business statistics: a decision-making approach, …chap 9-3 what is a hypothesis? we perform of...

Business Statistics

Department of Quantitative Methods & Information Systems

Dr. Mohammad Zainal QMIS 220

Chapter 9

Introduction to

Hypothesis Testing

Chap 9-2

Chapter Goals

After completing this chapter, you should be

able to:

Formulate null and alternative hypotheses for applications involving a single population mean, proportion and variance

Formulate a decision rule for testing a hypothesis

Know how to use the test statistic, critical value, and p-value approaches to test the null hypothesis

Know what Type I and Type II errors are

Compute the probability of a Type II error

QMIS 220, by Dr. M. Zainal

Chap 9-3

What is a Hypothesis?

We perform of hypothesis only when we are

making a decision about a population

parameter based the value of a sample statistic.

Consider an example of a person who has been

indicted for committing a crime and is being

tried in a court.

Based on the available evidence, the judge will

make one of the following decisions:

The person is not guilty (null hypothesis, Ho).

The person is guilty (alternative hypothesis, H1).


Chap 9-4


A null hypothesis is a claim about a population

parameter that is assumed to be true until it is

declared false.

An alternative hypothesis is a claim about a

population parameter that will be true if the null

hypothesis is false.

The alternative hypothesis is often what the test

is attempting to establish.


Chap 9-5


The research hypothesis should be expressed

as the alternative hypothesis

Manufacturers’ claims are usually given the

benefit of the doubt and stated as the null

hypothesis.


Chap 9-6


In the court example, we start the prosecution

by assuming the person is not guilty.

The prosecutors collect all possible evidences

to prove the null hypothesis is false (guilty).

Rejection and nonrejection regions

Not enough evidence to declare the

person guilty and, hence, the null

hypothesis is not rejected in this region

Enough evidence to declare the person

guilty and, hence, the null hypothesis is

rejected in this region

Nonrejection region Rejection region

C

Critical

point

Level of

evidence 0


Chap 9-7


Actual Situation

The person is not guilty

The person is guilty

Court’s decision

The person is not guilty

Correct decision Type II or

error

The person is guilty Type I or

error Correct decision


Chap 9-8


Remember: 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 = .68


Chap 9-9

The Null Hypothesis, H0

States the assumption (numerical) to be

tested

Example: 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

3μ:H0

3μ:H0 3x:H0


Chap 9-10

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 status quo

Always contains “=” , “≤” or “” sign

May or may not be rejected

(continued)


Chap 9-11

The Alternative Hypothesis, HA

Is the opposite of the null hypothesis

e.g.: The average number of TV sets in U.S.

homes is less than 3 ( HA: µ < 3 )

Challenges the status quo

Never contains the “=” , “≤” or “” sign

May or may not be accepted

Is generally the hypothesis that is believed

(or needs to be supported) by the

researcher – a research hypothesis


Chap 9-12

Formulating Hypotheses

Left-Tailed

Test

Right-Tailed

Test

Two-Tailed

Test

Sign in the null

hypothesis Ho = or = or =

Sign in the alternative

hypothesis HA < >

Critical region In the left tail In the right tail In both tails


Chap 9-13


Example 1: Ford motor company has

worked to reduce road noise inside the cab

of the redesigned F150 pickup truck. It

would like to report in its advertising that

the truck is quieter. The average of the

prior design was 68 decibels at 60 mph.

What is the appropriate hypothesis test?


Chap 9-14


Example 1: Ford motor company has worked to reduce road noise inside

the cab of the redesigned F150 pickup truck. It would like to report in its

advertising that the truck is quieter. The average of the prior design was

68 decibels at 60 mph.

What is the appropriate test?

H0: µ ≥ 68 (the truck is not quieter) status quo

HA: µ < 68 (the truck is quieter) wants to support

If the null hypothesis is rejected, Ford has sufficient evidence to support that the truck is now quieter.


Chap 9-15


Example 2: The average annual income of

buyers of Ford F150 pickup trucks is

claimed to be $65,000 per year. An

industry analyst would like to test this

claim.

