1. Slide Slide 1 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Lecture Slides Elementary Statistics Tenth Edition and the Triola Statistics Series by Mario F. Triola

2. Slide Slide 2 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. 11-1 Overview 11-2 Multinomial Experiments: Goodness-of-fit 11-3 Contingency Tables: Independence and Homogeneity 11-4 McNemar’s Test for Matched Pairs Chapter 11 Multinomial Experiments and Contingency Tables

3. Slide Slide 3 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Section 11-1 & 11-2 Overview and Multinomial Experiments: Goodness of Fit Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Jim Zimmer, Chattanooga State, Chattanooga, TN

4. Slide Slide 4 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Overview  We focus on analysis of categorical (qualitative or attribute) data that can be separated into different categories (often called cells).  Use the  2 (chi-square) test statistic (Table A- 4).  The goodness-of-fit test uses a one-way frequency table (single row or column).  The contingency table uses a two-way frequency table (two or more rows and columns).

5. Slide Slide 5 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Key Concept Given data separated into different categories, we will test the hypothesis that the distribution of the data agrees with or “fits” some claimed distribution. The hypothesis test will use the chi-square distribution with the observed frequency counts and the frequency counts that we would expect with the claimed distribution.

6. Slide Slide 6 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Multinomial Experiment This is an experiment that meets the following conditions: 1. The number of trials is fixed. 2. The trials are independent. 3. All outcomes of each trial must be classified into exactly one of several different categories. 4. The probabilities for the different categories remain constant for each trial. Definition

7. Slide Slide 7 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Last Digits of Weights When asked, people often provide weights that are somewhat lower than their actual weights. So how can researchers verify that weights were obtained through actual measurements instead of asking subjects?

8. Slide Slide 8 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Test the claim that the digits in Table 11-2 do not occur with the same frequency. Table 11-2 summarizes the last digit of weights of 80 randomly selected students. Example: Last Digits of Weights

9. Slide Slide 9 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. 1. The number of trials (last digits) is the fixed number 80. 2. The trials are independent, because the last digit of any individual’s weight does not affect the last digit of any other weight. 3. Each outcome (last digit) is classified into exactly 1 of 10 different categories. The categories are 0, 1, …, 9. 4. Finally, in testing the claim that the 10 digits are equally likely, each possible digit has a probability of 1/10, and by assumption, that probability remains constant for each subject. Verify that the four conditions of a multinomial experiment are satisfied. Example: Last Digits of Weights

10. Slide Slide 10 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Definition Goodness-of-fit Test A goodness-of-fit test is used to test the hypothesis that an observed frequency distribution fits (or conforms to) some claimed distribution.

11. Slide Slide 11 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. O represents the observed frequency of an outcome. E represents the expected frequency of an outcome. k represents the number of different categories or outcomes. n represents the total number of trials. Goodness-of-Fit Test Notation

12. Slide Slide 12 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Expected Frequencies If all expected frequencies are equal: the sum of all observed frequencies divided by the number of categories n E = k

13. Slide Slide 13 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. If expected frequencies are not all equal: Each expected frequency is found by multiplying the sum of all observed frequencies by the probability for the category. E = n p Expected Frequencies

14. Slide Slide 14 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Goodness-of-Fit Test in Multinomial Experiments 1.The data have been randomly selected. 2.The sample data consist of frequency counts for each of the different categories. 3.For each category, the expected frequency is at least 5. (The expected frequency for a category is the frequency that would occur if the data actually have the distribution that is being claimed. There is no requirement that the observed frequency for each category must be at least 5.) Requirements

15. Slide Slide 15 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Goodness-of-Fit Test in Multinomial Experiments Critical Values 1. Found in Table A- 4 using k – 1 degrees of freedom, where k = number of categories. 2. Goodness-of-fit hypothesis tests are always right-tailed.  2 =  ( O – E ) 2 E Test Statistics

16. Slide Slide 16 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley.  A large disagreement between observed and expected values will lead to a large value of  2 and a small P -value.  A significantly large value of  2 will cause a rejection of the null hypothesis of no difference between the observed and the expected.  A close agreement between observed and expected values will lead to a small value of  2 and a large P -value. Goodness-of-Fit Test in Multinomial Experiments

