 = .05/2 = .05

/2 = .05/2 = .05


v t.10 t.05 t.025


Critical Values of t Table Critical Values of t Table (Portion)(Portion)

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

df = n df = n -- 1 = 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 3.182t0

/2 = /2 = .05.05

/2 = .05/2 = .05


v t.10 t.05 t.025


Critical Values of t Table Critical Values of t Table (Portion)(Portion)

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

df = n df = n -- 1 = 21 = 2

8 8 -- 124124

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

/2 = .05/2 = .05

/2 = .05/2 = .05


TwoTwo--Tailed t TestTailed t TestExample Example

Does an average box of Does an average box of cereal contain cereal contain 368368grams of cereal? A grams of cereal? A random sample of random sample of 3636

8 8 -- 125125

random sample of random sample of 3636boxes 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.


TwoTwo--Tailed t Test Tailed t Test SolutionSolution

HH00: :

HHaa: :

= = df = df =


8 8 -- 126126

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

Decision:Decision:




HH00: : = 368= 368HHaa: : 368368 = = df = df =


8 8 -- 127127

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

Decision:Decision:




HH00: : = 368= 368HHaa: : 368368 = = .05.05df = df = 36 36 -- 1 = 351 = 35


8 8 -- 128128

df = df = 36 36 -- 1 = 351 = 35Critical Value(s):Critical Value(s):

Decision:Decision:




HH00: : = 368= 368HHaa: : 368368 = = .05.05df = df = 36 36 -- 1 = 351 = 35


8 8 -- 129129


Decision:Decision:


t0 2.0301-2.0301

.025

Reject H0 Reject H0

.025



HH00: : = 368= 368HHaa: : 368368 = = .05.05df = df = 36 36 -- 1 = 351 = 35


tX

Sn

372 5 3681236

2 25.

.

8 8 -- 130130


Decision:Decision:


t0 2.0301-2.0301

.025

Reject H0 Reject H0

.025

n 36



HH00: : = 368= 368HHaa: : 368368 = = .05.05df = df = 36 36 -- 1 = 351 = 35


tX

Sn

372 5 3681236

2 25.

.

8 8 -- 131131


Decision:Decision:


t0 2.0301-2.0301

.025

Reject H0 Reject H0

.025

n 36




HH00: : = 368= 368HHaa: : 368368 = = .05.05df = df = 36 36 -- 1 = 351 = 35


tX

Sn

372 5 3681236

2 25.

.

8 8 -- 132132


Decision:Decision:


t0 2.0301-2.0301

.025

Reject H0 Reject H0

.025

n 36


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


TwoTwo--Tailed t TestTailed 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

8 8 -- 133133

of detergent is of detergent is 3.253.25 lb. You lb. You take a random sample of take a random sample of 6464containers. You calculate the containers. You calculate the sample average to be sample average to be 3.2383.238lb. 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.


TwoTwo--Tailed t Test Tailed t Test Solution*Solution*

HH00: :

HHaa: :

df df


8 8 -- 134134

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

Decision:Decision:




HH00: : = 3.25= 3.25HHaa: : 3.253.25 df df


8 8 -- 135135

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

Decision:Decision:




HH00: : = 3.25= 3.25HHaa: : 3.253.25 .01.01df df 64 64 -- 1 = 631 = 63


8 8 -- 136136

df df 64 64 -- 1 = 631 = 63Critical Value(s):Critical Value(s):

Decision:Decision:




HH00: : = 3.25= 3.25HHaa: : 3.253.25 .01.01df df 64 64 -- 1 = 631 = 63


8 8 -- 137137


Decision:Decision:


t0 2.6561-2.6561

.005

Reject H0 Reject H0

.005



HH00: : = 3.25= 3.25HHaa: : 3.253.25 .01.01df df 64 64 -- 1 = 631 = 63


tX

Sn

3 238 3 2511764

82. .

..

8 8 -- 138138


Decision:Decision:


t0 2.6561-2.6561

.005

Reject H0 Reject H0

.005

n 64



HH00: : = 3.25= 3.25HHaa: : 3.253.25 .01.01df df 64 64 -- 1 = 631 = 63


tX

Sn

3 238 3 2511764

82. .

..

8 8 -- 139139


Decision:Decision:


t0 2.6561-2.6561

.005

Reject H0 Reject H0

.005

n 64




HH00: : = 3.25= 3.25HHaa: : 3.253.25 .01.01df df 64 64 -- 1 = 631 = 63


tX

Sn

3 238 3 2511764

82. .

