Chapter 10 Chi-Square Tests and the F-Distribution Larson/Farber 4th ed Larson/Farber 4th ed

Chapter Outline 10.1 Goodness of Fit 10.2 Independence 10.3 Comparing Two Variances 10.4 Analysis of Variance Larson/Farber 4th ed

Section 10.1 Goodness of Fit Larson/Farber 4th ed

Section 10.1 Objectives Use the chi-square distribution to test whether a frequency distribution fits a claimed distribution Larson/Farber 4th ed

Multinomial Experiments A probability experiment consisting of a fixed number of trials in which there are more than two possible outcomes for each independent trial. A binomial experiment had only two possible outcomes. The probability for each outcome is fixed and each outcome is classified into categories. Larson/Farber 4th ed

Multinomial Experiments Example: A radio station claims that the distribution of music preferences for listeners in the broadcast region is as shown below. Distribution of music Preferences Classical 4% Oldies 2% Country 36% Pop 18% Gospel 11% Rock 29% Each outcome is classified into categories. The probability for each possible outcome is fixed. Larson/Farber 4th ed

Chi-Square Goodness-of-Fit Test Used to test whether a frequency distribution fits an expected distribution. The null hypothesis states that the frequency distribution fits the specified distribution. The alternative hypothesis states that the frequency distribution does not fit the specified distribution. Larson/Farber 4th ed

Chi-Square Goodness-of-Fit Test Example: To test the radio station’s claim, the executive can perform a chi-square goodness-of-fit test using the following hypotheses. H0: The distribution of music preferences in the broadcast region is 4% classical, 36% country, 11% gospel, 2% oldies, 18% pop, and 29% rock. (claim) Ha: The distribution of music preferences differs from the claimed or expected distribution. Larson/Farber 4th ed

Chi-Square Goodness-of-Fit Test To calculate the test statistic for the chi-square goodness-of-fit test, the observed frequencies and the expected frequencies are used. The observed frequency O of a category is the frequency for the category observed in the sample data. Larson/Farber 4th ed

Chi-Square Goodness-of-Fit Test The expected frequency E of a category is the calculated frequency for the category. Expected frequencies are obtained assuming the specified (or hypothesized) distribution. The expected frequency for the ith category is Ei = npi where n is the number of trials (the sample size) and pi is the assumed probability of the ith category. Larson/Farber 4th ed

Example: Finding Observed and Expected Frequencies A marketing executive randomly selects 500 radio music listeners from the broadcast region and asks each whether he or she prefers classical, country, gospel, oldies, pop, or rock music. The results are shown at the right. Find the observed frequencies and the expected frequencies for each type of music. Survey results (n = 500) Classical 8 Country 210 Gospel 72 Oldies 10 Pop 75 Rock 125 Larson/Farber 4th ed

Solution: Finding Observed and Expected Frequencies Observed frequency: The number of radio music listeners naming a particular type of music Survey results (n = 500) Classical 8 Country 210 Gospel 72 Oldies 10 Pop 75 Rock 125 observed frequency Larson/Farber 4th ed

Solution: Finding Observed and Expected Frequencies Expected Frequency: Ei = npi Type of music % of listeners Observed frequency Expected frequency Classical 4% 8 Country 36% 210 Gospel 11% 72 Oldies 2% 10 Pop 18% 75 Rock 29% 125 500(0.04) = 20 500(0.36) = 180 500(0.11) = 55 500(0.02) = 10 500(0.18) = 90 500(0.29) = 145 n = 500 Larson/Farber 4th ed

Chi-Square Goodness-of-Fit Test For the chi-square goodness-of-fit test to be used, the following must be true. The observed frequencies must be obtained by using a random sample. Each expected frequency must be greater than or equal to 5. Larson/Farber 4th ed

Chi-Square Goodness-of-Fit Test Chapter 10 Chi-Square Goodness-of-Fit Test If these conditions are satisfied, then the sampling distribution for the goodness-of-fit test is approximated by a chi-square distribution with k – 1 degrees of freedom, where k is the number of categories. The test statistic for the chi-square goodness-of-fit test is where O represents the observed frequency of each category and E represents the expected frequency of each category. The test is always a right-tailed test. Larson/Farber 4th ed Larson/Farber 4th ed

