Chapter Hypothesis Testing with One Sample 1 of 101 7 © 2012 Pearson Education, Inc. All rights reserved.

ChapterHypothesis Testing with One Sample

1 of 101

7

© 2012 Pearson Education, Inc.All rights reserved.

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

© 2012 Pearson Education, Inc. All rights reserved. 2 of 101

Section 7.1

Introduction to Hypothesis Testing


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 p-value

• Make and interpret a decision based on the results of a statistical test

• Write a claim for a hypothesis test


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 (such as the mean).

• 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 H0 read “H sub-zero” or “H naught.” (or H –o)

Alternative hypothesis

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

• Must be true if H0 is false.

• Denoted Ha read “H sub-a.” (or H – a)

complementary statements


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.


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

Equality condition

Complement of H0

H0:

Ha:

(Claim)p = 0.61

p ≠ 0.61

Solution:


μ ≥ 15 minutes


Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim.2. A car dealership announces that the mean time for an oil change is less than 15 minutes.

Inequality condition

Complement of HaH0:

Ha:(Claim)μ < 15 minutes

Solution:


μ ≤ 18 years


Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim.3. A company advertises that the mean life of its furnaces is more than 18 years

Inequality condition

Complement of HaH0:

Ha:(Claim)μ > 18 years

Solution:


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. (NOTE: This may or may not be the claim)

• At the end of the test, one of two decisions will be made: reject the null hypothesis fail to reject the null hypothesis (FTR)

• Because your decision is based on a sample, there is the possibility of making the wrong decision.


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.

• Hint: 1 comes before 2, True comes before 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


KNOW THESE

Types of Errors

• Another way of looking at a Type 1 and Type 2 error:

• Type 1 Error –Rejecting a true “claim”

• Type 2 Error –Accepting (Failing To Reject, or FTR) a false “claim”

Larson/Farber 5th ed. 14

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)


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.



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)



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.


Level of Significance

Level of significance • Your maximum allowable probability of making a

type I error. Denoted by α, 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: α = 0.10 α = 0.05 α = 0.01

• P(type II error) = β (beta) We won’t cover this much, it’s higher level statistics.


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


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.


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.

• In other words, we’re testing the alternative hypothesis


Left-tailed Test

• The alternative hypothesis Ha contains the less-than inequality symbol (<).

z

0 1 2 3–3 –2 –1

Test statistic

H0: μ ≥ k

Ha: μ < kP is the area to the left of the standardized test statistic.


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

Right-tailed Test

H0: μ ≤ k

Ha: μ > k

Test statistic


z

0 1 2 3–3 –2 –1

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

Two-tailed Test

• The alternative hypothesis Ha contains the not-equal-to symbol (≠). Each tail has an area of ½P.

z

0 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 standardized test statistic.

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


Example: Identifying The Nature of a TestFor 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 school publicizes that the proportion of its students who are involved in at least one extracurricular activity is 61%.

H0:Ha:

p = 0.61p ≠ 0.61

Two-tailed test

Solution:


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

H0:Ha:

Left-tailed testz

0-z

P-value area

μ ≥ 15 minμ < 15 min

Solution:


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

H0:Ha:

Right-tailed testz

0 z

P-value areaμ ≤ 18 yr

μ > 18 yr

Solution:


Making a Decision

Decision Rule Based on P-value• Compare the P-value with α.

If P ≤ α , then reject H0.

If P > α, then fail to reject H0.

Claim

Decision Claim is H0 Claim is Ha

Reject H0

Fail to reject H0

There is enough evidence to reject the claim

There is not enough evidence to reject the claim

There is enough evidence to support the claim

There is not enough evidence to support the claim


Remember, α is your maximum allowable probability of making a type I error.

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 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 H0.

Solution: Interpreting a Decision

• If you reject H0, 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 H0, 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%.”


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

Solution:

• The claim is represented by Ha.

• H0 is “the mean time for an oil change is greater than or equal to 15 minutes.”


Solution: Interpreting a Decision

• If you reject H0, 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 H0, then you should conclude “there is not enough evidence to support the dealership’s claim that the mean time for an oil change is less than 15 minutes.”


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 sketch its graph.

4. Calculate the test statisticand its corresponding standardized test statistic.Add it to your sketch.

z0

Standardized test statistic

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


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


Remember: P-Value is 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.

So, if my sample is more extreme than my level of significance, I have to reject the null hypothesis

What you evaluating is the AREA in the tail

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 p-value

• Made and interpreted a decision based on the results of a statistical test

• Wrote a claim for a hypothesis test


Assignment

• Page 367 5-9,11-16,21-24


Section 7.2

Hypothesis Testing for the Mean (Large Samples)



• 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


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 α.

