Chapter 8Hypothesis Testing Review In Chapters 2 and 3 we used “descriptive statistics” when we summarized data using tools such as graphs, and statistics such as the mean and standard deviation. Methods of inferential statistics use sample data to make an inference or conclusion about a population. The two main activities of inferential statistics are using sample data to (1) estimate a population parameter (such as estimating a population parameter with a confidence interval), and (2) test a hypothesis or claim about a population parameter. In Chapter 7 we presented methods for estimating a population parameter with a confidence interval, and in this chapter we present the method of hypothesis testing. Bluman, Chapter 8, 08/2011 2 Hypothesis Testing Researchers are interested in answering many types of questions. For example, Is the earth warming up? Does a new medication lower blood pressure? Does the public prefer a certain color in a new fashion line? Is a new teaching technique better than a traditional one? Do seat belts reduce the severity of injuries? These types of questions can be addressed through statistical hypothesis testing, which is a decision-making process for evaluating claims about a population. A hypothesis is a claim or statement about a property of a population (μ, σ and p). Bluman, Chapter 8, 08/2011 3 Why need hypothesis testing? Bluman, Chapter 8, 08/2011 4 Hypothesis Testing Three methods used to test hypotheses: 1. The traditional method 2. The P-value method 3. The confidence interval method Bluman, Chapter 8, 08/2011 5 8.1 Steps in Hypothesis TestingTraditional Method A statistical hypothesis is a conjecture about a population parameter. This conjecture may or may not be true. The null hypothesis, symbolized by H0 (Hn or HN), is a statistical hypothesis that states that there is no difference between a parameter and a specific value, or that there is no difference between two parameters, use =. Bluman, Chapter 8, 08/2011 6 <. 08/2011 7 . Chapter 8.Steps in Hypothesis TestingTraditional Method The alternative hypothesis. The symbolic form of the alternative hypothesis must use one of these symbols: . Bluman. is a statistical hypothesis that states the existence of a difference between a parameter and a specific value. symbolized by H1 (Ha or HA). or states that there is a difference between two parameters. >. Chapter 8. the claim must be worded so that it becomes the alternative hypothesis. Bluman. 08/2011 8 .Note about Forming Your Own Claims (Hypotheses) If you are conducting a study and want to use a hypothesis test to support your claim. Note about Identifying H0 and H1 Bluman. Chapter 8. 08/2011 9 . One-tailed test By H1 No difference = = with equal sign Researcher’ s interest Note: In this book. 08/2011 10 . the null hypothesis is always stated using the equals sign There is no overlapping between H0 and H1 Bluman. Chapter 8. Situation A A medical researcher is interested in finding out whether a new medication will have any undesirable side effects. Bluman. or remain unchanged after a patient takes the medication? The researcher knows that the mean pulse rate for the population under study is 82 beats per minute. The hypotheses for this situation in symbolic form are H 0 : 82 H1 : 82 This is called a two-tailed hypothesis test. decrease. Will the pulse rate increase. 08/2011 11 . The researcher is particularly concerned with the pulse rate of the patients who take the medication. Chapter 8. The hypotheses for this situation are H 0 : 36 H1 : 36 This is called a right-tailed hypothesis test. the null hypothesis is always stated using the equals sign.Situation B A chemist invents an additive to increase the life of an automobile battery. The mean lifetime of the automobile battery without the additive is 36 months. In this book. Bluman. 08/2011 12 . Chapter 8. Chapter 8. 08/2011 13 .Situation C A contractor wishes to lower heating bills by using a special type of insulation in houses. her hypotheses about heating costs with the use of insulation are The hypotheses for this situation are H 0 : 78 H1 : 78 This is called a left-tailed hypothesis test. If the average of the monthly heating bills is $78. Bluman. or research hypothesis. Statistical evidence can be used to reject the claim if the claim is the null hypothesis. however. can be stated as either the null hypothesis or the alternative hypothesis. Bluman. the statistical evidence can only support the claim if it is the alternative hypothesis. A claim. These facts are important when you are stating the conclusion of a statistical study. though.Claim When a researcher conducts a study. Chapter 8. the claim should be stated as the alternative hypothesis. he or she is generally looking for evidence to support a claim. 