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Chapter

Hypothesis Testing with One Sample

1 of 101

Chapter Outline

• • • • • 7.1 Introduction to Hypothesis Testing 7.2 Hypothesis Testing for the Mean (Large Samples) 7.3 Hypothesis Testing for the Mean (Small Samples) 7.4 Hypothesis Testing for Proportions 7.5 Hypothesis Testing for Variance and Standard Deviation 2 of 101 © 2012 Pearson Education, Inc. All rights reserved.

Section 7.1

Introduction to Hypothesis Testing

3 of 101

Section 7.1 Objectives

• • • • • State a null hypothesis and an alternative hypothesis Identify type I and type II errors and interpret the level of significance Determine whether to use a one-tailed or two-tailed statistical test and find a

Hypothesis Tests

• •

Hypothesis test

A process that uses sample statistics to test a claim about the value of a population parameter.

For example:

An automobile manufacturer advertises that its new hybrid car has a mean mileage of 50 miles per gallon. To test this claim, a sample would be taken. If the sample mean differs enough from the advertised mean, you can decide the advertisement is wrong.

Hypothesis Tests

• • •

Statistical hypothesis

A statement, or claim, about a population parameter.

Need a pair of hypotheses • one that represents the claim • the other, its complement When one of these hypotheses is false, the other must be true.

Stating a Hypothesis

• •

Null hypothesis

A statistical hypothesis that contains a statement of equality such as ≤, =, or ≥.

Denoted

read “H sub-zero” or “H naught.” • • •

Alternative hypothesis

A statement of strict inequality such as >, ≠, or <.

Must be true if false.

0 is Denoted

H a

7 of 101

Stating a Hypothesis

• • To write the null and alternative hypotheses, translate the claim made about the population parameter from a verbal statement to a mathematical statement.

Then write its complement.

0 :

μ H a

≤

k H

0 :

μ H a

≥

k H

0 :

μ H a

≠

• Regardless of which pair of hypotheses you use, you always assume

and examine the sampling distribution on the basis of this assumption.

Example: Stating the Null and Alternative Hypotheses

Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim.

A school publicizes that the proportion of its students who are involved in at least one extracurricular activity is 61%.

Solution:

0 :

= 0.61

H a

≠ 0.61

Equality condition (Claim) Complement of

Example: Stating the Null and Alternative Hypotheses

Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim.

A car dealership announces that the mean time for an oil change is less than 15 minutes.

Solution:

0 :

≥ 15 minutes

H a

Complement of

H a

Inequality condition (Claim) 10 of 101

Example: Stating the Null and Alternative Hypotheses

Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim.

A company advertises that the mean life of its furnaces is more than 18 years

Solution:

0 :

μ H a

Complement of

H a

Inequality condition (Claim) 11 of 101

Types of Errors

• No matter which hypothesis represents the claim, always begin the hypothesis test

assuming that the equality condition in the null hypothesis is true

• At the end of the test, one of two decisions will be made:  reject the null hypothesis  fail to reject the null hypothesis • Because your decision is based on a sample, there is the possibility of making the wrong decision.

Types of Errors

Decision

Do not reject

0 Reject

Actual Truth of H 0

0 is true Correct Decision

0 is false

Type II Error Type I Error

Correct Decision • • A

type I error

occurs if the null hypothesis is rejected when it is true.

type II error

occurs if the null hypothesis is not rejected when it is false.

Example: Identifying Type I and Type II Errors

The USDA limit for salmonella contamination for chicken is 20%. A meat inspector reports that the chicken produced by a company exceeds the USDA limit. You perform a hypothesis test to determine whether the meat inspector’s claim is true. When will a type I or type II error occur? Which is more serious?

(Source: United States Department of Agriculture)

Solution: Identifying Type I and Type II Errors

Let

represent the proportion of chicken that is contaminated.

Hypotheses:

0 :

≤ 0.2

H a

> 0.2

(Claim) Chicken meets USDA limits.

0 :

≤ 0.20

0.16

0.18

Chicken exceeds USDA limits.