What is the appropriate hypothesis test?


Chap 9-16


Example 1: The average annual income of buyers of Ford F150 pickup trucks is claimed to be $65,000 per year. An industry analyst would like to test this claim.

What is the appropriate test?

H0: µ = 65,000 (income is as claimed) status quo

HA: µ ≠ 65,000 (income is different than claimed)

The analyst will believe the claim unless sufficient evidence is found to discredit it.


Population

Claim: the population mean age is 50.

Null Hypothesis:

REJECT

Suppose the sample

mean age is 20:

x = 20

Sample

Null Hypothesis

Is x = 20

likely if

µ = 50?

Hypothesis Testing Process

If not likely,

Now select a random sample:

H0: µ = 50

Chap 9-17 QMIS 220, by Dr. M. Zainal

Chap 9-18

Sampling Distribution of x

μ = 50 If 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 were

the population mean…

x


Chap 9-19

Errors in Making Decisions

Type I Error

Reject a true null hypothesis

Considered a serious type of error

The probability of Type I Error is

Called level of significance of the test

Set by researcher in advance


Chap 9-20

Errors in Making Decisions

Type II Error

Fail to reject a false null hypothesis

The probability of Type II Error is β

β is a calculated value, the formula is

discussed later in the chapter

(continued)


Chap 9-21

Outcomes and Probabilities

State of Nature

Decision

Do Not

Reject

H 0

No error

(1 - )

Type II Error

( β )

Reject

H 0

Type I Error

( )

Possible Hypothesis Test Outcomes

H0 False H0 True

Key:

Outcome

(Probability) No Error

( 1 - β )


Chap 9-22

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 )


Chap 9-23

Factors Affecting Type II Error

All else equal,

β when the difference between

hypothesized parameter and its true value

β when

β when σ

β when n

The formula used to

compute the value of β

is discussed later in the

chapter


If you increase the significance level, you reduce

the region of acceptance. As a result, you are

more likely to reject the null hypothesis. This

means you are less likely to accept the null

hypothesis when it is false

http://stattrek.com/Help/Glossary.aspx?Target=Region of acceptance

Chap 9-24

Level of Significance,

Defines unlikely values of sample statistic if

null hypothesis is true

Defines rejection region of the sampling

distribution

Is designated by , (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


Chap 9-25

Hypothesis Tests for the Mean

σ Known σ Unknown

Hypothesis

Tests for

Assume first that the population

standard deviation σ is known


Chap 9-26

1. Specify population parameter of interest

2. Formulate the null and alternative hypotheses

3. Specify the desired significance level, α

4. Define the rejection region

5. Take a random sample and determine

whether or not the sample result is in the

rejection region

6. Reach a decision and draw a conclusion

Process of Hypothesis Testing


Chap 9-27

Level of Significance and the Rejection Region

H0: μ ≥ 3

HA: μ < 3

0

H0: μ ≤ 3

HA: μ > 3

H0: μ = 3

HA: μ ≠ 3

/2

Lower tail test

Level of significance =

0

/2

Upper tail test Two tailed test

0

-zα zα -zα/2 zα/2

Reject H0 Reject H0 Reject H0 Reject H0 Do not

reject H0

Do not

reject H0

Do not

reject H0

Example: Example: Example:


Chap 9-28

Reject H0 Do not reject H0

The cutoff value, or ,

is called a critical value

-zα

xα

-zα xα

0

µ=3

H0: μ ≥ 3

HA: μ < 3

n

σzμx

Critical Value for Lower Tail Test


Chap 9-29


Critical Value for Upper Tail Test

zα

xα

0

H0: μ ≤ 3

HA: μ > 3

n

σzμx

µ=3

The cutoff value, or ,

is called a critical value

zα xα


Chap 9-30

Do not reject H0 Reject H0 Reject H0

There are two cutoff

values (critical values):

or

Critical Values for Two Tailed Tests

/2

-zα/2

xα/2

± zα/2

xα/2

0

H0: μ = 3

HA: μ 3

zα/2

xα/2

n

σzμx /2/2

Lower

Upper xα/2

Lower Upper

/2

µ=3


Chap 9-31

The Rejection Region

H0: μ ≥ 3

HA: μ < 3

0

H0: μ ≤ 3

HA: μ > 3

H0: μ = 3

HA: μ ≠ 3

/2

Lower tail test

0

/2

Upper tail test Two tailed test

0

-zα zα -zα/2 zα/2

Reject H0 Reject H0 Reject H0 Reject H0 Do not

reject H0

Do not

reject H0

Do not

reject H0

Example: Example: Example:

Reject H0 if z < -zα

i.e., if x < xα

αx αx α/2(L)x α/2(U)x

Reject H0 if z > zα

i.e., if x > xα

Reject H0 if z < -zα/2 or z > zα/2

i.e., if x < xα/2(L) or x > xα/2(U)


Chap 9-32

z-units: For given , find the critical z value(s):

-zα , zα ,or ±zα/2

Convert the sample mean x to a z test statistic:

Reject H0 if z is in the rejection region,

otherwise do not reject H0

x units: Given , calculate the critical value(s)

xα , or xα/2(L) and xα/2(U)

The sample mean is the test statistic. Reject H0 if x is in the

rejection region, otherwise do not reject H0

Two Equivalent Approaches to Hypothesis Testing

n

σ

μxz


Chap 9-33

Hypothesis Testing Example

Test the claim that the true mean # of TV

sets in US homes is at least 3.

(Assume σ = 0.8)

1. Specify the population value of interest

The mean number of TVs in US homes

2. Formulate the appropriate null and alternative

hypotheses

H0: μ 3 HA: μ < 3 (This is a lower tail

test)

3. Specify the desired level of significance

Suppose that = .05 is chosen for this test QMIS 220, by Dr. M. Zainal

Chap 9-34


4. Determine the rejection region

= .05

-zα= -1.645 0

This is a one-tailed test with = .05.

Since σ is known, the cutoff value is a z value:

Reject H0 if z < z = -1.645 ; otherwise do not reject H0

Hypothesis Testing Example (continued)


Chap 9-35

5. Obtain sample evidence and compute the

test statistic

Suppose a sample is taken with the following

results: n = 100, x = 2.84 ( = 0.8 is assumed known)

Then the test statistic is:

2.0.08

.16

100

0.8

32.84

n

σ

μxz

Hypothesis Testing Example


Chap 9-36


= .05

-1.645 0

6. Reach a decision and interpret the result

-2.0

Since z = -2.0 < -1.645, we reject the null

hypothesis that the mean number of TVs in US

homes is at least 3. There is sufficient evidence

that the mean is less than 3.


z


Chap 9-37

Reject H0

= .05

2.8684

Do not reject H0

3

An alternate way of constructing rejection region:

2.84

Since x = 2.84 < 2.8684, we reject the null hypothesis


x

Now

expressed

in x, not z

units

2.8684100

0.81.6453

n

σzμx αα


Chap 9-38

Convert Sample Statistic ( ) to Test Statistic

(a z value, if σ is known)

Determine the p-value from a table or

computer

Compare the p-value with

If p-value < , reject H0

If p-value , do not reject H0

x


p-Value Approach to Testing

Chap 9-39

p-Value 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

(continued)


Chap 9-40

p-value =.0228

= .05

p-value example

Example: How likely is it to see a sample mean

of 2.84 (or something further below the mean) if

the true mean is = 3.0?

2.8684 3

2.84

x .02282.0)P(z

1000.8

3.02.84zP

3.0)μ|2.84xP(

0 -1.645 -2.0

z


Chap 9-41

Compare the p-value with

If p-value < , reject H0

If p-value , do not reject H0

Here: p-value = .0228 = .05

Since .0228 < .05, we reject

the null hypothesis

(continued)

p-value example

p-value =.0228

= .05

2.8684 3

2.84 QMIS 220, by Dr. M. Zainal

Chap 9-42

Example: Upper Tail z Test for Mean ( Known)

A phone industry manager thinks that

customer monthly cell phone bill have

increased, and now average over $52 per

month. The company wishes to test this

claim. (Assume = 10 is known)