..

8 8 -- 140140


Decision:Decision:


t0 2.6561-2.6561

.005

Reject H0 Reject H0

.005

n 64


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


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

8 8 -- 141141

of Mean (of Mean ( Unknown)Unknown)


OneOne--Tailed t TestTailed t TestExample Example

Is the average capacity of Is the average capacity of batteries batteries at least 140 at least 140 ampereampere--hours? A random hours? A random sample of sample of 2020 batteries had batteries had

8 8 -- 142142

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 .05.05level.level.


OneOne--Tailed t Test Tailed t Test SolutionSolution

HH00: :

HHaa: :

==df =df =


8 8 -- 143143

df =df =





HH00: : = 140= 140HHaa: : < 140< 140 = = df = df =


8 8 -- 144144

df = df =





HH00: : = 140= 140HHaa: : < 140< 140 = = .05.05df = df = 20 20 -- 1 = 191 = 19


8 8 -- 145145

df = df = 20 20 -- 1 = 191 = 19





HH00: : = 140= 140HHaa: : < 140< 140 = = .05.05df = df = 20 20 -- 1 = 191 = 19


8 8 -- 146146

t0-1.7291

.05

Reject

df = df = 20 20 -- 1 = 191 = 19





HH00: : = 140= 140HHaa: : < 140< 140 = = .05.05df = df = 20 20 -- 1 = 191 = 19


tX

Sn

138 47 1402 66

20

2 57.

..

8 8 -- 147147

t0-1.7291

.05

Reject

df = df = 20 20 -- 1 = 191 = 19



n 20



HH00: : = 140= 140HHaa: : < 140< 140 = = .05.05df = df = 20 20 -- 1 = 191 = 19


tX

Sn

138 47 1402 66

20

2 57.

..

8 8 -- 148148

t0-1.7291

.05

Reject

df = df = 20 20 -- 1 = 191 = 19



n 20




HH00: : = 140= 140HHaa: : < 140< 140 = = .05.05df = df = 20 20 -- 1 = 191 = 19


tX

Sn

138 47 1402 66

20

2 57.

..

8 8 -- 149149

t0-1.7291

.05

Reject

df = df = 20 20 -- 1 = 191 = 19



n 20


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


OneOne--Tailed t TestTailed t TestThinking ChallengeThinking Challenge

Youre a marketing analyst for Youre a marketing analyst for WalWal--Mart. WalMart. Wal--Mart had teddy 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

8 8 -- 150150

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)?


OneOne--Tailed t Test Tailed t Test Solution*Solution*

HH00: :

HHaa: :

= = df = df =


8 8 -- 151151

df = df =





HH00: : = 5= 5HHaa: : > 5> 5 = = df =df =


8 8 -- 152152

df =df =





HH00: : = 5= 5HHaa: : > 5> 5 = = .05.05df = df = 10 10 -- 1 = 91 = 9


8 8 -- 153153

df = df = 10 10 -- 1 = 91 = 9





HH00: : = 5= 5HHaa: : > 5> 5 = = .05.05df = df = 10 10 -- 1 = 91 = 9


8 8 -- 154154

t0 1.8331

.05

Reject

df = df = 10 10 -- 1 = 91 = 9





HH00: : = 5= 5HHaa: : > 5> 5 = = .05.05df = df = 10 10 -- 1 = 91 = 9


tX

Sn

6 4 53 373

10

131..

.

8 8 -- 155155

t0 1.8331

.05

Reject

df = df = 10 10 -- 1 = 91 = 9



n 10



HH00: : = 5= 5HHaa: : > 5> 5 = = .05.05df = df = 10 10 -- 1 = 91 = 9


tX

Sn

6 4 53 373

10

131..

.

8 8 -- 156156

t0 1.8331

.05

Reject

df = df = 10 10 -- 1 = 91 = 9



n 10




HH00: : = 5= 5HHaa: : > 5> 5 = = .05.05df = df = 10 10 -- 1 = 91 = 9


tX

Sn

6 4 53 373

10

131..

.