0 :

> 0.20

0.20

0.22

0.24

Solution: Identifying Type I and Type II Errors

Hypotheses:

0 :

H a

≤ 0.2

> 0.2

(Claim) A type I error is rejecting

0 when it is true. The actual proportion of contaminated chicken is less than or equal to 0.2, but you decide to reject

0 .

A type II error is failing to reject

0 when it is false. The actual proportion of contaminated chicken is greater than 0.2, but you do not reject

0 .

Solution: Identifying Type I and Type II Errors

Hypotheses:

0 :

a :

≤ 0.2

> 0.2

(Claim) • • • With a type I error, you might create a health scare and hurt the sales of chicken producers who were actually meeting the USDA limits. With a type II error, you could be allowing chicken that exceeded the USDA contamination limit to be sold to consumers. A type II error could result in sickness or even death.

Level of Significance

• • • •

Level of significance

Your maximum allowable probability of making a type I error.  Denoted by

, the lowercase Greek letter alpha.

By setting the level of significance at a small value, you are saying that you want the probability of rejecting a true null hypothesis to be small.

Commonly used levels of significance: 

= 0.10

= 0.05

= 0.01

(type II error) =

Statistical Tests

• • After stating the null and alternative hypotheses and specifying the level of significance, a random sample is taken from the population and sample statistics are calculated.

The statistic that is compared with the parameter in the null hypothesis is called the

test statistic

Population parameter

Test statistic

x p σ

p ˆ

Standardized test statistic

(Section 7.2

≥ 30) (Section 7.3

< 30)

(Section 7.4)

2 (Section 7.5) 19 of 101

P-values

• •

P-value

(or

probability value

) The probability, if the null hypothesis is true, of obtaining a sample statistic with a value as extreme or more extreme than the one determined from the sample data.

Nature of the Test

• • • Three types of hypothesis tests  left-tailed test  right-tailed test  two-tailed test The type of test depends on the region of the sampling distribution that favors a rejection of

0 .

This region is indicated by the alternative hypothesis.

Left-tailed Test

• The alternative hypothesis

H a

inequality symbol (<).

contains the less-than

0 :

μ H a

≥ <

k k P

is the area to the left of the standardized test statistic.

0 1 2 3 22 of 101

Right-tailed Test

• The alternative hypothesis

H a

than inequality symbol (>).

contains the greater-

0 :

μ H a

≤

k P

is the area to the right of the standardized test statistic.

–1 0 1 Test statistic 2 3 23 of 101

Two-tailed Test

• The alternative hypothesis

H a

contains the not-equal to symbol (≠). Each tail has an area of ½

0 :

k H a

μ ≠ k P

is twice the is twice the area to area to the right the left of the of the positive negative standardized standardized test test statistic.

statistic.

0 1 Test statistic 2 3 24 of 101

Example: Identifying The Nature of a Test

For each claim, state

0 and

H a

. Then determine whether the hypothesis test is a left-tailed, right-tailed, or two-tailed test. Sketch a normal sampling distribution and shade the area for the

-value.

A school publicizes that the proportion of its students who are involved in at least one extracurricular activity is 61%.

Solution:

H a

: :

p p

= 0.61

≠ 0.61

Example: Identifying The Nature of a Test

For each claim, state

0 and

H a

. Then determine whether the hypothesis test is a left-tailed, right-tailed, or two-tailed test. Sketch a normal sampling distribution and shade the area for the

-value.

A car dealership announces that the mean time for an oil change is less than 15 minutes.

Solution:

0 :

≥ 15 min

H a

-value area -z 0

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Example: Identifying The Nature of a Test

For each claim, state

0 and

H a

. Then determine whether the hypothesis test is a left-tailed, right-tailed, or two-tailed test. Sketch a normal sampling distribution and shade the area for the

-value.

A company advertises that the mean life of its furnaces is more than 18 years.

Solution:

0 :

≤ 18 yr

H a

z 0

-value area 27 of 101

Making a Decision

•

Decision Rule Based on P-value

Compare the

-value with

  If

≤

, then reject

0 .

