Statistics for Economics not reject H0 Reject H0 0 zα μ0 H0: μ=μ0 H1: μ> μ0 Critical value Z...

Chapter 9 9-1

Statistics for Business and Economics, 7/e Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall

Statistics for Economics

Department of Economics

Dr. Mohammad ZainalECON 506

Chapter 9Hypothesis Testing:Single Population

Chapter Goals

After completing this chapter, you should be able to:

Formulate null and alternative hypotheses for applications involving a single population mean from a normal distribution a single population proportion (large samples) the variance of a normal distribution

Formulate a decision rule for testing a hypothesis

Know how to use the critical value and p-value approaches to test the null hypothesis (for both mean and proportion problems)

Know what Type I and Type II errors are

Assess the power of a testEcon 506, by Dr. M. Zainal Ch. 9-2

Chapter 9 9-2


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).

Econ 506, by Dr. M. Zainal Ch. 9-3


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.


Chapter 9 9-3



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.



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 theperson guilty and, hence, the nullhypothesis is not rejected in this region

Enough evidence to declare the personguilty and, hence, the null hypothesis isrejected in this region

Nonrejection region Rejection region

C

Critical point

Level of evidence0


Chapter 9 9-4



Actual Situation

The person is not guilty

The person is guilty

Court’s decision

The person is not guilty

Correct decisionType II or

error

The person is guiltyType I or

errorCorrect decision



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

9.1


Chapter 9 9-5


The Null Hypothesis, H0

States the assumption (numerical) 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

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 untilproven guilty

Refers to the status quo

Always contains “=” , “≤” or “” sign

May or may not be rejected

(continued)


Chapter 9 9-6


The Alternative Hypothesis, H1

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

homes is not equal to 3 ( H1: μ ≠ 3 )

Challenges the status quo

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

May or may not be supported

Is generally the hypothesis that the researcher is trying to support


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 Process

If not likely,

Now select a random sample

H0: μ = 50 )

X


Chapter 9 9-7


Reason for Rejecting H0

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.

20

... if in fact this werethe population mean…

X


Level of Significance,

Defines the unlikely values of the sample statistic if the 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


Chapter 9 9-8


Level of Significance and the Rejection Region

H0: μ ≥ 3 H1: μ < 3

0

H0: μ ≤ 3 H1: μ > 3

Representscritical value

Lower-tail test

Level of significance =

0Upper-tail test

Two-tail test

Rejection region is shaded

/2

0

/2H0: μ = 3 H1: μ ≠ 3


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


Chapter 9 9-9


Errors in Making Decisions

Type II Error

Fail to reject a false null hypothesis

The probability of Type II Error is β

(continued)


Outcomes and Probabilities

Actual Situation

Decision

Do NotReject

H0

No Error(1 - )

Type II Error( β )

RejectH0

Type I Error( )

Possible Hypothesis Test Outcomes

H0 FalseH0 True

Key:Outcome

(Probability) No Error( 1 - β )


Chapter 9 9-10


Type I & II Error Relationship

Type I and Type II errors can not happen atthe 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 )


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


Chapter 9 9-11


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



Example 1: A manufacturing company has worked to reduce the noise of one of their tools. They would like to show in its advertising that the tool is quieter. The average of the prior design was 150 decibels.

What is the appropriate hypothesis test?


Chapter 9 9-12



What is the appropriate test?

H0: µ ≥ 150 (the tool is not quieter) status quoHA: µ < 150 (the tool is quieter) wants to support

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



Example 2: The average annual income of buyers of Ford 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?


Chapter 9 9-13



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

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


Power of the Test

The power of a test is the probability of rejecting a null hypothesis that is false

i.e., Power = P(Reject H0 | H1 is true)

Power of the test increases as the sample size increases


Chapter 9 9-14


Hypothesis Tests for the Mean

Known Unknown

Hypothesis Tests for


Test of Hypothesisfor the Mean (σ Known)

Convert sample result ( ) to a z value

The decision rule is:

α0

0 z

n

σμx

z if H Reject

σ Known σ Unknown


Consider the test

00 μμ:H

01 μμ:H

(Assume the population is normal)

x

9.2


Chapter 9 9-15


Decision Rule

Reject H0Do not reject H0

zα0

μ0

H0: μ = μ0

H1: μ > μ0

Critical value

Z

α0

0 z

n

σμx

z if H Reject

nσ/Zμx if H Reject α00

n

σzμ α0

Alternate rule:

x

cxEcon 506, by Dr. M. Zainal Ch. 9-29

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


Chapter 9 9-16


p-Value Approach to Testing

Convert sample result (e.g., ) to test statistic (e.g., z statistic )

Obtain the p-value For an upper

tail test:

