# chapter 8 hypothesis testing ( 假设检验 )

59
Chapter 8 Hypothesis Testing ( 假假假假 ) The Null and Alternative The Null and Alternative Hypotheses and Errors in Hypotheses and Errors in Hypothesis Testing. Hypothesis Testing. z z Tests about a Population Tests about a Population Mean ( known): Mean ( known): t t Tests about a Population Tests about a Population Mean ( unknown) Mean ( unknown)

Post on 03-Jan-2016

77 views

Category:

## Documents

DESCRIPTION

Chapter 8 Hypothesis Testing ( 假设检验 ). The Null and Alternative Hypotheses and Errors in Hypothesis Testing. z Tests about a Population Mean ( known): t Tests about a Population Mean ( unknown). Chapter 8 Hypothesis Testing ( 假设检验 ). z Tests about a Population Proportion - PowerPoint PPT Presentation

TRANSCRIPT

Chapter 8 Hypothesis Testing(假设检验 )

The Null and Alternative Hypotheses The Null and Alternative Hypotheses

and Errors in Hypothesis Testing. and Errors in Hypothesis Testing.

( known): ( known):

( unknown)( unknown)

Type‖Error Probabilities and Sample Type‖Error Probabilities and Sample

Size DeterminationSize Determination

The Chi-Square DistributionThe Chi-Square Distribution

Statistical Inference for a Population Statistical Inference for a Population

VarianceVariance

Chapter 8 Hypothesis Testing(假设检验 )

Learning Objectives

In this chapter, you learn: The basic principles of hypothesis testing

How to use hypothesis testing to test a mean, proportion and population variance.

The assumptions of each hypothesis-testing procedure, how to evaluate them, and the consequences if they are seriously violated

What is a Hypothesis(假设 )?

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

population mean

population proportion

Example: The mean monthly cell phone bill of this city is μ = \$42

Example: The proportion of adults in this city with cell phones is p = 0.68

States the claim or assertion to be tested

Example: The average number of TV sets in

U.S. Homes is equal to three ( )

The Null Hypothesis(零假设 ), H0

3μ:H0

3μ:H0 3X:H0

The Null Hypothesis, H0

Begin with the assumption that the null hypothesis is true Similar to the notion of innocent until

proven guilty Refers to the Always contains “=” , “≤” or “” sign May or may not be rejected

(continued)

existing state

The Alternative Hypothesis(备择假设 ) Ha

Is the opposite of the null hypothesis e.g., The average number of TV sets in U.S.

homes is not equal to 3 ( Ha: μ ≠ 3 )

Challenges the existing state Never contains the “=” , “≤” or “” sign May or may not be proven Is generally the hypothesis that the

researcher is trying to prove

Example 8.1Example 8.1 Trash Bag Case

• Tests show that the current trash bag has a mean breaking strength close to but not exceeding 50 lbs

• The null hypothesis H0 is that the new bag has a mean breaking strength that is 50 lbs or less

• The new bag’s mean breaking strength is not known and is in question, but it is hoped that the new bag is stronger than the current one

• The alternative hypothesis Ha is that the new bag has a mean breaking strength that exceeds 50 lbs

H0: 50 vs. Ha: > 50

Example 8.2Example 8.2 Payment Time Case

• With a new billing system, the mean bill paying time is hoped to be less than 19.5 days

• The alternative hypothesis Ha is that the new billing system has a mean payment time that is less than 19.5 days

• With the old billing system, the mean bill paying time was close to but not less than 39 days

• The null hypothesis H0 is that the new billing system has a mean payment time close to but not less than 19.5 days

H0: 19.5 vs. Ha: < 19.5

As a result of testing H0 vs. Ha, will have to decide either of the following decisions for the null hypothesis H0:

Do not reject H0

OR Reject H0

Types of Decisions

Population

Claim: thepopulationmean age is 50.(Null Hypothesis:

REJECT

Supposethe samplemean age is 20: X = 20

SampleNull Hypothesis

20 likely if μ = 50?Is

Hypothesis Testing(假设检验 ) Process

If not likely,

Now select a random sample

H0: μ = 50 )

X

Sampling Distribution of X

μ = 50If H0 is true

If it is unlikely that we would get a sample mean of this value ...

... then we reject the null

hypothesis that μ = 50.

