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Chapter 10 Chi-Square Tests and the F-Distribution § 10.1 Goodness of Fit Multinomial Experiments A multinomial experiment is a probability experiment consisting of a fixed number of trials in which there are more than two possible outcomes for each independent trial. (Unlike the binomial experiment in which there were only two possible outcomes.) Example: A researcher claims that the distribution of favorite pizza toppings among teenagers is as shown below. Each outcome is classified into categories. Topping Cheese Pepperoni Sausage Mushrooms Onions Frequency, f 41% 25% 15% 10% 9% The probability for each possible outcome is fixed. Larson & Farber, Elementary Statistics: Picturing the World, 3e 3 1 Chi-Square Goodness-of-Fit Test A Chi-Square Goodness-of-Fit Test is used to test whether a frequency distribution fits an expected distribution. 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. 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, Elementary Statistics: Picturing the World, 3e 4 Observed and Expected Frequencies Example: 200 teenagers are randomly selected and asked what their favorite pizza topping is. The results are shown below. Find the observed frequencies and the expected frequencies. Topping Cheese Pepperoni Sausage Mushrooms Onions Results (n = 200) 78 52 30 25 15 % of teenagers 41% 25% 15% 10% 9% Observed Frequency 78 52 30 25 15 Expected Frequency 200(0.41) = 82 200(0.25) = 50 200(0.15) = 30 200(0.10) = 20 200(0.09) = 18 Larson & Farber, Elementary Statistics: Picturing the World, 3e 5 Chi-Square Goodness-of-Fit Test For the chi-square goodness-of-fit test to be used, the following must be true. 1. 2. The observed frequencies must be obtained by using a random sample. Each expected frequency must be greater than or equal to 5. The Chi-Square Goodness-of-Fit Test If the conditions listed above 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 χ2 =∑ (O − E )2 The test is always a right- test. E where O represents the observed frequencytailed of each category and E represents the expected frequency of each category. Larson & Farber, Elementary Statistics: Picturing the World, 3e 6 2 Chi-Square Goodness-of-Fit Test Performing a Chi-Square Goodness-of-Fit Test In Words In Symbols 1. Identify the claim. State the null and alternative hypotheses. State H0 and Ha. 2. Specify the level of significance. Identify α. 3. Identify the degrees of freedom. d.f. = k – 1 4. Determine the critical value. Use Table 6 in Appendix B. 5. Determine the rejection region. Continued. Larson & Farber, Elementary Statistics: Picturing the World, 3e 7 Chi-Square Goodness-of-Fit Test Performing a Chi-Square Goodness-of-Fit Test In Words In Symbols 6. Calculate the test statistic. 2 χ 2 = ∑ (O − E ) E 7. Make a decision to reject or fail to reject the null hypothesis. If χ2 is in the rejection region, reject H0. Otherwise, fail to reject H0. 8. Interpret the decision in the context of the original claim. Larson & Farber, Elementary Statistics: Picturing the World, 3e 8 Chi-Square Goodness-of-Fit Test Example: A researcher claims that the distribution of favorite pizza toppings among teenagers is as shown below. 200 randomly selected teenagers are surveyed. Topping Cheese Pepperoni Sausage Mushrooms Onions Frequency, f 39% 26% 15% 12.5% 7.5% Using α = 0.01, and the observed and expected values previously calculated, test the surveyor’s claim using a chi-square goodnessof-fit test. Continued. Larson & Farber, Elementary Statistics: Picturing the World, 3e 9 3 Chi-Square Goodness-of-Fit Test Example continued: H0: The distribution of pizza toppings is 39% cheese, 26% pepperoni, 15% sausage, 12.5% mushrooms, and 7.5% onions. (Claim) Ha: The distribution of pizza toppings differs from the claimed or expected distribution. Because there are 5 categories, the chi-square distribution has k – 1 = 5 – 1 = 4 degrees of freedom. With d.f. = 4 and α = 0.01, the critical value is χ20 = 13.277. Continued. Larson & Farber, Elementary Statistics: Picturing the World, 3e 10 Chi-Square Goodness-of-Fit Test Example continued: Rejection region α = 0.01 X2 χ20 = 13.277 χ2 =∑ Topping Cheese Pepperoni Sausage Mushrooms Onions Observed Frequency 78 52 30 25 15 Expected Frequency 82 50 30 20 18 (O − E )2 = (78 − 82)2 + (52 − 50)2 + (30 − 30)2 + (25 − 20)2 + (15 − 18)2 82 50 30 20 18 E ≈ 2.025 Fail to reject H0. There is not enough evidence at the 1% level to reject the surveyor’s claim. Larson & Farber, Elementary Statistics: Picturing the World, 3e 11 § 10.2 Independence 4 Contingency