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Chapter 8 Hypothesis Testing ( 假设检验 )PowerPoint Presentation

Chapter 8 Hypothesis Testing ( 假设检验 )

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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
- Type‖Error Probabilities and Sample Size Determination
- The Chi-Square Distribution
- Statistical Inference for a Population Variance

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

The Null Hypothesis(零假设), H0

- States the claim or assertion to be tested
Example: The average number of TV sets in U.S. Homes is equal to three ( )

- Is always about a population parameter, not about a sample statistic

The Null Hypothesis, H0

(continued)

- 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

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 8.1

- Tests show that the current trash bag has a mean breaking strength m 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 8.2

- With a new billing system, the mean bill paying time m 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 m 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

Hypothesis Testing(假设检验) Process

Claim:the

population

mean age is 50.

(Null Hypothesis:

Population

H0: μ = 50 )

Now select a random sample

X

=

likely if μ = 50?

Is

20

Suppose

the sample

If not likely,

REJECT

mean age

is 20: X = 20

Sample

Null Hypothesis

Reason for Rejecting H0

Sampling Distribution of X

X

20

μ= 50

IfH0 is true

... then we reject the null hypothesis that μ = 50.

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

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

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

a

Level of significance =

Represents

critical value

a

a

H0: μ = 3 Ha: μ ≠ 3

/2

/2

Rejection region is shaded

Two-tail test

3

H0: μ ≤ 50 Ha: μ > 50

a

50

Upper-tail test

H0: μ ≥ 19.5 Ha: μ < 19.5

a

Lower-tail test

19.5

Errors in Making Decisions

- 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 test
- Set by the researcher in advance

Errors in Making Decisions

(continued)

- 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

- Failing to reject the null hypothesis when it is false

Outcomes and Probabilities

Possible Hypothesis Test Outcomes

Actual Situation

Decision

H0 True

H0 False

Do Not

No error

(1 - )

Type II Error

( β )

Reject

Key:

Outcome

(Probability)

a

H

0

Reject

Type I Error

( )

No Error

( 1 - β )

H

a

0

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 )

Z Test of Hypothesis for the Mean (σ Known)

X

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

Hypothesis

Tests for

Known

σ Known

Unknown

σ Unknown

(Z test)

(t test)

The test statistic is:

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

Two-Tail Tests(双边检验)

H0: μ = 3 Ha: μ¹ 3

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

/2

/2

X

3

Reject H0

Do not reject H0

Reject H0

Z

-Z

+Z

0

Lower critical value

Upper critical value

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

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

H0: μ ≤ 50

Ha: μ > 50

H0: μ ≥ 19.5

Ha: μ < 19.5

- There is only one critical value, since the rejection area is in only one tail

a

Reject H0

Do not reject H0

Z

-Z

0

μ

X

Critical value

H0: μ ≤ 50

Ha: μ > 50

- There is only one critical value, since the rejection area is in only one tail

a

Do not reject H0

Reject H0

Z

Zα

0

_

μ

X

Critical value

6 Steps in Hypothesis Testing

- State the null hypothesis, H0 and the alternative hypothesis, Ha
- Choose the level of significance, , and the sample size, n
- Determine the appropriate test statistic and sampling distribution
- Determine the critical values that divide the rejection and non-rejection regions

6 Steps in Hypothesis Testing

(continued)

- Collect data and compute the value of the test statistic
- 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

Hypothesis Testing Example

Example 8.3

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

Hypothesis Testing Example

(continued)

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:

Hypothesis Testing Example

(continued)

- 6. Is the test statistic in the rejection region?

= 0.05/2

= 0.05/2

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

Reject H0

Do not reject H0

Reject H0

-Z= -1.96

0

+Z= +1.96

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

Hypothesis Testing Example

(continued)

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

= 0.05/2

= 0.05/2

Reject H0

Do not reject H0

Reject H0

-Z= -1.96

0

+Z= +1.96

-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

Trash Bag Case ( is known)

Example 8.4

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

Form hypothesis test:

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.

Case Study: Find Rejection Region

(continued)

- Suppose that = 0.05 is chosen for this test
Find the rejection region:

Reject H0

= 0.05

Do not reject H0

Reject H0

1.645

0

Reject H0 if Z > 1.645

Review:One-Tail Critical Value

Standardized Normal Distribution Table (Portion)

What is Z given a = 0.05?

0.45

.04

Z

.03

.05

a= 0.05

1.5

.4370

.4382

.4394

.4484

.4495

.4505

1.6

z

0

1.645

.4582

1.7

.4591

.4599

Critical Value

= 1.645

(continued)

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:

(continued)

Reach a decision and interpret the result:

Reject H0

= 0.05

Do not reject H0

Reject H0

1.645

0

Z = 2.20

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

Discussion Question

- Using Critical Value Approach
- In this payment time case, we assume that s is known and s = 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 ?

