9 Chapter Hypothesis Testing © 2012 Pearson Education, Inc.

Chapter Outline 9.1 Introduction to Statistical Tests
9.2 Testing the Mean µ 9.3 Testing a Proportion p © 2012 Pearson Education, Inc. All rights reserved. 2 of 101

Introduction to Statistical Tests
Section 9.1 Introduction to Statistical Tests © 2012 Pearson Education, Inc. All rights reserved. 3 of 101

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 © 2012 Pearson Education, Inc. All rights reserved. 4 of 101

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. © 2012 Pearson Education, Inc. All rights reserved. 5 of 101

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 © 2012 Pearson Education, Inc. All rights reserved. 6 of 101

complementary statements
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 © 2012 Pearson Education, Inc. All rights reserved. 7 of 101

Stating a Hypothesis To write the null and alternative hypotheses, translate the claim made about the population parameter from a verbal statement to a mathematical statement. Then write its complement. H0: μ = k H1: μ > k H0: μ = k H1: μ < k H0: μ = k H1: 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. © 2012 Pearson Education, Inc. All rights reserved. 8 of 101

Example: Stating the Null and Alternative Hypotheses
Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim. A school publicizes that the proportion of its students who are involved in at least one extracurricular activity is 61%. Solution: H0: H1: p = 0.61 Equality condition (Claim) p ≠ 0.61 Complement of H0 © 2012 Pearson Education, Inc. All rights reserved. 9 of 101

Exercise 1: Stating the Null and Alternative Hypotheses
Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim. A car dealership announces that the mean time for an oil change is less than 15 minutes. Solution: H0: H1: μ = 15 minutes Equality Condition μ < 15 minutes Inequality condition (Claim) © 2012 Pearson Education, Inc. All rights reserved. 10 of 101

Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim. A company advertises that the mean life of its furnaces is more than 18 years. Solution: H0: H1: μ = 18 years Equality Condition μ > 18 years Inequality condition (Claim) © 2012 Pearson Education, Inc. All rights reserved. 11 of 101

Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim. A company that manufactures ball bearings claims the average diameter is 6 mm. Solution: H0: H1: μ = 6 mm Complement of H1 (Claim) μ≠ 6 mm Inequality condition © 2012 Pearson Education, Inc. All rights reserved. 12 of 101

Types of Errors No matter which hypothesis represents the claim, always begin the hypothesis test assuming that the equality condition in the null hypothesis is true. At the end of the test, one of two decisions will be made: reject the null hypothesis fail to reject the null hypothesis Because your decision is based on a sample, there is the possibility of making the wrong decision. © 2012 Pearson Education, Inc. All rights reserved. 13 of 101

Types of Errors 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 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. © 2012 Pearson Education, Inc. All rights reserved. 14 of 101

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) © 2012 Pearson Education, Inc. All rights reserved. 15 of 101

Solution: Identifying Type I and Type II Errors
Let p represent the proportion of chicken that is contaminated. H0: H1: Hypotheses: p = 0.20 p > 0.20 (Claim) 0.16 0.18 0.20 0.22 0.24 p H0: p = 0.20 H1: p > 0.20 Chicken meets USDA limits. Chicken exceeds USDA limits. © 2012 Pearson Education, Inc. All rights reserved. 16 of 101

Type I Error H (TRUE) 𝑝=0.2 H1 𝑝> (Decision) A type I error is rejecting H0 when it is true. The actual proportion of contaminated chicken is less than or equal to 0.2, but you decide to reject H0. © 2012 Pearson Education, Inc. All rights reserved. 17 of 101

Type II Error H (FALSE) 𝑝= (Decision) H1 𝑝> 0.2 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. © 2012 Pearson Education, Inc. All rights reserved. 18 of 101

Type I Error H (TRUE) 𝑝=0.2 H1 𝑝> (Decision) 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. © 2012 Pearson Education, Inc. All rights reserved. 19 of 101

