chapter 7

Chapter 7

Hypothesis Testing with One Sample

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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

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Section 7.1

Introduction to Hypothesis Testing

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Section 7.1 Objectives

• State a null hypothesis and an alternative hypothesis• Identify type I and type I errors and interpret the level

of significance• Determine whether to use a one-tailed or two-tailed

statistical test and find a p-value• Make and interpret a decision based on the results of

a statistical test• Write a claim for a hypothesis test

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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.

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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.

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Stating a Hypothesis

Null hypothesis • A statistical hypothesis

that contains a statement of equality such as , =, or .

• Denoted H0 read “H subzero” or “H naught.”

Alternative hypothesis • A statement of

inequality such as >, , or <.

• Must be true if H0 is false.

• Denoted Ha read “H sub-a.”

complementary statements

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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.

H0: μ ≤ kHa: μ > k

H0: μ ≥ kHa: μ < k

H0: μ = kHa: μ ≠ k

• Regardless of which pair of hypotheses you use, you always assume μ = k and examine the sampling distribution on the basis of this assumption.

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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.1.A university publicizes that the proportion of its students who graduate in 4 years is 82%.

Equality condition

Complement of H0

H0:

Ha:

(Claim)p = 0.82

p ≠ 0.82

Solution:

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μ ≥ 2.5 gallons per minute


Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim.2.A water faucet manufacturer announces that the mean flow rate of a certain type of faucet is less than 2.5 gallons per minute.

Inequality condition

Complement of HaH0:

Ha:(Claim)μ < 2.5 gallons per minute

Solution:

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μ ≤ 20 ounces


Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim.3.A cereal company advertises that the mean weight of the contents of its 20-ounce size cereal boxes is more than 20 ounces.

Inequality condition

Complement of HaH0:

Ha:(Claim)μ > 20 ounces

Solution:

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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.

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Types of Errors

• A type I error occurs if the null hypothesis is rejected when it is true.

• A type II error occurs if the null hypothesis is not rejected when it is false.

Actual Truth of H0

Decision H0 is true H0 is false

Do not reject H0 Correct Decision Type II Error

Reject H0 Type I Error Correct Decision

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Hypothesis testing is sometimes compared to the legal system used in the United States. Under this system, the following steps are used.

1. A carefully worded accusation is written.

2. The defendant is assumed innocent ( H0) until proven guilty. The burden of proof lies with the prosecution. If the evidence is not strong enough, there is no conviction; a “not guilty” verdict does not prove that a defendant is innocent.

3. The evidence needs to be conclusive beyond a reasonable doubt. The system assumes that more harm is done by convicting the innocent (type I error) than by not convicting the guilty (type II error).

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)

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Let p represent the proportion of chicken that is contaminated.

Solution: Identifying Type I and Type II Errors

H0:

Ha:

p ≤ 0.2

p > 0.2

Hypotheses:

(Claim)

0.16 0.18 0.20 0.22 0.24p

H0: p ≤ 0.20 H0: p > 0.20

Chicken meets USDA limits.

Chicken exceeds USDA limits.

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A type I error is rejecting H0 when it is true. The actual proportion of contaminated chicken is lessthan or equal to 0.2, but you decide to reject H0.

A type II error is failing to reject H0 when it is false. The actual proportion of contaminated chicken is greater than 0.2, but you do not reject H0.

H0:Ha:

p ≤ 0.2p > 0.2

Hypotheses:(Claim)

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H0:Ha:

p ≤ 0.2p > 0.2

Hypotheses:(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.

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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

• P(type II error) = β (beta)

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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.

σ2

x

χ2 (Section 7.5)s2

z (Section 7.4)pt (Section 7.3 n < 30)z (Section 7.2 n 30)μ

Standardized test statistic

Test statisticPopulation parameter

p̂

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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.

• Depends on the nature of the test.

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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 H0.

• This region is indicated by the alternative hypothesis.

