1. Slide 1 Copyright © 2004 Pearson Education, Inc.

2. Slide 2 Copyright © 2004 Pearson Education, Inc. Chapter 10 Multinomial Experiments and Contingency Tables 10-1 Overview 10-2 Multinomial Experiments: Goodness-of-fit 10-3 Contingency Tables: Independence and Homogeneity

3. Slide 3 Copyright © 2004 Pearson Education, Inc. Created by Erin Hodgess, Houston, Texas Section 10-1 & 10-2 Overview and Multinomial Experiments: Goodness of Fit

4. Slide 4 Copyright © 2004 Pearson Education, Inc. 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 5 Copyright © 2004 Pearson Education, Inc. 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

6. Slide 6 Copyright © 2004 Pearson Education, Inc. Example: Last Digit Analysis In 2001, Barry Bonds hit 73 home runs. Table 10-2 summarizes the last digit of those home run distances. Verify that the four conditions of a multinomial experiment are satisfied.

7. Slide 7 Copyright © 2004 Pearson Education, Inc. Example: Last Digit Analysis 1. The number of trials (last digits) is the fixed number 73. 2. The trials are independent, because the last digit of the length of a home run does not affect the last digit of the length of any other home run. 3. Each outcome (last digit) is classified into exactly 1 of 10 different categories. The categories are 0, 1, …, 9. 4. Finally, if we assume that the home run distances are measured, the last digits should be equally likely, so that each possible digit has a probability of 1/10. In 2001, Barry Bonds hit 73 home runs. Table 10-2 summarizes the last digit of those home run distances. Verify that the four conditions of a multinomial experiment are satisfied.

8. Slide 8 Copyright © 2004 Pearson Education, Inc. 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.

9. Slide 9 Copyright © 2004 Pearson Education, Inc. 0 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

10. Slide 10 Copyright © 2004 Pearson Education, Inc. Expected Frequencies If all expected frequencies are equal: the sum of all observed frequencies divided by the number of categories n E = k

11. Slide 11 Copyright © 2004 Pearson Education, Inc. If all 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

12. Slide 12 Copyright © 2004 Pearson Education, Inc. Goodness-of-fit Test in Multinomial Experiments Test Statistic 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

13. Slide 13 Copyright © 2004 Pearson Education, Inc.  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.

14. Slide 14 Copyright © 2004 Pearson Education, Inc. Figure 10-3Relationships Among Components in Goodness-of-Fit Hypothesis Test

15. Slide 15 Copyright © 2004 Pearson Education, Inc. Example: Last Digit Analysis In 2001, Barry Bonds hit 73 home runs. Table 10-2 summarizes the last digit of those home run distances. Test the claim that the digits 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

16. Slide 16 Copyright © 2004 Pearson Education, Inc. Example: Last Digit Analysis In 2001, Barry Bonds hit 73 home runs. Table 10-2 summarizes the last digit of those home run distances. Test the claim that the digits do not occur with the same frequency.

17. Slide 17 Copyright © 2004 Pearson Education, Inc. Example: Last Digit Analysis In 2001, Barry Bonds hit 73 home runs. Table 10-2 summarizes the last digit of those home run distances. Test the claim that the digits do not occur with the same frequency. The test statistic is  2 = 251.521. Since the critical value is 16.919, we reject the null hypothesis. There is sufficient evidence to support the claim that the last digits do not occur with the same relative frequency.

18. Slide 18 Copyright © 2004 Pearson Education, Inc. Example: Last Digit Analysis In 2001, Barry Bonds hit 73 home runs. Table 10-2 summarizes the last digit of those home run distances. Test the claim that the digits do not occur with the same frequency.

19. Slide 19 Copyright © 2004 Pearson Education, Inc. Example: Detecting Fraud In the Chapter Problem, it was noted that statistics can be used to detect fraud. Table 10-1 list the percentages for leading digits. 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. 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

20. Slide 20 Copyright © 2004 Pearson Education, Inc. Example: Detecting Fraud In the Chapter Problem, it was noted that statistics can be used to detect fraud. Table 10-1 list the percentages for leading digits. 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.

21. Slide 21 Copyright © 2004 Pearson Education, Inc. Example: Detecting Fraud In the Chapter Problem, it was noted that statistics can be used to detect fraud. Table 10-1 list the percentages for leading digits. 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. 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.

22. Slide 22 Copyright © 2004 Pearson Education, Inc. Example: Detecting Fraud In the Chapter Problem, it was noted that statistics can be used to detect fraud. Table 10-1 list the percentages for leading digits. 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.

23. Slide 23 Copyright © 2004 Pearson Education, Inc. Example: Detecting Fraud In the Chapter Problem, it was noted that statistics can be used to detect fraud. Table 10-1 list the percentages for leading digits. 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 24 Copyright © 2004 Pearson Education, Inc. Created by Erin Hodgess, Houston, Texas Section 10-3 Contingency Tables: Independence and Homogeneity

25. Slide 25 Copyright © 2004 Pearson Education, Inc.  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

26. Slide 26 Copyright © 2004 Pearson Education, Inc.

27. Slide 27 Copyright © 2004 Pearson Education, Inc.  Test of Independence This method tests the null hypothesis that the row variable and column variable in a contingency table are not related. (The null hypothesis is the statement that the row and column variables are independent.) Definition

28. Slide 28 Copyright © 2004 Pearson Education, Inc. Assumptions 1. The sample data are randomly selected. 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.)