Type II Error H (FALSE) 𝑝= (Decision) H1 𝑝> 0.2 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. © 2012 Pearson Education, Inc. All rights reserved. 20 of 101

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) © 2012 Pearson Education, Inc. All rights reserved. 21 of 101

Standardized test statistic
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. z p t (n < 30) z (n ≥ 30) μ Standardized test statistic Test statistic Population parameter © 2012 Pearson Education, Inc. All rights reserved. 22 of 101

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) © 2012 Pearson Education, Inc. All rights reserved. 23 of 101

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 © 2012 Pearson Education, Inc. All rights reserved. 24 of 101

Left-tailed Test The alternative hypothesis H1 contains the less-than inequality symbol (<). H0: μ=k H1: μ < k P is the area to the left of the standardized test statistic. z 1 2 3 –3 –2 –1 Test statistic © 2012 Pearson Education, Inc. All rights reserved. 25 of 101

Right-tailed Test The alternative hypothesis H1 contains the greater-than inequality symbol (>). H0: μ = k H1: μ > k P is the area to the right of the standardized test statistic. z 1 2 3 –3 –2 –1 Test statistic © 2012 Pearson Education, Inc. All rights reserved. 26 of 101

Two-tailed Test The alternative hypothesis H1 contains the not-equal-to symbol (≠). Each tail has an area of ½P. H0: μ = k H1: μ ≠ k P is twice the area to the right of the positive standardized test statistic. P is twice the area to the left of the negative standardized test statistic. z 1 2 3 –3 –2 –1 Test statistic Test statistic © 2012 Pearson Education, Inc. All rights reserved. 27 of 101

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%. Solution: H0: H1: p = 0.61 p ≠ 0.61 Two-tailed test © 2012 Pearson Education, Inc. All rights reserved. 28 of 101

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. Solution: z -z P-value area H0: H1: μ =15 min μ < 15 min Left-tailed test © 2012 Pearson Education, Inc. All rights reserved. 29 of 101

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. Solution: z P-value area H0: H1: μ = 18 yr μ > 18 yr Right-tailed test © 2012 Pearson Education, Inc. All rights reserved. 30 of 101

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 H1 Reject H0 Fail to reject H0 There is enough evidence to reject the claim There is enough evidence to support the claim There is not enough evidence to reject the claim There is not enough evidence to support the claim © 2012 Pearson Education, Inc. All rights reserved. 31 of 101

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. © 2012 Pearson Education, Inc. All rights reserved. 32 of 101

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 𝑝= (Claim) H1 𝑝≠0.61 © 2012 Pearson Education, Inc. All rights reserved. 33 of 101

H0 𝑝= (Claim) H1 𝑝≠0.61 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%.” © 2012 Pearson Education, Inc. All rights reserved. 34 of 101

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 𝜇=15 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 H1 𝜇<15 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 (Claim) © 2012 Pearson Education, Inc. All rights reserved. 36 of 101

H0 𝜇=15 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 H1 𝜇<15 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 (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. © 2012 Pearson Education, Inc. All rights reserved. 37 of 101

Steps for Hypothesis Testing
State the claim mathematically and verbally. Identify the null and alternative hypotheses. H0: ? H1: ? Specify the level of significance. α = ? Determine the standardized sampling distribution and sketch its graph. Calculate the test statistic and its corresponding standardized test statistic. Add it to your sketch. This sampling distribution is based on the assumption that H0 is true. z z Standardized test statistic © 2012 Pearson Education, Inc. All rights reserved. 38 of 101

Steps for Hypothesis Testing
Find the P-value. Use the following decision rule. 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? No Fail to reject H0. Yes Reject H0. © 2012 Pearson Education, Inc. All rights reserved. 39 of 101

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. For a left-tailed test, P = (Area in left tail). For a right-tailed test, P = (Area in right tail). For a two-tailed test, P = 2(Area in tail of test statistic). © 2012 Pearson Education, Inc. All rights reserved. 40 of 101