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Left-tailed Test• The alternative hypothesis Ha contains the less-than

inequality symbol (<).

z0 1 2 3-3 -2 -1

Test statistic

H0: μ k

Ha: μ < k

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

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• The alternative hypothesis Ha contains the greater-than inequality symbol (>).

z0 1 2 3-3 -2 -1

Right-tailed Test

H0: μ ≤ k

Ha: μ > k

Test statistic

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

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Two-tailed Test• The alternative hypothesis Ha contains the not equal

inequality symbol (≠). Each tail has an area of ½P.

z0 1 2 3-3 -2 -1

Test statistic

Test statistic

H0: μ = k

Ha: μ kP is twice the area to the left of the negative test statistic.

P is twice the area to the right of the positive test statistic.

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Example: Identifying The Nature of a Test

For each claim, state H0 and Ha. 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 P-value.1.A university publicizes that the proportion of its students who graduate in 4 years is 82%.

H0:Ha:

p = 0.82p ≠ 0.82

Two-tailed test z0-z z

½ P-value area

½ P-value area

Solution:

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For each claim, state H0 and Ha. 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 P-value.2.A water faucet manufacturer announces that the mean flow rate of a certain type of faucet is less than 2.5 gallons per minute.

H0:Ha:

Left-tailed testz0-z

P-value area

μ ≥ 2.5 gpmμ < 2.5 gpm

Solution:

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For each claim, state H0 and Ha. 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 P-value.3.A cereal company advertises that the mean weight of the contents of its 20-ounce size cereal boxes is more than 20 ounces.

H0:Ha:

Right-tailed testz0 z

P-value areaμ ≤ 20 oz

μ > 20 oz

Solution:

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Making a DecisionDecision Rule Based on P-value• Compare the P-value with .

If P , then reject H0. If P > , then fail to reject H0.

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Example: Interpreting a Decision

You perform a hypothesis test for the following claim. How should you interpret your decision if you reject H0? If you fail to reject H0?

1.H0 (Claim): A university publicizes that the proportion of its students who graduate in 4 years is 82%.

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Solution: Interpreting a Decision

• The claim is represented by H0.

• If you reject H0 you should conclude “there is sufficient evidence to indicate that the university’s claim is false.”

• If you fail to reject H0, you should conclude “there is insufficient evidence to indicate that the university’s claim (of a four-year graduation rate of 82%) is false.”

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Example: Interpreting a Decision

You perform a hypothesis test for the following claim. How should you interpret your decision if you reject H0? If you fail to reject H0?

2.Ha (Claim): Consumer Reports states that the mean stopping distance (on a dry surface) for a Honda Civic is less than 136 feet.Solution:• The claim is represented by Ha.

• H0 is “the mean stopping distance…is greater than or equal to 136 feet.”

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Solution: Interpreting a Decision

• If you reject H0 you should conclude “there is enough evidence to support Consumer Reports’ claim that the stopping distance for a Honda Civic is less than 136 feet.”

• If you fail to reject H0, you should conclude “there is not enough evidence to support Consumer Reports’ claim that the stopping distance for a Honda Civic is less than 136 feet.”

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z0

Steps for Hypothesis Testing

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

H0: ? Ha: ?2. Specify the level of significance.

α = ?3. Determine the standardized

sampling distribution and draw its graph.

4. Calculate the test statisticand its standardized value.Add it to your sketch. z0

Test statistic

This sampling distribution is based on the assumption that H0 is true.

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Steps for Hypothesis Testing

5. Find the P-value.6. Use the following decision rule.

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

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

Fail to reject H0.

Yes

Reject H0.

No

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Section 7.1 Summary

• Stated a null hypothesis and an alternative hypothesis• Identified type I and type I errors and interpreted the

level of significance• Determined whether to use a one-tailed or two-tailed

statistical test and found a p-value• Made and interpreted a decision based on the results

of a statistical test• Wrote a claim for a hypothesis test

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Tests of Significance

http://www.learner.org/courses/againstallodds/unitpages/unit25.html


Section 7.2

Hypothesis Testing for the Mean (Large Samples)

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• Find P-values and use them to test a mean μ• Use P-values for a z-test• Find critical values and rejection regions in a normal

distribution• Use rejection regions for a z-test

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Using P-values to Make a Decision

Decision Rule Based on P-value• To use a P-value to make a conclusion in a hypothesis

test, compare the P-value with .1. If P , then reject H0.

