Statistics-Fundamentals of Hypothesis Testing

8/3/2019 Statistics-Fundamentals of Hypothesis Testing

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Basic Business Statistics(8th Edition)

Chapter 9

Fundamentals of HypothesisTesting: One-Sample Tests


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Chapter Topics

Hypothesis testing methodology

Z test for the mean ( known)

P-value approach to hypothesis testing Connection to confidence interval estimation

One-tail tests

T test for the mean ( unknown)

Z test for the proportion

Potential hypothesis-testing pitfalls and ethicalconsiderations

W

W


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What is a Hypothesis?

A hypothesis is aclaim (assumption)

about the populationparameter

Examples of parametersare population mean

or proportion The parameter must

be identified beforeanalysis

I claim the mean GPA of

this class is 3.5!

1984-1994 T/Maker Co.

Q !


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The Null Hypothesis, H0

States the assumption (numerical) to betested

e.g.: The average number of TV sets in U.S.Homes is at least three ( )

Is always about a population parameter( ), not about a sample

statistic ( )

0: 3H Q u

0 : 3H Q u

0 : 3H X u


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The Null Hypothesis, H0

Begins with the assumption that the nullhypothesis is true

Similar to the notion of innocent untilproven guilty

Refers to the status quo

Always contains the = 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 less than 3 ( )

Challenges the status quo

Never contains the = sign

May or may not be accepted

Is generally the hypothesis that isbelieved (or needed to be proven) to betrue by the researcher

1: 3H Q


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Hypothesis Testing Process

Identify the Population

Assume thepopulation

mean age is 50.

( )

REJECT

Take a Sample

Null Hypothesis

No, not likely!

X 20 likely ifIs ?Q! !

0: 50H Q !

20X !


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Sampling Distribution of

= 50

It is unlikely that

we would get asample mean ofthis value ...

... Therefore,

we reject thenull hypothesisthat m = 50.

Reason for Rejecting H0

Q20

If H0 is true

X

... if in fact this werethe population mean.

X


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Level of Significance,

Defines unlikely values of sample statistic ifnull hypothesis is true

Called rejection region of the sampling distribution Is designated by , (level of significance)

Typical values are .01, .05, .10

Is selected by the researcher at the beginning

Provides the critical value(s) of the test

E

E


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Level of Significanceand the Rejection Region

H0: Q u3

H1:Q < 3

0

0

0

H0:Q e 3

H1:Q > 3

H0:Q !3

H1:Q { 3

E

E

E/2

Critical

Value(s)

RejectionRegions


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Errors in Making Decisions

Type I Error Rejects a true null hypothesis

Has serious consequences

The probability of Type I Error is Called level of significance

Set by researcher

Type II Error Fails to reject a false null hypothesis

The probability of Type II Error is

The power of the test is

E

1 FF


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Errors in Making Decisions

Probability of not making Type I Error

Called the confidence coefficient

1 E

(continued)


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Result ProbabilitiesH0: Innocent

The Truth The Truth

Verdict Innocent Guilty Decision H0 True H0 False

Innocent Correct ErrorDo Not

Reject

H0

1 - EType II

Error (F )

Guilty Error CorrectReject

H0

Type IError(E )

Power

(1 - F )

Jury Trial Hypothesis Test


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Type I & II Errors Have anInverse Relationship

E

F

If you reduce the probability of one

error, the other one increases so that

everything else is unchanged.


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Factors Affecting Type II Error

True value of population parameter Increases when the difference between

hypothesized parameter and its true valuedecrease

Significance level Increases when decreases

Population standard deviation Increases when increases

Sample size Increases when n decreases

F

F

E

F W

F

E

F

n

F

F W


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How to Choose betweenType I and Type II Errors

Choice depends on the cost of the errors

Choose smaller Type I Error when the cost ofrejecting the maintained hypothesis is high A criminal trial: convicting an innocent person

The Exxon Valdez: causing an oil tanker to sink

Choose larger Type I Error when you have aninterest in changing the status quo A decision in a startup company about a new piece

of software

A decision about unequal pay for a covered group


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Critical ValuesApproach to Testing

Convert sample statistic (e.g.: ) to teststatistic (e.g.: Z, t or F statistic)