Chi-Square Goodness-of-Fit Test Chapter 10 Chi-Square Goodness-of-Fit Test In Words In Symbols Identify the claim. State the null and alternative hypotheses. Specify the level of significance. Identify the degrees of freedom. Determine the critical value. State H0 and Ha. Identify . d.f. = k – 1 Use Table 6 in Appendix B. Larson/Farber 4th ed Larson/Farber 4th ed

Chi-Square Goodness-of-Fit Test Chapter 10 Chi-Square Goodness-of-Fit Test In Words In Symbols Determine the rejection region. Calculate the 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 χ2 is in the rejection region, reject H0. Otherwise, fail to reject H0. Larson/Farber 4th ed Larson/Farber 4th ed

Example: Performing a Goodness of Fit Test Use the music preference data to perform a chi-square goodness-of-fit test to test whether the distributions are different. Use α = 0.01. Distribution of music preferences Classical 4% Country 36% Gospel 11% Oldies 2% Pop 18% Rock 29% Survey results (n = 500) Classical 8 Country 210 Gospel 72 Oldies 10 Pop 75 Rock 125 Larson/Farber 4th ed

Solution: Performing a Goodness of Fit Test H0: Ha: α = d.f. = Rejection Region music preference is 4% classical, 36% country, 11% gospel, 2% oldies, 18% pop, and 29% rock music preference differs from the claimed or expected distribution 0.01 6 – 1 = 5 Test Statistic: Decision: Conclusion: 0.01 χ2 15.086 Larson/Farber 4th ed

Solution: Performing a Goodness of Fit Test Chapter 10 Solution: Performing a Goodness of Fit Test Type of music Observed frequency Expected frequency Classical 8 20 Country 210 180 Gospel 72 55 Oldies 10 Pop 75 90 Rock 125 145 Larson/Farber 4th ed Larson/Farber 4th ed

Solution: Performing a Goodness of Fit Test H0: Ha: α = d.f. = Rejection Region music preference is 4% classical, 36% country, 11% gospel, 2% oldies, 18% pop, and 29% rock music preference differs from the claimed or expected distribution 0.01 6 – 1 = 5 Test Statistic: Decision: χ2 = 22.713 Reject H0 0.01 χ2 15.086 There is enough evidence to conclude that the distribution of music preferences differs from the claimed distribution. 22.713 Larson/Farber 4th ed

Example: Performing a Goodness of Fit Test The manufacturer of M&M’s candies claims that the number of different-colored candies in bags of dark chocolate M&M’s is uniformly distributed. To test this claim, you randomly select a bag that contains 500 dark chocolate M&M’s. The results are shown in the table on the next slide. Using α = 0.10, perform a chi-square goodness-of-fit test to test the claimed or expected distribution. What can you conclude? (Adapted from Mars Incorporated) Larson/Farber 4th ed

Example: Performing a Goodness of Fit Test Solution: The claim is that the distribution is uniform, so the expected frequencies of the colors are equal. To find each expected frequency, divide the sample size by the number of colors. E = 500/6 ≈ 83.3 Color Frequency Brown 80 Yellow 95 Red 88 Blue 83 Orange 76 Green 78 n = 500 Larson/Farber 4th ed

Solution: Performing a Goodness of Fit Test H0: Ha: α = d.f. = Rejection Region Distribution of different-colored candies in bags of dark chocolate M&Ms is uniform Distribution of different-colored candies in bags of dark chocolate M&Ms is not uniform 0.10 6 – 1 = 5 Test Statistic: Decision: Conclusion: 0.10 χ2 9.236 Larson/Farber 4th ed

Solution: Performing a Goodness of Fit Test Chapter 10 Solution: Performing a Goodness of Fit Test Color Observed frequency Expected frequency Brown 80 83.3 Yellow 95 Red 88 Blue 83 Orange 76 Green 78 Larson/Farber 4th ed Larson/Farber 4th ed

Solution: Performing a Goodness of Fit Test H0: Ha: α = d.f. = Rejection Region Distribution of different-colored candies in bags of dark chocolate M&Ms is uniform Distribution of different-colored candies in bags of dark chocolate M&Ms is not uniform 0.01 6 – 1 = 5 Test Statistic: Decision: χ2 = 3.016 Fail to Reject H0 0.10 χ2 9.236 There is not enough evidence to dispute the claim that the distribution is uniform. 3.016 Larson/Farber 4th ed