2. If P > , then fail to reject H0.

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Example: Interpreting a P-valueThe P-value for a hypothesis test is P = 0.0237. What is your decision if the level of significance is1.0.05?

2.0.01?

Solution:Because 0.0237 < 0.05, you should reject the null hypothesis.

Solution:Because 0.0237 > 0.01, you should fail to reject the null hypothesis.

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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 P-value.a.For a left-tailed test, P = (Area in left tail).b.For a right-tailed test, P = (Area in right tail).c.For a two-tailed test, P = 2(Area in tail of test statistic).

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Example: Finding the P-value

Find the P-value for a left-tailed hypothesis test with a test statistic of z = -2.23. Decide whether to reject H0 if the level of significance is α = 0.01.

z0-2.23

P = 0.0129

Solution:For a left-tailed test, P = (Area in left tail)

Because 0.0129 > 0.01, you should fail to reject H0

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z0 2.14

Example: Finding the P-value

Find the P-value for a two-tailed hypothesis test with a test statistic of z = 2.14. Decide whether to reject H0 if the level of significance is α = 0.05.

Solution:For a two-tailed test, P = 2(Area in tail of test statistic)

Because 0.0324 < 0.05, you should reject H0

0.9838

1 – 0.9838 = 0.0162 P = 2(0.0162)

= 0.0324

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Z-Test for a Mean μ

• Can be used when the population is normal and is known, or for any population when the sample size n is at least 30.

• The test statistic is the sample mean • The standardized test statistic is z

• When n 30, the sample standard deviation s can be substituted for .

xzn

standard error xn

x

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


2. Specify the level of significance.

3. Determine the standardized test statistic.

4. Find the area that corresponds to z.

State H0 and Ha.

Identify .

Use Table 4 in Appendix B.

xzn

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In Words In Symbols

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

Reject H0 if P-value is less than or equal to . Otherwise, fail to reject H0.

5. Find the P-value.a. For a left-tailed test, P = (Area in left tail).b. For a right-tailed test, P = (Area in right tail).c. For a two-tailed test, P = 2(Area in tail of test

statistic).6. Make a decision to reject or

fail to reject the null hypothesis.

7. Interpret the decision in the context of the original claim.

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In Words In Symbols

Example: Hypothesis Testing Using P-values

In an advertisement, a pizza shop claims that its mean delivery time is less than 30 minutes. A random selection of 36 delivery times has a sample mean of 28.5 minutes and a standard deviation of 3.5 minutes. Is there enough evidence to support the claim at = 0.01? Use a P-value.

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Solution: Hypothesis Testing Using P-values

• H0:

• Ha: • = • Test Statistic:

μ ≥ 30 minμ < 30 min0.01

28.5 303.5 362.57

xzn

• Decision:

At the 1% level of significance, you have sufficient evidence to conclude the mean delivery time is less than 30 minutes.

z0-2.57

0.0051

• P-value

0.0051 < 0.01Reject H0

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Example: Hypothesis Testing Using P-values

You think that the average franchise investment information shown in the graph is incorrect, so you randomly select 30 franchises and determine the necessary investment for each. The sample mean investment is $135,000 with astandard deviation of $30,000. Is there enough evidence to support your claim at = 0.05? Use a P-value.

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Solution: Hypothesis Testing Using P-values

• H0:

• Ha: • = • Test Statistic:

μ = $143,260μ ≠ $143,2600.05

135,000 143, 26030,000 30

1.51

xzn

• Decision:

At the 5% level of significance, there is not sufficient evidence to conclude the mean franchise investment is different from $143,260.

• P-valueP = 2(0.0655) = 0.1310

0.1310 > 0.05

Fail to reject H0

z0-1.51

0.0655

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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 z0 separates the rejection region from

the nonrejection region.

