Statistics for Business and Economics Chapter 9 Categorical

Statistics for Business and Economics Chapter 9 Categorical Data Analysis

Learning Objectives 1. Explain 2 Test for Proportions 2. Explain 2 Test of Independence 3. Solve Hypothesis Testing Problems • More Than Two Population Proportions • Independence

Data Types Data Quantitative Discrete Continuous Qualitative

Qualitative Data • Qualitative random variables yield responses that classify – Example: gender (male, female) • Measurement reflects number in category • Nominal or ordinal scale • Examples – What make of car do you drive? – Do you live on-campus or off-campus?

Hypothesis Tests Qualitative Data 1 pop. Proportion More than 2 pop. Independence 2 pop. Z Test c 2 Test

Chi-Square ( 2) Test for k Proportions

Hypothesis Tests Qualitative Data 1 pop. Proportion More than 2 pop. Independence 2 pop. Z Test c 2 Test

Multinomial Experiment • • • n identical trials k outcomes to each trial Constant outcome probability, pk Independent trials Random variable is count, nk Example: ask 100 people (n) which of 3 candidates (k) they will vote for

2) ( Chi-Square Test for k Proportions • Tests equality (=) of proportions only – Example: p 1 =. 2, p 2=. 3, p 3 =. 5 • One variable with several levels • Uses one-way contingency table

One-Way Contingency Table Shows number of observations in k independent groups (outcomes or variable levels) Outcomes (k = 3) Candidate Tom Bill Mary Total 35 20 45 100 Number of responses

Conditions Required for a Valid Test: One-way Table 1. A multinomial experiment has been conducted 2. The sample size n is large: E(ni) is greater than or equal to 5 for every cell

2 Test for k Proportions Hypotheses & Statistic 1. Hypotheses H 0: p 1 = p 1, 0, p 2 = p 2, 0, . . . , pk = pk, 0 Hypothesized probability Ha: At least one pi is different from above 2. Test Statistic Observed count Expected count: E(ni) = npi, 0 3. Degrees of Freedom: k – 1 Number of outcomes

2 Test Basic Idea 1. Compares observed count to expected count assuming null hypothesis is true 2. Closer observed count is to expected count, the more likely the H 0 is true • Measured by squared difference relative to expected count — Reject large values

Finding Critical Value Example What is the critical 2 value if k = 3, and =. 05? If ni = E(ni), 2 = 0. Do not reject H 0 Reject H 0 =. 05 df = k - 1 = 2 0 2 Table (Portion) DF. 995 1. . . 2 0. 010 5. 991 c 2 Upper Tail Area …. 95 … … 0. 004 … … 0. 103 … . 05 3. 841 5. 991

2 Test for k Proportions Example As personnel director, you want to test the perception of fairness of three methods of performance evaluation. Of 180 employees, 63 rated Method 1 as fair, 45 rated Method 2 as fair, 72 rated Method 3 as fair. At the. 05 level of significance, is there a difference in perceptions?

2 • • • Test for k Proportions Solution H 0: p 1 = p 2 = p 3 = 1/3 Test Statistic: Ha: At least 1 is different =. 05 n 1 = 63 n 2 = n 3 =72 45 Critical Value(s): Decision: Reject H 0 =. 05 0 5. 991 c 2 Conclusion:

2 Test for k Proportions Solution

2 • • • Test for k Proportions Solution H 0: p 1 = p 2 = p 3 = 1/3 Test Statistic: 2 = 6. 3 Ha: At least 1 is different =. 05 n 1 = 63 n 2 = n 3 =72 45 Critical Value(s): Decision: Reject H 0 Reject at =. 05 0 5. 991 c 2 Conclusion: There is evidence of a difference in proportions

Contingency Tables • Useful in situations involving multiple population proportions • Used to classify sample observations according to two or more characteristics • Also called a cross-classification table.