Section 10.1 Summary Used the chi-square distribution to test whether a frequency distribution fits a claimed distribution Larson/Farber 4th ed

Section 10.2 Independence Larson/Farber 4th ed

Section 10.2 Objectives Use a contingency table to find expected frequencies Use a chi-square distribution to test whether two variables are independent Larson/Farber 4th ed

Contingency Tables r  c contingency table Shows the observed frequencies for two variables. The observed frequencies are arranged in r rows and c columns. The intersection of a row and a column is called a cell. Larson/Farber 4th ed

Contingency Tables Example: The contingency table shows the results of a random sample of 550 company CEOs classified by age and size of company.(Adapted from Grant Thornton LLP, The Segal Company) Age Company size 39 and under 40 - 49 50 - 59 60 - 69 70 and over Small / Midsize 42 69 108 60 21 Large 5 18 85 120 22 Larson/Farber 4th ed

Finding the Expected Frequency Assuming the two variables are independent, you can use the contingency table to find the expected frequency for each cell. The expected frequency for a cell Er,c in a contingency table is Larson/Farber 4th ed

Example: Finding Expected Frequencies Find the expected frequency for each cell in the contingency table. Assume that the variables, age and company size, are independent. Age Company size 39 and under 40 - 49 50 - 59 60 - 69 70 and over Total Small / Midsize 42 69 108 60 21 300 Large 5 18 85 120 22 250 47 87 193 180 43 550 marginal totals Larson/Farber 4th ed

Solution: Finding Expected Frequencies Age Company size 39 and under 40 - 49 50 - 59 60 - 69 70 and over Total Small / Midsize 42 69 108 60 21 300 Large 5 18 85 120 22 250 47 87 193 180 43 550 Larson/Farber 4th ed

Solution: Finding Expected Frequencies Age Company size 39 and under 40 - 49 50 - 59 60 - 69 70 and over Total Small / Midsize 42 69 108 60 21 300 Large 5 18 85 120 22 250 47 87 193 180 43 550 Larson/Farber 4th ed

Solution: Finding Expected Frequencies Chapter 10 Solution: Finding Expected Frequencies Age Company size 39 and under 40 - 49 50 - 59 60 - 69 70 and over Total Small / Midsize 42 69 108 60 21 300 Large 5 18 85 120 22 250 47 87 193 180 43 550 Larson/Farber 4th ed Larson/Farber 4th ed

Chi-Square Independence Test Used to test the independence of two variables. Can determine whether the occurrence of one variable affects the probability of the occurrence of the other variable. Larson/Farber 4th ed

Chi-Square Independence Test For the chi-square independence test to be used, the following must be true. The observed frequencies must be obtained by using a random sample. Each expected frequency must be greater than or equal to 5. Larson/Farber 4th ed

Chi-Square Independence Test If these conditions are satisfied, then the sampling distribution for the chi-square independence test is approximated by a chi-square distribution with (r – 1)(c – 1) degrees of freedom, where r and c are the number of rows and columns, respectively, of a contingency table. The test statistic for the chi-square independence test is where O represents the observed frequencies and E represents the expected frequencies. The test is always a right-tailed test. Larson/Farber 4th ed

Chi-Square Independence Test Chapter 10 Chi-Square Independence Test In Words In Symbols Identify the claim. State the null and alternative hypotheses. Specify the level of significance. Identify the degrees of freedom. Determine the critical value. State H0 and Ha. Identify . d.f. = (r – 1)(c – 1) Use Table 6 in Appendix B. Larson/Farber 4th ed Larson/Farber 4th ed

Chi-Square Independence Test Chapter 10 Chi-Square Independence Test In Words In Symbols Determine the rejection region. Calculate the 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 χ2 is in the rejection region, reject H0. Otherwise, fail to reject H0. Larson/Farber 4th ed Larson/Farber 4th ed