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Rejection Regions and Critical ValuesFinding Critical Values in a Normal Distribution1. Specify the level of significance .2. Decide whether the test is left-, right-, or two-tailed.3. Find the critical value(s) z0. If the hypothesis test is

a. left-tailed, find the z-score that corresponds to an area of ,

b. right-tailed, find the z-score that corresponds to an area of 1 – ,

c. two-tailed, find the z-score that corresponds to ½ and 1 – ½.

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

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Example: Finding Critical Values

Find the critical value and rejection region for a two-tailed test with = 0.05.

z0 z0z0

½α = 0.025 ½α = 0.025

1 – α = 0.95

The rejection regions are to the left of -z0 = -1.96 and to the right of z0 = 1.96.

z0 = 1.96-z0 = -1.96

Solution:

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Decision Rule Based on Rejection Region

To use a rejection region to conduct a hypothesis test, calculate the standardized test statistic, z. If the standardized test statistic1. is in the rejection region, then reject H0.2. is not in the rejection region, then fail to reject H0.

z0z0

Fail to reject H0.

Reject H0.

Left-Tailed Testz < z0

z0 z0

Reject Ho.

Fail to reject Ho.

z > z0 Right-Tailed Test

z0z0

Two-Tailed Testz0z < -z0 z > z0

Reject H0

Fail to reject H0

Reject H0

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



3. Sketch the sampling distribution.

4. Determine the critical value(s).

5. Determine the rejection region(s).

State H0 and Ha.

Identify .


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In Words In Symbols

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

6. Find the standardized test statistic.

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


or if 30

use

xz nns

.If z is in the rejection region, reject H0. Otherwise, fail to reject H0.

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In Words In Symbols

Example: Testing with Rejection Regions

Employees in a large accounting firm claim that the mean salary of the firm’s accountants is less than that of its competitor’s, which is $45,000. A random sample of 30 of the firm’s accountants has a mean salary of $43,500 with a standard deviation of $5200. At α = 0.05, test the employees’ claim.

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Solution: Testing with Rejection Regions

• H0:

• Ha: • = • Rejection Region:

μ ≥ $45,000μ < $45,0000.05

43,500 45,0005200 30

1.58

xzn

• Decision:At the 5% level of significance, there is not sufficient evidence to support the employees’ claim that the mean salary is less than $45,000.

• Test Statistic

z0-1.645

0.05

-1.58-1.645

Fail to reject H0

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Example: Testing with Rejection Regions

The U.S. Department of Agriculture reports that the mean cost of raising a child from birth to age 2 in a rural area is $10,460. You believe this value is incorrect, so you select a random sample of 900 children (age 2) and find that the mean cost is $10,345 with a standard deviation of $1540. At α = 0.05, is there enough evidence to conclude that the mean cost is different from $10,460? (Adapted from U.S. Department of Agriculture Center for Nutrition Policy and Promotion)

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Solution: Testing with Rejection Regions

• H0:


μ = $10,460μ ≠ $10,4600.05

10,345 10, 4601540 900

2.24

xzn

• Decision:At the 5% level of significance, you have enough evidence to conclude the mean cost of raising a child from birth to age 2 in a rural area is significantly different from $10,460.

• Test Statistic

z0-1.96

0.025

1.96

0.025

-1.96 1.96

-2.24

Reject H0

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Section 7.2 Summary

• Found P-values and used them to test a mean μ• Used P-values for a z-test• Found critical values and rejection regions in a

normal distribution• Used rejection regions for a z-test

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Section 7.3

Hypothesis Testing for the Mean (Small Samples)

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• Find critical values in a t-distribution• Use the t-test to test a mean μ• Use technology to find P-values and use them with a

t-test to test a mean μ

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Finding Critical Values in a t-Distribution1. Identify the level of significance .2. Identify the degrees of freedom d.f. = n – 1.3. Find the critical value(s) using Table 5 in Appendix B in the

row with n – 1 degrees of freedom. If the hypothesis test isa. left-tailed, use “One Tail, ” column with a negative

sign,b. right-tailed, use “One Tail, ” column with a positive

sign,c. two-tailed, use “Two Tails, ” column with a negative

and a positive sign.