Obtain critical value(s) for a specifiedfrom a table or computer

If the test statistic falls in the critical region,reject H0

Otherwise do not reject H0

X

E


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p-Value Approach to Testing

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

Obtain the p-value from a table or computer

p-value: Probability of obtaining a test statisticmore extreme ( or ) than the observedsample value given H0 is true

Called observed level of significance

Smallest value of that an H0 can be rejected

Compare the p-value with

If p-value , do not reject H0

If p-value , reject H0

X

e u

e

u EE

E

E


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General Steps inHypothesis Testing

e.g.: Test the assumption that the true mean number of ofTV sets in U.S. homes is at least three ( Known)W

1. State the H0

2. State the H1

3. Choose4. Choose n

5. Choose Test

0

1

: 3

: 3

=.05

100

Z

H

H

n

test

Q

Q

E

u

!E


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100 households surveyed

Computed test stat =-2,

p-value = .0228

Reject null hypothesis

The true mean number of TV

sets is less than 3

(continued)RejectH0

E

-1.645 Z

6. Set up critical value(s)

7. Collect data

8. Compute test statistic

and p-value9. Make statistical decision

10. Express conclusion

General Steps inHypothesis Testing


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One-tail Z Test for Mean( Known)

Assumptions

Population is normally distributed

If not normal, requires large samples Null hypothesis has or sign only

Z test statistic

W

e u

/

X

X

X XZn

Q QW W ! !


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Rejection Region

Z0

RejectH0

Z0

RejectH0

H0: QuQ

0

H1: Q < Q0

H0: QeQ

0

H1: Q > Q0

Z Must Be SignificantlyBelow 0 to reject H0

Small values of Z dontcontradict H0

Dont Reject H0!

EE


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Example: One Tail Test

Q. Does an average box of

cereal contain more than

368 grams of ce

real? A

random sample of 25

boxes showed = 372.5.

The company has

specified W to be1

5 grams.

Test at the E!0.05 level.

368 gm.

H0: Qe368

H1: Q> 368

X


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Finding Critical Value: One Tail

Z .04 .06

1.6 .9495 .9505 .9515

1.7 .9591 .9599 .960

8

1.8 .9671 .9678 .9686

.9738 .9750

Z0 1.645

.05

1.9 .9744

Standardized CumulativeNormal Distribution Table

(Portion)What is Z given E = 0.05?

E = .05

Critical Value

= 1.645

.95

1ZW !


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Example Solution: One Tail Test

E = 0.5

n= 25

Critical Value:1.645

Test Statistic:

Decision:

Conclusion:Do Not Reject atE = .05

No evidence that true

mean is more than 368

Z0 1.645

.05

Reject

H0:Qe368

H1:Q > 368

1.50

X

Z

n

Q

W

! !

1.50


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p -Value Solution

Z0 1.50

P-Value =.0668

ZValue of Sample

Statistic

FromZTable:

Lookup 1.50 to

Obtain .9332

Use the

alternative

hypothesis

to find the

direction of

the rejectionregion.

1.0000

- .9332

.0

668

p-Value isP(Zu1.50) = 0.0668


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p -Value Solution(continued)

01.50

Z

Reject

(p-Value = 0.0668) u (E = 0.05)

Do Not Reject.

p Value = 0.0668

E = 0.05

Test Statistic 1.50 is in the Do Not Reject Region

1.645


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One-tail Z Test for Mean( Known) in PHStat

PHStat | one-sample tests | Z test for themean, sigma known

Example in excel spreadsheet

W

Microsoft Excel

Worksheet


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Example: Two-Tail Test

Q. Does an average box

of cereal contain 368

grams of ce

real? A

random sample of 25

boxes showed =

372.5. The company

has specified W to be15 grams. Test at the

E!0.05 level.

368 gm.

H0:Q !368

H1:Q { 368

X


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372.5 3681.5015

25

X

Zn

Q

W

! ! !E = 0

.0

5n= 25

Critical Value: 1.96

Example Solution: Two-Tail Test

Test Statistic:

Decision:

Conclusion:Do Not Reject atE = .05

No Evidence that TrueMean is Not 368Z0 1.96

.025

Reject

-1.96

.025

H0:Q!368

H1:Q{ 368

1.50


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p-Value Solution

(p Value = 0.1336) u (E = 0.05)

Do Not Reject.