08/2011 14 . Therefore. (e. students who attend class more have better grades) -Give the significance level. -Reach a conclusion. (e. test score) -Perform the calculations required for the statistical test. MDC students) -State the particular hypotheses that will be investigated. (e.g. (e. class attendance. (e. Bluman. α = 0. Chapter 8. 08/2011 15 .05) -Select a sample from the population.g.Hypothesis Testing A decision-making process for evaluating claims about a population. The researcher must: -Define the population under study. randomly select 2 classes) -Collect the data.g.g.g. Bluman. The numerical value obtained from a statistical test is called the test value. Chapter 8.Hypothesis Testing A statistical test uses the data obtained from a sample to make a decision about whether the null hypothesis should be rejected. there are four possible outcomes. In the hypothesis-testing situation. 08/2011 16 . and a decision is made to reject or not to reject it on the basis of the data obtained from a sample. Chapter 8. 08/2011 17 . the null hypothesis may or may not be true. A type I error (a error: false positive) occurs if one rejects the null hypothesis when it is true. Bluman.Hypothesis Testing In reality. A type II error (b error: false negative) occurs if one does not reject the null hypothesis when it is false. 08/2011 18 . however. α and β are related in that decrease one increases the other. Bluman. false negative) β can not easily be computed. Chapter 8.Hypothesis Testing (α error. false positive) (β error. Chapter 8. the more sensitive the test. The power is 1 – β.Power of a Statistical Test The power of a test measures the sensitivity of the test to detect a real difference in parameters if one actually exists. The higher the power. Bluman. 08/2011 19 . That is. Chapter 8. P(type I error) = a. This probability is symbolized by a (alpha). 08/2011 20 . P(type II error) = b (beta).Hypothesis Testing The level of significance is the maximum probability of committing a type I error and is the probability that the test statistic will fall in the critical region when the null hypothesis is actually true. Bluman. Likewise. when a = 0. Bluman. there is a 5% chance of rejecting a true null hypothesis. and 0. 08/2011 21 .10.Hypothesis Testing Typical significance levels are: 0. Chapter 8. 0.05.01 For example.05. V. C. The noncritical or nonrejection region is the range of values of the test value that indicates that the difference was probably due to chance and that the null hypothesis should not be rejected. 08/2011 22 .. Chapter 8. The critical or rejection region is the range of values of the test value that indicates that there is a significant difference and that the null hypothesis should be rejected. separates the critical region from the noncritical region. Bluman.Hypothesis Testing The critical value. 08/2011 23 .Bluman. Chapter 8. Two-tailed Test H0: = H1: a is divided equally between the two tails of the critical region Means less than or greater than Bluman. 08/2011 24 . Chapter 8. Left-tailed Test H0: = a the left tail H1: < Points Left Bluman. 08/2011 25 . Chapter 8. Chapter 8.Right-tailed Test H0: = H1: > Points Right Bluman. 08/2011 26 . 33 for α = 0. Chapter 8. 08/2011 27 .Hypothesis Testing Finding the Critical Value for α = 0.01 (Right-Tailed Test) Bluman.01 (Right-Tailed Test) z = 2. 08/2011 28 .33 for α = 0. Chapter 8.Hypothesis Testing Finding the Critical Value for α = 0. z = -2.01 (Left-Tailed Test) z Because of symmetry.01 (Left-Tailed Test) Bluman. Chapter 8.58 Bluman.01 (Two-Tailed Test) z = ±2. 08/2011 29 .Hypothesis Testing Finding the Critical Value for α = 0. 08/2011 30 . Chapter 8.= = Bluman. c. with an area equal to α. If the test is right-tailed. a. b. will be on the left side of the mean. 08/2011 31 . If the test is left-tailed. onehalf of the area will be to the right of the mean. will be on the right side of the mean. If the test is two-tailed. and one-half will be to the left of the mean. Chapter 8.Procedure Table Finding the Critical Values for Specific α Values. α must be divided by 2. Bluman. Using Table E Step 1 Draw the figure and indicate the appropriate area. with an area equal to α. the critical region. the critical region. For the right value. a. 08/2011 32 . Bluman. use the z value that corresponds to α / 2 for the left value. It will be negative. Using Table E Step 2 Find the z value in Table E.Procedure Table Finding the Critical Values for Specific α Values. For a left-tailed test. For a two-tailed test. It will be positive. For a right-tailed test. Chapter 8. c. b. use the z value that corresponds to the area equivalent to α in Table E. use the z value that corresponds to the area equivalent to 1 – α. use the z value that corresponds to the area equivalent to 1 – α / 2. Chapter 8 Hypothesis Testing Section 8-1 Example 8-2 Page #410 Bluman. 