1. 1 Power 14 Goodness of Fit & Contingency Tables

2. 2 Outline u I. Parting Shots On the Linear Probability Model u II. Goodness of Fit & Chi Square u III.Contingency Tables

3. 3 The Vision Thing u Discriminating BetweenTwo Populations u Decision Theory and the Regression Line

4. 4 income education  x = a,  x 2 >  y 2  y = b  x, y > 0 mean income non Mean educ. non Mean Educ Players Mean income Players Players Non-players Discriminating line

5. 5 Expected Costs of Misclassification u E C MC = C(n/p)*P(n/p)*P(p) + C(p/n)*P(n/p)*P(p)  where P(n) = 23/100  u Suppose C(n/p) = C(p/n) u then E C MC = C*P(n/p)*3/4 + C*P(p/n)*1/4 u And the two costs of misclassification will be balanced if P(p/n) =3/4 = Bern

6. 6 The Regression Line- Discriminant Function u Bern = 3/4 u Bern = c + b 1 *educ + b 2 *income u Bern = 3/4 = 1.39 - 0.0216*educ -0.0105* income, or u 0.0216*educ =0.64 - 0.0105*income u Educ = 29.63 - 0.486*income, u the regression line

7. 7 Lottery: Players and Non-Players Vs. Education & Income 0 5 10 15 20 25 0102030405060708090100 Income ($000) Education (Years) Discriminant Function or Decision Rule: Bern = ¾ = 1.39 – 0.0216*education – 0.0105*income Legend: Non-Players Players Mean- Nonplayers Mean-Players

8. 8 II. Goodness of Fit & Chi Square u Rolling a Fair Die u The Multinomial Distribution u Experiment: 600 Tosses

9. 9 The Expected Frequencies

10. 10 The Expected Frequencies & Empirical Frequencies Empirical Frequency

11. 11 Hypothesis Test u Null H 0 : Distribution is Multinomial u Statistic: (O i - E i ) 2 /E i, : observed minus expected squared divided by expected u Set Type I Error @ 5% for example u Distribution of Statistic is Chi Square P(n 1 =1, n 2 n 3 =0, n 4 =0, n 5 =0, n 6 =0) = n!/ P(n 1 =1, n 2 =0, n 3 =0, n 4 =0, n 5 =0, n 6 =0) = n!/ P(n 1 =1, n 2 n 3 =0, n 4 =0, n 5 =0, n 6 =0)= 1!/1!0!0!0!0!0!(1/6) 1 (1/6) 0 P(n 1 =1, n 2 =0, n 3 =0, n 4 =0, n 5 =0, n 6 =0)= 1!/1!0!0!0!0!0!(1/6) 1 (1/6) 0 (1/6) 0 (1/6) 0 (1/6) 0 (1/6) 0 One Throw, side one comes up: multinomial distribution

12. 12

13. 13 Chi Square: x 2 =  (O i - E i ) 2 = 6.15

14. Chi Square Density for 5 degrees of freedom 11.07 5 %

15. 15 Contingency Table Analysis u Tests for Association Vs. Independence For Qualitative Variables

16. 16 Does Consumer Knowledge Affect Purchases? Frost Free Refrigerators Use More Electricity

17. 17 Marginal Counts

18. 18 Marginal Distributions, f(x) & f(y)

19. 19 Joint Disribution Under Independence f(x,y) = f(x)*f(y)

20. 20 Expected Cell Frequencies Under Independence

21. 21 Observed Cell Counts

22. 22 Contribution to Chi Square: (observed-Expected) 2 /Expected Chi Sqare = 0.31 + 0.93 + 0.46 +1.39 = 3.09 (m-1)*(n-1) = 1*1=1 degrees of freedom Upper Left Cell: (314-324) 2 /324 = 100/324 =0.31

23. 5% 5.02

24. 24 Using Goodness of Fit to Choose Between Competing Proabaility Models u Men on base when a home run is hit

25. 25 Men on base when a home run is hit

26. 26 Conjecture u Distribution is binomial

27. 27 Average # of men on base Sum of products = n*p = 0.298+0.250+0.081 = 0.63

28. 28 Using the binomial k=men on base, n=# of trials u P(k=0) = [3!/0!3!] (0.21) 0 (0.79) 3 = 0.493 u P(k=1) = [3!/1!2!] (0.21) 1 (0.79) 2 = 0.393 u P(k=2) = [3!/2!1!] (0.21) 2 (0.79) 1 = 0.105 u P(k=3) = [3!/3!0!] (0.21) 3 (0.79) 0 = 0.009

29. 29 Goodness of Fit

30. Chi Square, 3 degrees of freedom 5% 7.81

31. 31 Conjecture: Poisson where  np = 0.63 u P(k=3) = 1- P(k=2)-P(k=1)-P(k=0)  P(k=0) = e -   k /k! = e -0.63 (0.63) 0 /0! = 0.5326  P(k=1) = e -   k /k! = e -0.63 (0.63) 1 /1! = 0.3355  P(k=2) = e -   k /k! = e -0.63 (0.63) 2 /2! = 0.1057

32. 32 Goodness of Fit