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. If P ≤ α, then reject H0. If P > α, then fail to reject H0. © 2012 Pearson Education, Inc. All rights reserved. 41 of 101

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. Solution: For a left-tailed test, P = (Area in left tail) z -2.23 P = Because > 0.01, you should fail to reject H0. © 2012 Pearson Education, Inc. All rights reserved. 42 of 101

Exercise 11: Finding the P-value
Find the P-value for a two-tailed hypothesis test with a test statistic of z = 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) z 2.14 1 – = P = 2(0.0162) = 0.9838 Because < 0.05, you should reject H0. © 2012 Pearson Education, Inc. All rights reserved. 43 of 101

Checkpoint: Interpreting a P-value
The P-value for a hypothesis test is P = What is your decision if the level of significance is α = 0.05? α = 0.01? Solution: Because < 0.05, you should reject the null hypothesis. Solution: Because > 0.01, you should fail to reject the null hypothesis. © 2012 Pearson Education, Inc. All rights reserved. 44 of 101

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 at The sample mean 𝑥 is 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? © 2012 Pearson Education, Inc. All rights reserved. 45 of 101

Let 𝜇 equal the population mean (beats/minute) over Rosie’s lifetime. 𝑃𝑣𝑎𝑙=𝑃 𝑥 <105.0 =𝑃 𝑧< −2.04 =𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 −𝐼, −2.04 =0.0207 Since 𝑃𝑣𝑎𝑙<0.05, we reject 𝐻 0 and conclude that there is sufficient evidence to support the claim that Rosie’s heart rate is indeed slowing down. H0 𝜇=115 H1 𝜇< (claim) © 2012 Pearson Education, Inc. All rights reserved. 46 of 101

𝑃𝑣𝑎𝑙=𝑃 𝑥 <105.0 =𝑃 𝑧< −2.04 =𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 −𝐼, −2.04 =0.0207 Since 𝑃𝑣𝑎𝑙<0.05, we reject 𝐻 0 and conclude that there is sufficient evidence to support the claim that Rosie’s heart rate is indeed slowing down. z z = -2.04 P-value = © 2012 Pearson Education, Inc. All rights reserved. 47 of 101

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). © 2012 Pearson Education, Inc. All rights reserved. 48 of 101

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 The sample mean is about 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? © 2012 Pearson Education, Inc. All rights reserved. 49 of 101

Exercise 13(contd.) Perform Hypothesis Test
𝜇=2.3 H1 μ≠ (claim) Population Normally distributed. Case: σ is known Test Type Two-tailed test Test Statistic 𝑥 ≈2.51 Standardized Test Statistic 𝑧= 𝑥 − 𝜇 𝜎 𝑛 ≈1.98 α = 0.01 Pval = 𝑃 𝑧< −1.98 𝑂𝑅 𝑧>1.98 =2𝑃 𝑧>1.98 =2 ⋅𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 1.98,𝐼 ≈0.0478 Conclusion 𝑃𝑣𝑎𝑙> 𝛼 ==> 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. © 2012 Pearson Education, Inc. All rights reserved. 50 of 101

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 © 2012 Pearson Education, Inc. All rights reserved. 51 of 101

Hypothesis Testing for the Mean Case: 𝝁 𝐢𝐬 𝐊𝐧𝐨𝐰𝐧
Section 9.2 Hypothesis Testing for the Mean Case: 𝝁 𝐢𝐬 𝐊𝐧𝐨𝐰𝐧 © 2012 Pearson Education, Inc. All rights reserved. 52 of 101

Section 9.2 Objectives 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 © 2012 Pearson Education, Inc. All rights reserved. 53 of 101

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. © 2012 Pearson Education, Inc. All rights reserved. 54 of 101

