hypothesis testing

ChapterHypothesis Testing

1 of 101

9

© 2012 Pearson Education, Inc.All rights reserved.

Chapter Outline

• 9.1 Introduction to Statistical Tests• 9.2 Testing the Mean • 9.3 Testing a Proportion p

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

Section 9.1

Introduction to Statistical Tests


Section 9.1 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. • 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

• Denoted H0 read “H sub-zero” or “H naught.”

Alternative hypothesis • A statement of strict

inequality such as >, ≠, or <.

• Must be true if H0 is false.

• Denoted H1 read “H sub-1.”

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 = kH1: μ > k

H0: μ = kH1: μ < k

H0: μ = kH1: 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.

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:

H1:

(Claim)p = 0.61

p ≠ 0.61

Solution:


μ = 15 minutes

Exercise 1: Stating the Null and Alternative Hypotheses


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

Inequality condition

Equality ConditionH0:

H1:(Claim)μ < 15 minutes

Solution:


μ = 18 years



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


Equality ConditionH0:

H1: (Claim)μ > 18 years

Solution:


μ = 6 mm



A company that manufactures ball bearings claims the average diameter is 6 mm.


Complement of H1H0:

H1:

(Claim)Solution:


6 mm

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

• 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 falseDo not reject H0 Correct Decision Type II ErrorReject H0 Type I Error Correct Decision


Exercise 4: 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:

H1:

p = 0.20

p > 0.20

Hypotheses:

(Claim)

0.16 0.18 0.20 0.22 0.24p

H0: p = 0.20 H1: 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.


Type I Error

H0 (TRUE)

H1 (Decision)


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.


Type II Error

H0 (FALSE) (Decision)

H1


• 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.


Type I Error

H0 (TRUE)

H1 (Decision)


• 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.


Type II Error

H0 (FALSE) (Decision)

H1

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)


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.

x

z pt (n < 30)z (n ≥ 30)μ

Standardized test statistic

Test statisticPopulation parameter

p̂


P-values

• The P-value is the probability of obtaining a sample statistic with a value as extreme (or more extreme) than the one determined from the sample data

• The P-value is computed based on the assumption that H0 is true

• Smaller P-values are stronger evidence against the null hypotheses

• The calculation of the P-value depends on the nature of the test (i.e. left-tailed, right tailed, or two-tailed)


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


Left-tailed Test• The alternative hypothesis H1 contains the less-than

inequality symbol (<).

z0 1 2 3–3 –2 –1

Test statistic

H0: μ=k H1: μ < kP is the area to the left of the standardized test statistic.


• The alternative hypothesis H1 contains the greater-than inequality symbol (>).

Right-tailed Test

H0: μ = k H1: μ > k

Test statistic


z0 1 2 3–3 –2 –1

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

Two-tailed Test• The alternative hypothesis H1 contains the not-equal-

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

z0 1 2 3–3 –2 –1

Test statistic

Test statistic

H0: μ = k H1: μ ≠ k

P 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.


Exercise 5: Identifying The Nature of a Test

For each claim, state H0 and H1. 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.A school publicizes that the proportion of its students who are involved in at least one extracurricular activity is 61%.

H0:H1:

p = 0.61p ≠ 0.61

Two-tailed test

Solution:



For each claim, state H0 and H1. 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.

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

H0:H1:

Left-tailed testz0-z

P-value area

μ =15 minμ < 15 min

Solution:



For each claim, state H0 and H1. 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.

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

H0:H1:

Right-tailed testz 0 z

P-value areaμ = 18 yr

μ > 18 yr

Solution:


Making a DecisionDecision 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

ClaimDecision Claim is H0 Claim is H1

Reject H0

Fail to reject H0

There is enough evidence to reject the claimThere is not enough evidence to reject the claim

There is enough evidence to support the claimThere is not enough evidence to support the claim


Exercise 8: 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?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%.”

• We believe that the proportion of students involved in extracurricular activities is different from 61%.


H0 (Claim)

H1


• 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%.”


H0 (Claim)

H1

Exercise 9: 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?H1 (Claim): A car dealership announces that the mean time for an oil change is less than 15 minutes.

Solution:• The claim is represented by H1. • H0 is “the mean time for an oil change is greater than

or equal to 15 minutes.”© 2012 Pearson Education, Inc. All rights reserved. 35 of 101


• 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.”


H0

H1 (Claim)


• 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.” We believe that the mean time is at least 15 minutes.