1. If P ≤ α, then reject H0.

2. If P > α, then fail to reject H0.


Example: Interpreting a P-value

The P-value for a hypothesis test is P = 0.0237. What is your decision if the level of significance is

1. α = 0.05?

2. α = 0.01?

Solution:Because 0.0237 < 0.05, you should reject the null hypothesis. –This would be outside the limits of the tail

Solution:Because 0.0237 > 0.01, you should fail to reject the null hypothesis. -This would be inside our limits


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).


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.


On a Calculator

• Calculate P-Value 2nd Vars normalcdf For a left tail test enter (–EE99,Zscore) For a right tail test enter (Zscore, EE99) For a 2 tail test, do one or the other and multiply

your answer times 2 (for 2 tails)


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


Z-Test for a Mean μ

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

xzn

standard error xn

x


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 α.

Use Table 4 in Appendix B.

xzn

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 α. 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.

In Words In Symbols


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

α = 0.01? Use a P-value.


Solution: Hypothesis Testing Using P-values

• H0:

• Ha:

• α = • Test Statistic:

μ ≥ 13 sec

μ < 13 sec (Claim)

0.01

• Decision:

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.

• P-value

0.0014 < 0.01Reject H0 .


12.9 13

0.19 32

2.98

xz

n

On a Calculator

• Stats Tests Z-Test

• Input Stats

• Enter the mean, the standard deviation, the sample mean, and n

• Here were our 2 equations:

• We will test the claim, so chose < μ0

• We get a p-value of .00145, which is less than .01


• H0:

• Ha:

• α =

μ ≥ 13 sec

μ < 13 sec (Claim)

0.01 (alpha)

Therefore, we should reject the null hypothesis, because our sample value is outside the limit we set for ourselves

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 α = 0.05? Use a P-value. (Adapted from National Institute of Diabetes and Digestive and Kidney Diseases)


Solution: Hypothesis Testing Using P-values

• H0:

• Ha:

• α = • Test Statistic:

μ = $22,500

μ ≠ 22,500 (Claim)

0.05

• Decision:

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.

• P-value

0.0836 > 0.05

Fail to reject H0 .


21,545 22,500

3015 30

1.73

xz

n

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.


Rejection Regions and Critical Values

Finding Critical Values in a Normal Distribution

1. Specify the level of significance α.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 α,

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

c. two-tailed, find the z-score that corresponds to ½α and1 – ½α.

4. 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 α = 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:


Calculator: Again, we can use invNorm(.025) and select either or both tails

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 Test

z < z0 z

0 z0

Reject Ho.

Fail to reject Ho.

z > z0

Right-Tailed Test

z0–z0

Two-Tailed Testz0z < –z0 z > z0

Reject H0

Fail to reject H0

Reject H0


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



3. Determine the critical value(s).

4. Determine the rejection region(s).

State H0 and Ha.

Identify α.


In Words In Symbols


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

5. Find the standardized test statistic.



or if 30

use

xz nn

s

.

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

In Words In Symbols


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 α = 0.05, test the employees’ claim.


Solution: Testing with Rejection Regions

• H0:

• Ha:

• α = • Rejection Region:

μ ≥ $68,000

μ < $68,000 (Claim)

0.05

• Decision: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.

• Test Statistic

Fail to reject H0 .


66,900 68,000

5500 30

1.10

xz

n

1.10z

On a Calculator

• Stats Tests Z-Test

• Input Stats

• Enter the mean, the standard deviation, the sample mean, and n

• Here were our 2 equations:

• We will test the claim, so chose < μ0

• We get a z-score of -1.09, which is inside our desired area

62

• H0:

• Ha:

• α =

Therefore, we should not reject the null hypothesis, because our sample value is inside the limit we set for ourselves

μ ≥ $68,000

μ < $68,000 (Claim)

0.05

Note: You could also compare P-Value to alpha. The p-value is not less than alpha, so we do not reject the null hypothesis

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. Atα = 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

• H0:

• Ha:


μ = $13,120 (Claim)

μ ≠ $13,120

0.10

• Decision: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.

• Test Statistic

Reject H0 .


12,925 13,120

1745 500

2.50

xz

n

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


Assignment

• P. 381 3-12

• P. 382 15-22, 25-34


Chapter 7 Quiz 1

• Using data from question 40 on page 384:

1. Write a null hypothesis and an alternate hypothesis (identify which is the claim)

2. Calculate the critical Z-value

3. Calculate the sample standard deviation

4. Calculate the p-value

5. Using a confidence level α = .09, is there enough information to reject the company’s claim?

4 points each question


Section 7.3

Hypothesis Testing for the Mean (Small Samples)



• 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


Finding Critical Values in a t-Distribution

1. Identify the level of significance α.

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 is

a. left-tailed, use “One Tail, α ” column with a negative sign,

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

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


Example: Finding Critical Values for t

Find the critical value t0 for a left-tailed test givenα = 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, α” column. • Because the test is left-

tailed, the critical value is negative.

t0t0 = –1.725

0.05


On the Calculator

• 2nd Vars InvT(area,D.F.)