Example: Performing a χ2 Independence Test Using the age/company size contingency table, can you conclude that the CEOs ages are related to company size? Use α = 0.01. Expected frequencies are shown in parentheses. Age Company size 39 and under 40 - 49 50 - 59 60 - 69 70 and over Total Small / Midsize 42 (25.64) 69 (47.45) 108 (105.27) 60 (98.18) 21 (23.45) 300 Large 5 (21.36) 18 (39.55) 85 (87.73) 120 (81.82) 22 (19.55) 250 47 87 193 180 43 550 Larson/Farber 4th ed

Solution: Performing a Goodness of Fit Test H0: Ha: α = d.f. = Rejection Region CEOs’ ages are independent of company size CEOs’ ages are dependent on company size 0.01 (2 – 1)(5 – 1) = 4 Test Statistic: Decision: 0.01 χ2 13.277 Larson/Farber 4th ed

Solution: Performing a Goodness of Fit Test Chapter 10 Solution: Performing a Goodness of Fit Test Larson/Farber 4th ed Larson/Farber 4th ed

Solution: Performing a Goodness of Fit Test H0: Ha: α = d.f. = Rejection Region CEOs’ ages are independent of company size CEOs’ ages are dependent on company size 0.01 (2 – 1)(5 – 1) = 4 Test Statistic: Decision: χ2 = 77.9 Reject H0 0.01 χ2 13.277 There is enough evidence to conclude CEOs’ ages are dependent on company size. 77.9 Larson/Farber 4th ed

Section 10.2 Summary Used a contingency table to find expected frequencies Used a chi-square distribution to test whether two variables are independent Larson/Farber 4th ed

Comparing Two Variances Section 10.3 Comparing Two Variances Larson/Farber 4th ed

Section 10.3 Objectives Interpret the F-distribution and use an F-table to find critical values Perform a two-sample F-test to compare two variances Larson/Farber 4th ed

F-Distribution Let represent the sample variances of two different populations. If both populations are normal and the population variances are equal, then the sampling distribution of is called an F-distribution. Larson/Farber 4th ed

Properties of the F-Distribution The F-distribution is a family of curves each of which is determined by two types of degrees of freedom: The degrees of freedom corresponding to the variance in the numerator, denoted d.f.N The degrees of freedom corresponding to the variance in the denominator, denoted d.f.D F-distributions are positively skewed. The total area under each curve of an F-distribution is equal to 1. Larson/Farber 4th ed

Properties of the F-Distribution Chapter 10 Properties of the F-Distribution F-values are always greater than or equal to 0. For all F-distributions, the mean value of F is approximately equal to 1. d.f.N = 1 and d.f.D = 8 F 1 2 3 4 d.f.N = 8 and d.f.D = 26 d.f.N = 16 and d.f.D = 7 d.f.N = 3 and d.f.D = 11 Larson/Farber 4th ed Larson/Farber 4th ed

Critical Values for the F-Distribution Chapter 10 Critical Values for the F-Distribution Specify the level of significance . Determine the degrees of freedom for the numerator, d.f.N. Determine the degrees of freedom for the denominator, d.f.D. Use Table 7 in Appendix B to find the critical value. If the hypothesis test is one-tailed, use the  F-table. two-tailed, use the ½ F-table. Larson/Farber 4th ed Larson/Farber 4th ed

Example: Finding Critical F-Values Find the critical F-value for a right-tailed test when α = 0.05, d.f.N = 6 and d.f.D = 29. Solution: The critical value is F0 = 2.43. Larson/Farber 4th ed

Example: Finding Critical F-Values Find the critical F-value for a two-tailed test when α = 0.05, d.f.N = 4 and d.f.D = 8. Solution: When performing a two-tailed hypothesis test using the F-distribution, you need only to find the right-tailed critical value. You must remember to use the ½α table. Larson/Farber 4th ed

Solution: Finding Critical F-Values ½α = 0.025, d.f.N = 4 and d.f.D = 8 The critical value is F0 = 5.05. Larson/Farber 4th ed

Two-Sample F-Test for Variances To use the two-sample F-test for comparing two population variances, the following must be true. The samples must be randomly selected. The samples must be independent. Each population must have a normal distribution. Larson/Farber 4th ed

Two-Sample F-Test for Variances Test Statistic where represent the sample variances with The degrees of freedom for the numerator is d.f.N = n1 – 1 where n1 is the size of the sample having variance The degrees of freedom for the denominator is d.f.D = n2 – 1, and n2 is the size of the sample having variance Larson/Farber 4th ed