, then fail to reject

0 .

Decision

Reject

0 Fail to reject

Claim

Claim is

0 There is enough evidence to reject the claim There is not enough evidence to reject the claim Claim is

Example: Interpreting a Decision

You perform a hypothesis test for the following claim. How should you interpret your decision if you reject

0 ? If you fail to reject

0 ?

0 (Claim) : A school publicizes that the proportion of its students who are involved in at least one extracurricular activity is 61%.

•

Solution:

The claim is represented by

H 0

Solution: Interpreting a Decision

• If you reject

0 , then you should conclude “there is enough evidence to reject the school’s claim that the proportion of students who are involved in at least one extracurricular activity is 61%.” • If you fail to reject

0 , then you should conclude “there is not enough evidence to reject the school’s claim that proportion of students who are involved in at least one extracurricular activity is 61%.” 30 of 101 © 2012 Pearson Education, Inc. All rights reserved.

Example: Interpreting a Decision

You perform a hypothesis test for the following claim. How should you interpret your decision if you reject

0 ? If you fail to reject

0 ?

a (Claim) : A car dealership announces that the mean time for an oil change is less than 15 minutes.

• •

Solution:

The claim is represented by

H a

Solution: Interpreting a Decision

• If you reject

0 , then you should conclude “there is enough evidence to support

the dealership’s

claim that the mean time for an oil change is less than 15 minutes.” • If you fail to reject

dealership’s H

0 , then you should conclude “there is not enough evidence to support the claim that the mean time for an oil change is less than 15 minutes.” 32 of 101 © 2012 Pearson Education, Inc. All rights reserved.

Steps for Hypothesis Testing

State the claim mathematically and verbally. Identify the null and alternative hypotheses.

0 : ?

a : ?

Specify the level of significance.

= ?

This sampling distribution is based on the assumption that

0 is true.

Determine the standardized sampling distribution and sketch its graph.

Calculate the test statistic and its corresponding standardized test statistic.

Add it to your sketch.

Steps for Hypothesis Testing

Find the

-value.

Use the following decision rule.

Is the

-value less than or equal to the level of significance?

Yes No

Fail to reject

0 . Reject

0 .

Write a statement to interpret the decision in the context of the original claim.

34 of 101

Section 7.1 Summary

• • • • • Stated a null hypothesis and an alternative hypothesis Identified type I and type II errors and interpreted the level of significance Determined whether to use a one-tailed or two-tailed statistical test and found a

Section 7.2

Hypothesis Testing for the Mean (Large Samples)

36 of 101

Section 7.2 Objectives

• • • • Find

-values and use them to test a mean μ Use

-values for a

-test Find critical values and rejection regions in a normal distribution Use rejection regions for a

Using P-values to Make a Decision

•

Decision Rule Based on P-value

To use a

-value to make a conclusion in a hypothesis test, compare the

-value with

≤

, then reject

0 .

, then fail to reject

0 .

Example: Interpreting a P-value

The

-value for a hypothesis test is

= 0.0237. What is your decision if the level of significance is 1.

α = 0.05?

Solution:

Because 0.0237 < 0.05, you should

reject the null hypothesis

α = 0.01?

Solution:

Because 0.0237 > 0.01, you should

fail to reject the null hypothesis

Finding the P-value

After determining the hypothesis test’s standardized test statistic and the test statistic’s corresponding area, do one of the following to find the

-value.

For a left-tailed test,

= (Area in left tail).

For a right-tailed test,

= (Area in right tail).

For a two-tailed test,

= 2(Area in tail of test statistic).

Example: Finding the P-value

Find the

-value for a left-tailed hypothesis test with a test statistic of

= –2.23. Decide whether to reject

0 the level of significance is

= 0.01.

Solution:

For a left-tailed test,

= (Area in left tail)

P = 0.0129

-2.23

0 Because 0.0129 > 0.01, you should

fail to reject H 0

41 of 101

Example: Finding the P-value

Find the

-value for a two-tailed hypothesis test with a test statistic of

= 2.14. Decide whether to reject

0 the level of significance is

= 0.05.

