8 - 1 © 2003 pearson prentice hall statistical methods

8 - 8 - 11

© 2003 Pearson Prentice Hall© 2003 Pearson Prentice Hall

Statistical MethodsStatistical Methods

StatisticalMethods

DescriptiveStatistics

InferentialStatistics

EstimationHypothesis

Testing

StatisticalMethods

DescriptiveStatistics

InferentialStatistics

EstimationHypothesis

Testing

8 - 8 - 22


Hypothesis Testing Hypothesis Testing ConceptsConcepts

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Hypothesis TestingHypothesis Testing

8 - 8 - 44



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.

Reject hypothesis! Not close.

Random Random samplesample

8 - 8 - 88


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!

© 1984-1994 T/Maker Co.

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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

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

4.4. Specified as HSpecified as H00: : Some Numeric Some Numeric

Value Value Specified with = Sign Even if Specified with = Sign Even if , or , or Example, HExample, H00: : 3 3

8 - 8 - 1010


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 will lead towill lead to two-sided tests two-sided tests <, > will lead to one-sided tests<, > will lead to one-sided tests

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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)

8 - 8 - 1212


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: :

1212

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: :

1212

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

Value

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion


1 - 1 -

Level of ConfidenceLevel of Confidence

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HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion


1 - 1 -


Observed sample statisticObserved sample statistic

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HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValueCritical

Value

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

RejectionRegion

NonrejectionRegion

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion


1 - 1 -


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HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion


1 - 1 -


Observed sample statisticObserved sample statistic

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HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion


1 - 1 -


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HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion


1 - 1 -


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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

ZX X

n

x

x

Z

X X

n

x

x

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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: :

nn

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

Test Statistic: Test Statistic:

Decision:Decision:

Conclusion:Conclusion:

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HH00: : = 368 = 368

HHaa: : 368 368

nn



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

.025

Reject H 0 Reject H 0

.025

Z0 1.96-1.96

.025


.025

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HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 2525



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025

ZX

n

372 5 368

1525

150.

.ZX

n

372 5 368

1525

150.

.

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HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 2525



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025

ZX

n

372 5 368

1525

150.

.ZX

n

372 5 368

1525

150.

.

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

.025


.025

Z0 1.96-1.96

.025


.025

ZX

n

372 5 368

1525

150.

.ZX

n

372 5 368

1525

150.

.


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

.025


.025

Z0 1.96-1.96

.025


.025

8 - 8 - 5353



HH00: : = 70 = 70

HHaa: : 70 70

= = .05.05

nn = = 3636



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025

ZX

n

69 7 70

3 536

51.

..Z

X

n

69 7 70

3 536

51.

..

8 - 8 - 5454



HH00: : = 70 = 70

HHaa: : 70 70

= = .05.05

nn = = 3636



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025

ZX

n

69 7 70

3 536

51.

..Z

X

n

69 7 70

3 536

51.

..


8 - 8 - 5555



HH00: : = 70 = 70

HHaa: : 70 70

= = .05.05

nn = = 3636



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025

ZX

n

69 7 70

3 536

51.

..Z

X

n

69 7 70

3 536

51.

..


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

ZX X

n

x

x

Z

X X

n

x

x

8 - 8 - 5959


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

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Z0

Reject H 0

Z0

Reject H 0


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

Must be Must be significantlysignificantly below below

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Z0

Reject H 0

Z0

Reject H 0

Z0

Reject H 0

Z0

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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Z0

= 1

Z0

= 1


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

= .025= .025

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Z0

= 1

Z0

= 1


.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

Z0

= 1

Z0

= 1


.500 .500 -- .025.025

.475.475

.06

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

Z0

= 1

1.96 Z0

= 1

1.96


.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:


8 - 8 - 7070



HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 2525



Decision:Decision:


8 - 8 - 7171



HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 2525



Decision:Decision:


Z0 1.645

.05

Reject

Z0 1.645

.05

Reject

8 - 8 - 7272



HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 2525



Decision:Decision:


Z0 1.645

.05

Reject

Z0 1.645

.05

Reject

ZX

n

372 5 368

1525

150.

.ZX

n

372 5 368

1525

150.

.

8 - 8 - 7373



HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 2525



Decision:Decision:


Z0 1.645

.05

Reject

Z0 1.645

.05

Reject

ZX

n

372 5 368

1525

150.

.ZX

n

372 5 368

1525

150.

.


8 - 8 - 7474



HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 2525



Decision:Decision:


Z0 1.645

.05

Reject

Z0 1.645

.05

Reject

ZX

n

372 5 368

1525

150.

.ZX

n

372 5 368

1525

150.

.


