hypothesistesting1sample(8)

8/8/2019 hypothesistesting1sample(8)

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Statistics for Managers Using Microsoft Excel, 4e 2004 Prentice-Hall, Inc. Chap 8-1

Chapter 9

Fundamentals of HypothesisTesting: One-Sample Tests

Statistics for ManagersUsing Microsoft Excel

5th Edition


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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 = $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 equal to three ( )

Is always about a population parameter,

not about a sample statistic

3:H0 !

3: ! 3X: !


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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, H1

Is the opposite of the null hypothesis

e.g., The average number of TV sets in U.S.homes is not equal to 3 ( H1: 3 )

Challenges the status quo

Never contains the = , or u sign

May or may not be accepted

Is generally the hypothesis that theresearcher is trying to show


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Statistics for Managers Using Microsoft Excel, 4e 2004 Prentice-Hall, Inc.

Population

Claim: thepopulationmean age is 50.(Null Hypothesis:

REJECT

Supposethe samplemean ageis 20: X = 20

SampleNull Hypothesis

20 likely if = 50?!Is

Hypothesis Testing Process

If not likely,

Now select arandom sample

H0: = 50 )

X


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Sampling Distribution ofX

= 50If H0 is true

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

... then wereject the null

hypothesis that = 50.

Reason for Rejecting H0

20

... if in fact this werethe population mean

X


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Level of Significance, E

Defines the unlikely values of the sample

statistic if the 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: 3

H1: < 3

0

H0: 3

H1: > 3

E

E

Representscritical value

Lower-tail test

Level of significance = E

0Upper-tail test

Two-tail test

Rejection

region isshaded

/2

0

E/2EH0: = 3

H1: 3


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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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Outcomes and Probabilities

Actual

SituationDecision

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

For two tailed test for the mean, known:

Convert sample statistic ( ) to test statistic (Z

statistic )

Determine the critical Z values for a specifiedlevel of significance E from a table orcomputer

Decision Rule: If the test statistic falls in therejection region, reject H0 ; otherwise do not

reject H0

X


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Do not reject H0 Reject H0Reject H0

There are twocutoff values(critical values),

defining theregions ofrejection

Two-Tail Tests

E/2

-Z 0

H0: = 3

H1: { 3

+Z

E/2

Lowercritical

value

Uppercritical

value

3

Z

X


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Review: 10 Steps inHypothesis Testing

1. State the null hypothesis, H0

2. State the alternative hypotheses, H1

3. Choose the level of significance,

4. Choose the sample size, n

5. Determine the appropriate statisticaltechnique and the test statistic to use

6. Find the critical values and determine therejection region(s)


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Review: 10 Steps inHypothesis Testing

7. Collect data and compute the test statisticfrom the sample result

8. Compare the test statistic to the criticalvalue to determine whether the test statisticsfalls in the region of rejection

9. Make the statistical decision: Reject H0 if the

test statistic falls in the rejection region 10.Express the decision in the context of the

problem


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Hypothesis Testing Example

Test the claim that the true mean # of TVsets in US homes is equal to 3.

(Assume = 0.8)

1-2. State the appropriate null and alternativehypotheses

H0: = 3 H1: 3 (This is a two tailed test)

3. Specify the desired level of significance

Suppose that E = .05 is chosen for this test

4. Choose a sample size

Suppose a sample of size n = 100 is selected


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2 .0.08

.

00

0.8

2.84

n

!

!

!

!

Hypothesis Testing Example

5. Determine the appropriate technique is known so this is a Z test

6. Set up the critical values

ForE = .05 the critical Z values are 1.96

7. Collect the data and compute the test statistic

Suppose the sample results are

n = 100, X = 2.84 ( = 0.8 is assumed known)

So the test statistic is:

(continued)


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Reject H0 Do not reject H0

8. Is the test statistic in the rejection region?

E = .05/2

-Z= -1.96 0Reject H0 ifZ < -1.96 or

Z > 1.96;otherwisedo notreject H0

Hypothesis Testing Example(continued)

E = .05/2

Reject H0

+Z= +1.96

Here, Z = -2.0 < -1.96, so thetest statistic is in the rejectionregion


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9-10. Reach a decision and interpret the result

-2.0

Since Z = -2.0 < -1.96, we reject the null hypothesisand conclude that there is sufficient evidence that themean number of TVs in US homes is not equal to 3

Hypothesis Testing Example(continued)


E = .05/2

-Z= -1.96 0

E = .05/2

Reject H0

+Z= +1.96


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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


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p-Value Approach to Testing

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

Obtain the p-value from a table or computer

Compare the p-value with E

If p-value < E , reject H0

If p-value u E , do not reject H0

X

(continued)


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.0228

E/2 = .025

p-Value Example

Example: How likely is it to see a sample mean of2.84 (or something further from the mean, in eitherdirection) if the true mean is Q = 3.0?

-1.96 0

-2.0

.02282.0)P(Z

.02282.0)P(Z

!"

!

Z1.96

2.0

X = 2.84 is translatedto a Z score of Z = -2.0

p-value

=.0228 + .0228 = .0456

.0228

E/2 = .025


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Compare the p-value with E

If p-value < E , reject H0

If p-value u E , do not reject H0

Here: p-value = .0456E = .05

Since .0456 < .05, wereject the nullhypothesis

(continued)

p-Value Example

.0228

E/2 = .025

-1.96 0

-2.0

Z1.96

2.0

.0228

E/2 = .025


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Connection to Confidence Intervals

ForX = 2.84, = 0.8 and n = 100, the 95%confidence interval is:

2.6832 2.9968

Since this interval does not contain the hypothesizedmean (3.0), we reject the null hypothesis at E = .05

000.( . ).t

000.( . )-.