H0

H1 (Claim)

z0

Steps for Hypothesis Testing

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

H0: ? H1: ?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.

z0Standardized 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


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).


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.


Exercise 10: 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.


z0 2.14

Exercise 11: 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


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

2. α = 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.


Exercise 12: Perform Hypothesis Test

Let x be a random variable that represents the heart rate in beats per minute of Rosie, and old sheep dog. From past experience the vet knows that x is normally distributed with a mean of 115 bpm and standard deviation of 12 bpm. Over the past several weeks Rosie’s heart rate (beats / min) was measured at93 109 110 89 112 117The sample mean is 105.0. The vet is concerned that Rosie’s heart rate may be slowing. At a 5% level of significance, do the data indicate that this is the case?



• Let equal the population mean (beats/minute) over Rosie’s lifetime.

• Since we reject and conclude that there is sufficient evidence to support the claim that Rosie’s heart rate is indeed slowing down.


H0

H1 (claim)


• Since we reject and conclude that there is sufficient evidence to support the claim that Rosie’s heart rate is indeed slowing down.


z0z = -2.04

P-value = 0.0207


The environmental Protection Agency has been studying Miller Creek regarding ammonia nitrogen concentration. For many years, the concentration has been 2.3 mg/l. However, a new golf course and housing developments re raising concern that the concentration may have changed because of lawn fertilizer. Any change (either an increase or a decrease) in the ammonia nitrogen concentration can affect plant and animal life in and around the creek. Let x be a random variable representing ammonia nitrogen concentration (in mg/l).


Exercise 13(contd.) Perform Hypothesis Test

Based on recent studies of Miller Creek, we may assume that x has a normal distribution with population standard deviation 0.3. Recently, a random sample of eight water tests from the creek gave the following x values.

2.1 2.5 2.2 2.8 3.0 2.2 2.4 2.9The sample mean is about 2.51. Construct a statistical test to examine the claim that the concentration of ammonia nitrogen has changed from 2.3 ml/g. Use 0.01 as our level of significance. State the null hypothesis and alternate hypothesis. What is your conclusion?


Exercise 13(contd.) Perform Hypothesis Test


H0

H1 2.3 (claim)Population Normally distributed. Case: σ is knownTest Type Two-tailed testTest StatisticStandardized Test Statistic

α = 0.01Pval =

Conclusion ==> Fail to Reject H0.

Interpretation At the 1% level of significance, there is insufficient evidence to support the claim that the ammonia nitrogen concentration has changed in Miller Creek.

Section 9.1 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


Section 9.2

Hypothesis Testing for the MeanCase:



• Find P-values and use them to test a mean μ• Test µ when σ is known

Use P-values for a z-test• Test µ when σ is unknown

Use P-values for a t-test• Find critical values and rejection regions in a normal

distribution• Use rejection regions for a z-test, or t-test


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 H1.

Identify α.

Use Table 5 in Appendix II.

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


Exercise 1: 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. Assume that the population standard deviation is 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: • H1: • α = • Test Statistic:

μ = 13 secμ < 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 .


12.9 130.19 322.98

xzn

Exercise 2: 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. Assume that 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 = $22,500μ ≠ 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 22,5003015 30

1.73

xzn

Section 9.2

Hypothesis Testing for the MeanCase:σ



• 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


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 Student’s t Distribution(Table 6) to Estimate P-Values

Suppose we calculate t = 2.22 for a one-tailed test from a sample size of 6.Thus, df = n – 1 = 5.We obtain: 0.025 < P-Value < 0.050

Exercise 3: Testing µ, σ Unknown: PvalueThe drug 6-mP (6-mercoptopurine) is used to treat leukemia. The following data represent the remission times (in weeks) for a random sample of 21 patients using 6-mP.


10 7 32 23 22 6 16 34 32 2511 20 19 6 17 35 6 13 9 610

The sample mean is 17.1 weeks with a sample standard deviation of 10.0 weeks. Let x be a random variable representing the remission times (in weeks) for all patients. Assume the x-distribution is mound-shaped and symmetric. A previous drug treatment had a remission time of 12.5 weeks. At a 1% level of significance do the data indicate the mean remission time for 6-mP is different (either way)?

Exercise 3: Testing µ, σ Unknown: Pvalue


H0

H1 (claim)Population Distribution is mound-shaped symmetric. Case: σ is unknown ,

Test Type Two-Tailed TestTest Statistic

212010.0

Standardized Test Statistic

α = 0.01Pval =Conclusion ==> Fail to Reject H0.