Reason for Rejecting H0

20

... if in fact this was the population mean…

X

Level of Significance(显著水平 ),

Defines the unlikely values of the sample statistic if the null hypothesis is true

Defines rejection region(拒绝域 ) of the sampling distribution

Is designated by , (level of significance)

Typical values are 0.01, 0.05, or 0.10

Is selected by the researcher at the beginning

Provides the critical value(s) of the test

Level of Significance and the Rejection Region

H0: μ ≥ 19.5

Ha: μ < 19.519.5

H0: μ ≤ 50

Ha: μ > 50

Represents critical value

Lower-tail test

Level of significance =

50Upper-tail test

Two-tail test

/2

3

/2H0: μ = 3

Ha: μ ≠ 3

Type I Error(第一类误差 ) Rejecting null hypothesis when it is true Considered a serious type of error

The probability of Type I Error is

Called level of significance of the testSet by the researcher in advance

Errors in Making Decisions

0H

Type II Error(第二类误差 ) Failing to reject the null hypothesis when it is

false

The probability of Type II Error is β

Errors in Making Decisions(continued)

0H

Outcomes and Probabilities

Actual SituationDecision

Do NotReject

H0

No error (1 - )

Type II Error ( β )

RejectH0

Type I Error( )

Possible Hypothesis Test Outcomes

H0 False H0 True

Key:Outcome

(Probability) No Error ( 1 - β )

Type I & II Error Relationship

Type I and Type II errors cannot happen at the 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 ( ) , then

Type II error probability ( β )

Hypothesis Tests for the Mean

Known Unknown

Hypothesis Tests for

(Z test) (t test)

Z Test of Hypothesis for the Mean (σ Known)

Convert sample statistic ( ) to a Z test statistic(Z检验统计量 )

X

The test statistic is:

n

σμX

Z

σ Known σ Unknown

Hypothesis Tests for

Known Unknown(Z test) (t test)

Critical Value Approach to Testing

Convert sample statistic ( ) to test statistic (Z statistic )

Determine the critical Z values for a specifiedlevel of significance from a table or computer

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

reject H0

X

Do not reject H0 Reject H0Reject H0

There are two cutoff values (critical values), defining the regions of rejection

Two-Tail Tests(双边检验 )

/2

-Z 0

H0: μ = 3

Ha: μ

3

+Z

/2

Lower critical value

Upper critical value

3

Z

X

One-Tail Tests(单边检验 )

In many cases, the alternative hypothesis focuses on a particular direction

H0: μ ≥ 19.5

Ha: μ < 19.5

H0: μ ≤ 50

Ha: μ > 50

This is a lower-tail test since the alternative hypothesis is focused on the lower tail below the mean of 19.5

This is an upper-tail test since the alternative hypothesis is focused on the upper tail above the mean of 50

Reject H0 Do not reject H0

There is only one

critical value, since

the rejection area is

in only one tail

Lower-Tail Tests

-Z 0

μ

H0: μ ≥ 19.5

Ha: μ < 19.5

Z

X

Critical value

Reject H0Do not reject H0

Upper-Tail Tests

Zα0

μ

H0: μ ≤ 50

Ha: μ > 50 There is only one

critical value, since

the rejection area is

in only one tail

Critical value

Z

X_

6 Steps in Hypothesis Testing

1. State the null hypothesis, H0 and the alternative hypothesis, Ha

2. Choose the level of significance, , and the sample size, n

3. Determine the appropriate test statistic and sampling distribution

4. Determine the critical values that divide the rejection and non-rejection regions

6 Steps in Hypothesis Testing

5. Collect data and compute the value of the test statistic

6. Make the statistical decision and state the managerial conclusion. If the test statistic falls into the non-rejection region, do not reject the null hypothesis H0. If the test statistic falls into the rejection region, reject the null hypothesis. Express the managerial conclusion in the context of the problem

(continued)

Hypothesis Testing Example

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

(Assume σ = 0.8)

1. State the appropriate null and alternative hypotheses

H0: μ = 3 Ha: μ ≠ 3 (This is a two-tail test)

2. Specify the desired level of significance and the sample size Suppose that = 0.05 and n = 100 are chosen

for this test

Example 8.3Example 8.3

2.0.08

.16

100

0.832.84

n

σμX

Z

Hypothesis Testing Example

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

4. Determine the critical values For = 0.05 the critical Z values are ±1.96

5. Collect the data and compute the test statistic

Suppose the sample results are

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

So the test statistic is:

(continued)

Reject H0 Do not reject H0

6. Is the test statistic in the rejection region?

= 0.05/2

-Z= -1.96 0Reject H0 if Z < -1.96 or Z > 1.96; otherwise do not reject H0

Hypothesis Testing Example(continued)