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:



Decision:Decision:


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HH00: : = 32 = 32

HHaa: : < 32 < 32

= =

nn = =



Decision:Decision:



Decision:Decision:


8 - 8 - 7878



HH00: : = 32 = 32

HHaa: : < 32 < 32

== .01 .01

nn = = 6060



Decision:Decision:



Decision:Decision:


8 - 8 - 7979



HH00: : = 32 = 32

HHaa: : < 32 < 32

= .01= .01

nn = 60= 60



Decision:Decision:



Decision:Decision:


Z0-2.33

.01

Reject

Z0-2.33

.01

Reject

8 - 8 - 8080



HH00: : = 32 = 32

HHaa: : < 32 < 32

= .01= .01

nn = 60= 60



Decision:Decision:



Decision:Decision:


Z0-2.33

.01

Reject

Z0-2.33

.01

Reject

ZX

n

30 7 32

3 860

2 65.

..Z

X

n

30 7 32

3 860

2 65.

..

8 - 8 - 8181



HH00: : = 32 = 32

HHaa: : < 32 < 32

= .01= .01

nn = 60= 60



Decision:Decision:



Decision:Decision:


Z0-2.33

.01

Reject

Z0-2.33

.01

Reject

ZX

n

30 7 32

3 860

2 65.

..Z

X

n

30 7 32

3 860

2 65.

..

Reject at Reject at = .01 = .01

8 - 8 - 8282



HH00: : = 32 = 32

HHaa: : < 32 < 32

= .01= .01

nn = 60= 60



Decision:Decision:



Decision:Decision:


Z0-2.33

.01

Reject

Z0-2.33

.01

Reject

ZX

n

30 7 32

3 860

2 65.

..Z

X

n

30 7 32

3 860

2 65.

..


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

8 - 8 - 8383


Decision Making RisksDecision Making Risks

8 - 8 - 8484


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)

8 - 8 - 8585


Jury Trial H0 Test

Actual Situation Actual Situation

Verdict Innocent Guilty Decision H0 True H0

False

Innocent Correct ErrorDo NotReject

H0

1 - Type IIError

()

Guilty Error Correct RejectH0

Type IError ()

Power(1 - )

Jury Trial H0 Test



False

Innocent Correct ErrorDo NotReject

H0

1 - Type IIError

()


Type IError ()

Power(1 - )

Decision ResultsDecision Results

HH00: Innocent: Innocent

8 - 8 - 8686


Jury Trial H0 Test



False

Innocent Correct Error AcceptH0

1 - Type IIError

()


Type IError ()

Power(1 - )

Jury Trial H0 Test



False

Innocent Correct Error AcceptH0

1 - Type IIError

()


Type IError ()

Power(1 - )

Decision ResultsDecision Results

HH00: Innocent: Innocent

8 - 8 - 8787


& & Have an Have an Inverse RelationshipInverse Relationship

You can’t reduce both errors simultaneously!

8 - 8 - 8888


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

8 - 8 - 8989


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

8 - 8 - 9090



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 - 9191


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 - 9292


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

tX

Sn

tX

Sn

8 - 8 - 9393


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

ValuesValues

8 - 8 - 9494


t0 t0


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

8 - 8 - 9595


t0 t0


ValuesValues

/2 = .05/2 = .05

/2 = .05/2 = .05

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

8 - 8 - 9696


t0 t0


ValuesValues

/2 = .05/2 = .05

/2 = .05/2 = .05

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

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

8 - 8 - 9797


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 - 9898


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

/2 = .05/2 = .05

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

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

8 - 8 - 9999


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 - 100100


Two-Tailed t Test Two-Tailed t Test SolutionSolution

HH00: :

HHaa: :

= =

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


Decision:Decision:


8 - 8 - 101101



HH00: : = 368 = 368

HHaa: : 368 368

= =

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


Decision:Decision:


8 - 8 - 102102



HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05

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


Decision:Decision:


8 - 8 - 103103



HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05



Decision:Decision:


t0 2.0301-2.0301

.025


.025

t0 2.0301-2.0301

.025


.025

8 - 8 - 104104



HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05



Decision:Decision:


t0 2.0301-2.0301

.025


.025

t0 2.0301-2.0301

.025


.025

tX

Sn

372 5 368

1236

2 25.

.tX

Sn

372 5 368

1236

2 25.

.

8 - 8 - 105105



HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05



Decision:Decision:


t0 2.0301-2.0301

.025


.025

t0 2.0301-2.0301

.025


.025

tX

Sn

372 5 368

1236

2 25.

.tX

Sn

372 5 368

1236

2 25.

.


8 - 8 - 106106



HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05



Decision:Decision:


t0 2.0301-2.0301

.025


.025

t0 2.0301-2.0301

.025


.025

tX

Sn

372 5 368

1236

2 25.

.tX

Sn

372 5 368

1236

2 25.

.


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

8 - 8 - 107107


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 - 108108


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

HH00: :

HHaa: :

df df



Decision:Decision:


8 - 8 - 109109



HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

df df



Decision:Decision:


8 - 8 - 110110



HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

.01.01

df df 64 - 1 = 6364 - 1 = 63



Decision:Decision:


8 - 8 - 111111



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

.005


.005

t0 2.6561-2.6561

.005


.005

8 - 8 - 112112



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

.005


.005

t0 2.6561-2.6561

.005


.005

tX

Sn

3 238 3 25

11764

82. .

..t

XSn

3 238 3 25

11764

82. .

..