01.50

Z

Reject

E = 0.05

1.96

p Value = 2 x 0.0668


Reject


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PHStat | one-sample tests | Z test for themean, sigma known


Two-tail Z Test for Mean( Known) in PHStatW

Microsoft ExcelW

orksheet


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For 372.5, 15 and 25,

the 95% confidence interval is:

372.5 1.96 15 / 25 372.5 1.96 15 / 25

or

366.62 378.38

If this interval contains the hypothesized mean (368),

we donot reject the null hypothesis.

I

X nW

Q

Q

! ! !

e e

e e

t does. Donot reject.

Connection toConfidence Intervals


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t Test: Unknown

Assumption

Population is normally distributed

If not normal, requires a large sample T test statistic with n-1 degrees of freedom

W

/

Xt

S n

Q!


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Example: One-Tail t Test

Does an average box of

cereal contain more than

368 grams of cereal? Arandom sample of 36

boxes showed X = 372.5,

and s! 15. Test at the E!

0.01 level.

368 gm.

H0:Qe 368

H1: Q" 368

W is not given


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Example Solution: One-Tail

E = 0

.01

n= 36, df = 35

Critical Value: 2.4377

Test Statistic:

Decision:

Conclusion:

Do Not Reject at E = .01

No evidence that true

mean is more than 368t350 2.437

7

.01

Reject

H0:Qe368

H1:Q" 368

372.5 3681.80

1536

Xt

Sn

Q ! ! !

1.80


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p -Value Solution

01.80

t35

Reject

(p Value is between .025 and .05) u (E = 0.01).

Do Not Reject.

p Value = [.025, .05]

E

=0.01


2.4377


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PHStat | one-sample tests | t test for themean, sigma known


t Test: Unknown in PHStatW

Microsoft ExcelW

orksheet


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Proportion

Involves categorical values

Two possible outcomes

Success (possesses a certain characteristic) andFailure (does not possesses a certaincharacteristic)

Fraction or proportion of population in the

success category is denoted by p


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Proportion

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

(continued)

Numberof SuccessesSample Size

s

Xpn

! !

sp pQ !

(1 )s

p

p p

nW

!


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Example: Z Test for Proportion

Q. A marketing companyclaims that it receives

4% responses from itsmailing. To test thisclaim, a randomsample of 500 were

surveyed with 25responses. Test at theE = .05 significancelevel.

Check:

500 .04 20

5

1 500 1 .04

480 5

np

n p

! !

u

!

! u


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.05 .041.14

1 .04 1 .04

500

Sp p

Zp p

n

$ ! !

Z Test for Proportion: Solution

E = .05

n = 500

Do not reject atE = .05

H0:p !.04

H1:p { .04

Critical Values:s 1.96

Test Statistic:

Decision:

Conclusion:

Z0

Reject Reject

.025.025

1.96-1.96

1.14

We do not have sufficientevidence to reject thecompanys claim of 4%

response rate.


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p -Value Solution

(p Value = 0.2542) u (E = 0.05).

Do Not Reject.

01.14

Z

Reject

E = 0.05

1.96

p Value = 2 x .1271


Reject


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Z Test for Proportion in PHStat

PHStat | one-sample tests | z test for theproportion


Microsoft Excel

Worksheet


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Potential Pitfalls andEthical Considerations

Randomize data collection method to reduce

selection biases

Do not manipulate the treatment of human

subjects without informed consent

Do not employ data snooping to choose

between one-tail and two-tail test, or to

determine the level of significance


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Potential Pitfallsand Ethical Considerations

Do not practice data cleansing to hide

observations that do not support a stated

hypothesis

Report all pertinent findings

(continued)


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Chapter Summary

Addressed hypothesis testing methodology

Performed Z Test for the mean ( Known)

Discussed p Value approach to hypothesis

testing

Made connection to confidence interval

estimation

W


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Chapter Summary

Performed one-tail and two-tail tests

Performed t test for the mean ( unknown) Performed Z test for the proportion

Discussed potential pitfalls and ethical

considerations

(continued)

W

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Statistics-Fundamentals of Hypothesis Testing

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