© 1999 Prentice-Hall, Inc. Chap. 8 - 1 Chapter Topics Hypothesis Testing Methodology Z Test for the...

© 1999 Prentice-Hall, Inc. Chap. 8 - 1

Chapter Topics•Hypothesis Testing Methodology

•Z Test for the Mean (Known)

• p-Value Approach to Hypothesis Testing

•Connection to Confidence Interval Estimation

•One Tail Test

• t Test of Hypothesis for the Mean

•Z Test of Hypothesis for the Proportion


A hypothesis is an assumption about the population parameter.

A parameter is a Population mean or proportion

The parameter must be identified before analysis.

I assume the mean GPA of this class is 3.5!

© 1984-1994 T/Maker Co.

What is a Hypothesis?


• States the Assumption (numerical) to be tested

e.g. The average # TV sets in US homes is at least 3 (H0: 3)

• Begin with the assumption that the null hypothesis is TRUE.

(Similar to the notion of innocent until proven guilty)

The Null Hypothesis, H0

•Refers to the Status Quo•Always contains the ‘ = ‘ sign

•The Null Hypothesis may or may not be rejected.


• Is the opposite of the null hypothesise.g. The average # TV sets in US homes

is less than 3 (H1: < 3)

• Challenges the Status Quo

• Never contains the ‘=‘ sign

• The Alternative Hypothesis may or may not be accepted

The Alternative Hypothesis, H1


Steps: State the Null Hypothesis (H0: 3) State its opposite, the Alternative

Hypothesis (H1: < 3)Hypotheses are mutually exclusive &

exhaustiveSometimes it is easier to form the

alternative hypothesis first.

Identify the Problem


Population

Assume thepopulationmean age is 50.(Null Hypothesis)

REJECT

The SampleMean Is 20

SampleNull Hypothesis

50?20 XIs

Hypothesis Testing Process

No, not likely!


Sample Mean = 50

Sampling DistributionIt is unlikely that we would get a sample mean of this value ...

... if in fact this were the population mean.

... Therefore, we reject the null

hypothesis that = 50.

20H0

Reason for Rejecting H0


• Defines Unlikely Values of Sample Statistic if Null Hypothesis Is True Called Rejection Region of Sampling

Distribution

• Designated (alpha) Typical values are 0.01, 0.05, 0.10

• Selected by the Researcher at the Start

• Provides the Critical Value(s) of the Test

Level of Significance,


Level of Significance, and the Rejection Region

H0: 3

H1: < 30

0

0

H0: 3

H1: > 3

H0: 3

H1: 3

/2

Critical Value(s)

Rejection Regions


• Type I Error Reject True Null Hypothesis Has Serious Consequences Probability of Type I Error Is

Called Level of Significance

• Type II Error Do Not Reject False Null Hypothesis Probability of Type II Error Is (Beta)

Errors in Making Decisions


H0: Innocent

Jury Trial Hypothesis 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 - )

Result Possibilities


Reduce probability of one error and the other one goes up.

& Have an Inverse Relationship


• Convert Sample Statistic (e.g., ) to Standardized Z Variable

• Compare to Critical Z Value(s) If Z test Statistic falls in Critical Region, Reject

H0; Otherwise Do Not Reject H0

Z-Test Statistics (Known)

Test Statistic

X

n

XXZ

X

X


• Probability of Obtaining a Test Statistic More Extreme or ) than Actual Sample Value Given H0 Is True

• Called Observed Level of Significance Smallest Value of a H0 Can Be Rejected

• Used to Make Rejection Decision If p value Do Not Reject H0

If p value <, Reject H0

p Value Test


1. State H0 H0 : 3

2. State H1 H1 :

3. Choose = .05

4. Choose n n = 100

5. Choose Test: Z Test (or p Value)

Hypothesis Testing: Steps

Test the Assumption that the true mean # of TV sets in US homes is at least 3.


6. Set Up Critical Value(s) Z = -1.645

7. Collect Data 100 households surveyed

8. Compute Test Statistic Computed Test Stat.= -2

9. Make Statistical Decision Reject Null Hypothesis

10. Express Decision The true mean # of TV set is less than 3 in the US households.

Hypothesis Testing: Steps

Test the Assumption that the average # of TV sets in US homes is at least 3.