8 - 8 - 113113



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

.005


.005

t0 2.6561-2.6561

.005


.005

tX

Sn

3 238 3 25

11764

82. .

..t

XSn

3 238 3 25

11764

82. .

..


8 - 8 - 114114



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

.005


.005

t0 2.6561-2.6561

.005


.005

tX

Sn

3 238 3 25

11764

82. .

..t

XSn

3 238 3 25

11764

82. .

..


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

8 - 8 - 115115


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

8 - 8 - 116116


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 - 117117


One-Tailed t Test One-Tailed t Test SolutionSolution

HH00: :

HHaa: :

==

df =df =



Decision:Decision:


8 - 8 - 118118



HH00: : = 140 = 140

HHaa: : < 140 < 140

= =

df = df =



Decision:Decision:


8 - 8 - 119119



HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

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



Decision:Decision:


8 - 8 - 120120


t0-1.7291

.05

Reject

t0-1.7291

.05

Reject


HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

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



Decision:Decision:


8 - 8 - 121121


t0-1.7291

.05

Reject

t0-1.7291

.05

Reject


HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

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



Decision:Decision:


tX

Sn

138 47 140

2 6620

2 57.

..t

XSn

138 47 140

2 6620

2 57.

..

8 - 8 - 122122


t0-1.7291

.05

Reject

t0-1.7291

.05

Reject


HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

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



Decision:Decision:


tX

Sn

138 47 140

2 6620

2 57.

..t

XSn

138 47 140

2 6620

2 57.

..


8 - 8 - 123123


t0-1.7291

.05

Reject

t0-1.7291

.05

Reject


HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

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



Decision:Decision:


tX

Sn

138 47 140

2 6620

2 57.

..t

XSn

138 47 140

2 6620

2 57.

..


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

8 - 8 - 124124


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 - 125125


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

HH00: :

HHaa: :

= =

df = df =



Decision:Decision:


8 - 8 - 126126



HH00: : = 5 = 5

HHaa: : > 5 > 5

= =

df =df =



Decision:Decision:


8 - 8 - 127127



HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

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



Decision:Decision:


8 - 8 - 128128


t0 1.8331

.05

Reject

t0 1.8331

.05

Reject


HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

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



Decision:Decision:


8 - 8 - 129129


t0 1.8331

.05

Reject

t0 1.8331

.05

Reject


HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

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



Decision:Decision:


tX

Sn

6 4 5

3 37310

131..

.tX

Sn

6 4 5

3 37310

131..

.

8 - 8 - 130130


t0 1.8331

.05

Reject

t0 1.8331

.05

Reject


HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

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



Decision:Decision:


tX

Sn

6 4 5

3 37310

131..

.tX

Sn

6 4 5

3 37310

131..

.


8 - 8 - 131131


t0 1.8331

.05

Reject

t0 1.8331

.05

Reject


HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

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



Decision:Decision:


tX

Sn

6 4 5

3 37310

131..

.tX

Sn

6 4 5

3 37310

131..

.


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

8 - 8 - 132132



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 - 133133


Confidence Intervals, Confidence Intervals, Hypothesis Tests, and p-Hypothesis Tests, and p-

valuesvaluesAll 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 1- is the confidence level1- is the confidence level

P-ValueP-Value Assume true mean Assume true mean Pr( > measured distance) = pPr( > measured distance) = p

For one-sided value, no absolute valueFor one-sided value, no absolute value

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

X

X 2/zX

X

2/z

8 - 8 - 134134


Calculating Type II Calculating Type II Error ProbabilitiesError Probabilities

8 - 8 - 135135


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 - 136136


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 - 137137


XX11 = 360= 360

XX00 = 368= 368


Finding PowerFinding PowerSteps 2 & 3Steps 2 & 3


HH11: : 00 < 368 < 368

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

= .05= .05

n =n =15/15/2525

DrawDraw

DrawDraw

SpecifySpecify

1-1-

8 - 8 - 138138


XX11 = 360= 360 363.065363.065

XX00 = 368= 368




HH11: : 00 < 368 < 368


065.363

25

1564.13680

n

ZX L

065.363

25

1564.13680

n

ZX L

= .05= .05

n =n =15/15/2525

DrawDraw

DrawDraw

SpecifySpecify

8 - 8 - 139139


XX11 = 360= 360 363.065363.065

XX00 = 368= 368




HH11: : 00 < 368 < 368


= .05= .05

n =n =15/15/2525

= .154= .154

1-1- =.846 =.846

DrawDraw

DrawDraw

SpecifySpecify

Z TableZ Table

065.363

25

1564.13680

n

ZX L

065.363

25

1564.13680

n

ZX L

8 - 8 - 140140


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 - 141141


ConclusionConclusion

1.1. Distinguished Types of Hypotheses Distinguished Types of Hypotheses

2.2. Described Hypothesis Testing ProcessDescribed Hypothesis Testing Process

3.3. Explained p-Value ConceptExplained p-Value Concept

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

5.5. Explained Power of a TestExplained Power of a Test

End of Chapter

Any blank slides that follow are blank intentionally.

8 - 1 © 2003 pearson prentice hall statistical methods

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