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Chapter 10

Chapter 10. Hypothesis testing: Categorical Data Analysis. Learning Objectives. Comparison of binomial proportion using Z and  2 Test. Explain  2 Test for Independence of 2 variables Explain The Fisher’s test for independence McNemar’s tests for correlated data Kappa Statistic

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Chapter 10

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  1. Chapter 10 Hypothesis testing: Categorical Data Analysis EPI809/Spring 2008

  2. Learning Objectives • Comparison of binomial proportion using Z and 2 Test. • Explain 2 Test for Independence of 2 variables • Explain The Fisher’s test for independence • McNemar’s tests for correlated data • Kappa Statistic • Use of SAS Proc FREQ EPI809/Spring 2008

  3. Data Types EPI809/Spring 2008

  4. Qualitative Data • Qualitative Random Variables Yield Responses That Can Be Put In Categories. Example: Gender (Male, Female) • Measurement or Count Reflect # in Category • Nominal (no order) or Ordinal Scale (order) • Data can be collected as continuous but recoded to categorical data. Example (Systolic Blood Pressure - Hypotension, Normal tension, hypertension ) EPI809/Spring 2008

  5. Hypothesis Tests Qualitative Data EPI809/Spring 2008

  6. Z Test for Differences in Two Proportions EPI809/Spring 2008

  7. Hypotheses for Two Proportions EPI809/Spring 2008

  8. Hypotheses for Two Proportions EPI809/Spring 2008

  9. Hypotheses for Two Proportions EPI809/Spring 2008

  10. Hypotheses for Two Proportions EPI809/Spring 2008

  11. Hypotheses for Two Proportions EPI809/Spring 2008

  12. Hypotheses for Two Proportions EPI809/Spring 2008

  13. Z Test for Difference in Two Proportions 1. Assumptions • Populations Are Independent • Populations Follow Binomial Distribution • Normal Approximation Can Be Used for large samples (All Expected Counts  5) • Z-Test Statistic for Two Proportions EPI809/Spring 2008

  14. Sample Distribution for Difference Between Proportions EPI809/Spring 2008

  15. Z Test for Two Proportions Thinking Challenge MA • You’re an epidemiologist for the US Department of Health and Human Services. You’re studying the prevalence of disease X in two states (MA and CA). In MA, 74 of 1500people surveyed were diseased and in CA, 129 of 1500 were diseased. At .05 level, does MA have a lower prevalence rate? CA EPI809/Spring 2008

  16. Z Test for Two Proportions Solution* EPI809/Spring 2008

  17. Z Test for Two Proportions Solution* H0: Ha:  = nMA = nCA= Critical Value(s): Test Statistic: Decision: Conclusion: EPI809/Spring 2008

  18. Z Test for Two Proportions Solution* H0: pMA - pCA = 0 Ha: pMA - pCA < 0 = nMA = nCA= Critical Value(s): Test Statistic: Decision: Conclusion: EPI809/Spring 2008

  19. Z Test for Two Proportions Solution* H0: pMA - pCA = 0 Ha: pMA - pCA < 0 = .05 nMA= 1500 nCA= 1500 Critical Value(s): Test Statistic: Decision: Conclusion: EPI809/Spring 2008

  20. Z Test for Two Proportions Solution* H0: pMA - pCA = 0 Ha: pMA - pCA < 0 = .05 nMA= 1500 nCA= 1500 Critical Value(s): Test Statistic: Decision: Conclusion: EPI809/Spring 2008

  21. Z Test for Two Proportions Solution* EPI809/Spring 2008

  22. Z Test for Two Proportions Solution* H0: pMA - pCA = 0 Ha: pMA - pCA < 0 = .05 nMA= 1500 nCA= 1500 Critical Value(s): Test Statistic: Decision: Conclusion: Z = -4.00 EPI809/Spring 2008

  23. Z Test for Two Proportions Solution* H0: pMA - pCA = 0 Ha: pMA - pCA < 0 = .05 nMA= 1500 nCA= 1500 Critical Value(s): Test Statistic: Decision: Conclusion: Z = -4.00 Reject at  = .05 EPI809/Spring 2008

  24. Z Test for Two Proportions Solution* H0: pMA - pCA = 0 Ha: pMA - pCA < 0 = .05 nMA= 1500 nCA= 1500 Critical Value(s): Test Statistic: Decision: Conclusion: Z = -4.00 Reject at  = .05 There is evidence MA is less than CA EPI809/Spring 2008