17. Slide Slide 17 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Relationships Among the  2 Test Statistic, P-Value, and Goodness-of-Fit Figure 11-3

18. Slide Slide 18 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Last Digit Analysis Test the claim that the digits in Table 11-2 do not occur with the same frequency. H 0 : p 0 = p 1 =  = p 9 H 1 : At least one of the probabilities is different from the others.  = 0.05 k – 1 = 9  2.05, 9 = 16.919

19. Slide Slide 19 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Last Digit Analysis Test the claim that the digits in Table 11-2 do not occur with the same frequency. Because the 80 digits would be uniformly distributed through the 10 categories, each expected frequency should be 8.

20. Slide Slide 20 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Last Digit Analysis Test the claim that the digits in Table 11-2 do not occur with the same frequency.

21. Slide Slide 21 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Last Digit Analysis From Table 11-3, the test statistic is  2 = 156.500. Since the critical value is 16.919, we reject the null hypothesis of equal probabilities. There is sufficient evidence to support the claim that the last digits do not occur with the same relative frequency. Test the claim that the digits in Table 11-2 do not occur with the same frequency.

22. Slide Slide 22 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Detecting Fraud Unequal Expected Frequencies In the Chapter Problem, it was noted that statistics can be used to detect fraud. Table 11-1 lists the percentages for leading digits from Benford’s Law.

23. Slide Slide 23 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Detecting Fraud Observed Frequencies and Frequencies Expected with Benford’s Law Test the claim that there is a significant discrepancy between the leading digits expected from Benford’s Law and the leading digits from the 784 checks.

24. Slide Slide 24 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Detecting Fraud H 0 : p 1 = 0.301, p 2 = 0.176, p 3 = 0.125, p 4 = 0.097, p 5 = 0.079, p 6 = 0.067, p 7 = 0.058, p 8 = 0.051 and p 9 = 0.046 H 1 : At least one of the proportions is different from the claimed values.  = 0.01 k – 1 = 8  2.01,8 = 20.090 Test the claim that there is a significant discrepancy between the leading digits expected from Benford’s Law and the leading digits from the 784 checks.

25. Slide Slide 25 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Detecting Fraud The test statistic is  2 = 3650.251. Since the critical value is 20.090, we reject the null hypothesis. There is sufficient evidence to reject the null hypothesis. At least one of the proportions is different than expected. Test the claim that there is a significant discrepancy between the leading digits expected from Benford’s Law and the leading digits from the 784 checks.

26. Slide Slide 26 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Detecting Fraud Test the claim that there is a significant discrepancy between the leading digits expected from Benford’s Law and the leading digits from the 784 checks. Figure 11-5

27. Slide Slide 27 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Detecting Fraud Figure 11-6 Comparison of Observed Frequencies and Frequencies Expected with Benford’s Law

28. Slide Slide 28 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Recap In this section we have discussed: Multinomial Experiments: Goodness-of-Fit - Equal Expected Frequencies - Unequal Expected Frequencies Test the hypothesis that an observed frequency distribution fits (or conforms to) some claimed distribution.

29. Slide Slide 29 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Section 11-3 Contingency Tables: Independence and Homogeneity Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Jim Zimmer, Chattanooga State, Chattanooga, TN

30. Slide Slide 30 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Key Concept In this section we consider contingency tables (or two-way frequency tables), which include frequency counts for categorical data arranged in a table with a least two rows and at least two columns. We present a method for testing the claim that the row and column variables are independent of each other. We will use the same method for a test of homogeneity, whereby we test the claim that different populations have the same proportion of some characteristics.