Using P-values for a z-Test for Mean μ
In Words In Symbols State the claim mathematically and verbally. Identify the null and alternative hypotheses. Specify the level of significance. Determine the standardized test statistic. Find the area that corresponds to z. State H0 and H1. Identify α. Use Table 5 in Appendix II. © 2012 Pearson Education, Inc. All rights reserved. 55 of 101

Using P-values for a z-Test for Mean μ
In Words In Symbols Find the P-value. For a left-tailed test, P = (Area in left tail). For a right-tailed test, P = (Area in right tail). For a two-tailed test, P = 2(Area in tail of test statistic). Make a decision to reject or fail to reject the null hypothesis. Interpret the decision in the context of the original claim. Reject H0 if P-value is less than or equal to α. Otherwise, fail to reject H0. © 2012 Pearson Education, Inc. All rights reserved. 56 of 101

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. © 2012 Pearson Education, Inc. All rights reserved. 57 of 101

Solution: Hypothesis Testing Using P-values
α = Test Statistic: μ = 13 sec μ < 13 sec (Claim) 0.01 < 0.01 Reject H0 . 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. © 2012 Pearson Education, Inc. All rights reserved. 58 of 101

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 𝜎=$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) © 2012 Pearson Education, Inc. All rights reserved. 59 of 101

Solution: Hypothesis Testing Using P-values
Ha: α = Test Statistic: μ = $22,500 μ ≠ 22,500 (Claim) 0.05 Decision: > 0.05 Fail to reject H0 . 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. © 2012 Pearson Education, Inc. All rights reserved. 60 of 101

Hypothesis Testing for the Mean Case:σ 𝐢𝐬 𝐔𝐧𝐤𝐧𝐨𝐰𝐧
Section 9.2 Hypothesis Testing for the Mean Case:σ 𝐢𝐬 𝐔𝐧𝐤𝐧𝐨𝐰𝐧 © 2012 Pearson Education, Inc. All rights reserved. 61 of 101

Section 9.2 Objectives 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 © 2012 Pearson Education, Inc. All rights reserved. 62 of 101

t-Test for a Mean μ (n < 30, σ Unknown)
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. © 2012 Pearson Education, Inc. All rights reserved. 63 of 101

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: < P-Value < 0.050

Exercise 3: Testing µ, σ Unknown: Pvalue
The 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 25 11 20 19 17 35 13 9 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)? © 2012 Pearson Education, Inc. All rights reserved. 65 of 101

H0 𝜇=12.5 H1 𝜇≠12.5 (claim) Population Distribution is mound-shaped symmetric. Case: σ is unknown , 𝑛< 30 Test Type Two-Tailed Test Test Statistic 𝑥 ≈17.1 𝑛= 21 𝑑𝑓=𝑛−1= 20 𝑠= 10.0 Standardized Test Statistic 𝑡= 𝑥 − 𝜇 𝑠 𝑛 = 17.1− ≈ 2.108 α = 0.01 Pval = 2𝑃 𝑡< −2.108 =2 ⋅𝑡𝑐𝑑𝑓 −𝐼, −2.108, 20 ≈0.0478 Conclusion 𝑃𝑣𝑎𝑙> 𝛼 ==> 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. © 2012 Pearson Education, Inc. All rights reserved. 66 of 101

If using Table 6, Appendix II, note that the sample statistic 𝑡=2.108 falls between and The P-value for the sample t falls between the corresponding two-tail areas and This means 0.020<𝑃𝑣𝑎𝑙𝑢𝑒< The entire range is greater than α. This means the specific P-value is greater than α, so we cannot reject 𝐻 0 . © 2012 Pearson Education, Inc. All rights reserved. 67 of 101

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. © 2012 Pearson Education, Inc. All rights reserved. 68 of 101