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Example: Finding Critical Values for t

Find the critical value t0 for a left-tailed test given = 0.05 and n = 21.Solution:• The degrees of freedom are

d.f. = n – 1 = 21 – 1 = 20.• Look at α = 0.05 in the

“One Tail, ” column. • Because the test is left-

tailed, the critical value is negative.

t0-1.725

0.05

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Example: Finding Critical Values for t

Find the critical values t0 and -t0 for a two-tailed test given = 0.05 and n = 26.Solution:• The degrees of freedom are

d.f. = n – 1 = 26 – 1 = 25.• Look at α = 0.05 in the

“Two Tail, ” column. • Because the test is two-

tailed, one critical value is negative and one is positive.

t0-2.060

0.025

2.060

0.025

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t-Test for a Mean μ (n < 30, Unknown)

t-Test for a Mean • A statistical test for a population mean. • The t-test can be used when the population is normal

or nearly normal, is unknown, and n < 30. • The test statistic is the sample mean • The standardized test statistic is t.

• The degrees of freedom are d.f. = n – 1.

xts n

x

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



3. Identify the degrees of freedom and sketch the sampling distribution.

4. Determine any critical value(s).

State H0 and Ha.

Identify .


d.f. = n – 1.

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In Words In Symbols

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

5. Determine any rejection region(s).




xts n

If t is in the rejection region, reject H0. Otherwise, fail to reject H0.

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In Words In Symbols

Example: Testing μ with a Small Sample

A used car dealer says that the mean price of a 2005 Honda Pilot LX is at least $23,900. You suspect this claim is incorrect and find that a random sample of 14 similar vehicles has a mean price of $23,000 and a standard deviation of $1113. Is there enough evidence to reject the dealer’s claim at α = 0.05? Assume the population is normally distributed. (Adapted from Kelley Blue Book)

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Solution: Testing μ with a Small Sample

• H0:

• Ha: • α = • df = • Rejection Region:

• Test Statistic:

• Decision:

μ ≥ $23,900μ < $23,9000.0514 – 1 = 13

23,000 23,900 3.0261113 14

xts n

t0-1.771

0.05

-3.026-1.771

At the 0.05 level of significance, there is enough evidence to reject the claim that the mean price of a 2005 Honda Pilot LX is at least $23,900

Reject H0

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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 α = 0.05? Assume the population is normally distributed.

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Solution: Testing μ with a Small Sample

• H0:


• Test Statistic:

• Decision:

μ = 6.8μ ≠ 6.80.0519 – 1 = 18

6.7 6.8 1.8160.24 19

xts n

At the 0.05 level of significance, there is not enough evidence to reject the claim that the mean pH is 6.8.t0-2.101

0.025

2.101

0.025

-2.101 2.101

-1.816

Fail to reject H0

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Example: Using P-values with t-Tests

The American Automobile Association claims that the mean daily meal cost for a family of four traveling on vacation in Florida is $118. A random sample of 11 such families has a mean daily meal cost of $128 with a standard deviation of $20. Is there enough evidence to reject the claim at α = 0.10? Assume the population is normally distributed. (Adapted from American Automobile Association)

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Solution: Using P-values with t-Tests• H0:

• Ha:

• Decision:

μ = $118μ ≠ $118

Fail to reject H0. At the 0.10 level of significance, there is not enough evidence to reject the claim that the mean daily meal cost for a family of four traveling on vacation in Florida is $118.

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0.128 > 0.10

Section 7.3 Summary

• Found critical values in a t-distribution• Used the t-test to test a mean μ• Used technology to find P-values and used them with

a t-test to test a mean μ

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Small Sample Inference for One Mean



Section 7.4

Hypothesis Testing for Proportions

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• Use the z-test to test a population proportion p

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z-Test for a Population Proportion

z-Test for a Population Proportion• A statistical test for a population proportion. • Can be used when a binomial distribution is given such that

np 5 and nq 5.• The test statistic is the sample proportion . • The standardized test statistic is z.

