Ch08 Hypothesis Testing

8/6/2019 Ch08 Hypothesis Testing

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Business Statistics: A Decision-Making Approach, 6e 2005 Prentice-Hall, Inc. Chap 8-1

Business Statistics:A Decision-Making Approach6th Edition

Chapter 8Introduction to

Hypothesis Testing


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Chapter Goals

After completing this chapter, you should be

able to:

Formulate null and alternative hypotheses forapplications involving a single population mean orproportion

Formulate a decision rule for testing a hypothesis

Know how to use the test statistic, critical value, andp-value approaches to test the null hypothesis

Know what Type I and Type II errors are

Compute the probability of a Type II error


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What is a Hypothesis?

A hypothesis is a claim(assumption) about apopulation parameter:

population mean

population proportion

Example: The mean monthly cell phone billof this city is Q = $42

Example: The proportion of adults in thiscity with cell phones is p = .68


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The Null Hypothesis, H0

States the assumption (numerical) to betested

Example: The average number of TV sets inU.S. Homes is at least three ( )

Is always about a population parameter,

not about a sample statistic

3:H0 u

3:0 u 3x:0 u


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The Null Hypothesis, H0

Begin with the assumption that the nullhypothesis is true

Similar to the notion of innocent untilproven guilty

Refers to the status quo

Always contains = , or u sign May or may not be rejected

(continued)


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The Alternative Hypothesis, HA

Is the opposite of the null hypothesis

e.g.: The average number of TV sets in U.S.homes is less than ( HA: Q < )

Challenges the status quo

Never contains the = , or u sign

May or may not be accepted

Is generally the hypothesis that is believed(or needs to be supported) by theresearcher


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Business Statistics: A Decision-Making Approach, 6e 2005 Prentice-Hall, Inc.

Population

Claim: thepopulationmean age is 50.(Null Hypothesis:

REJECT

Supposethe samplemean ageis 20: x = 20

SampleNull Hypothesis

20 likely if Q = 50?!Is

Hypothesis Testing Process

If not likely,

Now select arandom sample

H0: Q = 50 )

x


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Sampling Distribution of x

Q = 50IfH0 is true

If it is unlikely thatwe would get asample mean ofthis value ...

... then wereject the null

hypothesis thatQ = 50.

Reason for Rejecting H0

20

... if in fact this werethe population mean

x


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Level of Significance, E

Defines unlikely values of sample statistic if

null hypothesis is true

Defines rejection region of the samplingdistribution

Is designated by E , (level of significance)

Typical values are .01, .05, or .10 Is selected by the researcher at the beginning

Provides the critical value(s) of the test


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Level of Significanceand the Rejection Region

H0:

HA:

H0: =

HA:

E

E

/2

Representscritical value

Lower tail test

Level of significance = E

0

0

E/2E

Upper tail test

Two tailed test

Rejection

region isshaded


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Errors in Making Decisions

Type IError

Reject a true null hypothesis

Considered a serious type of error

The probability of Type I Error is E

Called level of significance of the test Set by researcher in advance


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Errors in Making Decisions

Type IIError

Fail to reject a false null hypothesis

The probability of Type II Error is

(continued)


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Outcomes and Probabilities

State of Nature

Decision

Do NotReject

H0

No error(1 - )E

Type IIError( )

RejectH0

Type IError( )E

Possible Hypothesis Test Outcomes

H0 FalseH0 True

Key:

Outcome(Probability) No Error( 1 - )


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Type I & II Error Relationship

Type I and Type II errors can not happen atthe same time

Type I error can only occur if H0 is true

Type II error can only occur if H0 is false

If Type I error probability ( E ) , then

Type II error probability ( )


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Factors Affecting Type II Error

All else equal,

when the difference between

hypothesized parameter and its true value

when E

when when n


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Critical ValueApproach to Testing

Convert sample statistic (e.g.: ) to test

statistic ( Z or t statistic )

Determine the critical value(s) for a specifiedlevel of significance E from a table or

computer

If the test statistic falls in the rejection region,reject H0 ; otherwise do not reject H0

x


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Reject H0 Do not reject H0

The cutoff value,

or , is called a

critical value

Lower Tail Tests

E

-z

x

-z x

0

H0: 3

HA: < 3

n

zx

EE!


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Reject H0Do not reject H0

The cutoff value,

or , is called a

critical value

Upper Tail Tests

E

z

x

z x

0

H0: 3

HA: > 3

n

zx EE !