H0 𝜇=1.75 H1 μ> (claim) Population Distribution unknown. Case: σ is unknown. However, since 𝑛>30, 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 Test Test Statistic 𝑥 ≈2.05 Standardized Test Statistic 𝑧= 𝑥 − 𝜇 𝜎 𝑛 = 2.05− ≈ 2.48 α = 0.01 Pval = 𝑃 𝑧> 2.48 =𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 2.48,𝐼 ≈0.0066 Conclusion 𝑃𝑣𝑎𝑙≤ 𝛼 ==> 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. © 2012 Pearson Education, Inc. All rights reserved. 69 of 101

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) © 2012 Pearson Education, Inc. All rights reserved. 70 of 101

Solution: Testing μ via Pvalue: (n < 30)
H0 𝜇=20,500 (claim) H1 μ<20,500 Population Normal distribution. Case: σ is unknown. n<30 Test Type Left-Tail Test Test Statistic 𝑥 ≈19,850 𝑛= 14 𝑑𝑓=𝑛−1= 13 𝑠= 1084.0 Standardized Test Statistic 𝑡= 𝑥 − 𝜇 𝑠 𝑛 = − ≈ −2.244 α = 0.05 Pval = 𝑃 𝑡< −2.244 =𝑡𝑐𝑑𝑓 −𝐼, −2.244, 23 ≈0.0174 Conclusion 𝑃𝑣𝑎𝑙≤ 𝛼 ==> 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. © 2012 Pearson Education, Inc. All rights reserved. 71 of 101

Hypothesis Testing for the Mean Critical Region Method
Section 9.2 Hypothesis Testing for the Mean Critical Region Method © 2012 Pearson Education, Inc. All rights reserved. 72 of 101

Rejection Regions and Critical Values
Rejection region (or critical region) The range of values for which the null hypothesis is not probable. If a test statistic falls in this region, the null hypothesis is rejected. A critical value z0 separates the rejection region from the nonrejection region. © 2012 Pearson Education, Inc. All rights reserved. 73 of 101

Testing µ via the Critical Region Method
The values of 𝑥 that will result in the rejection of the null hypothesis are called the critical region of the 𝑥 distribution When we use a predetermined significance level α, the Critical Value Method and the P-Value Method are logically equivalent

Rejection Regions and Critical Values
Finding Critical Values in a Normal Distribution Specify the level of significance α. Decide whether the test is left-, right-, or two-tailed. Find the critical value(s) z0. If the hypothesis test is left-tailed, find the z-score that corresponds to an area of α, right-tailed, find the z-score that corresponds to an area of 1 – α, two-tailed, find the z-score that corresponds to ½α and 1 – ½α. Sketch the standard normal distribution. Draw a vertical line at each critical value and shade the rejection region(s). © 2012 Pearson Education, Inc. All rights reserved. 75 of 101

Critical Regions for H0: µ = k Left-Tailed Test

Critical Regions for H0: µ = k Right-Tailed Test

Critical Regions for H0: µ = k Two-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 statistic is in the rejection region, then reject H0. is not in the rejection region, then fail to reject H0. z z0 Fail to reject H0. Reject H0. Left-Tailed Test z < z0 z z0 Reject Ho. Fail to reject Ho. z > z0 Right-Tailed Test z –z0 Two-Tailed Test z0 z < –z0 z > z0 Reject H0 Fail to reject H0 © 2012 Pearson Education, Inc. All rights reserved. 79 of 101

Exercise 1: Find Critical Value 𝒛 𝟎
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 𝒛 𝟎
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 −𝒛 𝟎 , +𝒛 𝟎
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
In Words In Symbols State the claim mathematically and verbally. Identify the null and alternative hypotheses. Specify the level of significance. Determine the critical value(s). Determine the rejection region(s). State H0 and H1. Identify α. Use Table 5 in Appendix II. © 2012 Pearson Education, Inc. All rights reserved. 83 of 101

Using Rejection Regions for a z-Test for a Mean μ
In Words In Symbols Find the standardized test statistic. Make a decision to reject or fail to reject the null hypothesis. Interpret the decision in the context of the original claim. If z is in the rejection region, reject H0. Otherwise, fail to reject H0. © 2012 Pearson Education, Inc. All rights reserved. 84 of 101