Solution:

For a two-tailed test,

= 2(Area in tail of test statistic) 0.9838

0 1 – 0.9838 = 0.0162

z P

= 2(0.0162) =

0.0324

2.14

Because 0.0324 < 0.05, you should

reject H 0

Z-Test for a Mean

• • • • Can be used when the population is normal and

is known, or for any population when the sample size

is at least 30. The

test statistic

The

standardized test statistic

 

 



 standard error  

When

≥ 30, the sample standard deviation

substituted for

σ.

Using P-values for a z-Test for Mean μ

In Words

State the claim mathematically and verbally. Identify the null and alternative hypotheses.

Specify the level of significance.

Determine the standardized test statistic.

Find the area that corresponds to

In Symbols

State

0 and

H a

. Identify

 

 

Use Table 4 in Appendix B.

44 of 101

Using P-values for a z-Test for Mean

In Words In Symbols

Find the

-value.

For a left-tailed test,

= (Area in left tail).

For a right-tailed test, For a two-tailed test,

P P

= (Area in right tail).

= 2(Area in tail of test statistic).

Make a decision to reject or fail to reject the null hypothesis.

Reject

0 if

-value is less than or equal to

α.

Otherwise, fail to reject

0 .

Interpret the decision in the context of the original claim.

45 of 101

Example: Hypothesis Testing Using P values

In auto racing, a pit crew claims that its mean pit stop time (for 4 new tires and fuel) is less than 13 seconds. A random selection of 32 pit stop times has a sample mean of 12.9 seconds and a standard deviation of 0.19 second. Is there enough evidence to support the claim at

= 0.01? Use a

-value.

Solution: Hypothesis Testing Using P values

•

P-value

• • • •

0 : μ ≥ 13 sec

H a



= μ < 13 sec (Claim) 0.01

Test Statistic:

 

 

 0.19

  2.98

•

Decision: Reject H 0 .

0.0014 < 0.01

At the 1% level of significance, you have sufficient evidence to support the claim that the mean pit stop time is less than 13 seconds.

47 of 101

Example: Hypothesis Testing Using P values

The National Institute of Diabetes and Digestive and Kidney Diseases reports that the average cost of bariatric (weight loss) surgery is $22,500. You think this information is incorrect. You randomly select 30 bariatric surgery patients and find that the average cost for their surgeries is $21,545 with a standard deviation of $3015. Is there enough evidence to support your claim at

= 0.05? Use a

-value.

(Adapted from National Institute of Diabetes and Digestive and Kidney Diseases)

• • • •

Solution: Hypothesis Testing Using P values

•

0 : μ = $22,500

H a

= μ ≠ 22,500 (Claim) 0.05

P-value Test Statistic:

 

 

 3015   1.73

•

Decision:

0.0836 > 0.05

Fail to reject H 0 .

At the 5% level of significance, there is not sufficient evidence to support the claim that the mean cost of bariatric surgery is different from $22,500.

49 of 101

Rejection Regions and Critical Values

• • •

Rejection region

(or

critical region

) The range of values for which the null hypothesis is not probable. If a test statistic falls in this region, the null hypothesis is rejected. A critical value

Rejection Regions and Critical Values

Finding Critical Values in a Normal Distribution

Specify the level of significance

Decide whether the test is left-, right-, or two-tailed.

Find the critical value(s)

0 . If the hypothesis test is

left-tailed

, find the

-score that corresponds to an area of

, 4.

right-tailed

, find the

-score that corresponds to an area of 1 –

two-tailed

, find the

-score that corresponds to ½

1 – ½

α .

and Sketch the standard normal distribution. Draw a vertical line at each critical value and shade the rejection region(s).

Example: Finding Critical Values

Find the critical value and rejection region for a two tailed test with

= 0.05.

Solution:

1 – α = 0.95

½α = 0.025

–

0 =

0 0

0 1.96

The rejection regions are to the left of

–z

0 and to the right of

0 = 1.96.