Two-Sample F-Test for Variances Chapter 10 Two-Sample F-Test for Variances In Words In Symbols Identify the claim. State the null and alternative hypotheses. Specify the level of significance. Identify the degrees of freedom. Determine the critical value. State H0 and Ha. Identify . d.f.N = n1 – 1 d.f.D = n2 – 1 Use Table 7 in Appendix B. Larson/Farber 4th ed Larson/Farber 4th ed

Two-Sample F-Test for Variances Chapter 10 Two-Sample F-Test for Variances In Words In Symbols Determine the rejection region. Calculate the 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 F is in the rejection region, reject H0. Otherwise, fail to reject H0. Larson/Farber 4th ed Larson/Farber 4th ed

Example: Performing a Two-Sample F-Test A restaurant manager is designing a system that is intended to decrease the variance of the time customers wait before their meals are served. Under the old system, a random sample of 10 customers had a variance of 400. Under the new system, a random sample of 21 customers had a variance of 256. At α = 0.10, is there enough evidence to convince the manager to switch to the new system? Assume both populations are normally distributed. Larson/Farber 4th ed

Solution: Performing a Two-Sample F-Test Because 400 > 256, H0: Ha: α = d.f.N= d.f.D= Rejection Region: σ12 ≤ σ22 σ12 > σ22 Test Statistic: Decision: 0.10 9 20 Fail to Reject H0 There is not enough evidence to convince the manager to switch to the new system. F 1.96 0.10 1.96 1.56 Larson/Farber 4th ed

Example: Performing a Two-Sample F-Test You want to purchase stock in a company and are deciding between two different stocks. Because a stock’s risk can be associated with the standard deviation of its daily closing prices, you randomly select samples of the daily closing prices for each stock to obtain the results. At α = 0.05, can you conclude that one of the two stocks is a riskier investment? Assume the stock closing prices are normally distributed. Stock A Stock B n2 = 30 n1 = 31 s2 = 3.5 s1 = 5.7 Larson/Farber 4th ed

Solution: Performing a Two-Sample F-Test Because 5.72 > 3.52, H0: Ha: ½α = d.f.N= d.f.D= Rejection Region: σ12 = σ22 σ12 ≠ σ22 Test Statistic: Decision: 0. 025 30 29 Reject H0 There is enough evidence to support the claim that one of the two stocks is a riskier investment. F 2.09 0.025 2.09 2.65 Larson/Farber 4th ed

Section 10.3 Summary Interpreted the F-distribution and used an F-table to find critical values Performed a two-sample F-test to compare two variances Larson/Farber 4th ed

Section 10.4 Analysis of Variance Larson/Farber 4th ed

Section 10.4 Objectives Use one-way analysis of variance to test claims involving three or more means Introduce two-way analysis of variance Larson/Farber 4th ed

One-Way ANOVA One-way analysis of variance Chapter 10 One-Way ANOVA One-way analysis of variance A hypothesis-testing technique that is used to compare means from three or more populations. Analysis of variance is usually abbreviated ANOVA. Hypotheses: H0: μ1 = μ2 = μ3 =…= μk (all population means are equal) Ha: At least one of the means is different from the others. Larson/Farber 4th ed Larson/Farber 4th ed

One-Way ANOVA In a one-way ANOVA test, the following must be true. Each sample must be randomly selected from a normal, or approximately normal, population. The samples must be independent of each other. Each population must have the same variance. Larson/Farber 4th ed

One-Way ANOVA The variance between samples MSB measures the differences related to the treatment given to each sample and is sometimes called the mean square between. The variance within samples MSW measures the differences related to entries within the same sample. This variance, sometimes called the mean square within, is usually due to sampling error. Larson/Farber 4th ed

One-Way Analysis of Variance Test If the conditions for a one-way analysis of variance are satisfied, then the sampling distribution for the test is approximated by the F-distribution. The test statistic is The degrees of freedom for the F-test are d.f.N = k – 1 and d.f.D = N – k where k is the number of samples and N is the sum of the sample sizes. Larson/Farber 4th ed