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. © 2012 Pearson Education, Inc. All rights reserved. 85 of 101

Solution: Testing μ via Critical Region Method: Case σ is Known
𝜇=13 H1 μ< (claim) Population Distribution unknown. Case: σ is known , 𝑛>30, Test Type Left-Tail Test Test Statistic 𝑥 ≈12.9 Standardized Test Statistic 𝑧= 𝑥 − 𝜇 𝜎 𝑛 ≈ −2.98 α = 0.01 𝑧 0 = −2.33 Conclusion 𝑧 ≤ 𝑧 0 ==> 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 © 2012 Pearson Education, Inc. All rights reserved. 86 of 101

Using the t-Test for a Mean μ (Critical Region Method)
In Words In Symbols State the claim mathematically and verbally. Identify the null and alternative hypotheses. Specify the level of significance. Identify the degrees of freedom. Determine the critical value(s). Determine the rejection region(s). State H0 and H1. Identify α. d.f. = n – 1. Use Table 6 in Appendix II © 2012 Pearson Education, Inc. All rights reserved. 87 of 101

Using the t-Test for a Mean μ (Critical Region Method)
In Words In Symbols Find the standardized test statistic and sketch the sampling distribution Make a decision to reject or fail to reject the null hypothesis. Interpret the decision in the context of the original claim. If t is in the rejection region, reject H0. Otherwise, fail to reject H0. © 2012 Pearson Education, Inc. All rights reserved. 88 of 101

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) © 2012 Pearson Education, Inc. All rights reserved. 90 of 101

Solution: Testing μ with a Small Sample
Test Statistic: Decision: H0: Ha: α = df = Rejection Region: μ = $20,500 (Claim) μ < $20,500 0.05 Reject H0 . 14 – 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. t ≈ –2.244 © 2012 Pearson Education, Inc. All rights reserved. 91 of 101

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. © 2012 Pearson Education, Inc. All rights reserved. 92 of 101

Solution: Testing μ with a Small Sample
α = df = Rejection Region: μ = 6.8 (Claim) μ ≠ 6.8 Test Statistic: Decision: 0.05 19 – 1 = 18 Fail to reject H0 . At the 5% level of significance, there is not enough evidence to reject the claim that the mean pH is 6.8. t –2.101 0.025 2.101 –2.101 2.101 –1.816 © 2012 Pearson Education, Inc. All rights reserved. 93 of 101

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. © 2012 Pearson Education, Inc. All rights reserved. 94 of 101

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 © 2012 Pearson Education, Inc. All rights reserved. 95 of 101

Objectives Identify the components needed for testing a proportion
Compute the sample test statistic Find the P-value and conclude the test

Testing a Proportion p 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 𝑢𝑠𝑒 𝑝 = 𝑟/𝑛 as our estimate for p, the population probability of success on each trial.

Testing a Proportion p 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

Testing a Proportion p

Testing a Proportion p The null and alternate hypotheses for tests of proportions are as follows: Left-Tailed Test Right-Tailed Test Two-Tailed Test H0: p = k Ho: p = k H0: p = k H1: p < k H1: p > k H1: p ≠ k

Exercise 1: Testing p A 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 – Solution H0 𝑝=0.30 H1 𝑝>0.30 (claim) Population
Binomial Distribution 𝑛= 225 𝑛𝑝= 225(0.30) = (>5) 𝑛𝑞= 225(0.70) = (>5) Test Type Right-tail Test Statistic 𝑝 = 𝑟 𝑛 = ≈0.39 Standardized Test Statistic 𝑧= 𝑝 − 𝑝 𝑝𝑞 𝑛 = − (0.30)(0.70) ≈ 2.95 α = 0.01 𝑃𝑣𝑎𝑙= 𝑃 𝑧>2.95 =𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 2.95, 𝐼 =0.0016 Conclusion 𝑃𝑣𝑎𝑙 ≤ 𝛼 ==> 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: P-value Area
Figure 8.10