8 8 -- 157157

t0 1.8331

.05

Reject

df = df = 10 10 -- 1 = 91 = 9



n 10


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


Z Test of ProportionZ Test of Proportion

8 8 -- 158158


Data TypesData Types

Data

8 8 -- 159159

Numerical Qualitative

Discrete Continuous


1.1. Approximated by Approximated by Normal DistributionNormal Distribution

Excludes 0 or nExcludes 0 or n

Sampling Distribution Sampling Distribution of Proportionof Proportion


.2.2

.3.3P(PP(P^^ )) 13 ppnpn

8 8 -- 160160

pp

Excludes 0 or nExcludes 0 or n

2.2. MeanMean

3.3. Standard ErrorStandard Error

P p

where where pp00 = Population Proportion= Population Proportionpp^^pp

nn 11

.0.0

.1.1

.2.2

.0.0 .2.2 .4.4 .6.6 .8.8 1.01.0

PP^^

00 00


Standardizing Sampling Standardizing Sampling Distribution of ProportionDistribution of Proportion

Sampling Sampling DistributionDistribution

Standardized Standardized Normal DistributionNormal Distribution

ZZpp pp pp

pp pp

nn

^^pp

pp

^^

^^

(( ))11

^^00

00 00

8 8 -- 161161

Z Z = 0= 0

zz= 1= 1

ZZ

DistributionDistribution Normal DistributionNormal Distribution

PP^^PP

PP

^^

^^



OnePopulation

8 8 -- 162162

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)


OneOne--Sample Z Test Sample Z Test for Proportionfor Proportion

8 8 -- 163163



1.1. AssumptionsAssumptions Two Categorical OutcomesTwo Categorical Outcomes

Population Follows Binomial DistributionPopulation Follows Binomial Distribution

8 8 -- 164164

Normal Approximation Can Be UsedNormal Approximation Can Be Used

Does Not Contain 0 or nDoes Not Contain 0 or n 13 ppnpn



1.1. AssumptionsAssumptions Two Categorical OutcomesTwo Categorical Outcomes

Population Follows Binomial DistributionPopulation Follows Binomial Distribution

8 8 -- 165165

Normal Approximation Can Be UsedNormal Approximation Can Be Used

Does Not Contain 0 or nDoes Not Contain 0 or n

2.2. ZZ--test statistic for proportiontest statistic for proportion

Zp p

p pn

( )0

0 01Hypothesized Hypothesized population proportionpopulation proportion

13 ppnpn


OneOne--Proportion Z Test 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

8 8 -- 166166

Using a new system, a Using a new system, a random sample of random sample of 200200boxes 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.


OneOne--Proportion Z Test Proportion Z Test SolutionSolution

HH00: :

HHaa: :

= = nn ==


8 8 -- 167167

nn ==





HH00: : pp = .10= .10

HHaa: : pp < .10< .10

= = nn = =


8 8 -- 168168

nn = =





HH00: : pp ==.10.10HHaa: : pp < .10< .10

= = .05.05nn = = 200200


8 8 -- 169169

nn = = 200200





HH00: : pp = .10= .10

HHaa: : pp < .10< .10

= = .05.05nn = = 200200


8 8 -- 170170

nn = = 200200



Z0-1.645

.05

Reject



HH00: : pp = .10= .10

HHaa: : pp < .10< .10

= = .05.05nn = = 200200


Zp p

p p

( )

.

. ( . ).0

0 01

11200

10

10 1 10212

8 8 -- 171171

nn = = 200200



Z0-1.645

.05

Reject

n0 0

200



HH00: : pp = .10= .10

HHaa: : pp < .10< .10

= = .05.05nn = = 200200


Zp p

p p

( )

.

. ( . ).0

0 01

11200

10

10 1 10212

8 8 -- 172172

nn = = 200200



Z0-1.645

.05

Reject Reject at Reject at = .05= .05

n0 0

200



HH00: : pp = .10= .10

HHaa: : pp < .10< .10

= = .05.05nn = = 200200


Zp p

p p

( )

.

. ( . ).0

0 01

11200

10

10 1 10212

8 8 -- 173173

nn = = 200200



Z0-1.645

.05

Reject Reject at Reject at = .05= .05

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

n0 0

200


OneOne--Proportion Z Test Proportion Z Test Thinking ChallengeThinking Challenge

Youre an accounting Youre an accounting manager. A yearmanager. A year--end audit end audit showed showed 4%4% of transactions of transactions had errors. You implement had errors. You implement

8 8 -- 174174

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 the at the .05.05 levellevel? ?