52 of 101

Decision Rule Based on Rejection Region

To use a rejection region to conduct a hypothesis test, calculate the standardized test statistic,

. If the standardized test statistic 1.

is in the rejection region, then reject

0 .

not

in the rejection region, then fail to reject

0 .

Fail to reject

Reject

0 0

Left-Tailed Test

Fail to reject

0 Reject

< –

0 –

0 0

Two-Tailed Test

Reject

z z

Reject

Right-Tailed Test

53 of 101

Using Rejection Regions for a z-Test for a Mean

μ

In Words

State the claim mathematically and verbally. Identify the null and alternative hypotheses.

Specify the level of significance.

Determine the critical value(s).

Determine the rejection region(s).

In Symbols

State

0 and

H a

. Identify

Use Table 4 in Appendix B.

Using Rejection Regions for a z-Test for a Mean μ

In Words

Find the standardized test statistic.

Make a decision to reject or fail to reject the null hypothesis.

In Symbols

 

 

or if

 30 use  

is in the rejection region, reject

0 . Otherwise, fail to reject

0 .

Interpret the decision in the context of the original claim.

Example: Testing with Rejection Regions

Employees at a construction and mining company claim that the mean salary of the company’s mechanical engineers is less than that of the one of its competitors, which is $68,000. A random sample of 30 of the company’s mechanical engineers has a mean salary of $66,900 with a standard deviation of $5500. At

= 0.05, test the employees’ claim.

Solution: Testing with Rejection Regions

• • • •

0 : μ ≥ $68,000

H a

= μ < $68,000 (Claim) 0.05

Rejection Region:

  1.10

•

Test Statistic

 

 

 5500   1.10

30 •

Decision: Fail to reject H 0 .

At the 5% level of significance, there is not sufficient evidence to support the employees’ claim that the mean salary is less than $68,000.

Example: Testing with Rejection Regions

The U.S. Department of Agriculture claims that the mean cost of raising a child from birth to age 2 by husband-wife families in the U.S. is $13,120. A random sample of 500 children (age 2) has a mean cost of $12,925 with a standard deviation of $1745. At

= 0.10, is there enough evidence to reject the claim?

(Adapted from U.S. Department of Agriculture Center for Nutrition Policy and Promotion)

Solution: Testing with Rejection Regions

• • • •

0 : μ = $13,120 (Claim)

H a

= μ ≠ $13,120 0.10

Rejection Region:

•

Test Statistic

 

 

 1745   2.50

500 •

Decision: Reject H 0 .

At the 10% level of significance, you have enough evidence to reject the claim that the mean cost of raising a child from birth to age 2 by husband-wife families in the U.S. is $13,120.

Section 7.2 Summary

• • • • Found

-values and used them to test a mean

Used

-values for a

-test Found critical values and rejection regions in a normal distribution Used rejection regions for a

60 of 101

Section 7.3

Hypothesis Testing for the Mean (Small Samples)

61 of 101

Section 7.3 Objectives

• • • Find critical values in a

-distribution Use the

-test to test a mean

Use technology to find

-values and use them with a

-test to test a mean

Finding Critical Values in a t-Distribution

Identify the level of significance

Identify the degrees of freedom d.f. =

– 1.

Find the critical value(s) using Table 5 in Appendix B in the row with

– 1 degrees of freedom. If the hypothesis test is

left-tailed

, use “One Tail,

” column with a negative sign,

right-tailed

, use “One Tail,

” column with a positive sign,

two-tailed

, use “Two Tails,

” column with a negative and a positive sign.

63 of 101

Example: Finding Critical Values for t

Find the critical value

= 0.05 and

= 21.

for a left-tailed test given • • •

Solution:

The degrees of freedom are d.f. =

– 1 = 21 – 1 = 20 .

Look at α = 0.05 in the “One Tail,

α”

column. Because the test is left tailed, the critical value is negative.