08/2011 33 . Chapter 8. Example 8-2: Using Table E Using Table E in Appendix C, find the critical value(s) for each situation and draw the appropriate figure, showing the critical region. a. A left-tailed test with α = 0.10. Step 1 Draw the figure and indicate the appropriate area. Step 2 Find the area closest to 0.1000 in Table E. In this case, it is 0.1003. The z value is -1.28. Bluman, Chapter 8, 08/2011 34 Example 8-2: Using Table E Using Table E in Appendix C, find the critical value(s) for each situation and draw the appropriate figure, showing the critical region. b. A two-tailed test with α = 0.02. Step 1 Draw the figure with areas α /2 = 0.02/2 = 0.01. Step 2 Find the areas closest to 0.01 and 0.99. The areas are 0.0099 and 0.9901. The z values are -2.33 and 2.33. Bluman, Chapter 8, 08/2011 35 Example 8-2: Using Table E Using Table E in Appendix C, find the critical value(s) for each situation and draw the appropriate figure, showing the critical region. c. A right-tailed test with α = 0.005. Step 1 Draw the figure and indicate the appropriate area. Step 2 Find the area closest to 1 – α = 0.995. There is a tie: 0.9949 and 0.9951. Average the z values of 2.57 and 2.58 to get 2.575 or 2.58. Bluman, Chapter 8, 08/2011 36 Chapter 8. Step 3 Compute the test value. Step 4 Make the decision to reject or not reject the null hypothesis.Procedure Table Solving Hypothesis-Testing Problems (Traditional Method) Step 1 State the hypotheses and identify the claim. Bluman. 08/2011 37 . Step 2 Find the critical value(s) from the appropriate table in Appendix C. Step 5 Summarize the results. or when the population is normally distributed if n <30 and is known. It can be used when n 30.2 z Test for a Mean The z test is a statistical test for the mean of a population. 08/2011 38 . The formula for the z test is X z n where X = sample mean μ = hypothesized population mean = population standard deviation n = sample size Bluman.8. Chapter 8. Chapter 8. 08/2011 39 .Chapter 8 Hypothesis Testing Section 8-2 Example 8-3 Page #414 Bluman. 000 and H1: μ > $42.05.000 per year. Step 1: State the hypotheses and identify the claim. At α = 0.000 (claim) Step 2: Find the critical value.260. the critical value is z = 1. H0: μ = $42. The standard deviation of the population is $5230.05 and the test is a right-tailed test.Example 8-3: Professors’ Salaries A researcher reports that the average salary of assistant professors is more than $42. Since α = 0. A sample of 30 assistant professors has a mean salary of $43. Bluman. 08/2011 40 . Chapter 8.65. test the claim that assistant professors earn more than $42.000. At α = 0. Chapter 8.05. A sample of 30 assistant professors has a mean salary of $43.260.Example 8-3: Professors’ Salaries A researcher reports that the average salary of assistant professors is more than $42. The standard deviation of the population is $5230. 08/2011 41 .000. Step 3: Compute the test value. 43260 42000 X z 1.000 per year.32 5230 30 n Bluman. test the claim that assistant professors earn more than $42. 1. 08/2011 42 .32.Example 8-3: Professors’ Salaries Step 4: Make the decision. There is not enough evidence to support the claim that assistant professors earn more on average than $42. Chapter 8. the decision is to not reject the null hypothesis. Step 5: Summarize the results. is less than the critical value. Since the test value. and is not in the critical region.65. 1.000 per year. Bluman. Important Comments Even though in Example 8–3 the sample mean of $43,260 is higher than the hypothesized population mean of $42,000, it is not significantly higher. Hence, the difference may be due to chance. When the null hypothesis is not rejected, there is still a probability of a type II error, i.e., of not rejecting the null hypothesis when it is false. When the null hypothesis is not rejected, it cannot be accepted as true. There is merely not enough evidence to say that it is false. Bluman, Chapter 8, 08/2011 43 Chapter 8 Hypothesis Testing Section 8-2 Example 8-4 Page #415 Bluman, Chapter 8, 08/2011 44 Example 8-4: Cost of Men’s Shoes A researcher claims that the average cost of men’s athletic shoes is less than $80. He selects a random sample of 36 pairs of shoes from a catalog and finds the following costs (in dollars). (The costs have been rounded to the nearest dollar.) Is there enough evidence to support the researcher’s claim at α = 0.10? Assume = 19.2. 60 70 75 55 80 55 50 40 80 70 50 95 120 90 75 85 80 60 110 65 80 85 85 45 75 60 90 90 60 95 110 85 45 90 70 70 Step 1: State the hypotheses and identify the claim. H0: μ = $80 and H1: μ < $80 (claim) Bluman, Chapter 8, 08/2011 45 Since α = 0. Chapter 8.2. (The costs have been rounded to the nearest dollar.Example 8-4: Cost of Men’s Shoes A researcher claims that the average cost of men’s athletic shoes is less than $80. He selects a random sample of 36 pairs of shoes from a catalog and finds the following costs (in dollars).) Is there enough evidence to support the researcher’s claim at α = 0.10? Assume = 19.28. 