Interpretation At the 1% level of significance, there is NOT sufficient evidence to support the claim that the mean remission time is different from 12.5 weeks.



• If using Table 6, Appendix II, note that the sample statistic falls between 2.086 and 2.528. The P-value for the sample t falls between the corresponding two-tail areas 0.050 and 0.020. This means . The entire range is greater than α. This means the specific P-value is greater than α, so we cannot reject .

Exercise 4: Testing µ, σ Unknown via P-value Method


A random sample of 46 adult coyotes in a region of northern Minnesota showed the average age to be 2.05 years with sample standard deviation of 0.82 years. However, it is thought that the overall population mean age of coyotes is 1.75 years. Does the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use α =0.01.



H0

H1 (claim)Population Distribution unknown. Case: σ is unknown. However,

we may use a normal distribution, and use s to approximate σ. A more accurate result would be obtained from using the Student’s t-distribution.

Test Type Right-Tail TestTest StatisticStandardized Test Statistic

α = 0.01Pval =Conclusion ==> Reject H0.

Interpretation At the 1% level of significance, there is sufficient evidence to support the claim that the mean age of coyotes is greater than 1.75 years.

Exercise 5: Testing μ with a Small Sample via P-Value Method

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 via Pvalue: (n < 30)


H0 (claim)H1

Population Normal distribution. Case: σ is unknown. Test Type Left-Tail TestTest Statistic

14131084.0


α = 0.05Pval =Conclusion ==> Reject H0.

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

Section 9.2

Hypothesis Testing for the MeanCritical Region Method


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.


Testing µ via the Critical Region Method

• The values of that will result in the rejection of the null hypothesis are called the critical region of the

• When we use a predetermined significance level α, the Critical Value Method and the P-Value Method are logically equivalent

Rejection Regions and Critical ValuesFinding Critical Values in a Normal Distribution1. 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).


Critical Regions for H0: µ = kLeft-Tailed Test

Critical Regions for H0: µ = kRight-Tailed Test

Critical Regions for H0: µ = kTwo-Tailed Test

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

z0–z0

Two-Tailed Testz0z < –z0 z > z0

Reject H0

Fail to reject H0

Reject H0


Exercise 1: Find Critical Value

Find the critical value and sketch rejection region for a left-tailed z-test with α = 0.05 and α = 0.01

Exercise 2: Find Critical Value

Find the critical value and sketch rejection region for a right-tailed z-test with α = 0.05 and α = 0.01

Exercise 3: Find Critical Values ,

Find the critical values and sketch rejection region for a two-tailed z-test with α = 0.05 and α = 0.01

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 H1.

Identify α.

Use Table 5 in Appendix II.

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 nns

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

In Words In Symbols


Exercise 4: Testing μ via Critical Region Method: Case σ is Known

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. Assume that the population standard deviation is 0.19 second. Is there enough evidence to support the claim at α = 0.01? Use the critical region method.


Solution: Testing μ via Critical Region Method: Case σ is Known

H0

H1 (claim)Population Distribution unknown. Case: σ is known , Test Type Left-Tail Test

Test StatisticStandardized Test Statistic

α = 0.01

Conclusion ==> Reject H0.

Interpretation At the 1% level of significance, there is sufficient evidence to support the claim that the mean pit stop is less than 13 seconds


Using the t-Test for a Mean μ(Critical Region Method)



3. Identify the degrees of freedom.



State H0 and H1.

Identify α.

Use Table 6 in Appendix II

d.f. = n – 1.

In Words In Symbols


Using the t-Test for a Mean μ(Critical Region Method)

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: Find Critical Value

.


Exercise 5: Testing μ via Critical Region Method: Case σ is Unknown

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. Use the critical region method. (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 20,500 2.2441084 14

xts n

t ≈ –2.244


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. Use the critical region method.


Solution: Testing μ with a Small Sample

• H0: • H1: • α = • df = • Rejection Region:

• Test Statistic:

• Decision:

μ = 6.8 (Claim)μ ≠ 6.80.0519 – 1 = 18

6.7 6.8 1.8160.24 19

xts 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 .



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. Use the critical region method.


Section 9.2 Summary

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

normal distribution• Used rejection regions for a z-test and t-test


Section 9.3

Hypothesis Testing for Proportions



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


Objectives

• Identify the components needed for testing a proportion

• Compute the sample test statistic

• Find the P-value and conclude the test


• Throughout this section, we will assume that the situations we are dealing with satisfy the conditions underlying the binomial distribution.