H0: μ ≤ 52 the average is not over $52 per month

HA: μ > 52 the average is greater than $52 per month (i.e., sufficient evidence exists to support the manager’s claim)

Form hypothesis test:


Chap 9-43


Suppose that = .10 is chosen for this test

Find the rejection region:

= .10

zα=1.28 0

Reject H0

Reject H0 if z > 1.28

Example: Find Rejection Region (continued)


Chap 9-44

Review: Finding Critical Value - One Tail

Z .07 .09

1.1 .3790 .3810 .3830

1.2 .3980 .4015

1.3 .4147 .4162 .4177 z 0 1.28

.08

Standard Normal

Distribution Table (Portion) What is z given = 0.10?

= .10

Critical Value

= 1.28

.90

.3997

.10

.40 .50


Chap 9-45

Obtain sample evidence and compute the test

statistic

Suppose a sample is taken with the following

results: n = 64, x = 53.1 (=10 was assumed known)

Then the test statistic is:

0.88

64

10

5253.1

n

σ

μxz

Example: Test Statistic (continued)


Chap 9-46


Example: Decision

= .10

1.28 0

Reject H0

Do not reject H0 since z = 0.88 ≤ 1.28

i.e.: there is not sufficient evidence that the mean bill is over $52

z = .88

Reach a decision and interpret the result:

(continued)


Chap 9-47

Reject H0

= .10

Do not reject H0 1.28

0

Reject H0

z = .88

Calculate the p-value and compare to

(continued)

.1894

.3106.50.88)P(z

6410

52.053.1zP

52.0)μ|53.1xP(

p-value = .1894

p -Value Solution

Do not reject H0 since p-value = .1894 > = .10


Chap 9-48

Critical Value Approach to Testing

When σ is known, convert sample statistic ( ) to

a z test statistic

x

Known Unknown

Hypothesis

Tests for

The test statistic is:

n

σ

μxz


Chap 9-49

Critical Value Approach to Testing

When σ is unknown, convert sample statistic ( )

to a t test statistic

x

Known Unknown

Hypothesis

Tests for


n

s

μxt 1n

(The population must be

approximately normal)


Chap 9-50

Hypothesis Tests for μ, σ Unknown

1. Specify the population value of interest

2. Formulate the appropriate null and alternative hypotheses

3. Specify the desired level of significance

4. Determine the rejection region (critical values are from the t-distribution with n-1 d.f.)

5. Obtain sample evidence and compute the t test statistic

6. Reach a decision and interpret the result


Chap 9-51

Example: 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


Chap 9-52

= 0.05

n = 25

Critical Values:

t24 = ± 2.0639

is unknown, so

use a t statistic

Example Solution: Two-Tail Test

Do not reject H0: not sufficient evidence that

true mean cost is different than $168

Reject H0 Reject H0

/2=.025

-tα/2

Do not reject H0

0 tα/2

/2=.025

-2.0639 2.0639

1.46

25

15.40

168172.50

n

s

μxt 1n

1.46

H0: μ = 168

HA: μ 168


Chap 9-53

Hypothesis Tests for Proportions

Involves categorical values

Two possible outcomes

“Success” (possesses a certain characteristic)

“Failure” (does not possesses that characteristic)

Fraction or proportion of population in the

“success” category is denoted by P


Chap 9-54

Proportions

The sample proportion of successes is denoted

by 𝑝 :

When both nP and n(1- P) are at least 5, 𝑝 is

approximately normally distributed with mean

and standard deviation

(continued)

sizesample

sampleinsuccessesofnumber

n

xp̂

Pμ p̂ n

P)P(1σ p̂


Chap 9-55

The sampling

distribution of p is

normal, so the test

statistic is a z

value:

Hypothesis Tests for Proportions

n

P)P(1

Pp̂z

nP 5

and

n(1-P) 5

Hypothesis

Tests for P

nP < 5

or

n(1-P) < 5

Not discussed

in this chapter


Chap 9-56

Example: 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 = .05

significance level.