ˆ

ˆ

ˆ ˆp

p

p p pzpq n

p̂

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Using a z-Test for a Proportion p



3. Sketch the sampling distribution.


State H0 and Ha.

Identify .


Verify that np ≥ 5 and nq ≥ 5

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In Words In Symbols

Using a z-Test for a Proportion p





If z is in the rejection region, reject H0. Otherwise, fail to reject H0.

p̂ pzpq n

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In Words In Symbols

Example: Hypothesis Test for Proportions

Zogby International claims that 45% of people in the United States support making cigarettes illegal within the next 5 to 10 years. You decide to test this claim and ask a random sample of 200 people in the United States whether they support making cigarettes illegal within the next 5 to 10 years. Of the 200 people, 49% support this law. At α = 0.05 is there enough evidence to reject the claim?

Solution:•Verify that np ≥ 5 and nq ≥ 5.

np = 200(0.45) = 90 and nq = 200(0.55) = 11087

Solution: Hypothesis Test for Proportions

• H0:


p = 0.45p ≠ 0.450.05

ˆ 0.49 0.45(0.45)(0.55) 200

1.14

p pzpq n

• Decision:At the 5% level of significance, there is not enough evidence to reject the claim that 45% of people in the U.S. support making cigarettes illegal within the next 5 to 10 years.

• Test Statistic

z0-1.96

0.025

1.96

0.025

-1.96 1.96

1.14

Fail to reject H0

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Example: Hypothesis Test for Proportions

The Pew Research Center claims that more than 55% of U.S. adults regularly watch their local television news. You decide to test this claim and ask a random sample of 425 adults in the United States whether they regularly watch their local television news. Of the 425 adults, 255 respond yes. At α = 0.05 is there enough evidence to support the claim?

Solution:•Verify that np ≥ 5 and nq ≥ 5.

np = 425(0.55) ≈ 234 and nq = 425 (0.45) ≈ 191

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Solution: Hypothesis Test for Proportions

• H0:


p ≤ 0.55p > 0.550.05

ˆ 255 425 0.55(0.55)(0.45) 425

2.07

p pzpq n

• Decision:At the 5% level of significance, there is enough evidence to support the claim that more than 55% of U.S. adults regularly watch their local television news.

• Test Statistic

z0 1.645

0.05

2.071.645

Reject H0

91

Section 7.4 Summary

• Used the z-test to test a population proportion p

92

93

Inference for Proportions



Section 7.5

Hypothesis Testing for Variance and Standard Deviation

94


• Find critical values for a χ2-test• Use the χ2-test to test a variance or a standard

deviation

95

Finding Critical Values for the χ2-Test

1. Specify the level of significance .2. Determine the degrees of freedom d.f. = n – 1.3. The critical values for the χ2-distribution are found in Table 6

of Appendix B. To find the critical value(s) for aa. right-tailed test, use the value that corresponds to d.f. and

.b. left-tailed test, use the value that corresponds to d.f. and

1 – .c. two-tailed test, use the values that corresponds to d.f. and

½ and d.f. and 1 – ½.

96

Finding Critical Values for the χ2-Test

χ2

12

2L 2

R

12

1 – α

χ220

1 – α

χ2

20

1 – α

Right-tailed Left-tailed

Two-tailed

97

Example: Finding Critical Values for χ2

Find the critical χ2-value for a left-tailed test whenn = 11 and = 0.01.

Solution:•Degrees of freedom: n – 1 = 11 – 1 = 10 d.f. •The area to the right of the critical value is 1 – = 1 – 0.01 = 0.99.

From Table 6, the critical value is .

20 2.558

χ2

0.01

20 2.558

98

Example: Finding Critical Values for χ2

Find the critical χ2-value for a two-tailed test when n = 13 and = 0.01.