0.05

t 0 = –1.725

64 of 101

Example: Finding Critical Values for t

Find the critical values –

0 given

= 0.10 and

= 26.

and

0 for a two-tailed test • • •

Solution:

The degrees of freedom are d.f. =

– 1 = 26 – 1 = 25 .

Look at

= 0.10 in the “Two Tail,

α”

column. Because the test is two tailed, one critical value is negative and one is positive.

65 of 101

t-Test for a Mean μ (n < 30, σ Unknown)

• • • • •

t-Test for a Mean

A statistical test for a population mean. The

-test can be used when the population is normal or nearly normal,

is unknown, and

< 30. The

test statistic

The

standardized test statistic



s x

 

The degrees of freedom are d.f. =

– 1 .

Using the t-Test for a Mean μ (Small Sample)

In Words

State the claim mathematically and verbally. Identify the null and alternative hypotheses.

Specify the level of significance.

Identify the degrees of freedom.

Determine the critical value(s).

Determine the rejection region(s).

In Symbols

State

0 and

H a

. Identify

d.f. =

– 1.

Use Table 5 in Appendix B.

67 of 101

Using the t-Test for a Mean μ (Small Sample)

In Words

Find the standardized test statistic and sketch the sampling distribution 7.

Make a decision to reject or fail to reject the null hypothesis.

In Symbols



s x

 

is in the rejection region, reject

0 . Otherwise, fail to reject

0 .

Interpret the decision in the context of the original claim.

Example: Testing μ with a Small Sample

A used car dealer says that the mean price of a 2008 Honda CR-V is at least $20,500. You suspect this claim is incorrect and find that a random sample of 14 similar vehicles has a mean price of $19,850 and a standard deviation of $1084. Is there enough evidence to reject the dealer’s claim at

= 0.05? Assume the population is normally distributed.

(Adapted from Kelley Blue Book)

Solution: Testing μ with a Small Sample

• • • • •

0 : μ ≥ $20,500 (Claim)

H a

: μ < $20,500 α = 0.05

df = 14 – 1 = 13

•



s x

 

•

Test Statistic:



Decision:

1084 14   2.244

Reject H 0 .

Rejection Region:

At the 5% level of significance, there is enough evidence to reject the claim that the mean price of a 2008 Honda CR-V is at least $20,500.

≈ –2.244

Example: Testing μ with a Small Sample

An industrial company claims that the mean pH level of the water in a nearby river is 6.8. You randomly select 19 water samples and measure the pH of each. The sample mean and standard deviation are 6.7 and 0.24, respectively. Is there enough evidence to reject the company’s claim at

= 0.05? Assume the population is normally distributed.

Solution: Testing μ with a Small Sample

• • • • •

0 : μ = 6.8 (Claim)

H a

: α = μ ≠ 6.8

0.05

df = 19 – 1 = 18 Rejection Region:

0.025

0 –1.816

•

Test Statistic:



 

s n

 6.7

 6.8

0.24

19   1.816

•

Decision: Fail to reject H 0 .

At the 5% level of significance, there is not enough evidence to reject the claim that the mean pH is 6.8.

72 of 101

Example: Using P-values with t-Tests

A Department of Motor Vehicles office claims that the mean wait time is less than 14 minutes. A random sample of 10 people has a mean wait time of 13 minutes with a standard deviation of 3.5 minutes. At

Solution: Using P-values with t-Tests

• •

0 :

H a

≥ 14 min μ < 14 min (Claim)

TI-83/84 setup: Calculate: Draw:

P ≈ 0.1949

•

Decision: Since 0.1949 > 0.10

, fail to reject H

At the 10% level of significance, there is not enough evidence to support the office’s claim that the mean wait time is less than 14 minutes.

74 of 101

Section 7.3 Summary

• • • Found critical values in a

-distribution Used the

-test to test a mean

Used technology to find

-values and used them with a

-test to test a mean

Section 7.4

Hypothesis Testing for Proportions

76 of 101

Section 7.4 Objectives

• Use the

-test to test a population proportion

77 of 101

z-Test for a Population Proportion

• • • •

z-Test for a Population Proportion p

A statistical test for a population proportion

. Can be used when a binomial distribution is given such that

≥ 5 and

≥ 5.