• Area = α = 0.05

• n = 21 so D.F. = 20

• Solve InvT(.05,20) and get -1.7247 or -1.725

• NOTE: This is a left tailed test. For a right tailed test, use the “Positive” answer 1.725


Example: Finding Critical Values for t

Find the critical values –t0 and t0 for a two-tailed test given α = 0.10 and n = 26.

Solution:• The degrees of freedom are

d.f. = n – 1 = 26 – 1 = 25.• Look at α = 0.10 in the

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

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


On a Calculator

• 2nd Vars InvT(area,D.F.)

• Area = α = 0.10HOWEVER, this is the total area. So we need to divide this in half since InvT solves for one tail

• n = 26 so D.F. = 25

• Solve InvT(.05,25) and get -1.708

• NOTE: This is a two tailed test. Therefore, we must have 2 rejection regions -1.708 and 1.708

• This will give us our total area of .10


t-Test for a Mean μ (n < 30, σ 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, σ 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


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



3. Identify the degrees of freedom.



State H0 and Ha.

Identify α.


d.f. = n – 1.

In Words In Symbols


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

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



xts n

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

In Words In Symbols


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 α = 0.05? Assume the population is normally distributed. (Adapted from Kelley Blue Book)


Solution: Testing μ with a Small Sample

• H0:

• Ha:

• α =

• df = • Rejection Region:

• Test Statistic:

• Decision:

μ ≥ $20,500 (Claim)μ < $20,500

0.0514 – 1 = 13

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.

Reject H0 .


19,850 20,5002.244

1084 14

xt

s n

t ≈ –2.244

On the Calculator

• Stats Test Ttest

• You must test the ALTERNATE Hypothesis

• Which means you have to know what it is

• Enter data

• Test μ < $20,500

• You will get a t-test value of -2.244 If you had calculated the t-critical value, you could compare, and see that the test value is less than the critical value, and thus reject the null hypothesis

• You get a p-value of .02 This is less than the alpha of .05

• Thus, the probability of occurrence is in the rejection region, so we reject the null hypothesis


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.


Solution: Testing μ with a Small Sample

• H0:

• Ha:

• α =

• df = • Rejection Region:

• Test Statistic:

• Decision:

μ = 6.8 (Claim)μ ≠ 6.8

0.05

19 – 1 = 18

6.7 6.81.816

0.24 19

xt

s n

At the 5% 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 .


On the Calculator

• Stats Test Ttest

• You must test the ALTERNATE Hypothesis

• Which means you have to know what it is

• Enter data

• Test μ ≠ 6.8

• You will get a t-test value of -1.816

• Remember, this is a 2 tail test, so you also have a t-test value of 1.816

• If you had calculated the T-critical value you could compare these

• You get a p-value of .09 This is not less than the alpha of .05

• Thus, the probability of occurrence is in not in the rejection region, so we fail to reject the null hypothesis


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 α = 0.10, test the office’s claim. Assume the population is normally distributed.


Solution: Using P-values with t-Tests


• H0:

• Ha:

• Decision:

μ ≥ 14 minμ < 14 min (Claim)

TI-83/84 setup: Calculate: Draw:

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.

P ≈ 0.1949Since 0.1949 > 0.10, fail to reject H0.

Chapter 7 Quiz 2



2. Calculate the critical Z-value (based on α = .06)

3. Calculate the sample standard deviation


5. Using a confidence level α = .06, is there enough information to support the scientist’s claim?



Assignment

• P. 393 4-30 even


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


Chapter 7 Quiz 3




3. Compare the p-value to α =.05 and decide whether to reject or fail to reject the null hypothesis

4. Using a confidence level α = .05, is there enough information to support the university’s claim? (Either the claim is false, or the claim is true).



Section 7.4

Hypothesis Testing for Proportions



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


z-Test for a Population Proportion

z-Test for a Population Proportion p

• A statistical test for a population proportion p.

• 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


Remember: Binomial Distribution means 2 outcomes

Using a z-Test for a Proportion p





State H0 and Ha.

Identify α.


Verify that np ≥ 5 and nq ≥ 5.

In Words In Symbols


Using a z-Test for a Proportion p

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



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

p̂ pzpq n

In Words In Symbols


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 α = 0.01, is there enough evidence to support the researcher’s claim? (Adopted from Pew Research Center)

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

np = 100(0.50) = 50 and nq = 100(0.50) = 50© 2012 Pearson Education, Inc. All rights reserved. 95 of 101

Solution: Hypothesis Test for Proportions

• H0:

• Ha:


p ≥ 0.5

p ≠ 0.45

0.01

• Decision: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.