31. Slide Slide 31 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Contingency Table (or two-way frequency table) A contingency table is a table in which frequencies correspond to two variables. (One variable is used to categorize rows, and a second variable is used to categorize columns.) Contingency tables have at least two rows and at least two columns. Definition

32. Slide Slide 32 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Case-Control Study of Motorcycle Drivers Is the color of the motorcycle helmet somehow related to the risk of crash related injuries? 491 213 704 377 112 489 31 8 39 899 333 1232 BlackWhiteYellow/Orange Row Totals Controls (not injured) Cases (injured or killed) Column Totals

33. Slide Slide 33 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Test of Independence A test of independence tests the null hypothesis that there is no association between the row variable and the column variable in a contingency table. (For the null hypothesis, we will use the statement that “the row and column variables are independent.”) Definition

34. Slide Slide 34 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Requirements 1.The sample data are randomly selected and are represented as frequency counts in a two-way table. 2.The null hypothesis H 0 is the statement that the row and column variables are independent; the alternative hypothesis H 1 is the statement that the row and column variables are dependent. 3.For every cell in the contingency table, the expected frequency E is at least 5. (There is no requirement that every observed frequency must be at least 5. Also there is no requirement that the population must have a normal distribution or any other specific distribution.)

35. Slide Slide 35 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Test of Independence Test Statistic Critical Values 1. Found in Table A-4 using degrees of freedom = (r – 1)(c – 1) r is the number of rows and c is the number of columns 2. Tests of Independence are always right-tailed.  2 =  ( O – E ) 2 E

36. Slide Slide 36 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. (row total) (column total) (grand total) E = Total number of all observed frequencies in the table Expected Frequency

37. Slide Slide 37 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Test of Independence This procedure cannot be used to establish a direct cause-and-effect link between variables in question. Dependence means only there is a relationship between the two variables.

38. Slide Slide 38 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Expected Frequency for Contingency Tables E = grand total row total column total grand total E = (row total) (column total) (grand total) (probability of a cell) n p

39. Slide Slide 39 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. 491 213 704 377 112 489 31 8 39 899 333 1232 BlackWhiteYellow/Orange Row Totals Controls (not injured) Cases (injured or killed) Column Totals For the upper left hand cell: = 513.714 E = (899)(704) 1232 Case-Control Study of Motorcycle Drivers (row total) (column total) E = (grand total) 899 1232 704 899 1232

40. Slide Slide 40 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. 491 513.714 213 704 377 112 489 31 8 39 899 333 1232 BlackWhiteYellow/Orange Row Totals Controls (not injured) Expected Cases (injured or killed) Column Totals Case-Control Study of Motorcycle Drivers = 513.714 E = (899)(704) 1232 (row total) (column total) E = (grand total) Expected

41. Slide Slide 41 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. 491 513.714 213 704 377 112 489 31 8 39 899 333 1232 BlackWhiteYellow/Orange Row Totals Controls (not injured) Expected Cases (injured or killed) Column Totals 190.286 Calculate expected for all cells. To interpret this result for the upper left hand cell, we can say that although 491 riders with black helmets were not injured, we would have expected the number to be 513.714 if crash related injuries are independent of helmet color. 356.827 132.173 28.459 10.541 Case-Control Study of Motorcycle Drivers Expected

42. Slide Slide 42 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Using a 0.05 significance level, test the claim that group (control or case) is independent of the helmet color. H 0 : Whether a subject is in the control group or case group is independent of the helmet color. (Injuries are independent of helmet color.) H 1 : The group and helmet color are dependent. Case-Control Study of Motorcycle Drivers

43. Slide Slide 43 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. 491 513.714 213 704 377 112 489 31 8 39 899 333 1232 BlackWhiteYellow/Orange Row Totals Cases (injured or killed) Expected Column Totals Controls (not injured) Expected 190.286 356.827 132.173 28.459 10.541 Case-Control Study of Motorcycle Drivers

44. Slide Slide 44 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. H 0 : Row and column variables are independent. H 1 : Row and column variables are dependent. The test statistic is  2 = 8.775  = 0.05 The number of degrees of freedom are (r–1)(c–1) = (2–1)(3–1) = 2. The critical value (from Table A-4) is  2.05,2 = 5.991. Case-Control Study of Motorcycle Drivers

45. Slide Slide 45 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. We reject the null hypothesis. It appears there is an association between helmet color and motorcycle safety. Case-Control Study of Motorcycle Drivers Figure 11-4

46. Slide Slide 46 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Relationships Among Key Components in Test of Independence Figure 11-8

47. Slide Slide 47 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Definition Test of Homogeneity In a test of homogeneity, we test the claim that different populations have the same proportions of some characteristics.