Check:

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

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


Chap 9-57

Z Test for Proportion: Solution

= .05

n = 500, 𝑝 = .05

Reject H0 at = .05

H0: P = .08

HA: P .08

Critical Values: ± 1.96

Test Statistic:

Decision:

Conclusion:

z 0

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

P)P(1

Pp̂z

-1.96


Chap 9-58

Do not reject H0

Reject H0 Reject H0

/2 = .025

1.96 0

z = -2.47

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

(continued)

0.01362(.0068)

.4932)2(.5

2.47)P(x2.47)P(z

p-value = .0136:

p -Value Solution

Reject H0 since p-value = .0136 < = .05

z = 2.47

-1.96

/2 = .025

.0068 .0068


Chap 9-59

Reject

H0: μ 52

Do not reject H0 : μ 52

Type II Error

Type II error is the probability of

failing to reject a false H0

52 50

Suppose we fail to reject H0: μ 52

when in fact the true mean is μ = 50


Chap 9-60

Reject

H0: 52

Do not reject H0 : 52

Type II Error

Suppose we do not reject H0: 52 when in fact

the true mean is = 50

52 50

This is the true

distribution of x if = 50

This is the range of x where

H0 is not rejected

(continued)


Chap 9-61

Reject

H0: μ 52


Type II Error

Suppose we do not reject H0: μ 52 when

in fact the true mean is μ = 50

52 50

β

Here, β = P( x cutoff ) if μ = 50

(continued)


Chap 9-62

Reject

H0: μ 52


Suppose n = 64 , σ = 6 , and = .05

52 50

So β = P( x 50.766 ) if μ = 50

Calculating β

50.76664

61.64552

n

σzμxcutoff

(for H0 : μ 52)

50.766


Chap 9-63

Reject

H0: μ 52


.1539.3461.51.02)P(z

646

5050.766zP50)μ|50.766xP(

Suppose n = 64 , σ = 6 , and = .05

52 50

Calculating β (continued)

Probability of

type II error:

β = .1539


Single Population

Hypothesis Tests for Variances

Tests for a Single

Population Variance

Chi-Square test statistic

H0: σ2 = σ0

2

HA: σ2 ≠ σ02

H0: σ2 σ0

2

HA: σ2 < σ02

H0: σ2 ≤ σ0

2

HA: σ2 > σ02

* Two tailed test

Lower tail test

Upper tail test

QMIS 220, by Dr. M. Zainal Chap 11-64

Chi-Square Test Statistic

Hypothesis Tests for Variances

Tests for a Single

Population Variance

Chi-Square test statistic *

The chi-squared test statistic for

a Single Population Variance is:

2

22

σ

1)s(n

where

2 = standardized chi-square variable

n = sample size

s2 = sample variance

σ2 = hypothesized variance


The Chi-square Distribution

The chi-square distribution is a family of

distributions, depending on degrees of freedom

(Like the t distribution).

The chi-square distribution curve starts at the

origin and lies entirely to the right of the vertical

axis.

The chi-square distribution assumes

nonnegative values only, and these are denoted

by the symbol 2 (read as “chi-square”).


The Chi-square Distribution

The shape of a specific chi-square distribution

depends on the number of degrees of freedom.

peak of a 2 distribution curve with 1 or 2

degrees of freedom occurs at zero and for a

curve with 3 or more degrees of freedom at

(df−2).

0 4 8 12 16 20 24 28 0 4 8 12 16 20 24 28 0 4 8 12 16 20 24 28

d.f. = 1 d.f. = 5 d.f. = 15

2 2 2


Finding the Critical Value

The critical value, , is found from the

chi-square table

Do not reject H0 Reject H0

2

2

2

H0: σ2 ≤ σ0

2

HA: σ2 > σ02

Upper tail test:




Example: Find the value of 2 for 7 degrees of freedom and an

area of .10 in the right tail of the chi-square distribution curve.



Example: Find the value of 2 for 9 degrees of freedom and an

area of .05 in the left tail of the chi-square distribution curve.