Solution:•Degrees of freedom: n – 1 = 13 – 1 = 12 d.f. •The areas to the right of the critical values are

From Table 6, the critical values are and

02

051 .0

11 92

0.9 5 χ2

1 0.0052 1 0.0052

2 3.074L 2 28.299R 2 3.074L

2 28.299R 99

The Chi-Square Test

χ2-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 is s2. • The standardized test statistic

follows a chi-square distribution with degrees of freedom d.f. = n – 1.

22

2( 1)n s

100

Using the χ2-Test for a Variance or Standard Deviation



3. Determine the degrees of freedom and sketch the sampling distribution.


State H0 and Ha.

Identify .


d.f. = n – 1

101

In Words In Symbols

Using the χ2-Test for a Variance or Standard Deviation

22

2( 1)n s

If χ2 is in the rejection region, reject H0. Otherwise, fail to reject H0.





102

In Words In Symbols

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

103

Solution: Hypothesis Test for the Population Variance

• H0:


• Test Statistic:

• Decision:

σ2 ≤ 0.25σ2 > 0.250.0541 – 1 = 40

22

2

( 1) (41 1)(0.27)0.25

43.2

n s

Fail to Reject H0

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.

χ2

0.05

55.75855.75843.2

105

Example: Hypothesis Test for the Standard Deviation

A restaurant claims that the standard deviation in the length of serving times is less than 2.9 minutes. A random sample of 23 serving times has a standard deviation of 2.1 minutes. At α = 0.10, is there enough evidence to support the restaurant’s claim? Assume the population is normally distributed.

106

• H0:


• Test Statistic:

• Decision:

σ ≥ 2.9 min.σ < 2.9 min.0.1023 – 1 = 22

2 22

2 2

( 1) (23 1)(2.1)2.9

11.536

n s

Reject H0

At the 10% level of significance, there is enough evidence to support the claim that the standard deviation for the length of serving times is less than 2.9 minutes.11.536

14.042χ2

0.10

14.042

108

Solution: Hypothesis Test for the Standard Deviation:

Example: Hypothesis Test for the Population Variance

A sporting goods manufacturer claims that the variance of the strength in a certain fishing line is 15.9. A random sample of 15 fishing line spools has a variance of 21.8. At α = 0.05, is there enough evidence to reject the manufacturer’s claim? Assume the population is normally distributed.

109

5.629χ2

1 0.0252

26.119

• H0:


• Test Statistic:

• Decision:

σ2 = 15.9σ2 ≠ 15.90.0515 – 1 = 14

22

2

( 1) (15 1)(21.8)15.9

19.194

n s

Fail to Reject H0

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

5.62919.194

26.119

111

Solution: Hypothesis Test for the Population Variance:

Section 7.5 Summary

• Found critical values for a χ2-test• Used the χ2-test to test a variance or a standard

deviation

112

113 .

Elementary Statistics:Picturing the World

Fifth Edition

by Larson and Farber

Chapter 7: Hypothesis Testing with One Sample

114 .

State the null and alternative hypotheses.

A company claims the mean lifetime of its AA batteries is more than 16 hours.

A. H0: μ > 16 Ha: μ ≤ 16

B. H0: μ < 16 Ha: μ ≥ 16

C. H0: μ ≤ 16 Ha: μ > 16

D. H0: μ ≥ 16 Ha: μ < 16

115 .


A company claims the mean lifetime of its AA batteries is more than 16 hours.

A. H0: μ > 16 Ha: μ ≤ 16

B. H0: μ < 16 Ha: μ ≥ 16

C. H0: μ ≤ 16 Ha: μ > 16

D. H0: μ ≥ 16 Ha: μ < 16

116 .


A student claims the mean cost of a textbook is at least $125.

A. H0: μ > 125 Ha: μ ≤ 125

B. H0: μ < 125 Ha: μ ≥ 125

C. H0: μ ≤ 125 Ha: μ > 125

D. H0: μ ≥ 125 Ha: μ < 125

117 .


A student claims the mean cost of a textbook is at least $125.