Tables An 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. The following contingency table shows a random sample of 321 fatally injured passenger vehicle drivers by age and gender. (Adapted from Insurance Institute for Highway Safety) Age Gender 16 – 20 21 – 30 31 – 40 41 – 50 51 – 60 Male 32 51 52 43 28 61 and older 10 Female 13 22 33 21 10 6 Larson & Farber, Elementary Statistics: Picturing the World, 3e 13 Expected Frequency Assuming the two variables are independent, you can use the contingency table to find the expected frequency for each cell. Finding the Expected Frequency for Contingency Table Cells The expected frequency for a cell Er,c in a contingency table is E xpect ed fr equ en cy E r ,c = (Su m of r ow r ) × (Su m of colu m n c ) . Sa m ple size Larson & Farber, Elementary Statistics: Picturing the World, 3e 14 Expected Frequency Example: Find the expected frequency for each “Male” cell in the contingency table for the sample of 321 fatally injured drivers. Assume that the variables, age and gender, are independent. Age Gender 16 – 20 21 – 30 31 – 40 41 – 50 51 – 60 61 and older Total Male 32 51 52 43 28 10 216 Female 13 22 33 21 10 6 105 Total 45 73 85 64 38 16 321 Continued. Larson & Farber, Elementary Statistics: Picturing the World, 3e 15 5 Expected Frequency Example continued: Age Gender 16 – 20 21 – 30 31 – 40 41 – 50 51 – 60 Male Female Total 32 13 45 51 22 73 52 33 85 43 21 64 28 10 38 E xpect ed fr equ en cy E r ,c = 61 and older 10 6 16 Total 216 105 321 (Su m of r ow r ) × (Su m of colu m n c ) Sa m ple size E 1,1 = 216 ⋅ 45 ≈ 30.28 321 E 1,2 = 216 ⋅ 73 ≈ 49.12 321 E 1,3 = 216 ⋅ 85 ≈ 57.20 321 E 1,4 = 216 ⋅ 64 ≈ 43.07 321 E 1,5 = 216 ⋅ 38 ≈ 25.57 321 E 1,6 = 216 ⋅ 16 ≈ 10.77 321 Larson & Farber, Elementary Statistics: Picturing the World, 3e 16 Chi-Square Independence Test A chi-square independence test is used to test the independence of two variables. Using a chi-square test, you can determine whether the occurrence of one variable affects the probability of the occurrence of the other variable. For the chi-square independence test to be used, the following must be true. 1. The observed frequencies must be obtained by using a random sample. 2. Each expected frequency must be greater than or equal to 5. Larson & Farber, Elementary Statistics: Picturing the World, 3e 17 Chi-Square Independence Test The Chi-Square Independence Test If the conditions listed are satisfied, then the sampling distribution for the chi-square independence test is approximated by a chisquare 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 χ2 =∑ (O − E )2 E The test is always a righttailed test. where O represents the observed frequencies and E represents the expected frequencies. Larson & Farber, Elementary Statistics: Picturing the World, 3e 18 6 Chi-Square Independence Test Performing a Chi-Square Independence Test In Words In Symbols 1. Identify the claim. State the null and alternative hypotheses. State H0 and Ha. 2. Specify the level of significance. Identify α. 3. Identify the degrees of freedom. d.f. = (r – 1)(c – 1) 4. Determine the critical value. Use Table 6 in Appendix B. 5. Determine the rejection region. Continued. Larson & Farber, Elementary Statistics: Picturing the World, 3e 19 Chi-Square Independence Test Performing a Chi-Square Independence Test In Words In Symbols 6. Calculate the test statistic. 2 χ 2 = ∑ (O − E ) E 7. Make a decision to reject or fail to reject the null hypothesis. If χ2 is in the rejection region, reject H0. Otherwise, fail to reject H0. 8. Interpret the decision in the context of the original claim. Larson & Farber, Elementary Statistics: Picturing the World, 3e 20 Chi-Square Independence Test Example: The following contingency table shows a random sample of 321 fatally injured passenger vehicle drivers by age and gender. The expected frequencies are displayed in parentheses. At α = 0.05, can you conclude that the drivers’ ages are related to gender in such accidents? Age Gender 16 – 20 21 – 30 31 – 40 41 – 50 51 – 60 Male 32 (30.28) 13 (14.72) 45 51 (49.12) 22 (23.88) 73 52 (57.20) 33 (27.80) 85 43 (43.07) 21 (20.93) 64 28 (25.57) 10 (12.43) 38 Female 61 and Total older 216 10 (10.77) 6 (5.23) 105 Larson & Farber, Elementary Statistics: Picturing the World, 3e 16 321 21 7 Chi-Square Independence Test Example continued: Because each expected frequency is at least 5 and the drivers were randomly selected, the chi-square independence test can be used to test whether the variables are independent. H0: The drivers’ ages are independent of gender. Ha: The