• In particular, we will let r be a binomial random variable. This means that r is the number of successes out of n independent binomial trials.

• We will as our estimate for p, the population probability of success on each trial.


• The letter q again represents the population probability of failure on each trial, and so q = 1 – p. We also assume that the samples are large (i.e., np > 5 and nq > 5).

Testing a Proportion p cont’d


The null and alternate hypotheses for tests of proportions are as follows:

Left-Tailed Test Right-Tailed Test Two-Tailed TestH0: p = k Ho: p = k H0: p = kH1: p < k H1: p > k H1: p ≠ k

Exercise 1: Testing pA team of eye surgeons has developed a new technique for a risky eye operation to restore the sight of people blinded from a certain disease. Under the old method, it is known that only 30% of the patients who undergo this operation recover their eyesight.

Suppose that surgeons in various hospitals have performed a total of 225 operations using the new method and that 88 have been successful (i.e., the patients fully recovered their sight). Can we justify the claim that the new method is better than the old one? (Use a 1% level of significance.)

Exercise 1 – SolutionH0

H1 (claim)Population Binomial Distribution

225225(0.30) = 67.5 (>5)225(0.70) = 157.5 (>5)

Test Type Right-tailTest Statistic


α = 0.01

Conclusion ==> Reject H0.

Interpretation At the 1% level of significance, there is sufficient evidence to support the claim that the new surgical technique is better than the old one.

Example 6 – Solution

P-value of the test statistic:

Figure 8.10

P-value Area

Exercise 2: Hypothesis Test for Proportions – Pvalue


A botanist has produced a new variety of hybrid wheat that is better able to withstand drought than other varieties. The botanist knows that for the parent plants, the proportion of seeds germinating is 80%. The proportion of seeds germinating for the hybrid variety is unknown, but the botanist claims that it is 80%. To test this claim, 400 seeds from the hybrid plant are tested, and it is found that 312 germinate. Use a 5% level of significance to test the claim that the proportion germinating for the hybrid is 80%.

Exercise 2: Hypothesis Test for p – Pvalue Method


H0 (claim)H1 Population Binomial Distribution

400400(0.80) = 320 (>5)400(0.20) = 80 (>5)

Test Type Two-tailTest Statistic


α = 0.05

Conclusion ==> Fail to reject H0.

Interpretation At the 5% level of significance, there is insufficient evidence to support the claim that the botanist is wrong.

Using a z-Test for a Proportion p – Critical Region Method





State H0 and H1.

Identify α.

Use Table 4 in Appendix B.

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


Exercise 3: Hypothesis Test for Proportions – Critical Region Method


A botanist has produced a new variety of hybrid wheat that is better able to withstand drought than other varieties. The botanist knows that for the parent plants, the proportion of seeds germinating is 80%. The proportion of seeds germinating for the hybrid variety is unknown, but the botanist claims that it is 80%. To test this claim, 400 seeds from the hybrid plant are tested, and it is found that 312 germinate. Use a 5% level of significance to test the claim that the proportion germinating for the hybrid is 80%.



H0 (claim)H1 Population Binomial Distribution

400Test Type Two-tailTest Statistic


α = 0.05

Conclusion ==> Fail to Reject H0.

Interpretation At the 5% level of significance, there is insufficient evidence to conclude that the botanist is wrong.

Section 9.3 Summary

• Used the z-test to test a population proportion p via Pvalue method

• Used the z-test to test a population proportion p via critical region method


Section 9.1-3

Hypothesis Testing using Technology


117

Tcdf Function

Larson/Farber 5th ed.

Pval: Two-Tail Area, t=2.31, df=8

118

Tcdf Function


1. Pval: Two-Tail Area, t=2.31, df=82. Pval: Right-Tail Area, t=2.31, df =8

119

ZTest Function


120

T-Test Function


121

1- PropZTest Function


Section 9.1-3

Extra Practice


Example: Hypothesis Test for Proportions – Critical Region

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. 123 of 101

Solution: Hypothesis Test for Proportions

• H0: • Ha: • α = • Rejection Region:

p ≥ 0.5p ≠ 0.450.01

• 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 pzpq n

z = –2.2

Example: Hypothesis Test for Proportions – Critical Region Method

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: • α = • Rejection Region:

p = 0.25 (Claim)p ≠ 0.250.10

• 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 pzpq n

z ≈ –1.31

hypothesis testing

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