60 70 75 55 80 55 50 40 80 70 50 95 120 90 75 85 80 60 110 65 80 85 85 45 75 60 90 90 60 95 110 85 45 90 70 70 Step 2: Find the critical value. the critical value is z = -1. 08/2011 46 . Bluman.10 and the test is a left-tailed test. 08/2011 47 .) Is there enough evidence to support the researcher’s claim at α = 0. (The costs have been rounded to the nearest dollar.2.56 n 19.10? Assume = 19. we find X = 75. Using technology.Example 8-4: Cost of Men’s Shoes A researcher claims that the average cost of men’s athletic shoes is less than $80.2. 75 80 X z 1.2 36 Bluman. He selects a random sample of 36 pairs of shoes from a catalog and finds the following costs (in dollars).0 and = 19. Chapter 8. Step 3: Compute the test value. There is enough evidence to support the claim that the average cost of men’s athletic shoes is less than $80. Bluman. -1.56. Since the test value. 08/2011 48 . Chapter 8. falls in the critical region. the decision is to reject the null hypothesis. Step 5: Summarize the results.Example 8-4: Cost of Men’s Shoes Step 4: Make the decision. Chapter 8.Chapter 8 Hypothesis Testing Section 8-2 Example 8-5 Page #416 Bluman. 08/2011 49 . can it be concluded that the average cost of stroke rehabilitation at a particular hospital is different from $24. 08/2011 50 .226. H0: μ = $24. The standard deviation of the population is $3251. Chapter 8.01.672 and H1: μ $24.Example 8-5: Cost of Rehabilitation The Medical Rehabilitation Education Foundation reports that the average cost of rehabilitation for stroke victims is $24.672. To see if the average cost of rehabilitation is different at a particular hospital.672 (claim) Bluman.672? Step 1: State the hypotheses and identify the claim. At α = 0. a researcher selects a random sample of 35 stroke victims at the hospital and finds that the average cost of their rehabilitation is $25. Chapter 8.58. the critical values are z = ±2.Example 8-5: Cost of Rehabilitation The Medical Rehabilitation Education Foundation reports that the average cost of rehabilitation for stroke victims is $24. Since α = 0. The standard deviation of the population is $3251. At α = 0. can it be concluded that the average cost of stroke rehabilitation at a particular hospital is different from $24.226.672? Step 2: Find the critical value. To see if the average cost of rehabilitation is different at a particular hospital. Bluman.672. 08/2011 51 . a researcher selects a random sample of 35 stroke victims at the hospital and finds that the average cost of their rehabilitation is $25.01 and a two-tailed test.01. 672 X z 1. Chapter 8. 08/2011 52 .01 3251 35 n Bluman.672” “a random sample of 35 stroke victims” “the average cost of their rehabilitation is $25. 25. 226 24.Example 8-5: Cost of Rehabilitation “reports that the average cost of rehabilitation for stroke victims is $24.01” Step 3: Find the test value.226” “the standard deviation of the population is $3251” “α = 0. Bluman.Example 8-5: Cost of Rehabilitation Step 4: Make the decision. Chapter 8. There is not enough evidence to support the claim that the average cost of rehabilitation at the particular hospital is different from $24.672. since the test value falls in the noncritical region. Step 5: Summarize the results. 08/2011 53 . Do not reject the null hypothesis. 08/2011 54 54 .Bluman. Chapter 8. Chapter 8. 08/2011 55 . Bluman.Hypothesis Testing The P-value (or probability value) is the probability of getting a sample statistic (such as the mean) or a more extreme sample statistic in the direction of the alternative hypothesis when the null hypothesis is true. the traditional method for solving hypothesis-testing problems compares z-values: critical value test value The P-value method for solving hypothesistesting problems compares areas: alpha P-value Bluman. Chapter 8.Hypothesis Testing In this section. 08/2011 56 . 08/2011 57 .= Bluman. Chapter 8. Bluman. 08/2011 58 . Chapter 8. Bluman. 08/2011 59 . Step 3 Find the P-value. Step 5 Summarize the results. Step 4 Make the decision. Chapter 8.Procedure Table Solving Hypothesis-Testing Problems (P-Value Method) Step 1 State the hypotheses and identify the claim. Step 2 Compute the test value. Chapter 8. 08/2011 60 .Chapter 8 Hypothesis Testing Section 8-2 Example 8-6 Page #419 Bluman. The population standard deviation is $659. Is there evidence to support the claim at a 0. 