drivers’ ages are dependent on gender. (Claim) d.f. = (r – 1)(c – 1) = (2 – 1)(6 – 1) = (1)(5) = 5 With d.f. = 5 and α = 0.05, the critical value is χ20 = 11.071. Continued. Larson & Farber, Elementary Statistics: Picturing the World, 3e 22 Chi-Square Independence Test Example continued: Rejection region α = 0.05 X2 χ20 = 11.071 χ2 =∑ (O − E )2 = 2.84 E Fail to reject H0. O E O–E 32 51 52 43 28 10 13 22 33 21 10 6 30.28 49.12 57.20 43.07 25.57 10.77 14.72 23.88 27.80 20.93 12.43 5.23 1.72 1.88 −5.2 −0.07 2.43 −0.77 −1.72 −1.88 5.2 0.07 −2.43 0.77 2 (O – E)2 (O − E ) E 2.9584 3.5344 27.04 0.0049 5.9049 0.5929 2.9584 3.5344 27.04 0.0049 5.9049 0.5929 0.0977 0.072 0.4727 0.0001 0.2309 0.0551 0.201 0.148 0.9727 0.0002 0.4751 0.1134 There is not enough evidence at the 5% level to conclude that age is dependent on gender in such accidents. Larson & Farber, Elementary Statistics: Picturing the World, 3e 23 § 10.3 Comparing Two Variances 8 F-Distribution Let s 12 an d s 22represent the sample variances of two different populations. If both populations are normal and the population 2 variances are equal,σthen the σ 22sampling distribution of 1 an d F = s 12 is called an F-distribution. s 22 There are several properties of this distribution. 1. 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, and the degrees of freedom corresponding to the variance in the denominator, denoted d.f.D. Continued. Larson & Farber, Elementary Statistics: Picturing the World, 3e 25 F-Distribution Properties of the F-distribution continued: 2. 3. 4. 5. F-distributions are positively skewed. The total area under each curve of an F-distribution is equal to 1. 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 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 F 1 2 3 4 Larson & Farber, Elementary Statistics: Picturing the World, 3e 26 Critical Values for the F-Distribution Finding Critical Values for the F-Distribution 1. Specify the level of significance α. 2. Determine the degrees of freedom for the numerator, d.f.N. 3. Determine the degrees of freedom for the denominator, d.f.D. 4. Use Table 7 in Appendix B to find the critical value. If the hypothesis test is a. one-tailed, use the α F-table. 1 b. two-tailed, use the α 2F-table. Larson & Farber, Elementary Statistics: Picturing the World, 3e 27 9 Critical Values for the F-Distribution Example: Find the critical F-value for a right-tailed test when d.f.N = 5 and d.f.D = 28. α = 0.05, 1 2 1 161.4 18.51 Appendix B: Table 7: F-Distribution α = 0.05 d.f.N: Degrees of freedom, numerator 2 3 4 5 6 199.5 215.7 224.6 230.2 234.0 19.00 19.16 19.25 19.30 19.33 27 28 29 4.21 4.20 4.18 3.35 3.34 3.33 d.f.D: Degrees of freedom, denominator 2.96 2.95 2.93 2.73 2.71 2.70 2.57 2.56 2.55 2.46 2.45 2.43 The critical value is F0 = 2.56. Larson & Farber, Elementary Statistics: Picturing the World, 3e 28 Critical Values for the F-Distribution Example: 1 Find the critical F-value for a two-tailed test when 1 α = (0.10) = 0.05 2 2 α = 0.10, d.f.N = 4 and d.f.D = 6. d.f.D: Degrees of freedom, denominator 1 2 3 4 5 6 7 1 161.4 18.51 10.13 7.71 6.61 5.99 5.59 Appendix B: Table 7: F-Distribution α = 0.05 d.f.N: Degrees of freedom, numerator 2 3 4 5 6 199.5 215.7 224.6 230.2 234.0 19.00 19.16 19.25 19.30 19.33 9.55 9.28 9.12 9.01 8.94 6.94 6.59 6.39 6.26 6.16 5.79 5.41 5.19 5.05 4.95 5.14 4.76 4.53 4.39 4.28 4.74 4.35 4.12 3.97 3.87 The critical value is F0 = 4.53. Larson & Farber, Elementary Statistics: Picturing the World, 3e 29 Two-Sample F-Test for Variances Two-Sample F-Test for Variances A two-sample F-test is used to compare two population variances when sample σ 12aan d σ 22 is randomly selected from each population. The populations must be independent and normally distributed. The test statistic is s2 F = 12 s2 where s 12 an d s 22 represent the sample variances with The of freedom for the numerator is d.f.N = n1 – 1 and the s 12 ≥degrees s 22. degrees of freedom for the denominator is d.f.D = n2 – 1, where n1 is the size of the sample having variance and n2 is the size of the sample having variance s 12 s 22. Larson & Farber, Elementary Statistics: Picturing the World, 3e 30 10 Two-Sample F-Test for Variances Using a Two-Sample F-Test to Compare In Words In Symbols 1. Identify the claim. State the null and alternative hypotheses. State H0 and Ha. 2. Specify the level of significance. Identify α. 3. Identify the degrees of freedom. d.f.N = n1 – 1 d.f.D = n2 – 1 4. Determine the critical value. Use Table 7 in Appendix B. Continued. Larson & Farber, Elementary Statistics: Picturing the World, 3e 31 11