Decision rule: compare the p-value to

If p-value < , reject H0

If p-value , do not reject H0

(continued)

x

)μμ | nσ/

μ-x P(z

true) is H that given , nσ/

μ-x P(z value-p

00

00


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

H1: μ > 52 the average is greater than $52 per month(i.e., sufficient evidence exists to support the manager’s claim)

Form hypothesis test:


Chapter 9 9-17


Suppose that = .10 is chosen for this test

Find the rejection region:


= .10

1.280

Reject H0

Example: Find Rejection Region(continued)

1.28nσ/

μxz if H Reject 0

0


Example: Sample Results

Obtain sample and compute the test statistic

Suppose a sample is taken with the following results: n = 64, x = 53.1 ( = 10 was assumed known)

Using the sample results,

0.88

64

105253.1

n

σμx

z 0

(continued)


Chapter 9 9-18


Example: Decision

Reach a decision and interpret the result:


= .10

1.280

Reject H0

Do not reject H0 since z = 0.88 < 1.28

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

z = 0.88

(continued)


Example: p-Value Solution

Calculate the p-value and compare to (assuming that μ = 52.0)

Reject H0

= .10

Do not reject H0 1.28

0

Reject H0

Z = .88

(continued)

.1894

.810610.88)P(z

6410/

52.053.1zP

52.0) μ | 53.1xP(

p-value = .1894

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


Chapter 9 9-19


One-Tail Tests

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

H0: μ ≥ 3

H1: μ < 3

H0: μ ≤ 3

H1: μ > 3

This is a lower-tail test since the alternative hypothesis is focused on the lower tail below the mean of 3

This is an upper-tail test since the alternative hypothesis is focused on the upper tail above the mean of 3


Upper-Tail Tests


zα0

μ

H0: μ ≤ 3

H1: μ > 3 There is only one

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

Critical value

Z

x

cx


Chapter 9 9-20


Lower-Tail Tests

Reject H0 Do not reject H0

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

-z 0

μ

H0: μ ≥ 3

H1: μ < 3

Z

Critical value

x

cx


Two-Tail Tests

In some settings, the alternative hypothesis does not specify a unique direction

Do not reject H0 Reject H0Reject H0

There are two critical values, defining the two regions of rejection

/2

0

H0: μ = 3 H1: μ 3

/2

Lower critical value

Uppercritical value

3

z

x

-z/2 +z/2


Chapter 9 9-21


Hypothesis Testing Example

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

(Assume σ = 0.8)

State the appropriate null and alternativehypotheses

H0: μ = 3 , H1: μ ≠ 3 (This is a two tailed test)

Specify the desired level of significance

Suppose that = .05 is chosen for this test

Choose a sample size

Suppose a sample of size n = 100 is selected



2.0.08

.16

100

0.832.84

n

σμX

z 0

Determine the appropriate technique σ is known so this is a z test

Set up the critical values

For = .05 the critical z values are ±1.96

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)


Chapter 9 9-22



Is the test statistic in the rejection region?


= .05/2

-z = -1.96 0

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

(continued)

= .05/2

Reject H0

+z = +1.96

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



Reach a decision and interpret the result

-2.0

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

(continued)


= .05/2

-z = -1.96 0

= .05/2

Reject H0

+z = +1.96


Chapter 9 9-23


Example: 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?

.0228

/2 = .025

-1.96 0

-2.0

.02282.0)P(z

.02282.0)P(z

Z1.96

2.0

x = 2.84 is translated to a z score of z = -2.0

p-value

= .0228 + .0228 = .0456

.0228

/2 = .025


Example: p-Value

Compare the p-value with

If p-value < , reject H0

If p-value , do not reject H0

Here: p-value = .0456 = .05

Since .0456 < .05, we reject the null hypothesis

(continued)

.0228

/2 = .025

-1.96 0

-2.0

Z1.96

2.0

.0228

/2 = .025


Chapter 9 9-24


t Test of Hypothesis for the Mean (σ Unknown)

Convert sample result ( ) to a t test statistic

σ Known σ Unknown


x


α , 1-n0

0 t

n

sμx

t if H Reject

Consider the test

00 μμ:H

01 μμ:H

(Assume the population is normal)

9.3


t Test of Hypothesis for the Mean (σ Unknown)

For a two-tailed test:


α/2 , 1-n0

α/2 , 1-n0

0 t

n

sμx

t if or t

n

sμx

t if H Reject

Consider the test

00 μμ:H

01 μμ:H

(Assume the population is normal, and the population variance is unknown)

(continued)


Chapter 9 9-25


Example: Two-Tail Test( Unknown)

The average cost of a hotel room in Chicago 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 H1: μ 168