Example

A commercial freezer must hold the selected

temperature with little variation. Specifications call

for a standard deviation of no more than 4 degrees

(or variance of 16 degrees2). A sample of 16

freezers is tested and

yields a sample variance

of s2 = 24. Test to see

whether the standard

deviation specification

is exceeded. Use

= .05



Use the chi-square table to find the critical value:

Do not reject H0 Reject H0

= .05

2

2

2

= 24.9958

= 24.9958 ( = .05 and 16 – 1 = 15 d.f.)

22.516

1)24(16

σ

1)s(n2

22


Since 22.5 < 24.9958,

do not reject H0

There is not significant

evidence at the = .05 level

that the standard deviation

specification is exceeded


Lower Tail or Two Tailed Chi-square Tests

H0: σ2 = σ0

2

HA: σ2 ≠ σ02

H0: σ2 σ0

2

HA: σ2 < σ02

2/2

Do not reject H0 Reject

21-

2

Do not reject H0

Reject

/2

21-/2

2

/2

Reject

Lower tail test: Two tail test:

(2U) (2

L)


Confidence Interval Estimate for σ2

2/2

/2

21-/2

2

/2

(2U) (2

L)

The confidence interval estimate for σ2 is

2

L

22

2

U

2 1)s(nσ

1)s(n

χχ

Where 2L and 2

U are from the

2 distribution with n -1 degrees

of freedom


Example

A sample of 16 freezers yields a sample

variance of s2 = 24.

Form a 95% confidence interval for the

population variance.


Example (continued)

Problems


Suppose a sample of 30 ECC students are given an IQ test. If

the sample has a standard deviation of 12.23 points, find a

90% confidence interval for the population standard deviation.

Chap 11-78

Problems


Problems


Example: Consider Ho : = 23, HA: 23

o What type of error would you make if the null hypothesis is

actually false and you fail to reject it?

o What type of error would you make if the null hypothesis is

actually true and you reject it?

Problems


Example: Write the null and alternative hypotheses for each of

the following examples. Determine if each is a case of a two-

tailed, a left-tailed, or a right-tailed test.

o To test if the mean number of hours spent working per week

by college students who hold jobs is different from 20 hours.

o To test whether or not a bank's ATM is out of service for an

average of more than 10 hours per month.

o To test if the mean length of experience of airport security

guards is different from three years.

o To test if the mean credit card debt of college seniors is less

than $1000.

Problems


Problems


The management of Priority Health Club claims that its

members lose an average of 10 pounds or more within the first

month after joining the club. A consumer agency that wants to

check this claim took a random sample of 36 members of this

health club and found that they lost an average of 9.2 pounds

within the first month of membership with a standard deviation

of 2.4 pounds.

o Find the p-value for this test.

o What will your decision be if = .01? What if = .05?

Problems


Problems


Canon Inc. has introduced a copying machine that features

two-color copying capability in a compact system copier. The

average speed of the standard compact system copier is 27

copies per minute. Suppose the company wants to test

whether the new copier has the same average speed as its

standard compact copier. Conduct a test at an = 0.05 level

of significance.

Ho : = 27

HA : 27

n = 24

Problems


Problems


Direct Mailing Company sells computers and computer parts by

mail. The company claims that at least 90% of all orders are

mailed within 72 hours after they are received. The quality

control department at the company often takes samples to

check if this claim is valid. A recently taken sample of 150

orders showed that 129 of them were mailed within 72 hours.

Do you think the company's claim is true? Use a 2.5%

significance level.

Problems


Chap 9-92

Chapter Summary

Addressed hypothesis testing methodology

Performed z Test for the mean (σ known)

Discussed p–value approach to

hypothesis testing

Performed one-tail and two-tail tests . . .


Chap 9-93

Chapter Summary

Performed t test for the mean (σ

unknown)

Performed z test for the proportion

Discussed Type II error and computed its

probability

(continued)


Copyright

The materials of this presentation were mostly

taken from the PowerPoint files accompanied

Business Statistics: A Decision-Making Approach,

7e © 2008 Prentice-Hall, Inc.


business statistics: a decision-making approach, …chap 9-3 what is a hypothesis? we perform of...

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