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One-Tail Tests

In many cases, the alternative hypothesisfocuses on a particular direction

H0: 3

H1: < 3

H0: 3

H1: > 3

This is a lower-tail test since thealternative hypothesis is focused onthe lower tail below the mean of 3

This is an upper-tail test since thealternative hypothesis is focused onthe upper tail above the mean of 3


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There is only one

critical value, since

the rejection area isin only one tail

Lower-Tail Tests

E

-Z 0

H0: 3

H1: < 3

Z

X

Critical value


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Reject H0Do not reject H0

Upper-Tail Tests

E

Z0

H0: 3

H1: > 3 There is only one

critical value, since

the rejection area isin only one tail

Critical value

Z

X


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Example: Upper-Tail Z Testfor Mean (W Known)

A phone industry manager thinks thatcustomer monthly cell phone bill haveincreased, and now average over $52 per

month. The company wishes to test thisclaim. (Assume W = 10 is known)

H0: 52 the average is not over $52 per monthH1: > 52 the average is greater than $52 per month

(i.e., sufficient evidence exists to support themanagers claim)

Form hypothesis test:


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Suppose that E = .10 is chosen for this test

Find the rejection region:

E = .10

1.280

Reject H0

Reject H0 if Z > 1.28

Example: Find Rejection Region

(continued)


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Review:One-Tail Critical Value

Z .07 .09

1.1 .8790 .8810 .8830

1.2.8980 .9015

1.3 .9147 .9162 .9177z 0 1.28

.08

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

E = .10

Critical Value= 1.28

.90

.8997

.10

.90


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Obtain sample and compute the test statistic

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(assuming that = 52.0)

(continued)

.

.. )(

/

.3.

3. )X(

!

!u!

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 sampleof 25 hotels resulted inX = $172.50 and

S = $15.40. Test at the

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

H0: = 168

H1: { 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 n-1,/2Do not reject H0

0

E/2=.025

-2.0639 2.0639

1.46

25

15.40

168172.50

n

S

Xt

1n!

!

!

1.46

H0: = 168

H1: { 168

t n-1,/2


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Connection to Confidence Intervals

ForX = 172.5, S = 15.40 and n = 25, the 95%confidence interval is:

172.5 - (2.0639) 15.4/ 25 to 172.5 + (2.0639) 15.4/ 25

166.14 178.86

Since this interval contains the Hypothesized mean (168),we do not reject the null hypothesis at E = .05


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Hypothesis Tests for Proportions

Involves categorical variables

Two possible outcomes

Success (possesses a certain characteristic)

Failure (does not possesses that characteristic)

Fraction or proportion of the population in the

success category is denoted by p


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Proportions

Sample proportion in the success category isdenoted by ps

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

can be approximated by a normal distributionwith mean and standard deviation

sizesamplesampleinsuccessesofnumber

nXps !!

p sp !n

p)p(1

sp

!

(continued)


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The samplingdistribution of psis approximatelynormal, so the teststatistic is a Zvalue:

Hypothesis Tests for Proportions

n

)p(p

ppZ

s

! 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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An equivalent formto the last slide,but in terms of thenumber ofsuccesses, X:

Z Test for Proportionin Terms of Number of Successes

)p(npnpXZ

!

X u 5and

n-X u 5

HypothesisTests for X

X < 5or

n-X < 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, ps = .05

Reject H0 at E = .05

H0: p = .08

H1: 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

s!

!

!

-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)

0.01362(.0068)

2. )(2. )(

!!

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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Deregulation leads to lower air travel prices.

The university discriminates against women

faculty members Stock prices increase when a firm announces a

layoff.

The cost minimizing strategy is to hire high school

graduates via recommendations from our currentemployees.

Example Hypotheses:

How do you test them?


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Hiring Policy Hypotheses

A recruiter must decide who to hire. This is likeforming a hypothesis about whether or not theindividual predicts to be a good employee.

Assume that an employee is either good orpoor.

WHAT ARE THE NULL AND ALTERNATIVE

HYPOTHESES?


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NULL: THE INDIVIDUAL DOES NOT MEETTHE STANDARDS (MEANZ)


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WHAT ARE THE POSSIBLE ERRORS THATTHE RECRUITER CAN MAKE?

HOW DO THESE RELATE TO TYPE I ANDTYPE II ERRORS?


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FAILURE TO HIRE A GOOD EMPLOYEE(failure to reject a false null=type II error)

FAILURE TO REJECT A POOREMPLOYEE(rejecting a null when it is reallytrue is type I error)


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A positive decision is a decision to reject thenull. A false positive is therefore a type I error(hiring a poor person).

A negative decision is a failure to reject thenull. A false negative is therefore a type IIerror (not hiring a good person)


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The addition of more criteria should increaseyour ability to distinguish poor candidates =>type I error falls

However, more criteria mean that more goodemployees are cut accidentally => type IIerror increases.

When do you use more criteria?


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When would type II error start to be moreimportant?

What does this tell you about balancing oftype I and type II error and the criteria that areused in hiring different for kinds ofoccupations?

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