  25. 2 Test of Independence Between 2 Categorical Variables EPI809/Spring 2008

  26. Hypothesis Tests Qualitative Data EPI809/Spring 2008

  27. 2 Test of Independence 1. Shows If a Relationship Exists Between 2 Qualitative Variables, but does Not Show Causality 2. Assumptions Multinomial Experiment All Expected Counts  5 3. Uses Two-Way Contingency Table EPI809/Spring 2008

  28. 2 Test of Independence Contingency Table • 1. Shows # Observations From 1 Sample Jointly in 2 Qualitative Variables EPI809/Spring 2008

  29. 2 Test of Independence Contingency Table 1. Shows # Observations From 1 Sample Jointly in 2 Qualitative Variables Levels of variable 2 Levels of variable 1 EPI809/Spring 2008

  30. 2 Test of Independence Hypotheses & Statistic 1. Hypotheses • H0: Variables Are Independent • Ha: Variables Are Related (Dependent) EPI809/Spring 2008

  31. 2 Test of Independence Hypotheses & Statistic 1. Hypotheses H0: Variables Are Independent Ha: Variables Are Related (Dependent) 2. Test Statistic Observed count Expected count EPI809/Spring 2008

  32. 2 Test of Independence Hypotheses & Statistic 1. Hypotheses H0: Variables Are Independent Ha: Variables Are Related (Dependent) 2. Test Statistic Degrees of Freedom: (r - 1)(c - 1) Observed count Expected count Rows Columns EPI809/Spring 2008

  33. 2 Test of Independence Expected Counts 1. Statistical Independence Means Joint Probability Equals Product of Marginal Probabilities 2. Compute Marginal Probabilities & Multiply for Joint Probability 3. Expected Count Is Sample Size Times Joint Probability EPI809/Spring 2008

  34. Expected Count Example EPI809/Spring 2008

  35. Expected Count Example 112 160 Marginal probability = EPI809/Spring 2008

  36. Expected Count Example 112 160 Marginal probability = 78 160 Marginal probability = EPI809/Spring 2008

  37. 112 160 78 160 Expected Count Example 112 160 Joint probability = Marginal probability = 78 160 Marginal probability = EPI809/Spring 2008

  38. 112 160 78 160 112 160 78 160 Expected count = 160· Expected Count Example 112 160 Joint probability = Marginal probability = 78 160 Marginal probability = = 54.6 EPI809/Spring 2008

  39. Expected Count Calculation EPI809/Spring 2008

  40. Expected Count Calculation EPI809/Spring 2008

  41. Expected Count Calculation 112x78 160 112x82 160 48x78 160 48x82 160 EPI809/Spring 2008

  42. 2 Test of Independence Example on HIV • You randomly sample 286 sexually active individuals and collect information on their HIV status and History of STDs. At the .05 level, is there evidence of a relationship? EPI809/Spring 2008

  43. 2 Test of Independence Solution EPI809/Spring 2008

  44. 2 Test of Independence Solution H0: Ha:  = df = Critical Value(s): Test Statistic: Decision: Conclusion: EPI809/Spring 2008

  45. 2 Test of Independence Solution H0: No Relationship Ha: Relationship  = df = Critical Value(s): Test Statistic: Decision: Conclusion: EPI809/Spring 2008

  46. 2 Test of Independence Solution H0: No Relationship Ha: Relationship  = .05 df = (2 - 1)(2 - 1) = 1 Critical Value(s): Test Statistic: Decision: Conclusion: EPI809/Spring 2008

  47. 2 Test of Independence Solution H0: No Relationship Ha: Relationship  = .05 df = (2 - 1)(2 - 1) = 1 Critical Value(s): Test Statistic: Decision: Conclusion:  = .05 EPI809/Spring 2008

  48. 2 Test of Independence Solution  E(nij) 5 in all cells 116x132 286 154x116 286 170x132 286 170x154 286 EPI809/Spring 2008

  49. 2 Test of Independence Solution EPI809/Spring 2008

  50. 2 Test of Independence Solution H0: No Relationship Ha: Relationship  = .05 df = (2 - 1)(2 - 1) = 1 Critical Value(s): Test Statistic: Decision: Conclusion: 2 = 54.29  = .05 EPI809/Spring 2008

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