The

test statistic

p The

standardized test statistic

     

p pq n

78 of 101

Using a z-Test for a Proportion p

Verify that

≥ 5 and

≥ 5.

In Words

State the claim mathematically and verbally. Identify the null and alternative hypotheses.

Specify the level of significance.

Determine the critical value(s).

In Symbols

State

0 and

a . Identify

Use Table 4 in Appendix B.

Determine the rejection region(s).

79 of 101

Using a z-Test for a Proportion p

In Words

Find the standardized test statistic and sketch the sampling distribution.

Make a decision to reject or fail to reject the null hypothesis.

In Symbols

 

p pq n

is in the rejection region, reject

0 . Otherwise, fail to reject

0 .

Interpret the decision in the context of the original claim.

Example: Hypothesis Test for Proportions

A research center claims that less than 50% of U.S. adults have accessed the Internet over a wireless network with a laptop computer. In a random sample of 100 adults, 39% say they have accessed the Internet over a wireless network with a laptop computer. At

= 0.01, is there enough evidence to support the researcher’s claim?

(Adopted from Pew Research Center)

•

Solution:

Verify that

≥ 5 and

≥ 5.

= 100(0.50) = 50 and

81 of 101

Solution: Hypothesis Test for Proportions

• • • •

0 : p ≥ 0.5

H a

: α = p ≠ 0.45

0.01

Rejection Region:

= –2.2

•

Test Statistic

 

p pq n

 (0.5)(0.5) 100   2.2

•

Decision: Fail to reject H 0 .

At the 1% level of significance, there is not enough evidence to support the claim that less than 50% of U.S. adults have accessed the Internet over a wireless network with a laptop computer.

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Example: Hypothesis Test for Proportions

A research center claims that 25% of college graduates think a college degree is not worth the cost. You decide to test this claim and ask a random sample of 200 college graduates whether they think a college degree is not worth the cost. Of those surveyed, 21% reply yes. At

= 0.10 is there enough evidence to reject the claim?

•

Solution:

Verify that

≥ 5 and

≥ 5.

= 200(0.25) = 50 and

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Solution: Hypothesis Test for Proportions

• • • •

0 : p = 0.25 (Claim)

H a

: α = p ≠ 0.25

0.10

Rejection Region:

≈ –1.31

•

Test Statistic

 



pq n

(0.25)(0.75) 200   1.31

•

Decision: Fail to reject H 0 .

At the 10% level of significance, there is enough evidence to reject the claim that 25% of college graduates think a college degree is not worth the cost.

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Section 7.4 Summary

• Used the

-test to test a population proportion

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Section 7.5

Hypothesis Testing for Variance and Standard Deviation

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Section 7.5 Objectives

• • Find critical values for a

2 -test Use the

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Finding Critical Values for the

χ 2

-Test

Specify the level of significance

Determine the degrees of freedom d.f. =

– 1.

The critical values for the

2 -distribution are found in Table 6 in Appendix B. To find the critical value(s) for a

right-tailed test

, use the value that corresponds to d.f. and

α.

left-tailed test

, use the value that corresponds to d.f. and 1 –

α.

two-tailed test

, use the values that corresponds to d.f. and ½

α,

and d.f. and 1 – ½

α.

Finding Critical Values for the

χ 2

-Test

Right-tailed Left-tailed 1 –   0 2  Two-tailed   0 2 1   1 2  1 –  



2 1 2  89 of 101

Example: Finding Critical Values for

χ 2

Find the critical

2 -value for a left-tailed test when

= 11 and

= 0.01.

•

Solution:

Degrees of freedom:

– 1 = 11 – 1 = 10 d.f. • The area to the right of the critical value is 1 –

= 1 – 0.01 = 0.99

  0.01

χ 0 = 2.558

From Table 6, the critical value is .