Test Statistic for a One-Way ANOVA Chapter 10 Test Statistic for a One-Way ANOVA In Words In Symbols Find the mean and variance of each sample. Find the mean of all entries in all samples (the grand mean). Find the sum of squares between the samples. Find the sum of squares within the samples. Larson/Farber 4th ed Larson/Farber 4th ed

Test Statistic for a One-Way ANOVA Chapter 10 Test Statistic for a One-Way ANOVA In Words In Symbols Find the variance between the samples. Find the variance within the samples Find the test statistic. Larson/Farber 4th ed Larson/Farber 4th ed

Performing a One-Way ANOVA Test Chapter 10 Performing a One-Way ANOVA Test In Words In Symbols Identify the claim. State the null and alternative hypotheses. Specify the level of significance. Identify the degrees of freedom. Determine the critical value. State H0 and Ha. Identify . d.f.N = k – 1 d.f.D = N – k Use Table 7 in Appendix B. Larson/Farber 4th ed Larson/Farber 4th ed

Performing a One-Way ANOVA Test Chapter 10 Performing a One-Way ANOVA Test In Words In Symbols Determine the rejection region. Calculate the 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 F is in the rejection region, reject H0. Otherwise, fail to reject H0. Larson/Farber 4th ed Larson/Farber 4th ed

Chapter 10 ANOVA Summary Table A table is a convenient way to summarize the results in a one-way ANOVA test. d.f.D SSW Within d.f.N SSB Between F Mean squares Degrees of freedom Sum of squares Variation Larson/Farber 4th ed Larson/Farber 4th ed

Example: Performing a One-Way ANOVA A medical researcher wants to determine whether there is a difference in the mean length of time it takes three types of pain relievers to provide relief from headache pain. Several headache sufferers are randomly selected and given one of the three medications. Each headache sufferer records the time (in minutes) it takes the medication to begin working. The results are shown on the next slide. At α = 0.01, can you conclude that the mean times are different? Assume that each population of relief times is normally distributed and that the population variances are equal. Larson/Farber 4th ed

Example: Performing a One-Way ANOVA Medication 1 Medication 2 Medication 3 12 16 14 15 17 21 20 19 Solution: k = 3 (3 samples) N = n1 + n2 + n3 = 4 + 5 + 4 = 13 (sum of sample sizes) Larson/Farber 4th ed

Solution: Performing a One-Way ANOVA H0: Ha: α = d.f.N= d.f.D= Rejection Region: μ1 = μ2 = μ3 At least one mean is different Test Statistic: Decision: 0. 01 3 – 1 = 2 13 – 3 = 10 F 7.56 0.01 Larson/Farber 4th ed

Solution: Performing a One-Way ANOVA To find the test statistic, the following must be calculated. Larson/Farber 4th ed

Solution: Performing a One-Way ANOVA To find the test statistic, the following must be calculated. Larson/Farber 4th ed

Solution: Performing a One-Way ANOVA H0: Ha: α = d.f.N= d.f.D= Rejection Region: μ1 = μ2 = μ3 At least one mean is different Test Statistic: Decision: 0. 01 3 – 1 = 2 13 – 3 = 10 Fail to Reject H0 There is not enough evidence at the 1% level of significance to conclude that there is a difference in the mean length of time it takes the three pain relievers to provide relief from headache pain. F 7.56 0.01 1.50 Larson/Farber 4th ed

Example: Using the TI-83/84 to Perform a One-Way ANOVA Three airline companies offer flights between Corydon and Lincolnville. Several randomly selected flight times (in minutes) between the towns for each airline are shown on the next slide. Assume that the populations of flight times are normally distributed, the samples are independent, and the population variances are equal. At α = 0.01, can you conclude that there is a difference in the means of the flight times? Use a TI-83/84. Larson/Farber 4th ed

Example: Using the TI-83/84 to Perform a One-Way ANOVA Airline 1 Airline 2 Airline 3 122 119 120 135 133 158 126 143 155 131 149 125 114 147 116 124 164 134 108 151 142 140 113 136 141 Larson/Farber 4th ed

Solution: Using the TI-83/84 to Perform a One-Way ANOVA Ha: Store data into lists L1, L2, and L3 μ1 = μ2 = μ3 At least one mean is different Decision: P-value < α Reject H0 There is enough evidence to support the claim. You can conclude that there is a difference in the means of the flight times. Larson/Farber 4th ed