(continued)


• Assumptions Population Is Normally Distributed If Not Normal, use large samples Null Hypothesis Has or Sign Only

• Z Test Statistic:

One-Tail Z Test for Mean (Known)

n

xxz

x

x


Z0

Reject H0

Z0

Reject H0

H0: H1: < 0

H0: 0 H1: > 0

Must Be Significantly Below = 0

Small values don’t contradict H0

Don’t Reject H0!

Rejection Region


Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed X = 372.5. The company has specified to be 15 grams. Test at the 0.05 level.

368 gm.

Example: One Tail Test

H0: 368 H1: > 368

_


Z .04 .06

1.6 .5495 .5505 .5515

1.7 .5591 .5599 .5608

1.8 .5671 .5678 .5686

.5738 .5750

Z0

Z = 1

1.645

.50 -.05

.45

.05

1.9 .5744

Standardized Normal Probability Table (Portion)

What Is Z Given = 0.05?

= .05

Finding Critical Values: One Tail

Critical Value = 1.645


= 0.025

n = 25

Critical Value: 1.645

Test Statistic:

Decision:

Conclusion:

Do Not Reject at = .05

No Evidence True Mean Is More than 368Z0 1.645

.05

Reject

Example Solution: One Tail

H0: 368 H1: > 368 50.1

n

XZ


Z0 1.50

p Value.0668

Z Value of Sample Statistic

From Z Table: Lookup 1.50

.9332

Use the alternative hypothesis to find the direction of the test.

1.0000 - .9332 .0668

p Value is P(Z 1.50) = 0.0668

p Value Solution


0 1.50 Z

Reject

(p Value = 0.0668) ( = 0.05). Do Not Reject.

p Value = 0.0668

= 0.05

Test Statistic Is In the Do Not Reject Region

p Value Solution


Does an average box of cereal contains 368 grams of cereal? A random sample of 25 boxes showed X = 372.5. The company has specified to be 15 grams. Test at the 0.05 level.

368 gm.

Example: Two Tail Test

H0: 368

H1: 368


= 0.05

n = 25

Critical Value: ±1.96

Test Statistic:

Decision:

Conclusion:

Do Not Reject at = .05

No Evidence that True Mean Is Not 368Z0 1.96

.025

Reject

Example Solution: Two Tail

-1.96

.025

H0: 386

H1: 38650.1

2515

3685.372

n

XZ


Connection to Confidence Intervals

For X = 372.5oz, = 15 and n = 25,

The 95% Confidence Interval is:

372.5 - (1.96) 15/ 25 to 372.5 + (1.96) 15/ 25

or

366.62 378.38

If this interval contains the Hypothesized mean (368), we do not reject the null hypothesis.

It does. Do not reject.

_


Assumptions Population is normally distributed If not normal, only slightly skewed & a large

sample taken

Parametric test procedure

t test statistic

t-Test: Unknown

nSX

t


Example: One Tail t-Test

Does an average box of cereal contain more than 368 grams of cereal? A random sample of 36 boxes showed X = 372.5, and 15. Test at the 0.01 level.

368 gm.

H0: 368 H1: 368

is not given,


Example:Z Test for Proportion

•Problem: A marketing company claims that it receives 4% responses from its Mailing.

•Approach: To test this claim, a random sample of 500 were surveyed with 25 responses.

•Solution: Test at the = .05 significance level.


= .05

n = 500

Do not reject at Do not reject at = .05

Z Test for Proportion: Solution

H0: p .04

H1: p .04

Critical Values: 1.96

Test Statistic:

Decision:

Conclusion:We do not have sufficient

evidence to reject the company’s claim of 4% response rate.

Z p - p

p (1 - p)n

s=

.04 -.05.04 (1 - .04)

500

= 1.14

Z0

Reject Reject

.025.025

© 1999 Prentice-Hall, Inc. Chap. 8 - 1 Chapter Topics Hypothesis Testing Methodology Z Test for the...

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Transcript of © 1999 Prentice-Hall, Inc. Chap. 8 - 1 Chapter Topics Hypothesis Testing Methodology Z Test for the...