0  2.558

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Example: Finding Critical Values for

χ 2

Find the critical

2 -value for a two-tailed test when

= 9 and

= 0.05.

• •

Solution:

Degrees of freedom:

– 1 = 9 – 1 = 8 d.f. The areas to the right of the critical values are 1 2   0 .0

25 1  1 2   0 9 5 .

 2 From Table 6, the critical values are and

  2  17.535

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The Chi-Square Test

• • • •

2 -Test for a Variance or Standard Deviation

A statistical test for a population variance or standard deviation. Can be used when the population is normal.

The

test statistic

2 . The

standardized test statistic

 2  (

  1)

2 2 follows a chi-square distribution with degrees of freedom d.f. =

– 1 .

Using the

-Test for a Variance or Standard Deviation

In Words

State the claim mathematically and verbally. Identify the null and alternative hypotheses.

Specify the level of significance.

Determine the degrees of freedom.

Determine the critical value(s).

In Symbols

State

0 and

H a

. Identify

d.f. =

– 1 Use Table 6 in Appendix B.

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Using the

-Test for a Variance or Standard Deviation

In Symbols

In Words

Determine the rejection region(s).

Find the standardized test statistic and sketch the sampling distribution.

Make a decision to reject or fail to reject the null hypothesis.

 2  (

  1)

2 2 If

2 is in the rejection region, reject

0 . Otherwise, fail to reject

0 .

Interpret the decision in the context of the original claim.

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Example: Hypothesis Test for the Population Variance

A dairy processing company claims that the variance of the amount of fat in the whole milk processed by the company is no more than 0.25. You suspect this is wrong and find that a random sample of 41 milk containers has a variance of 0.27. At

Solution: Hypothesis Test for the Population Variance

• • • • •

0 :

2 ≤ 0.25 (Claim)

H a

2 > 0.25

α = 0.05

df = 41 – 1 = 40 Rejection Region:

  0.05

 2  43.2

 0 2  55.758

•

Test Statistic:

 2 

(



1)

2  2 

43.2



0.25

•

Decision: Fail to Reject H 0 .

At the 5% level of significance, there is not enough evidence to reject the company’s claim that the variance of the amount of fat in the whole milk is no more than 0.25.

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Example: Hypothesis Test for the Standard Deviation

A company claims that the standard deviation of the lengths of time it takes an incoming telephone call to be transferred to the correct office is less than 1.4 minutes. A random sample of 25 incoming telephone calls has a standard deviation of 1.1 minutes. At

= 0.10, is there enough evidence to support the company’s claim? Assume the population is normally distributed.

• • • • •

Solution: Hypothesis Test for the Standard Deviation

0 : σ ≥ 1.4 min.

•

Test Statistic:

H a

: σ < 1.4 min. (Claim) α = df = 0.10

25 – 1 = 24

 2  (

 1)

2  2  14.816

 1.4

2 2

Rejection Region:

•

Decision: Reject H 0 .

At the 10% level of significance, there is enough evidence to support the claim that the standard deviation of the lengths of time it takes an incoming telephone call to be transferred to the correct office is less than 1.4 minutes.

Example: Hypothesis Test for the Population Variance

A sporting goods manufacturer claims that the variance of the strengths of a certain fishing line is 15.9. A random sample of 15 fishing line spools has a variance of 21.8. At

= 0.05, is there enough evidence to reject the manufacturer’s claim? Assume the population is normally distributed.

Solution: Hypothesis Test for the Population Variance

• • • • •

0 :

H a

2 = 15.9 (Claim) ≠ 15.9

α = 0.05

df = 15 – 1 = 14 Rejection Region:

 2 • 

Test Statistic:

(



1)

2  2 

(15



19.195

•

Decision:



1)(21.8) 15.9

Fail to Reject H 0

1 2   0.025



2  5.629

  19.195

1 2   0.025



2  26.119

At the 5% level of significance, there is not enough evidence to reject the claim that the variance in the strengths of the fishing line is 15.9.

Section 7.5 Summary

• • Found critical values for a

2 -test Used the

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