= 0.05/2

Reject H0

+Z= +1.96

Here, Z = -2.0 < -1.96, so the test statistic is in the rejection region

6(continued). Reach a decision and interpret the result

-2.0

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

Hypothesis Testing Example(continued)

Reject H0 Do not reject H0

= 0.05/2

-Z= -1.96 0

= 0.05/2

Reject H0

+Z= +1.96

Tests show that the current trash bag has a mean breaking strength μ close to but not exceeding 50 lbs, The new bag’s mean breaking strength is not known and is in question, but it is hoped that the new bag is stronger than the current one

H0: μ ≤ 50 the average breaking strength is not over 50

lbs

Ha: μ > 50 the average breaking strength for new trash

bag is greater than 50 lbs.

Form hypothesis test:

Example 8.4Example 8.4 Trash Bag Case ( is known)

Reject H0Do not reject H0

Suppose that = 0.05 is chosen for this test

Find the rejection region:

= 0.05

1.6450

Reject H0

Reject H0 if Z > 1.645

Case Study: Find Rejection Region(continued)

Review:One-Tail Critical Value

Z .03 .05

1.5 .4370 .4382 .4394

1.6 .4484 .4505

1.7 .4582 .4591 .4599z 0 1.645

.04

Standardized Normal Distribution Table (Portion)What is Z given = 0.05?

= 0.05

Critical Value = 1.645

0.45

.4495

20.2

40

1.655050.575

n

σμX

Z

Obtain sample and compute the test statistic

Suppose a sample is taken with the following results: n = 40, X = 50.575 (=1.65 was assumed

known)

Then the test statistic is:

Case Study: Test Statistic(continued)

Reject H0Do not reject H0

Case Study: Decision

= 0.05

1.6450

Reject H0

Reject H0 since Z = 2.20 ≥1.645

i.e.: there is strong evidence that the mean breaking

strength for new trash bag is over 50lbs

Z = 2.20

Reach a decision and interpret the result:(continued)

Discussion Question

Using Critical Value Approach In this payment time case, we assume that is

known and = 4.2 days. A sample of n = 65,

= 18.1077 days Have we strong evidence that the mean

payment time for new billing system is less than 19.5 days at =0.01 significance level ?

x

p-Value (p 值 ) Approach to Testing

p-value: Probability of obtaining a test statistic more extreme ( ≤ or ) than the observed sample value given H0 is

true

Also called observed level of significance

Smallest value of for which H0 can be

rejected

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

Obtain the p-value from a table or computer based on test statistic.

Compare the p-value with

If p-value < , reject H0

If p-value , do not reject H0

p-Value Approach to Testing

X

(continued)

Calculate the test statistic and the corresponding p-value

Rate the strength of the conclusion about the null hypothesis H0 according to these rules:

If p < 0.10, then there is some evidence to reject H0

If p < 0.05, then there is strong evidence to reject H0

If p < 0.01, then there is very strong evidence to reject H0

If p < 0.001, then there is extremely strong evidence to reject H0

Weight of Evidence Against the Null

0.0228

/2 = 0.025

p-Value Example

How likely is it to see a sample mean of 2.84 (or something further from the mean, in either direction) if the true mean is = 3.0?

-1.96 0

-2.0

0.02282.0)P(Z

0.02282.0)P(Z

Z1.96

2.0

X = 2.84 is translated to a Z score of Z = -2.0

p-value

= 0.0228 + 0.0228 = 0.0456

0.0228

/2 = 0.025

Example 8.5Example 8.5

Compare the p-value with If p-value < , reject H0

If p-value , do not reject H0

Here: p-value = 0.0456 = 0.05

Since 0.0456 < 0.05, we reject the null hypothesis

(continued)

p-Value Example

0.0228

/2 = 0.025

-1.96 0

-2.0

Z1.96

2.0

0.0228

/2 = 0.025

Connection to Confidence Intervals

For X = 2.84, σ = 0.8 and n = 100, the 95% confidence interval is:

2.6832 ≤ μ ≤ 2.9968

Since this interval does not contain the hypothesized mean (3.0), we reject the null hypothesis at = 0.05

100

0.8 (1.96) 2.84 to

100

0.8 (1.96) - 2.84

Reject H0

= 0.05

Do not reject H0 1.645

0

Reject H0

Z = 2.20

Calculate the p-value and compare to (assuming that μ = 50)