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Do not reject H0 Reject H0Reject H0

There are two cutoff

values (critical values):

or

Two Tailed Tests

E/2

-z/2

x/2

z/2

x/2

00

H0: = 3

HA: { 3

z/2

x/2

n

zx /2/2 EE s!

Lower

Upperx/2

Lower Upper

E/2


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Critical ValueApproach to Testing

Convert sample statistic ( ) to a test statistic( Z or t statistic )

x

W Known

LargeSamples

W Unknown

HypothesisTests forQ

SmallSamples


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W Known

LargeSamples

W Unknown

HypothesisTests for

SmallSamples

The test statistic is:

Calculating the Test Statistic

!


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W Known

LargeSamples

W Unknown


SmallSamples



st 1

!

But is sometimesapproximated

using a z:

n

xz

!

(continued)


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W Known

LargeSamples

W Unknown


SmallSamples



st 1

!

(The population must beapproximately normal)

(continued)


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Review: Steps in Hypothesis Testing

1. Specify the population value of interest

2. Formulate the appropriate null andalternative hypotheses

. Specify the desired level of significance

4. Determine the rejection region

5. Obtain sample evidence and compute thetest statistic

6. Reach a decision and interpret the result


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Hypothesis Testing Example

Test the claim that the true mean # of TVsets in US homes is at least 3.

(Assume = 0.8)

1. Specify the population value of interest

The mean number of TVs in US homes

2. Formulate the appropriate null and alternativehypotheses

H0: u 3 HA:


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4. Determine the rejection region

E = .05

-z= -1.645 0

This is a one-tailed test with E = .05.Since is known, the cutoff value is a z value:

Reject H0 if z < zE = -1.645 ; otherwise do not reject H0


(continued)


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5. Obtain sample evidence and compute thetest statistic

Suppose a sample is taken with the following

results: n = 100, x = 2.84 (W = 0.8 is assumed known)

Then the test statistic is:

22

n

!

!

!

!



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E = .05

-1.645 0

6. Reach a decision and interpret the result

-2.0

Since z = -2.0< -1.645, we reject the nullhypothesis that the mean number of TVs in UShomes is at least 3


(continued)

z


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Reject H0

E = .05

2.8684

Do not reject H0

3

An alternate way of constructing rejection region:

2.84

Since x = 2.84


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p-Value Approach to Testing

Convert Sample Statistic (e.g. ) to TestStatistic ( Z or t statistic )

Obtain the p-value from a table or computer

Compare the p-value with E

If p-value


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p-Value Approach to Testing

p-value: Probability of obtaining a test

statistic more extreme ( oru ) than the

observed sample value given H0 is true

Also called observed level of significance

Smallest value of E for which H0 can berejected

(continued)


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p-value =.0228

E = .05

p-value example

Example: How likely is it to see a sample meanof2.84 (or something further below the mean) ifthe true mean is Q = 3.0?

2.8684 3

2.84

x

.0.0)(z

000.

3.0.

z

3.0)|.xP(

!!

!

!


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Compare the p-value with E

If p-value


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Suppose that E = .10 is chosen for this test

Find the rejection region:

E = .10

z=1.280

Reject H0

Reject H0 if z > 1.28

Example: Find Rejection Region

(continued)


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Review:Finding Critical Value - One Tail

Z .07 .09

1.1 .3790 .3810 .3830

1.2.3980 .4015

1.3 .4147 .4162 .4177z 0 1.28

.08

Standard NormalDistribution Table (Portion)What is z given E = 0.10?

E = .10

Critical Value= 1.28

.90

.3997

.10

.40.50


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Obtain sample evidence and compute the teststatistic

Suppose a sample is taken with the following

results: n = 64, x = 53.1 (W=10 was assumed known)

Then the test statistic is:

n

!

!

!

Example: Test Statistic

(continued)


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Example: Decision

E = .10

1.280

Reject H0

Do not reject H0 since z = 0.88 1.28

i.e.: there is not sufficient evidence that themean bill is over $52

z = .88

Reach a decision and interpret the result:(continued)


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Reject H0

E = .10

Do not reject H0 1.28

0

Reject H0

z = .88

Calculate the p-value and compare to E

(continued)

. 9

.3 0.0. )P(z

0

.03.zP

5 .0)|53.xP(

!!u!

!