• Test Statistic

Fail to reject H0 .


ˆ 0.39 0.5

(0.5)(0.5) 100

2.2

p pz

pq n

z = –2.2

On a Calculator

• Choose Stats Tests 1-PropZTest

• P0 = .50 (null hypothesis)

• X = number of successes. They tell us success happened 39% of the time. There were 100 sampled, so x = .39 x 100 = 39

• n = 100

• Prop = “not equal” remember, we are testing the Alternative Hypothesis

• Z test score = -2.2 –we could compare this to a Z critical score if we calculated it

• P = .02

• This is not less than .01, so it is not in the tail, so we do not reject the null hypothesis


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. Atα = 0.10 is there enough evidence to reject the claim?

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

np = 200(0.25) = 50 and nq = 200 (0.75) = 150


Solution: Hypothesis Test for Proportions

• H0:

• Ha:


p = 0.25 (Claim)

p ≠ 0.25

0.10

• Decision: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.

• Test Statistic

Fail to reject H0 .


ˆ 0.21 0.25

(0.25)(0.75) 200

1.31

p pz

pq n

z ≈ –1.31

On the Calculator

100

• Choose Stats Tests 1-PropZTest

• P0 = .25 (null hypothesis

• X = number of successes. They tell us success happened 21% of the time. There were 200 sampled, so x = .21 x 200 = 42

• NOTE: You can actually type in .21 x 200 for this entry

• n = 200

• Prop = “not equal” remember, we are testing the Alternative Hypothesis (Binomial is ALWAYS not equal)

• Z test score = -1.30 –we could compare this to a Z critical score if we calculated it

• P = .19

• This is not less than .10, so it is not in the tail(s), so we do not reject the null hypothesis

Section 7.4 Summary

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


Assignment

• Page 401 4-16 Even



Section 7.5

Hypothesis Testing for Variance and Standard Deviation



• Find critical values for a χ2-test

• Use the χ2-test to test a variance or a standard deviation


Finding Critical Values for the χ2-Test

1. Specify the level of significance α.

2. Determine the degrees of freedom d.f. = n – 1.

3. The critical values for the χ2-distribution are found in Table 6 in Appendix B. To find the critical value(s) for a

a. right-tailed test, use the value that corresponds to d.f. and α.

b. left-tailed test, use the value that corresponds to d.f. and 1 – α.

c. two-tailed test, use the values that corresponds to d.f. and ½α, and d.f. and 1 – ½α.


Finding Critical Values for the χ2-Test


20

α1 – α

Right-tailed

Two-tailed

1 – α 12α

2L 2

R

12α

1 α

α

20

Left-tailed

Example: Finding Critical Values for χ2

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

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

From Table 6, the critical value is .20 2.558


0.01α

χ0 = 2.558

Example: Finding Critical Values for χ2

Find the critical χ2-value for a two-tailed test when n = 9 and α = 0.05.

Solution:• Degrees of freedom: n – 1 = 9 – 1 = 8 d.f. • The areas to the right of the critical values are

From Table 6, the critical values are and .

02

251

.0α

01

1 . 7 .52

9α

2 2.180L 2 17.535R


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


Using the χ2-Test for a Variance or Standard Deviation



3. Determine the degrees of freedom.


State H0 and Ha.

Identify α.


d.f. = n – 1

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.


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



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.


Solution: Hypothesis Test for the Population Variance

• H0:

• Ha:

• α =

• df =

• Rejection Region:

• Test Statistic:

• Decision:

σ2 ≤ 0.25 (Claim)

σ2 > 0.25

0.0541 – 1 = 40

22

2

( 1) (41 1)(0.27)

0.2543.2

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.


0.05α

20 55.758 2 43.2

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 α = 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

• H0:

• Ha:

• α =

• df =


• Test Statistic:

• Decision:

σ ≥ 1.4 min.

σ < 1.4 min. (Claim)0.10

25 – 1 = 24

Reject H0 .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.


2 22

2 2

( 1) (25 1)(1.1)

1.414.816

n s

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 α = 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

• H0:

• Ha:

• α =

• df =


• Test Statistic:

• Decision:

σ2 = 15.9 (Claim)

σ2 ≠ 15.9

0.0515 – 1 = 14

2 (n 1)s2

2

(15 1)(21.8)

15.919.195

Fail to Reject H0

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.


10.025

2α

10.025

2α

2 5.629L 2 26.119R 19.195

Section 7.5 Summary

• Found critical values for a χ2-test

• Used the χ2-test to test a variance or a standard deviation


Chapter Hypothesis Testing with One Sample 1 of 101 7 © 2012 Pearson Education, Inc. All rights reserved.

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