Contingency Table Example Left-Handed vs. Gender Dominant Hand: Left vs. Right Gender: Male vs. Female § 2 categories for each variable, so called a 2 x 2 table § Suppose we examine a sample of 300 children

Contingency Table Example (continued) Sample results organized in a contingency table: sample size = n = 300: 120 Females, 12 were left handed 180 Males, 24 were left handed Hand Preference Gender Left Right Female 12 108 120 Male 24 156 180 36 264 300

2 Test for the Difference Between Two Proportions H 0: π1 = π2 (Proportion of females who are left handed is equal to the proportion of males who are left handed) H 1: π1 ≠ π2 (The two proportions are not the same hand preference is not independent of gender) • If H 0 is true, then the proportion of left-handed females should be the same as the proportion of left-handed males • The two proportions above should be the same as the proportion of left-handed people overall

The Chi-Square Test Statistic The Chi-square test statistic is: • where: fo = observed frequency in a particular cell fe = expected frequency in a particular cell if H 0 is true (Assumed: each cell in the contingency table has expected frequency of at least 5)

Decision Rule The test statistic approximately follows a chisquared distribution with one degree of freedom Decision Rule: If , reject H 0, otherwise, do not reject H 0 0 Do not reject H 0 Reject H 0 2α 2

Computing the Average Proportion The average proportion is: 120 Females, 12 were left handed Here: 180 Males, 24 were left handed i. e. , of all the children the proportion of left handers is 0. 12, that is, 12%

Finding Expected Frequencies • To obtain the expected frequency for left handed females, multiply the average proportion left handed (p) by the total number of females • To obtain the expected frequency for left handed males, multiply the average proportion left handed (p) by the total number of males If the two proportions are equal, then P(Left Handed | Female) = P(Left Handed | Male) =. 12 i. e. , we would expect (. 12)(120) = 14. 4 females to be left handed (. 12)(180) = 21. 6 males to be left handed

Observed vs. Expected Frequencies Gender Hand Preference Left Right Female Observed = 12 Expected = 14. 4 Observed = 108 Expected = 105. 6 120 Male Observed = 24 Expected = 21. 6 Observed = 156 Expected = 158. 4 180 36 264 300

The Chi-Square Test Statistic Gender Female Male The test statistic is: Hand Preference Left Right Observed = 12 Observed = 108 Expected = 14. 4 Expected = 105. 6 Observed = 24 Observed = 156 Expected = 21. 6 Expected = 158. 4 36 264 120 180 300

Decision Rule: If > 3. 841, reject H 0, otherwise, do not reject H 0 Here, 0. 05 0 Do not reject H 0 Reject H 0 20. 05 = 3. 841 2 = 0. 7576< = 3. 841, so we do not reject H 0 and conclude that there is not sufficient evidence that the two proportions are different at = 0. 05

2 Test for Differences Among More Than Two Proportions • Extend the 2 test to the case with more than two independent populations: H 0: π1 = π2 = … = πc H 1: Not all of the πj are equal (j = 1, 2, …, c)

The Chi-Square Test Statistic The Chi-square test statistic is: • Where: fo = observed frequency in a particular cell of the 2 x c table fe = expected frequency in a particular cell if H 0 is true (Assumed: each cell in the contingency table has expected frequency of at least 1)

Computing the Overall Proportion The overall proportion is: • Expected cell frequencies for the c categories are calculated as in the 2 x 2 case, and the decision rule is the same: Decision Rule: If , reject H 0, otherwise, do not reject H 0 Where is from the chisquared distribution with c – 1 degrees of freedom

The Marascuilo Procedure • Used when the null hypothesis of equal proportions is rejected • Enables you to make comparisons between all pairs • Start with the observed differences, pj – pj’, for all pairs (for j ≠ j’). . . • . . . then compare the absolute difference to a calculated critical range

2 Test of Independence

Hypothesis Tests Qualitative Data 1 pop. Proportion More than 2 pop. Independence 2 pop. Z Test c 2 Test

2 Test of Independence • Shows if a relationship exists between two qualitative variables – One sample is drawn – Does not show causality • Uses two-way contingency table

2 Test of Independence Contingency Table Shows number of observations from 1 sample jointly in 2 qualitative variables Levels of variable 2 Levels of variable 1

Conditions Required for a 2 Test: Independence Valid 1. Multinomial experiment has been conducted 2. The sample size, n, is large: Eij is greater than or equal to 5 for every cell