!u

p-value = .1894

p -Value Solution

Do not reject H0 since p-value = .1894 > E = .10


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Example: Two-Tail Test(W Unknown)

The average cost of ahotel room in New York

is said to be $168 pernight. A random sampleof25 hotels resulted inx = $172.50 and

s = $15.40. Test at the

E = 0.05 level.(Assume the population distribution is normal)

H0: = 168

HA: { 168


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E = 0.05

n = 25

W is unknown, so

use a t statistic Critical Value:

t24 = 2.0639

Example Solution: Two-Tail Test

Do not reject H0: not sufficient evidence thattrue mean cost is different than $168

Reject H0Reject H0

E/2=.025

-t/2Do not reject H0

0t/2

E/2=.025

-2.0639 2.0639

1.46

2515.40

168172.50

ns

xt

1n

!

!

!

1.46

H0: = 168

HA: { 168


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Hypothesis Tests for Proportions

Involves categorical values

Two possible outcomes

Success (possesses a certain characteristic)

Failure (does not possesses that characteristic)

Fraction or proportion of population in the

success category is denoted by p


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Proportions

Sample proportion in the success category isdenoted by p

When both np and n(1-p) are at least 5, p can

be approximated by a normal distribution withmean and standard deviation

(continued)

sizesample sampleinsuccessesofnumbernxp!!

pP!

n

p)p(1

p

!


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The samplingdistribution of p isnormal, so the teststatistic is a zvalue:

Hypothesis Tests for Proportions

n

)p(p

pp

z

! 1

np u 5and

n(1-p) u 5

HypothesisTests for p

np < 5or

n(1-p) < 5

Not discussedin this chapter


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Example: z Test for Proportion

A marketing companyclaims that it receives8% responses from itsmailing. To test thisclaim, a random sampleof 500 were surveyedwith 25 responses. Testat the E = .05significance level.

Check:

np = (500)(.08) = 40

n(1-p) = (500)(.92) = 460


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Z Test for Proportion: Solution

E = .05

n = 500, p = .05

Reject H0 at E = .05

H0: p = .08

HA: p { .08

Critical Values: 1.96

Test Statistic:

Decision:

Conclusion:

z0

Reject Reject

.025.025

1.96

-2.47

There is sufficientevidence to reject thecompanys claim of 8%

response rate.

2.47

500

.08).08(1

.08.05

n

p)p(1

ppz !

!

!

-1.96


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Do not reject H0Reject H0Reject H0

E/2 = .025

1.960

z = -2.47

Calculate the p-value and compare to E(For a two sided test the p-value is always two sided)

(continued)

0.0 3(.00 )

. 93 )(.5

. 7)P(x. 7)P(z

!!

!

ue

p-value = .0136:

p -Value Solution

Reject H0 since p-value = .0136 < E = .05

z = 2.47

-1.96

E/2 = .025

.0068.0068


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RejectH0: u 52

Do not rejectH0: u 52

Type II Error

Type II error is the probability of

failing to reject a false H0

5250

Suppose we fail to reject H0: u 52

when in fact the true mean is = 50

E


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RejectH0: Q u 52

Do not rejectH0: Q u 52

Type II Error

Suppose we do not reject H0: Q u 52 when in factthe true mean is Q = 50

5250

This is the truedistribution of x ifQ = 50

This is the range of x whereH0 is not rejected

(continued)


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RejectH0: u 52


Type II Error

Suppose we do not reject H0: u 52 whenin fact the true mean is = 50

E

5250

Here, = P( x u cutoff ) if = 50

(continued)


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RejectH0: u 52


Suppose n = 64 , = 6 , and E = .05

E

5250

So = P( x u 50.766 ) if = 50

Calculating

50.7. 55

n

zxc t ff !!!! EE

(for H0: u 52)

50.766


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RejectH0: u 52


. 539.3.5.02)P(z5050.7

zP50)|50.7xP( !!u!

u!!u

Suppose n = 64 , = 6 , and E = .05

E

5250

Calculating

(continued)

Probability oftype II error:

= .1539


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Using PHStat

Options


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Sample PHStat Output

Input

Output


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Chapter Summary

Addressed hypothesis testing methodology

Performed z Test for the mean ( known)

Discussed pvalue approach to

hypothesis testing

Performed one-tail and two-tail tests . . .


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Business Statistics: A Decision Making Approach 6e 2005 Prentice Hall Inc Chap 8 56

Chapter Summary

Performed t test for the mean (

unknown)

Performed z test for the proportion

Discussed type II error and computed its

probability

(continued)

Ch08 Hypothesis Testing

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