08/2011 61 . H0: μ = $5700 and H1: μ > $5700 (claim) Bluman.05? Use the P-value method. Chapter 8. Step 1: State the hypotheses and identify the claim.Example 8-6: Cost of College Tuition A researcher wishes to test the claim that the average cost of tuition and fees at a four-year public college is greater than $5700. She selects a random sample of 36 four-year public colleges and finds the mean to be $5950. Step 2: Compute the test value. The population standard deviation is $659.28 659 36 n Bluman. 08/2011 62 . X 5950 5700 z 2.05? Use the P-value method. Is there evidence to support the claim at a 0. Chapter 8. She selects a random sample of 36 four-year public colleges and finds the mean to be $5950.Example 8-6: Cost of College Tuition A researcher wishes to test the claim that the average cost of tuition and fees at a four-year public college is greater than $5700. Step 3: Find the P-value.Example 8-6: Cost of College Tuition A researcher wishes to test the claim that the average cost of tuition and fees at a four-year public college is greater than $5700.05? Use the P-value method.0113. Using Table E. Hence. Bluman. She selects a random sample of 36 four-year public colleges and finds the mean to be $5950. The population standard deviation is $659. the P-value is 1. The area is 0.9887. Is there evidence to support the claim at a 0.28.9887 = 0. 08/2011 63 . Chapter 8. Subtract from 1.0000 – 0.0000 to find the area of the tail. find the area for z = 2. 08/2011 64 . Note: If α = 0. There is enough evidence to support the claim that the tuition and fees at four-year public colleges are greater than $5700.01. the null hypothesis would not be rejected. Bluman. Since the P-value is less than 0.Example 8-6: Cost of College Tuition Step 4: Make the decision.05. the decision is to reject the null hypothesis. Chapter 8. Step 5: Summarize the results. Chapter 8 Hypothesis Testing Section 8-2 Example 8-7 Page #420 Bluman. 08/2011 65 . Chapter 8. Chapter 8. is there enough evidence to reject the claim? Use the P-value method.2 miles per hour.2 8 X 1.6 32 Bluman. At α = 0.6 mile per hour.Example 8-7: Wind Speed (option) A researcher claims that the average wind speed in a certain city is 8 miles per hour. The standard deviation of the population is 0. A sample of 32 days has an average wind speed of 8.05. Step 1: State the hypotheses and identify the claim. 08/2011 66 .89 z n 0. H0: μ = 8 (claim) and H1: μ 8 Step 2: Compute the test value. 8. 9706. Bluman. The area for z = 1.05. A sample of 32 days has an average wind speed of 8.0294) = 0.0294 must be doubled to get the P-value. 08/2011 67 . Chapter 8. is there enough evidence to reject the claim? Use the P-value method. Step 3: Find the P-value.Example 8-7: Wind Speed A researcher claims that the average wind speed in a certain city is 8 miles per hour. Subtract: 1.0000 – 0.0294.0588. the area of 0. The standard deviation of the population is 0. At α = 0. The P-value is 2(0. Since this is a two-tailed test.9706 = 0.2 miles per hour.89 is 0.6 mile per hour. 08/2011 68 . There is not enough evidence to reject the claim that the average wind speed is 8 miles per hour. Step 5: Summarize the results. The decision is to not reject the null hypothesis. since the P-value is greater than 0. Bluman. Chapter 8.Example 8-7: Wind Speed Step 4: Make the decision.05. Procedure for Finding P-Values Bluman. 08/2011 69 . Chapter 8. 10. reject the null hypothesis.01 but P-value 0. 08/2011 70 .01.05. If P-value > 0. reject the null hypothesis. Bluman. Chapter 8. The difference is highly significant. If P-value > 0.05 but P-value 0. do not reject the null hypothesis. The difference is not significant.Guidelines for P-Values With No α If P-value 0. If P-value > 0. The difference is significant. consider the consequences of type I error before rejecting the null hypothesis.10. When the null hypothesis is rejected at a specific significance level. 08/2011 71 . the results may not have any practical significance. it can be concluded that the difference is probably not due to chance and thus is statistically significant. Chapter 8. However. It is up to the researcher to use common sense when interpreting the results of a statistical test. Bluman.Significance The researcher should distinguish between statistical significance and practical significance. σ is unknown.8. This is a conservative approach. in this textbook. Chapter 8.3 t Test for a Mean The t test is a statistical test for the mean of a population and is used when the population is normally or approximately normally distributed. however. Bluman.f. use d. s n Note: When the degrees of freedom are above 30. For example. = n – 1.f. 08/2011 72 .f. you should round down to the nearest table value. if d. The formula for the t test is X t The degrees of freedom are d. = 59. some textbooks will tell you to use the nearest table value. = 55 to find the critical value or values. Chapter 8 Hypothesis Testing Section 8-3 Example 8-8 Page #428 Bluman. 