A. H0: μ > 125 Ha: μ ≤ 125

B. H0: μ < 125 Ha: μ ≥ 125

C. H0: μ ≤ 125 Ha: μ > 125

D. H0: μ ≥ 125 Ha: μ < 125

118 .

True or false:

Testing the claim that at least 88% of students have a cell phone would be a right-tail test.

A. True

B. False

119 .

True or false:

Testing the claim that at least 88% of students have a cell phone would be a right-tail test.

A. True

B. False

0 : 0.88: 0.88a

H pH p

0

120 .

You are testing the claim that the mean cost of a new car is more than $25,200. How should you interpret a decision that rejects the null hypothesis?

A. There is enough evidence to reject the claim.

B. There is enough evidence to support the claim.

C. There is not enough evidence to reject the claim.

D. There is not enough evidence to support the claim.

You are testing the claim that the mean cost of a new car is more than $25,200. How should you interpret a decision that rejects the null hypothesis?

A. There is enough evidence to reject the claim.

B. There is enough evidence to support the claim.

C. There is not enough evidence to reject the claim.

D. There is not enough evidence to support the claim.

122 .

True or false:

Given H0: μ = 40 Ha: μ ≠ 40 and P = 0.0436. You would reject the null hypothesis at the 0.05 level of significance.

A. True

B. False

123 .

True or false:

Given H0: μ = 40 Ha: μ ≠ 40 and P = 0.0436. You would reject the null hypothesis at the 0.05 level of significance.

A. True

B. Falsep

124 .

Find the critical value, z0, for a left-tailed test at the 0.10 level of significance.

A. z0 = –1.645

B. z0 = 1.645

C. z0 = –1.28

D. z0 = 1.28

125 .

Find the critical value, z0, for a left-tailed test at the 0.10 level of significance.

A. z0 = –1.645

B. z0 = 1.645

C. z0 = –1.28

D. z0 = 1.28

126 .

Find the standardized test statistic z for the following situation:

Claim: μ >15; s = 3.4 n = 40

A. z = 2.60

B. z = –2.60

C. z = –0.07

D. z = 12.90

.x 13 6

127 .


Claim: μ >15; s = 3.4 n = 40

A. z = 2.60

B. z = –2.60

C. z = –0.07

D. z = 12.90

.x 13 6

( ) (13.6 15) 2.60423.4( )40

xz sn

128 .

Find the critical value(s), t0, for a two-tailed test, α = 0.05, and n = 8.

A. –t0 = –1.96 and t0 = 1.96

B. –t0 = –2.306 and t0 = 2.306

C. –t0 = –1.895 and t0 = 1.895

D. –t0 = –2.365 and t0 = 2.365

129 .

Find the critical value(s), t0, for a two-tailed test, α = 0.05, and n = 8.

130 .

Use technology to find the P-value for the following test:

H0: μ ≤ 20 Ha: μ > 20 s = 2.1 n = 16

A. 0.0128

B. 0.0257

C. 0.9872

D. 0.0066

.x 21 3

131 .

Use technology to find the P-value for the following test:

H0: μ ≤ 20 Ha: μ > 20 s = 2.1 n = 16

A. 0.0128

B. 0.0257

C. 0.9872

D. 0.0066

.x 21 3

132 .


Claim: p ≠ 0.23; x = 52 n = 200

A. z = 0.97

B. z = 1.01

C. z = 0.51

D. z = –1.01

133 .


Claim: p ≠ 0.23; x = 52 n = 200

A. z = 0.97

B. z = 1.01

C. z = 0.51

D. z = –1.01

134 .

Find the standardized test statistic χ2 for the following situation:

Claim: σ < 5.2; s = 4.47 n = 20

A. χ2 = 25.71

B. χ2 = 16.33

C. χ2 = 14.04

D. χ2 = 14.78

.

Find the standardized test statistic χ2 for the following situation: Claim: σ < 5.2; s = 4.47 n = 20

2 22

2 2

( 1) (20 1)(4.47) 14.0398(5.2)

n s

20

Don't confuse test statistic with

the critical value !

chapter 7

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