1. Chi-Square Procedures Chi-Square Test for Goodness of Fit, Independence of Variables, and Homogeneity of Proportions

2. The chi-square Goodness of Fit Test : you have only one set of data on a single characteristic, and you want to know if it matches an expected distribution based on the laws of probability(1 variable, 1population)

3. In a chi-square goodness of fit test, the null hypothesis is always H o : The data follow a specified distribution The alternative hypothesis is always H a : The data does not follow a specified distribution

4. The idea behind testing these types of claims is to compare actual counts to the counts we would expect if the null hypothesis were true. If a significant difference between the actual counts and expected counts exists, we would take this as evidence against the null hypothesis.

5. The method for obtaining the expected counts requires that we determine the number of observations within each cell under the assumption the null hypothesis is true.

6. Test Statistic for the Test of Goodness of Fit Let O i represent the observed number of counts in the ith cell, E i represent the expected number of counts in the ith cell. Then, approximately follows the chi-square distribution with (# of cells– 1) degrees of freedom in the contingency table

7. The Chi-Square Test for Goodness of Fit If a claim is made regarding the data following a certain distribution, we can use the following steps to test the claim provided 1. the data is randomly selected

8. The Chi-Square Test for Goodness of Fit If a claim is made regarding the data following a certain distribution, we can use the following steps to test the claim provided 1. the data is randomly selected 2. all expected frequencies are greater than or equal to 1.

9. The Chi-Square Test for Goodness of Fit If a claim is made regarding the data following a certain distribution, we can use the following steps to test the claim provided 1. the data is randomly selected 2. all expected frequencies are greater than or equal to 1. 3. 80% of the expected cell counts are greater than or equal to 5.

10. EXAMPLETesting for Goodness of Fit In consumer marketing, a common problem that any marketing manager faces is the selection of appropriate colors for package design. Assume that a marketing manager wishes to compare five different colors of package design. He is interested in knowing if there is a preference among the five colors so that it can be introduced in the market. A random sample of 400 consumers reveals the following. Do the consumer preferences for package colors show any significant difference? Package Color RedBlueGreenPinkOrangeTotal Costumers Preference 70106807074400

11. Step 1. A claim is made regarding the data fit to a certain distribution. H o : H a :

12. Step 1. A claim is made regarding the data fit to a certain distribution. H o : the number of customers who prefer each color are the same. H a : the number of customers who prefer each color are not the same.

13. Step 2: Calculate the expected frequencies (counts) for each cell in the contingency table.

14. Package Color RedBlueGreenPinkOrangeTotal Costumers Preference 70106807074400 Package Color RedBlueGreenPinkOrangeTotal Costumers Preference 80 400 Observed Counts Expected Counts

15. Step 3: Verify the requirements for the chi-square test for goodness of fit are satisfied. (1) data is randomly selected (2) all expected frequencies are greater than or equal to 1 (3) 80% of the expected cell counts are greater than or equal to 5. Step 4: Select a proper level of significance 

17. Step 5: Compute the test statistic and P-value P-value = cdf(min,max,df)

18. Step 5: Compute the test statistic and P-value P-value = 0.0224 11.4

19. If P-value < , reject null hypothesis

20. 11.4>9.49 and 0.0224<0.05. Therefore I would reject the null hypothesis. The data is statistically significant and I am led to believe that there is a difference in preference of package color

21. The chi-square independence test : you have two characteristics of a population, and you want to see if there is any association between the characteristics(2 variables, 1 population)

22. In a chi-square independence test, the null hypothesis is always H o : the variables are independent The alternative hypothesis is always H a : the variables are dependent

23. The idea behind testing these types of claims is to compare actual counts to the counts we would expect if the null hypothesis were true (if the variables are independent). If a significant difference between the actual counts and expected counts exists, we would take this as evidence against the null hypothesis.

24. The method for obtaining the expected counts requires that we determine the number of observations within each cell under the assumption the null hypothesis is true.

25. Expected Frequencies in a Chi-Square Independence Test To find the expected frequencies in a cell when performing a chi-square independence test, multiply the row total of the row containing the cell by the column total of the column containing the cell and divide this result by the table total. That is

26. Test Statistic for the Test of Independence Let O i represent the observed number of counts in the ith cell, E i represent the expected number of counts in the ith cell. Then, approximately follows the chi-square distribution with (r – 1)(c – 1) degrees of freedom where r is the number of rows and c is the number of columns in the contingency table

27. The Chi-Square Test for Independence If a claim is made regarding the association between (or independence of) two variables in a contingency table, we can use the following steps to test the claim provided 1. the data is randomly selected

28. The Chi-Square Test for Independence If a claim is made regarding the association between (or independence of) two variables in a contingency table, we can use the following steps to test the claim provided 1. the data is randomly selected 2. all expected frequencies are greater than or equal to 1.