2 Test of Independence Hypotheses & Statistic 1. Hypotheses • H 0: Variables are independent • Ha: Variables are related (dependent) 2. Test Statistic Observed count Expected count 3. Degrees of Freedom: (r – 1)(c – 1) Rows Columns

2 Test of Independence Expected Counts 1. Statistical independence means joint probability equals product of marginal probabilities 2. Compute marginal probabilities and multiply for joint probability 3. Expected count is sample size times joint probability

Expected Count Example Marginal probability = 112 160 House Style Location Urban Rural Obs. Total Split–Level 63 49 112 Ranch 15 33 48 Total 78 82 160

Expected Count Example Marginal probability = 112 160 House Style Location Urban Rural Obs. Total Split–Level 63 49 112 Ranch 15 33 48 Total 78 82 160 Marginal probability = 78 160

Expected Count Example Joint probability = House Style 112 78 160 Marginal probability = 112 160 Location Urban Rural Obs. Total Split–Level 63 49 112 Ranch 15 33 48 Total 78 82 160 Marginal probability = 78 160 112 78 Expected count = 160· 160 = 54. 6

Expected Count Calculation 112· 78 160 House Style House Location Urban Rural Obs. Exp. 112· 82 160 Total Split-Level 63 54. 6 49 57. 4 112 Ranch 15 23. 4 33 24. 6 48 Total 78 78 82 82 48· 78 160 48· 82 160

2 Test of Independence Example As a realtor you want to determine if house style and house location are related. At the. 05 level of significance, is there evidence of a relationship?

2 Test of Independence Solution • • • H 0: No Relationship Ha: Relationship =. 05 df = (2 - 1) = 1 Critical Value(s): Test Statistic: Decision: Reject H 0 =. 05 0 3. 841 c 2 Conclusion:

2 Test of Independence Solution Eij 5 in all cells 112· 78 160 112· 82 160 48· 78 160 48· 82 160

2 Test of Independence Solution

2 Test of Independence Solution • • • H 0: No Relationship Ha: Relationship =. 05 df = (2 - 1) = 1 Critical Value(s): Reject H 0 =. 05 0 3. 841 c 2 Test Statistic: 2 = 8. 41 Decision: Reject at =. 05 Conclusion: There is evidence of a relationship

2 Test of Independence Thinking Challenge You’re a marketing research analyst. You ask a random sample of 286 consumers if they purchase Diet Pepsi or Diet Coke. At the. 05 level of significance, is there evidence of a relationship? Diet Coke No Yes Total Diet Pepsi No Yes 84 32 48 122 132 154 Total 116 170 286

2 Test of Independence Solution* • • • H 0: No Relationship Ha: Relationship =. 05 df = (2 - 1) = 1 Critical Value(s): Test Statistic: Decision: Reject H 0 =. 05 0 3. 841 c 2 Conclusion:

2 Test of Independence Solution* Eij 5 in all cells 116· 132 286 154· 132 286 170· 154 286

2 Test of Independence Solution*

2 Test of Independence Solution* • • • H 0: No Relationship Ha: Relationship =. 05 df = (2 - 1) = 1 Critical Value(s): Reject H 0 =. 05 0 3. 841 c 2 Test Statistic: 2 = 54. 29 Decision: Reject at =. 05 Conclusion: There is evidence of a relationship

2 Test of Independence Thinking Challenge 2 There is a statistically significant relationship between purchasing Diet Coke and Diet Pepsi. So what do you think the relationship is? Aren’t they competitors? Diet Coke No Yes Total Diet Pepsi No Yes 84 32 48 122 132 154 Total 116 170 286

You Re-Analyze the Data High Income Diet Coke No Yes Total Diet Pepsi No Yes 4 30 40 2 44 32 Total 34 42 76 Diet Pepsi No Yes 80 2 8 120 88 122 Total 82 128 210 Low Income Diet Coke No Yes Total

True Relationships* Diet Coke Underlying causal relation Control or intervening variable (true cause) Apparent relation Diet Pepsi

Moral of the Story* Numbers don’t think - People do! © 1984 -1994 T/Maker

Conclusion 1. Explained 2 Test for Proportions 2. Explained 2 Test of Independence 3. Solved Hypothesis Testing Problems • • More Than Two Population Proportions Independence