48. Slide Slide 48 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. How to Distinguish Between a Test of Homogeneity and a Test for Independence: Were predetermined sample sizes used for different populations (test of homogeneity), or was one big sample drawn so both row and column totals were determined randomly (test of independence)?

49. Slide Slide 49 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Using Table 11-6 with a 0.05 significance level, test the effect of pollster gender on survey responses by men. Example: Influence of Gender

50. Slide Slide 50 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. H 0 : The proportions of agree/disagree responses are the same for the subjects interviewed by men and the subjects interviewed by women. H 1 : The proportions are different. Using Table 11-6 with a 0.05 significance level, test the effect of pollster gender on survey responses by men. Example: Influence of Gender

51. Slide Slide 51 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Using Table 11-6 with a 0.05 significance level, test the effect of pollster gender on survey responses by men. Example: Influence of Gender Minitab

52. Slide Slide 52 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Recap In this section we have discussed: Contingency tables where categorical data is arranged in a table with a least two rows and at least two columns. * Test of Independence tests the claim that the row and column variables are independent of each other. * Test of Homogeneity tests the claim that different populations have the same proportion of some characteristics.

53. Slide Slide 53 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Section 11-4 McNemar’s Test for Matched Pairs Created by Erin Hodgess, Houston, Texas Revised to accompany 10th Edition, Jim Zimmer, Chattanooga State, Chattanooga, TN

54. Slide Slide 54 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Key Concept The Contingency table procedures in Section 11-3 are based on independent data. For 2 x 2 tables consisting of frequency counts that result from matched pairs, we do not have independence, and for such cases, we can use McNemar’s test for matched pairs.

55. Slide Slide 55 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Table 11-9 is a general table summarizing the frequency counts that result from matched pairs.

56. Slide Slide 56 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. McNemar’s Test uses frequency counts from matched pairs of nominal data from two categories to test the null hypothesis that for a table such as Table 11-9, the frequencies b and c occur in the same proportion. Definition

57. Slide Slide 57 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Requirements 1.The sample data have been randomly selected. 2.The sample data consist of matched pairs of frequency counts. 3.The data are at the nominal level of measurement, and each observation can be classified two ways: (1) According to the category distinguishing values with each matched pair, and (2) according to another category with two possible values. 4.For tables such as Table 11-9, the frequencies are such that b + c ≥ 10.

58. Slide Slide 58 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Requirements Test Statistic (for testing the null hypothesis that for tables such as Table 11-9, the frequencies b and c occur in the same proportion): Where the frequencies of b and c are obtained from the 2 x 2 table with a format similar to Table 11-9. Critical values 1. The critical region is located in the right tail only. 2. The critical values are found in Table A- 4 by using degrees of freedom = 1.

59. Slide Slide 59 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Comparing Treatments Table 11-10 summarizes the frequency counts that resulted from the matched pairs using Pedacream on one foot and Fungacream on the other foot.

60. Slide Slide 60 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Comparing Treatments H 0 : The following two proportions are the same: The proportion of subjects with no cure on the Pedacream-treated foot and a cure on the Fungacream-treated foot. The proportion of subjects with a cure on the Pedacream-treated and no cure on the Fungacream-treated foot. Based on the results, does there appear to be a difference?

61. Slide Slide 61 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Comparing Treatments Note that the requirements have been met: The data consist of matched pairs of frequency counts from randomly selected subjects. Each observation can be categorized according to two variables. (One variable has values of “Pedacream” and “Fungacream,” and the other variable has values of “cured” and “not cured.”) b = 8 and c = 40, so b + c ≥ 10.

62. Slide Slide 62 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Comparing Treatments Test statistic Critical value from Table A-4 Reject the null hypothesis. It appears the creams produce different results.

63. Slide Slide 63 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Example: Comparing Treatments Note that the test did not use the categories where both feet were cured or where neither foot was cured. Only results from categories that are different were used. Definition. Discordant pairs of results come from pairs of categories in which the two categories are different (as in cure/no cure or no cure/cure).

64. Slide Slide 64 Copyright © 2007 Pearson Education, Inc Publishing as Pearson Addison-Wesley. Recap In this section we have discussed: McNemar’s test for matched pairs. Data are place in a 2 x 2 table where each observation is classified in two ways. The test only compares categories that are different (discordant pairs).