08/2011 73 . Chapter 8. Bluman.f.746. Chapter 8.05 column in the top row and 16 in the left-hand column. Find the 0. The critical t value is +1. = 16 for a right-tailed t test. 08/2011 74 .Example 8-8: Table F Find the critical t value for α = 0.05 with d. 08/2011 75 .Chapter 8 Hypothesis Testing Section 8-3 Example 8-9 & 8-10 Page #428 Bluman. Chapter 8. 734. = 18 for a twotailed t test.01 column in the One tail row. = 22 for a lefttailed test.f. and 18 in the d.f. The critical values are 1.10 with d. Find the 0.f.10 column in the Two tails row.01 with d. Find the 0.Example 8-9: Table F Find the critical t value for α = 0.508 since the test is a one-tailed left test. column. The critical value is t = -2. and 22 in the d.734 and -1. Chapter 8. Example 8-10: Table F Find the critical value for α = 0.f. 08/2011 76 . Bluman. column. 08/2011 77 . Chapter 8.Bluman. 08/2011 78 .Chapter 8 Hypothesis Testing Section 8-3 Example 8-12 Page #429 Bluman. Chapter 8. A random sample of 10 weeks had a mean number of 17.3 Step 2: Find the critical value.262 for α = 0. H0: μ = 16.8.3.262 and -2.7 infections. The sample standard deviation is 1.3 (claim) and H1: μ 16. Is there enough evidence to reject the investigator’s claim at α = 0.05? Step 1: State the hypotheses and identify the claim. The critical values are 2. = 9.Example 8-12: Hospital Infections A medical investigation claims that the average number of infections per week at a hospital in southwestern Pennsylvania is 16. Chapter 8. 08/2011 79 .f.05 and d. Bluman. Is there enough evidence to reject the investigator’s claim at α = 0.7 16. A random sample of 10 weeks had a mean number of 17.3 2. t 17. The sample standard deviation is 1.8 10 Bluman. Chapter 8.Example 8-12: Hospital Infections A medical investigation claims that the average number of infections per week at a hospital in southwestern Pennsylvania is 16.3.05? Step 3: Find the test value.8. 08/2011 80 .46 1.7 infections. 3. Reject the null hypothesis since 2.Example 8-12: Hospital Infections Step 4: Make the decision.262. Step 5: Summarize the results. There is enough evidence to reject the claim that the average number of infections is 16. Bluman. Chapter 8. 08/2011 81 .46 > 2. Chapter 8.Chapter 8 Hypothesis Testing Section 8-3 Example 8-13 Page #430 Bluman. 08/2011 82 . H0: μ = 60 and H1: μ < 60 (claim) Step 2: Find the critical value. the critical value is -1. A random sample of eight school districts is selected. Is there enough evidence to support the educator’s claim at α = 0. Pennsylvania. 08/2011 83 . At α = 0. Bluman. Chapter 8.415.10? 60 56 60 55 70 55 60 55 Step 1: State the hypotheses and identify the claim. and the daily salaries (in dollars) are shown. is less than $60 per day. = 7.f.Example 8-13: Substitute Salaries An educator claims that the average salary of substitute teachers in school districts in Allegheny County.10 and d. 58.1 8 s n Bluman. Using the Stats feature of the TI-84.9 60 X t 0.88 and s = 5. 08/2011 -0. we find X = 58.08. Chapter 8.61 5.Example 8-13: Substitute Salaries Step 3: Find the test value.624 84 . 08/2011 85 . There is not enough evidence to support the claim that the average salary of substitute teachers in Allegheny County is less than $60 per day. Step 5: Summarize the results. Bluman.624 falls in the noncritical region. Do not reject the null hypothesis since -0. Chapter 8.Example 8-13: Substitute Salaries Step 4: Make the decision. Chapter 8 Hypothesis Testing Section 8-3 Example 8-16: P-value method Page #432 Bluman. Chapter 8. 08/2011 86 . 6 milliliters per kilogram (ml/kg) and a standard deviation of 6 ml/kg. 08/2011 87 .7 2. is there enough evidence to support the physician’s claim at α = 0.517 t 6 15 s n Bluman.05? Step 1: State the hypotheses and identify the claim. X 40.7 (claim) Step 2: Compute the test value. Chapter 8. A sample of 15 joggers has a mean of 40. H0: μ = 36.Example 8-16: Jogger’s Oxygen Intake A physician claims that joggers’ maximal volume oxygen uptake is greater than the average of all adults.7 ml/kg. If the average of all adults is 36.6 36.7 and H1: μ > 36. 624.Example 8-16: Jogger’s Oxygen Intake A physician claims that joggers’ maximal volume oxygen uptake is greater than the average of all adults.01. A sample of 15 joggers has a mean of 40. Chapter 8. = 14 row.517 falls between 2.7 ml/kg. Bluman.145 and 2. is there enough evidence to support the physician’s claim at α = 0.6 milliliters per kilogram (ml/kg) and a standard deviation of 6 ml/kg. 08/2011 88 .01 and 0. the P-value is somewhere between 0. 2.025 and α = 0. since this is a one-tailed test. corresponding to α = 0. In the d.025.f.05? Step 3: Find the P-value. Thus. If the average of all adults is 36. There is enough evidence to support the claim that the joggers’ maximal volume oxygen uptake is greater than 36. 