Example Solution: Two-Tail Test

= 0.05

n = 25

is unknown, so use a t statistic

Critical Value:

t24 , .025 = ± 2.0639 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 H1: μ 168

t n-1,α/2


Chapter 9 9-26


Tests of the Population Proportion

Involves categorical variables

Two possible outcomes

“Success” (a certain characteristic is present)

“Failure” (the characteristic is not present)

Fraction or proportion of the population in the “success” category is denoted by P

Assume sample size is large

9.4


Proportions

Sample proportion in the success category is denoted by

When nP(1 – P) > 5, can be approximated by a normal distribution with mean and standard deviation

sizesample

sampleinsuccessesofnumber

n

xp ˆ

Pμ pn

P)P(1σ

p

(continued)

p

p


Chapter 9 9-27


Hypothesis Tests for Proportions

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

n)P(1P

Ppz

00

0

ˆ

nP(1 – P) > 5

Hypothesis Tests for P

Not discussed in this chapter

p

nP(1 – P) < 5


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: Our approximation for P is

= 25/500 = .05

nP(1 - P) = (500)(.05)(.95)= 23.75 > 5

p


Chapter 9 9-28


Z Test for Proportion: Solution

= .05

n = 500, = .05

Reject H0 at = .05

H0: P = .08 H1: P .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)P(1P

Ppz

00

0

ˆ

-1.96

p


p-Value Solution

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

Do not reject H0Reject H0Reject H0

/2 = .025

1.960

Z = -2.47

(continued)

0.01362(.0068)

2.47)P(Z2.47)P(Z

p-value = .0136:

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

Z = 2.47

-1.96

/2 = .025

.0068.0068


Chapter 9 9-29


Power of the Test

Recall the possible hypothesis test outcomes:Actual Situation

Decision

Do Not Reject H0

No error(1 - )

Type II Error( β )

Reject H0Type I Error

( )

H0 FalseH0 TrueKey:

Outcome(Probability)

No Error( 1 - β )

β denotes the probability of Type II Error

1 – β is defined as the power of the test

Power = 1 – β = the probability that a false null hypothesis is rejected

9.5


Type II Error

or


α0

0 zn/σ

μxz if H Reject

00 μμ:H

01 μμ:H

Assume the population is normal and the population variance is known. Consider the test

nσ/Zμxx if H Reject α0c0

If the null hypothesis is false and the true mean is μ*, then the probability of type II error is

n/σ

*μxzPμ*)μ|xxP(β c

c


Chapter 9 9-30


Type II Error Example

Type II error is the probability of failing to reject a false H0

Reject H0: μ 52

Do not reject H0 : μ 52

5250

Suppose we fail to reject H0: μ 52when in fact the true mean is μ* = 50

cx



Suppose we do not reject H0: μ 52 when in fact the true mean is μ* = 50

Reject H0: μ 52


5250

This is the true distribution of x if μ = 50

This is the range of x where H0 is not rejected

(continued)


Chapter 9 9-31



Suppose we do not reject H0: μ 52 when in fact the true mean is μ* = 50

Reject H0: μ 52


5250

β

Here, β = P( x ) if μ* = 50

(continued)

cx

cx


Calculating β

Suppose n = 64 , σ = 6 , and = .05

Reject H0: μ 52


5250

So β = P( x 50.766 ) if μ* = 50

50.76664

61.64552

n

σzμx α0c

(for H0 : μ 52)

50.766


Chapter 9 9-32


Calculating β

Suppose n = 64 , σ = 6 , and = .05

Reject H0: μ 52


.1539.3461.51.02)P(z

646

5050.766zP50)μ*|50.766xP(

5250

(continued)

Probability of type II error:

β = .1539


Power of the Test Example

If the true mean is μ* = 50,

The probability of Type II Error = β = 0.1539

The power of the test = 1 – β = 1 – 0.1539 = 0.8461

Actual Situation

Decision

Do Not Reject H0

No error1 - = 0.95

Type II Errorβ = 0.1539

Reject H0Type I Error = 0.05

H0 FalseH0 TrueKey:

Outcome(Probability)

No Error1 - β = 0.8461

(The value of β and the power will be different for each μ*)


Chapter 9 9-33


Hypothesis Tests of one Population Variance

If the population is normally distributed,

2

22

1n σ

1)s(n

has a chi-square distribution with (n – 1) degrees of freedom

Goal: Test hypotheses about thepopulation variance, σ2

9.6


Hypothesis Tests of one Population Variance

The test statistic for hypothesis tests about one population variance is

20

22

1n σ

1)s(nχ

(continued)


Chapter 9 9-34


Decision Rules: Variance

Population variance

Lower-tail test:

H0: σ2 σ02

H1: σ2 < σ02

Upper-tail test:

H0: σ2 ≤ σ02

H1: σ2 > σ02

Two-tail test:

H0: σ2 = σ02

H1: σ2 ≠ σ02

/2 /2

Reject H0 if Reject H0 if Reject H0 if

or

2, 1n χ

2,1 1n χ

2,1 1n 2/χ

2, 1n 2/χ

2,1 1n

21n χχ

2, 1n

21n χχ

2, 1n

21n 2/ χχ

2,1 1n

21n 2/ χχ


Proportions

The sample proportion of successes is denoted by p :

When both nπ and n(1- π) are at least 5, pis approximately normally distributed with mean and standard deviation

(continued)

sizesample

sampleinsuccessesofnumber

n

xp

πμp n

π)π(1σp


Chapter 9 9-35


Hypothesis Tests for Variances

Hypothesis Testsfor Variances

Tests for a SinglePopulation Variance

Tests for TwoPopulation Variances

Chi-Square test statistic F test statistic


Single Population



Chi-Square test statistic

H0: σ2 = σ02

HA: σ2 ≠ σ02

H0: σ2 σ02

HA: σ2 < σ02

H0: σ2 ≤ σ02

HA: σ2 > σ02

*Two tailed test

Lower tail test

Upper tail test


Chapter 9 9-36


Chi-Square Test Statistic



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”).


Chapter 9 9-37


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 22


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 ≤ σ02

HA: σ2 > σ02

Upper tail test:


Chapter 9 9-38





Example: Find the value of 2 for 7 degrees of freedom and anarea of .10 in the right tail of the chi-square distribution curve.


Chapter 9 9-39



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


Chapter 9 9-40



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

The test statistic is:

Since 22.5 < 24.9958, do not reject H0

There is not significant evidence at the = .05 levelthat the standard deviation specification is exceeded


Lower Tail or Two Tailed Chi-square Tests

H0: σ2 = σ02

HA: σ2 ≠ σ02

H0: σ2 σ02

HA: σ2 < σ02

2/2

Do not reject H0Reject

21-

2

Do not reject H0

Reject

/2

21-/2

2

/2

Reject

Lower tail test: Two tail test:

(2U)(2

L)


Chapter 9 9-41


Confidence Interval Estimatefor σ2

2/2

/2

21-/2

2

/2

(2U)(2

L)

The confidence interval estimate for σ2 is

2L

22

2U

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.


Chapter 9 9-42


Use the chi-square table to find 2L and 2

U :

6.2621

( = .05 and 16 – 1 = 15 d.f.)

Example(continued)

2.025

/2=.025

2.975

/2=.025

(2U)(2

L)27.4884 57.489σ13.096

6.2621

1)24(16σ

27.4884

1)24(16

1)s(nσ

1)s(n

2

2

2L

22

2U

2

χχ

We are 95% confident that the population variance is between 13.096 and 57.489 degrees2. (Taking the square root, we are 95% confident that the population standard deviation is between 3.619 and 7.582 degrees.)


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.


Chapter 9 9-43


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?


Chapter 9 9-44


Problems

Data:

Hypothesis:

Test statistic:

(a) P-value =

(b)

- Since the value of is smaller than the p-value, then we decide not to reject at %1 level of significance.

- Since the value of is greater than the p-value, then we decide to reject at %5 level of significance.

36, 9.2, 2.4n x s

0 1: 10 . : 10H vs H

0 9.2 102

2.4 / 6/c

xZ

s n

( 2) 0.02275P Z

0.01

0.05


Problems

Canon Inc. has introduced a copying machine that featurestwo-color copying capability in a compact system copier. Theaverage speed of the standard compact system copier is 27copies per minute. Suppose the company wants to testwhether the new copier has the same average speed as itsstandard compact copier. Conduct a test at an = 0.05 levelof significance.

Ho : = 27

HA : 27

n = 24


Chapter 9 9-45


Problems

For = 0.05 and df = 24 - 1 = 23, the critical values of t are ± 2.069

The rule is:

Do not reject Ho if: [-2.069 t 2.069]

The test statistic is:

Lower Rejection Region

Upper Rejection Region

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

.025 .025

.95

Nonrejection Region

t 2.069-2.069

The t Distribution

7.4

24.6 277.424

⇒

2.41.51

1.59


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.


Chapter 9 9-46


Problems


Chapter Summary

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

Performed t test for the mean (σ unknown)

Performed Z test for the proportion

Discussed type II error and power of the test

Performed a hypothesis test for the variance (χ2)


Statistics for Economics not reject H0 Reject H0 0 zα μ0 H0: μ=μ0 H1: μ> μ0 Critical value Z...

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