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

p-Value Approach to Testing

(continued)

X

- 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

Weight of Evidence Against the Null

- 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

p-Value Example

Example 8.5

- 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?

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

/2 = 0.025

/2 = 0.025

0.0228

0.0228

p-value

= 0.0228 + 0.0228 = 0.0456

-1.96

0

1.96

Z

-2.0

2.0

p-Value Example

(continued)

- 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

/2 = 0.025

/2 = 0.025

0.0228

0.0228

-1.96

0

1.96

Z

-2.0

2.0

Connection to Confidence Intervals

- ForX = 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

p -Value Solution for trash bag case

Example 8.6

Calculate the p-value and compare to

(assuming that μ = 50)

p-value = 0.0139

Reject H0

= 0.05

0

Do not reject H0

Reject H0

1.645

Z = 2.20

Reject H0 since p-value = 0.0139 ＜ = 0.05

- Using p-Value Approach
- In this payment time case, we assume that s is known and s = 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 ?

t Test of Hypothesis for the Mean (σ Unknown)

X

- Convert sample statistic ( ) to a ttest statistic

Hypothesis

Tests for

Known

σ Known

Unknown

σ Unknown

(Z test)

(t test)

The test statistic is:

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)

Two-Tail Test( Unknown)Example 8.7

H0: μ= 168 Ha: μ ¹ 168

a $168 per night. A random sample of 25 hotels resulted in X = $172.50 and = 0.05

n= 25

is unknown, so

use a t statistic

Critical Value:

t24 = ± 2.0639

Example Solution: Two-Tail TestH0: μ= 168 Ha: μ ¹ 168

a/2=.025

a/2=.025

Reject H0

Do not reject H0

Reject H0

t n-1,α/2

-t n-1,α/2

0

2.0639

-2.0639

1.46

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

Connection to Confidence Intervals $168 per night. A random sample of 25 hotels resulted in X = $172.50 and

- 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

t Test of Hypothesis for the Mean ( $168 per night. A random sample of 25 hotels resulted in X = $172.50 and σ Unknown) Continued

- ta, ta/2, and p-values are based on n – 1 degrees of freedom (for a sample of size n)
* either t > ta/2 or t < –ta/2

Hypothesis Tests for Proportions $168 per night. A random sample of 25 hotels resulted in X = $172.50 and

- 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

Proportions $168 per night. A random sample of 25 hotels resulted in X = $172.50 and

(continued)

- 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

Hypothesis Tests for Proportions $168 per night. A random sample of 25 hotels resulted in X = $172.50 and

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

Hypothesis

Tests for p

np 5

and

n(1-p) 5

np < 5

or

n(1-p) < 5

Not discussed in this chapter

Z Test for Proportion $168 per night. A random sample of 25 hotels resulted in X = $172.50 and in Terms of Number of Successes

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

Hypothesis

Tests for X

X 5

and

n-X 5

X < 5

or

n-X < 5

Not discussed in this chapter

Z Test for Proportion $168 per night. A random sample of 25 hotels resulted in X = $172.50 and

Example 8.8

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:

np= (500)(.08) = 40

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

Z Test for Proportion: Solution $168 per night. A random sample of 25 hotels resulted in X = $172.50 and

Test Statistic:

H0: p = 0.08 Ha: p¹ 0.08

a= 0.05

n = 500, = 0.05

Decision:

Critical Values: ± 1.96

Reject H0 at = 0.05

Reject

Reject

Conclusion:

.025

.025

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

z

-1.96

0

1.96

-2.47

p-Value Solution $168 per night. A random sample of 25 hotels resulted in X = $172.50 and

(continued)

Calculate the p-value and compare to

(For a two-tail test the p-value is always two-tail)

Do not reject H0

Reject H0

Reject H0

p-value = 0.0136:

/2= .025

/2= .025

0.0068

0.0068

0

-1.96

1.96

Z = -2.47

Z = 2.47

Reject H0 since p-value = 0.0136 < = 0.05

Reject H $168 per night. A random sample of 25 hotels resulted in X = $172.50 and 0 if:

p-value

Alternative

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|

*

Hypothesis Tests about aPopulation Proportion Continued

where the test statistic is:

* either z > za/2 or z < –za/2

Chapter Summary $168 per night. A random sample of 25 hotels resulted in X = $172.50 and

- Addressed hypothesis testing methodology
- Performed Z Test for the mean (σ known)
- Discussed critical value and p–value approaches to hypothesis testing
- Performed one-tail and two-tail tests

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