OneOne--Proportion Z Test Proportion Z Test Solution*Solution*

HH00: :

HHaa: :

= = nn = =


8 8 -- 175175

nn = =





HH00: : pp = .04= .04

HHaa: : pp .04.04 = = nn = =


8 8 -- 176176

nn = =





HH00: : pp = .04= .04

HHaa: : pp .04.04 = = .05.05nn = = 500500


8 8 -- 177177

nn = = 500500





HH00: : pp = .04= .04

HHaa: : pp .04.04 = = .05.05nn = = 500500


8 8 -- 178178

nn = = 500500



Z0 1.96-1.96

.025

Reject H0 Reject H0

.025



HH00: : pp = .04= .04

HHaa: : pp .04.04 = = .05.05nn = = 500500


Zp p

p p

( )

.

. ( . ).0

0 01

25500

04

04 1 04114

8 8 -- 179179

nn = = 500500



Z0 1.96-1.96

.025

Reject H0 Reject H0

.025

n0 0

500



HH00: : pp = .04= .04

HHaa: : pp .04.04 = = .05.05nn = = 500500


Zp p

p p

( )

.

. ( . ).0

0 01

25500

04

04 1 04114

8 8 -- 180180

nn = = 500500



Z0 1.96-1.96

.025

Reject H0 Reject H0

.025


n0 0

500



HH00: : pp = .04= .04

HHaa: : pp .04.04 = = .05.05nn = = 500500


Zp p

p p

( )

.

. ( . ).0

0 01

25500

04

04 1 04114

8 8 -- 181181

nn = = 500500



Z0 1.96-1.96

.025

Reject H0 Reject H0

.025


There is no evidence There is no evidence proportion has proportion has changed from 4% changed from 4%

n0 0

500



OnePopulation

8 8 -- 182182

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)


Confidence Intervals, HypothesisConfidence Intervals, HypothesisTests, and pTests, and p--valuesvalues

All Start with Known Sampling Distribution forAll Start with Known Sampling Distribution forConfidence IntervalConfidence Interval

Pr( > given distance from ) = Pr( > given distance from ) = Draw an interval of size around actualDraw an interval of size around actual

X

X 2/zX

2/z

8 8 -- 183183

Draw an interval of size around actualDraw an interval of size around actual 11-- is the confidence levelis the confidence level

PP--ValueValue Assume true mean Assume true mean Pr( > measured distance) = pPr( > measured distance) = p

For oneFor one--sided value, no absolute valuesided value, no absolute value

Hypothesis testHypothesis test Pick , If p < , reject the null hypothesisPick , If p < , reject the null hypothesis

X

X

2/z


Calculating Type II Error Calculating Type II Error ProbabilitiesProbabilities

8 8 -- 184184

ProbabilitiesProbabilities


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 --

8 8 -- 185185

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


RejectRejectDo NotDo NotRejectReject

Finding PowerFinding PowerStep 1Step 1

Hypothesis:Hypothesis:HH00: : 00 368368HH11: : 00 < 368< 368 = .05= .05

n =n =15/15/2525

DrawDraw

8 8 -- 186186

XX00 = 368= 368



Finding PowerFinding PowerSteps 2 & 3Steps 2 & 3


n =n =15/15/2525

DrawDraw

8 8 -- 187187 XX11 = 360= 360

XX00 = 368= 368True Situation:True Situation:11 = 360= 360

DrawDraw

SpecifySpecify

11--





n =n =15/15/2525

DrawDraw

8 8 -- 188188 XX11 = 360= 360 363.065363.065

XX00 = 368= 368True Situation:True Situation:11 = 360= 360

065.363

25

1564.13680

n

ZXL

DrawDraw

SpecifySpecify





n =n =15/15/2525

DrawDraw

8 8 -- 189189 XX11 = 360= 360 363.065363.065

XX00 = 368= 368True Situation:True Situation:11 = 360= 360 = .154= .154

11-- =.846=.846

DrawDraw

SpecifySpecify

Z TableZ Table

065.363

25

1564.13680

n

ZXL


Power CurvesPower Curves

PowerPower PowerPowerHH00: : 00 HH00: : 00

8 8 -- 190190

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

= 368 in = 368 in ExampleExample


ConclusionConclusion

1.1. Distinguished Types of Hypotheses Distinguished Types of Hypotheses

2.2. Described Hypothesis Testing ProcessDescribed Hypothesis Testing Process

3.3. Explained pExplained p--Value ConceptValue Concept

8 8 -- 191191

3.3. Explained pExplained p--Value ConceptValue 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.

hypos testing

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