08/2011 89 . The decision is to reject the null hypothesis.Example 8-16: Jogger’s Oxygen Intake Step 4: Make the decision. Chapter 8.05.7 ml/kg. Step 5: Summarize the results. since the P-value < 0. Bluman. Chapter 8.Whether to use z or t Bluman. 08/2011 90 . Chapter 8. The formula for the z test for a proportion is pˆ p z pq n where Bluman.8. the standard normal distribution can be used to test hypotheses for proportions.4 z Test for a Proportion Since a normal distribution can be used to approximate the binomial distribution when np 5 and nq 5. 08/2011 91 . Chapter 8.Chapter 8 Hypothesis Testing Section 8-4 Example 8-17 Page #438 Bluman. 08/2011 92 . H0: p = 0.60 Step 2: Find the critical value. 08/2011 93 .60 (claim) and H1: p 0.Example 8-17: Avoiding Trans Fats A dietitian claims that 60% of people are trying to avoid trans fats in their diets.05 and the test is a two-tailed test.05. Bluman. the critical value is z = ±1. Chapter 8. Since α = 0. is there enough evidence to reject the dietitian’s claim? Step 1: State the hypotheses and identify the claim. She randomly selected 200 people and found that 128 people stated that they were trying to avoid trans fats in their diets.96. At α = 0. 05. Chapter 8. X 128 pˆ 0. is there enough evidence to reject the dietitian’s claim? Step 3: Compute the test value. At α = 0. 08/2011 94 .64 n 200 Bluman. She randomly selected 200 people and found that 128 people stated that they were trying to avoid trans fats in their diets.Example 8-17: Avoiding Trans Fats A dietitian claims that 60% of people are trying to avoid trans fats in their diets. Example 8-17: Avoiding Trans Fats Step 4: Make the decision. Step 5: Summarize the results. Do not reject the null hypothesis since the test value falls outside the critical region. Chapter 8. 08/2011 95 . Bluman. There is not enough evidence to reject the claim that 60% of people are trying to avoid trans fats in their diets. Chapter 8.Chapter 8 Hypothesis Testing Section 8-4 Example 8-18 Page #439 Bluman. 08/2011 96 . 58.01 and the test is a two-tailed test.01.Example 8-18: Call-Waiting Service A telephone company representative estimates that 40% of its customers have call-waiting service. Chapter 8. she selected a sample of 100 customers and found that 37% had call waiting. Since α = 0.40 (claim) and H1: p 0. the critical value is z = ±2. H0: p = 0. At α = 0.40 Step 2: Find the critical value. Bluman. To test this hypothesis. is there enough evidence to reject the claim? Step 1: State the hypotheses and identify the claim. 08/2011 97 . Bluman. is there enough evidence to reject the claim? Step 3: Compute the test value. 08/2011 98 . Chapter 8.Example 8-18: Call-Waiting Service A telephone company representative estimates that 40% of its customers have call-waiting service. At α = 0.01. To test this hypothesis. she selected a sample of 100 customers and found that 37% had call waiting. There is not enough evidence to reject the claim that 40% of the telephone company’s customers have call waiting. Do not reject the null hypothesis since the test value falls outside the critical region. Bluman. 08/2011 99 . Step 5: Summarize the results. Chapter 8.Example 8-18: Call-Waiting Service Step 4: Make the decision. test the claim that 18% of all high school students smoke at least one pack of cigarettes a day. Chapter 8. At Wilson High School. 08/2011 100 . Bluman. a study found that 50 students smoked at least one pack of cigarettes a day.05. Use the P .Section 8-4: Z Test for a Proportion P-Value method Researchers suspect that 18% of all high school students smoke at least one pack of cigarettes a day.value method. At a = 0. not relevant to hypothesis test here. Hind: at least one pack of … is just a label. with an enrollment of 300 students. do not reject the null hypothesis.025 0 z=.2743) = .60 Table E: area = .18 = = – 0.pˆ = 50 = 0.1667 – 0.82) n 300 .1667 p = 0..60 pq (0. n = 300.18 pˆ – p 0. Bluman.2743. There is not enough evidence to reject the claim that 18% of all high school students smoke at least a pack of cigarettes a day. Chapter 8.05 H0 :p = 0.5486 Since P .18 (claim) z= H1 : p 0.18 q = 0. P-value = 2 (.18)(0.60 .025 z=.82 300 X = 50.05.2743 .value > 0. 08/2011 101 .2743 . α = . = n – 1 and n = sample size s2 = sample variance 2 = population variance Bluman.f. Chapter 8.8. The formula for the chi-square test for a variance is with degrees of freedom d. 08/2011 102 .5 c Test for a Variance or a Standard Deviation 2 The chi-square distribution is also used to test a claim about a single variance or standard deviation. The observations must be independent of one another. Bluman. 3. 08/2011 103 . The population must be normally distributed for the variable under study. Chapter 8. The sample must be randomly selected from the population. 