29. The Chi-Square Test for Independence If a claim is made regarding the association between (or independence of) two variables in a contingency table, we can use the following steps to test the claim provided 1. the data is randomly selected 2. all expected frequencies are greater than or equal to 1. 3. 80% of the expected cell counts are greater than or equal to 5.

30. EXAMPLETesting for Independence MoneyHealthLove Men82446355 Women46574273

31. Step 1. A claim is made regarding the independence of the data. H o : H a :

32. Step 1. A claim is made regarding the independence of the data. H o : there is not association between gender of lifestyle choice, the variables are independent H a : there is an association between gender of lifestyle choice, the variables are dependent

33. Step 2: Calculate the expected frequencies (counts) for each cell in the contingency table.

34. MoneyHealthLove Men63.64507.13312.23 Women64.36512.87315.77 MoneyHealthLove Men82446355 Women46574273 Observed Counts Expected Counts

35. Step 3: Verify the requirements for the chi-square test for independence are satisfied. (1) data is randomly selected (2) all expected frequencies are greater than or equal to 1 (3) 80% of the expected cell counts are greater than or equal to 5. Step 4: Select a proper level of significance 

36. Step 5: Compute the test statistic and P-Value P-value = cdf(min,max,df)

37. Step 5: Compute the test statistic and P-Value 36.84 P = 0.00000001

38. If P-value < , reject null hypothesis

39. 36.84>5.99 and 0.00000001<0.05. Therefore I would reject the null hypothesis. The data is statistically significant and I am led to believe that there is an association between gender and lifestyle choice and that these variables are dependent

40. In a chi-square test for homogeneity: you take samples from different populations, and you want to test to see if the proportions in various categories is the same for each population(1 variable, multiple populations)

41. In a chi-square homogeneity test, the null hypothesis is always H o : populations have the same proportion of individuals with some characteristic. The alternative hypothesis is always H a : populations have different proportion of individuals with some characteristic.

42. The idea behind testing these types of claims is to compare actual counts to the counts we would expect if the null hypothesis were true (proportions are equal). If a significant difference between the actual counts and expected counts exists, we would take this as evidence against the null hypothesis.

43. The method for obtaining the expected counts requires that we determine the number of observations within each cell under the assumption the null hypothesis is true.

44. Expected Frequencies in a Chi-Square Homogeneity Test To find the expected frequencies in a cell when performing a chi-square independence test, multiply the row total of the row containing the cell by the column total of the column containing the cell and divide this result by the table total. That is

45. Test Statistic for the Test of Homogeneity Let O i represent the observed number of counts in the ith cell, E i represent the expected number of counts in the ith cell. Then, approximately follows the chi-square distribution with (r – 1)(c – 1) degrees of freedom where r is the number of rows and c is the number of columns in the contingency table

46. The Chi-Square Test for Homogeneity If a claim is made regarding that different populations have the same proportion of individuals with some characteristic, we can use the following steps to test the claim provided 1. the data is randomly selected

47. The Chi-Square Test for Homogeneity If a claim is made regarding that different populations have the same proportion of individuals with some characteristic, we can use the following steps to test the claim provided 1. the data is randomly selected 2. all expected frequencies are greater than or equal to 1.

48. The Chi-Square Test for Homogeneity If a claim is made regarding that different populations have the same proportion of individuals with some characteristic, we can use the following steps to test the claim provided 1. the data is randomly selected 2. all expected frequencies are greater than or equal to 1. 3. 80% of the expected cell counts are greater than or equal to 5.

49. EXAMPLEA Test of Homogeneity of Proportions The following question was asked of a random sample of individuals in 1992, 1998, and 2001: “Would you tell me if you feel being a teacher is an occupation of very great prestige?” The results of the survey are presented below: 199219982001 Yes549539570 No522578599

50. Step 1. A claim is made regarding the homogeneity of the data. H o : H a :

51. Step 1. A claim is made regarding the homogeneity of the data. H o : the proportions of individuals who feel teaching is an occupation of very great prestige in each year are equal H a : the proportions of individuals who feel teaching is an occupation of very great prestige in each year are not equal

52. Step 2: Calculate the expected frequencies (counts) for each cell in the contingency table.

53. 199219982001 Yes528.96551.68577.36 No542.04565.32591.64 199219982001 Yes549539570 No522578599 Observed Counts Expected Counts

54. Step 3: Verify the requirements for the chi-square test for homogeneity are satisfied. (1) data is randomly selected (2) all expected frequencies are greater than or equal to 1 (3) 80% of the expected cell counts are greater than or equal to 5. Step 4: Select a proper level of significance 

55. Step 5: Compute the test statistic and P-Value P-value = cdf(min,max,df)

56. Step 5: Compute the test statistic and P-Value 2.26 P = 0.3228

57. If P-value < , reject null hypothesis

58. 2.26 0.01. Therefore I would fail to reject the null hypothesis. The data is not statistically significant and I can not conclude that the proportions of individuals who feel teaching is an occupation of very great prestige is different each year