0.0139

0.48610.52.20)P(Z

401.65/

5050.575ZP

50.575)XP(

p-value = 0.0139

p -Value Solution for trash bag case

Reject H0 since p-value = 0.0139 ＜ = 0.05

Example 8.6Example 8.6

Question

Using p-Value Approach In this payment time case, we assume that is

known and = 4.2 days. A sample of n = 65,

= 18.1077 days Are there strong evidence that the mean

payment time for new billing system is less than 19.5 days at =0.01 significance level ?

x

t Test of Hypothesis for the Mean (σ Unknown)

Convert sample statistic ( ) to a t test statistic X

The test statistic is:

n

sμX

t 1-n

Hypothesis Tests for

σ Known σ Unknown Known Unknown(Z test) (t test)

Two-Tail Test( Unknown)

The average cost of a hotel room in New York is said to be \$168 per night. A random sample of 25 hotels resulted in X = \$172.50 and

s = \$15.40. Test at the

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

H0: μ=

168 Ha:

μ 168

Example 8.7Example 8.7

= 0.05

n = 25

is unknown, so use a t statistic

Critical Value:

t24 = ± 2.0639

Example Solution: Two-Tail Test

Do not reject H0: not sufficient evidence that true mean cost is different than \$168

Reject H0Reject H0

/2=.025

-t n-1,α/2

Do not reject H0

0

/2=.025

-2.0639 2.0639

1.46

25

15.40168172.50

n

SμX

t 1n

1.46

H0: μ=

168 Ha:

μ 168t n-1,α/2

Connection to Confidence Intervals

For X = 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 ≤ μ ≤ 178.86

Since this interval contains the Hypothesized mean (168), we do not reject the null hypothesis at = 0.05

Alternative

Reject H0 if: p-value

Ha: > 0 t > t Area under t distribution to right of t

Ha: < 0 t < –t Area under t distribution to left of –t

Ha: 0 |t| > t /2 * Twice area under t distribution to right of |t|

t, t/2, and p-values are based on n – 1 degrees of freedom (for a sample of size n)

* either t > t/2 or t < –t/2

t Test of Hypothesis for the Mean (σ Unknown) Continued

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

Sample proportion in the success category is denoted by

When both np and n(1-p) are at least 5, can be approximated by a normal distribution with mean and standard deviation

Proportions

sizesample

sampleinsuccessesofnumber

n

Xp^

pp^μ

n

)(1σ ^

p

pp

(continued)

^

p

^

p

Hypothesis Tests for Proportions

The sampling distribution of is approximately normal, so the test statistic is a Z value:

n)(1

pZ

^

pp

p

np 5and

n(1-p) 5

Hypothesis Tests for p

np < 5or

n(1-p) < 5

Not discussed in this chapter

^

p

An equivalent form to the last slide, but in terms of the number of successes, X:

Z Test for Proportionin Terms of Number of Successes

)(1n

nXZ

pp

p

X 5and

n-X 5

Hypothesis Tests for X

X < 5or

n-X < 5

Not discussed in this chapter

Z Test for Proportion

A marketing company claims that it receives 8% responses from its mailing. To test this claim, a random sample of 500 were surveyed with 25 responses. Test at the = 0.05 significance level.

Check:

n p= (500)(.08) = 40

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

Example 8.8Example 8.8

Z Test for Proportion: Solution

= 0.05

n = 500, = 0.05

Reject H0 at = 0.05

H0: p = 0.08

Ha: p

0.08

Critical Values: ± 1.96

Test Statistic:

Decision:

Conclusion:

z0

Reject Reject

.025.025

1.96

-2.47

There is sufficient evidence to reject the company’s claim of 8% response rate.

2.47

500.08).08(1

.08.05

n)(1

pZ

^

pp

p

-1.96

^

p

Do not reject H0

Reject H0Reject H0

/2 = .025

1.960

Z = -2.47

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

(continued)

0.01362(0.0068)

2.47)P(Z2.47)P(Z

p-value = 0.0136:

p-Value Solution

Reject H0 since p-value = 0.0136 < = 0.05

Z = 2.47

-1.96

/2 = .025

0.00680.0068

Hypothesis Tests about aPopulation Proportion Continued

* either z > z/2 or z < –z/2

where the test statistic is: n

pp

pp̂z

00

0

1

0

0

0

:

:

:

ppH

ppH

ppH

a

a

a

2/zz

zz

zz

Alternative

Reject H0 if: p-value

Area under standard normal to the right of z

Area under standard normal to the left of –z

Twice the area under standard normal to the right of |z|

*

Chapter Summary