2.Assumptions for the c Test for a Variance or a Standard Deviation 2 1. Chapter 8 Hypothesis Testing Section 8-5 Example 8-21 Page #445 Bluman. 08/2011 104 . Chapter 8. 05 and the test is right-tailed.996 Bluman.Example 8-21: Table G Find the critical chi-square value for 15 degrees of freedom when α = 0. c 2 24. Chapter 8. 08/2011 105 . 08/2011 106 . Chapter 8.Chapter 8 Hypothesis Testing Section 8-5 Example 8-22 Page #446 Bluman. the α value must be subtracted from 1.05 and the test is left-tailed. Chapter 8.95. because the chi-square table gives the area to the right of the critical value. Bluman.05 = 0. 1 – 0. 08/2011 107 .Example 8-22: Table G Find the critical chi-square value for 10 degrees of freedom when α = 0. that is. When the test is left-tailed. The left side of the table is used. and the chisquare statistic cannot be negative. 95. Use Table G.Example 8-22: Table G Find the critical chi-square value for 10 degrees of freedom when α = 0. c 2 3. Chapter 8. 08/2011 108 . looking in row 10 and column 0.940 Bluman.05 and the test is left-tailed. Chapter 8 Hypothesis Testing Section 8-5 Example 8-23 Page #447 Bluman. 08/2011 109 . Chapter 8. 975 correspond to chi-square values of 36. The area to the right of the larger value is α /2.025. the area must be split. 08/2011 110 .975. or 0. or 0. Chapter 8. Bluman.05 and a two-tailed test is conducted.982. The area to the right of the smaller value is 1 – α /2. When the test is two-tailed. With 22 degrees of freedom.Example 8-23: Table G Find the critical chi-square value for 22 degrees of freedom when α = 0.025 and 0. areas 0.781 and 10. Chapter 8. 08/2011 111 .Bluman. Chapter 8.Chapter 8 Hypothesis Testing Section 8-5 Example 8-24 Page #448 Bluman. 08/2011 112 . Is there enough evidence to support the claim that the variation of the students is less than the population variance (2 =225) at α = 0.f.95.05? Assume that the scores are normally distributed. 08/2011 113 . Step 1: State the hypotheses and identify the claim. H0: 2 = 225 and H1: 2 < 225 (claim) Step 2: Find the critical value. table G The critical value is c 2= 12. Bluman.Example 8-24: Variation of Test Scores An instructor wishes to see whether the variation in scores of the 23 students in her class is less than the variance of the population.338. α = . d. =22. The variance of the class is 198. Chapter 8. Example 8-24: Variation of Test Scores An instructor wishes to see whether the variation in scores of the 23 students in her class is less than the variance of the population. Step 3: Compute the test value. Is there enough evidence to support the claim that the variation of the students is less than the population variance (2 =225) at α = 0. Bluman. Chapter 8. 08/2011 114 .05? Assume that the scores are normally distributed. The variance of the class is 198. There is not enough evidence to support the claim that the variation in test scores of the instructor’s students is less than the variation in scores of the population.36 falls in the noncritical region. Chapter 8. 08/2011 115 . Bluman. Step 5: Summarize the results.Example 8-24: Variation of Test Scores Step 4: Make the decision. Do not reject the null hypothesis since the test value 19. Chapter 8 Hypothesis Testing Section 8-5 Example 8-26 Page #450 Bluman. Chapter 8. 08/2011 116 . α/2 = . is there enough evidence to reject the manufacturer’s claim? Step 1: State the hypotheses and identify the claim.644 Step 2: Find the critical value. and assume that it is normally distributed.f = 19.Example 8-26: Nicotine Content A cigarette manufacturer wishes to test the claim that the variance of the nicotine content of its cigarettes is 0.05. d.025.907.025. α = . Bluman. A sample of 20 cigarettes has a standard deviation of 1. 08/2011 117 . Chapter 8. H0: 2 = 0. αright = . At α = 0.852 and 8. Nicotine content is measured in milligrams.644. αleft = .00 milligram.05.975 The critical values are 32.644 (claim) and H1: 2 0. 08/2011 08/2009 118 . Chapter Chapter 8.Bluman. Bluman.Zhao. Wei8. is there enough evidence to reject the manufacturer’s claim? Step 3: Compute the test value. Bluman.00 milligram.Example 8-26: Nicotine Content A cigarette manufacturer wishes to test the claim that the variance of the nicotine content of its cigarettes is 0. Chapter 8. Nicotine content is measured in milligrams.05. The standard deviation s must be squared in the formula. 08/2011 119 .644. A sample of 20 cigarettes has a standard deviation of 1. and assume that it is normally distributed. At α = 0. Chapter 8. 08/2011 120 .644. since the test value falls in the noncritical region. Bluman. Step 5: Summarize the results. There is not enough evidence to reject the manufacturer’s claim that the variance of the nicotine content of the cigarettes is 0. Do not reject the null hypothesis.Example 8-26: Nicotine Content Step 4: Make the decision.