29. Slide 29 Copyright © 2004 Pearson Education, Inc. 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

30. Slide 30 Copyright © 2004 Pearson Education, Inc. (row total) (column total) (grand total) E = Total number of all observed frequencies in the table

31. Slide 31 Copyright © 2004 Pearson Education, Inc. Tests of Independence H 0 : The row variable is independent of the column variable H 1 : The row variable is dependent (related to) the column variable 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.

32. Slide 32 Copyright © 2004 Pearson Education, Inc. 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

33. Slide 33 Copyright © 2004 Pearson Education, Inc. Observed and Expected Frequencies 332 1360 1692 318 104 422 29 35 64 27 18 45 706 1517 2223 Men Women Boys GirlsTotal Survived Died Total We will use the mortality table from the Titanic to find expected frequencies. For the upper left hand cell, we find: = 537.360 E = (706)(1692) 2223

34. Slide 34 Copyright © 2004 Pearson Education, Inc. 332 537.360 1360 1692 318 104 422 29 35 64 27 18 45 706 1517 2223 Men Women Boys GirlsTotal Survived Died Total Find the expected frequency for the lower left hand cell, assuming independence between the row variable and the column variable. = 1154.640 E = (1517)(1692) 2223 Observed and Expected Frequencies

35. Slide 35 Copyright © 2004 Pearson Education, Inc. 332 537.360 1360 1154.64 1692 318 134.022 104 287.978 422 29 20.326 35 43.674 64 27 14.291 18 30.709 45 706 1517 2223 Men Women Boys GirlsTotal Survived Died Total To interpret this result for the lower left hand cell, we can say that although 1360 men actually died, we would have expected 1154.64 men to die if survivablility is independent of whether the person is a man, woman, boy, or girl. Observed and Expected Frequencies

36. Slide 36 Copyright © 2004 Pearson Education, Inc. Example: Using a 0.05 significance level, test the claim that when the Titanic sank, whether someone survived or died is independent of whether that person is a man, woman, boy, or girl. H 0 : Whether a person survived is independent of whether the person is a man, woman, boy, or girl. H 1 : Surviving the Titanic and being a man, woman, boy, or girl are dependent.

37. Slide 37 Copyright © 2004 Pearson Education, Inc. Example: Using a 0.05 significance level, test the claim that when the Titanic sank, whether someone survived or died is independent of whether that person is a man, woman, boy, or girl.  2 = (332–537.36) 2 + (318–132.022) 2 + (29–20.326) 2 + (27–14.291) 2 537.36 134.022 20.326 14.291 + (1360–1154.64) 2 + (104–287.978) 2 + (35–43.674) 2 + (18–30.709) 2 1154.64 287.978 43.674 30.709  2 =78.481 + 252.555 + 3.702+11.302+36.525+117.536+1.723+5.260 = 507.084

38. Slide 38 Copyright © 2004 Pearson Education, Inc. Example: Using a 0.05 significance level, test the claim that when the Titanic sank, whether someone survived or died is independent of whether that person is a man, woman, boy, or girl. The number of degrees of freedom are (r–1)(c–1)= (2–1)(4–1)=3.  2.05,3 = 7.815. We reject the null hypothesis. Survival and gender are dependent.

39. Slide 39 Copyright © 2004 Pearson Education, Inc. Test Statistic  2 = 507.084 with  = 0.05 and ( r – 1) ( c– 1) = (2 – 1) (4 – 1) = 3 degrees of freedom Critical Value  2 = 7.815 (from Table A-4)

40. Slide 40 Copyright © 2004 Pearson Education, Inc. Relationships Among Components in X 2 Test of Independence Figure 10-8

41. Slide 41 Copyright © 2004 Pearson Education, Inc. Definition  Test of Homogeneity In a test of homogeneity, we test the claim that different populations have the same proportions of some characteristics.

42. Slide 42 Copyright © 2004 Pearson Education, Inc. 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)?

43. Slide 43 Copyright © 2004 Pearson Education, Inc. Example: Using Table 10-7 as seen below, with a 0.05 significance level, test the effect of pollster gender on survey responses by men.

44. Slide 44 Copyright © 2004 Pearson Education, Inc. Example: Using Table 10-7 as seen below, with a 0.05 significance level, test the effect of pollster gender on survey responses by men. 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.

45. Slide 45 Copyright © 2004 Pearson Education, Inc. Example: Using Table 10-7 as seen below, with a 0.05 significance level, test the effect of pollster gender on survey responses by men.

46. Slide 46 Copyright © 2004 Pearson Education, Inc. Example: Using Table 10-7 as seen below, with a 0.05 significance level, test the effect of pollster gender on survey responses by men. The Minitab display includes the test statistic of  2 = 6.529 and a P-value of 0.011. Using the P-value approach, we reject the null hypothesis of equal(homogeneous) proportions(because the P-value of 0.011 is less than 0.05. There is sufficient evidence to reject the claim of equal proportions.