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%. © 2012 Pearson Education, Inc. All rights reserved. 108 of 101

Exercise 2: Hypothesis Test for p – Pvalue Method
𝑝= (claim) H1 𝑝≠0.80 Population Binomial Distribution 𝑛= 400 𝑛𝑝= 400(0.80) = (>5) 𝑛𝑞= 400(0.20) = (>5) Test Type Two-tail Test Statistic 𝑝 = 𝑟 𝑛 = ≈ Standardized Test Statistic 𝑧= 𝑝 − 𝑝 𝑝𝑞 𝑛 = − (0.80)(0.20) ≈ −1.00 α = 0.05 𝑃𝑣𝑎𝑙= 2𝑃 𝑧< −1.00 =2𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 −𝐼,−1.00 =0.3174 Conclusion 𝑃𝑣𝑎𝑙> 𝛼 ==> Fail to reject H0. Interpretation At the 5% level of significance, there is insufficient evidence to support the claim that the botanist is wrong. © 2012 Pearson Education, Inc. All rights reserved. 109 of 101

Using a z-Test for a Proportion p – Critical Region Method
Verify that np ≥ 5 and nq ≥ 5. In Words In Symbols State the claim mathematically and verbally. Identify the null and alternative hypotheses. Specify the level of significance. Determine the critical value(s). Determine the rejection region(s). State H0 and H1. Identify α. Use Table 4 in Appendix B. © 2012 Pearson Education, Inc. All rights reserved. 110 of 101

Using a z-Test for a Proportion p
In Words In Symbols Find the standardized test statistic and sketch the sampling distribution. Make a decision to reject or fail to reject the null hypothesis. Interpret the decision in the context of the original claim. If z is in the rejection region, reject H0. Otherwise, fail to reject H0. © 2012 Pearson Education, Inc. All rights reserved. 111 of 101

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%. © 2012 Pearson Education, Inc. All rights reserved. 112 of 101

𝑝= (claim) H1 𝑝≠0.80 Population Binomial Distribution 𝑛= 400 Test Type Two-tail Test Statistic 𝑝 = 𝑟 𝑛 = ≈0.0.78 Standardized Test Statistic 𝑧= 𝑝 − 𝑝 𝑝𝑞 𝑛 = − (0.80)(0.20) ≈ −1.00 α = 0.05 Critical Values 𝑧 0 =1.96 Conclusion −𝑧 0 <𝑧< 𝑧 0 ==> Fail to Reject H0. Interpretation At the 5% level of significance, there is insufficient evidence to conclude that the botanist is wrong. © 2012 Pearson Education, Inc. All rights reserved. 113 of 101

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 © 2012 Pearson Education, Inc. All rights reserved. 115 of 101

Tcdf Function Pval: Two-Tail Area, t=2.31, df=8 Larson/Farber 5th ed.

Tcdf Function Pval: Two-Tail Area, t=2.31, df=8
Pval: Right-Tail Area, t=2.31, df =8 Larson/Farber 5th ed.

ZTest Function Larson/Farber 5th ed.

T-Test Function Larson/Farber 5th ed.

1- PropZTest Function Larson/Farber 5th ed.

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
Test Statistic H0: Ha: α = Rejection Region: p ≥ 0.5 p ≠ 0.45 0.01 Decision: Fail to reject H0 . 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. z = –2.2 © 2012 Pearson Education, Inc. All rights reserved. 124 of 101

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. At α = 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 © 2012 Pearson Education, Inc. All rights reserved. 125 of 101

Solution: Hypothesis Test for Proportions
Test Statistic H0: Ha: α = Rejection Region: p = (Claim) p ≠ 0.25 0.10 Decision: Fail to reject H0 . 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. z ≈ –1.31 © 2012 Pearson Education, Inc. All rights reserved. 126 of 101

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