Power 14 Goodness of Fit & Contingency Tables

11

Power 14Goodness of Fit

& Contingency Tables

22

Outline

I. Parting Shots On the Linear Probability I. Parting Shots On the Linear Probability ModelModel

II. Goodness of Fit & Chi SquareII. Goodness of Fit & Chi Square III.Contingency TablesIII.Contingency Tables

33

The Vision Thing

Discriminating BetweenTwo PopulationsDiscriminating BetweenTwo Populations Decision Theory and the Regression LineDecision Theory and the Regression Line

44

income

education

x = a, x2 > y

2

y = b

x, y > 0

mean income non

Meaneduc.non

MeanEduc

Players

Mean income Players

Players

Non-players Discriminatingline

55

Expected Costs of Misclassification

E CE CMCMC = C(n/p)*P(n/p)*P(p) + = C(n/p)*P(n/p)*P(p) +

C(p/n)*P(n/p)*P(p)C(p/n)*P(n/p)*P(p) where P(n) = 23/100where P(n) = 23/100 Suppose C(n/p) = C(p/n)Suppose C(n/p) = C(p/n) then E Cthen E CMC MC = C*P(n/p)*3/4 + C*P(p/n)*1/4 = C*P(n/p)*3/4 + C*P(p/n)*1/4

And the two costs of misclassification will And the two costs of misclassification will be balanced if P(p/n) =3/4 = Bernbe balanced if P(p/n) =3/4 = Bern

66

The Regression Line-Discriminant Function

Bern = 3/4Bern = 3/4 Bern = c + bBern = c + b1 1 *educ + b*educ + b2 2 *income*income

Bern = 3/4 = 1.39 - 0.0216*educ -0.0105* Bern = 3/4 = 1.39 - 0.0216*educ -0.0105* income, or income, or

0.0216*educ =0.64 - 0.0105*income0.0216*educ =0.64 - 0.0105*income Educ = 29.63 - 0.486*income, Educ = 29.63 - 0.486*income, the regression linethe regression line

77

Lottery: Players and Non-Players Vs. Education & Income

0

5

10

15

20

25

0 10 20 30 40 50 60 70 80 90 100

Income ($000)

Ed

uca

tio

n (

Yea

rs)

Discriminant Function or Decision Rule:Bern = ¾ = 1.39 – 0.0216*education – 0.0105*income

Legend: Non-Players Players

Mean- NonplayersMean- NonplayersMean-PlayersMean-Players

88

II. Goodness of Fit & Chi Square

Rolling a Fair DieRolling a Fair Die The Multinomial DistributionThe Multinomial Distribution Experiment: 600 TossesExperiment: 600 Tosses

99

Outcome Probability Expected Frequency1 1/6 1002 1/6 1003 1/6 1004 1/6 1005 1/6 1006 1/6 100

The Expected Frequencies The Expected Frequencies

1010

Outcome Expected Frequencies Expected Frequency1 100 1142 100 943 100 844 100 1015 100 1076 100 107

The Expected Frequencies & Empirical FrequenciesThe Expected Frequencies & Empirical Frequencies

Empirical FrequencyEmpirical Frequency

1111

Hypothesis Test

Null HNull H00: Distribution is Multinomial: Distribution is Multinomial

Statistic: (OStatistic: (Oii - E - Eii))22/E/Ei, i, : observed minus : observed minus

expected squared divided by expectedexpected squared divided by expected Set Type I Error @ 5% for exampleSet Type I Error @ 5% for example Distribution of Statistic is Chi SquareDistribution of Statistic is Chi Square

P(nP(n1 1 =1, n=1, n2 2 =0, nn3 3 =0, n =0, n4 4 =0, n=0, n5 5 =0, n=0, n6 6 =0) = n!/=0) = n!/

n

j

jnn

j

jpjn1

)(

1

)]([])(

P(nP(n1 1 =1, n=1, n2 2 =0, nn3 3 =0, n =0, n4 4 =0, n=0, n5 5 =0, n=0, n6 6 =0)= 1!/1!0!0!0!0!0!(1/6)=0)= 1!/1!0!0!0!0!0!(1/6)11(1/6)(1/6)00

(1/6)(1/6)0 0 (1/6)(1/6)0 0 (1/6)(1/6)0 0 (1/6)(1/6)00

One Throw, side one comes up: multinomial distributionOne Throw, side one comes up: multinomial distribution

1212

Outcome Expected Observed Oi - E i (Oi - E i)2

1 100 114 -14 196/1002 100 92 8 64/1003 100 84 16 256/1004 100 101 -1 1/1005 100 107 -7 49/1006 100 107 -7 49/100

Sum = 6.15

1313

Outcome Expected Observed Oi - E i (Oi - E i)2

1 100 114 -14 196/1002 100 92 8 64/1003 100 84 16 256/1004 100 101 -1 1/1005 100 107 -7 49/1006 100 107 -7 49/100

Sum = 6.15

Chi Square: xChi Square: x22 = = (O (Oii - E - Eii))2 2 = 6.15 = 6.15

0.00

0.05

0.10

0.15

0.20

0 5 10 15

CHI

DE

NS

ITY

Chi Square Density for 5 degrees of freedomChi Square Density for 5 degrees of freedom

11.0711.07

5 %5 %

1515

Contingency Table Analysis

Tests for Association Vs. Independence For Tests for Association Vs. Independence For Qualitative VariablesQualitative Variables

1616

Purchase Consumer Inform Cons. Not Inform . TotalsFrost FreeNot Frost FreeTotals

Does Consumer Knowledge Affect Purchases?Does Consumer Knowledge Affect Purchases?

Frost Free Refrigerators Use More ElectricityFrost Free Refrigerators Use More Electricity

1717

Purchase Consumer Inform Cons. Not Inform . TotalsFrost Free 432Not Frost Free 288Totals 540 180 720

Marginal CountsMarginal Counts

1818

Purchase Consumer Inform Cons. Not Inform . TotalsFrost Free 0.6Not Frost Free 0.4Totals 0.75 0.25 1

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

1919

Purchase Consumer Inform Cons. Not Inform . TotalsFrost Free 0.45 0.15 0.6Not Frost Free 0.3 0.1 0.4Totals 0.75 0.25 1

Joint Disribution Under IndependenceJoint Disribution Under Independencef(x,y) = f(x)*f(y)f(x,y) = f(x)*f(y)

2020

Purchase Consumer Inform Cons. Not Inform . TotalsFrost Free 324 108 432Not Frost Free 216 72 288Totals 540 180 720

Expected Cell Frequencies Under IndependenceExpected Cell Frequencies Under Independence

2121

Purchase Consumer Inform Cons. Not Inform . TotalsFrost Free 314 118Not Frost Free 226 62Totals

Observed Cell CountsObserved Cell Counts

2222

Purchase Consumer Inform Cons. Not Inform . TotalsFrost Free 0.31 0.93Not Frost Free 0.46 1.39Totals

Contribution to Chi Square: (observed-Expected)Contribution to Chi Square: (observed-Expected)22/Expected/Expected

Chi Sqare = 0.31 + 0.93 + 0.46 +1.39 = 3.09Chi Sqare = 0.31 + 0.93 + 0.46 +1.39 = 3.09(m-1)*(n-1) = 1*1=1 degrees of freedom (m-1)*(n-1) = 1*1=1 degrees of freedom

Upper Left Cell: (314-324)Upper Left Cell: (314-324)22/324 = 100/324 =0.31/324 = 100/324 =0.31

0.0

0.2

0.4

0.6

0.8

1.0

0 2 4 6 8 10 12 14

Chi-Square Variable

Figure 4: Chi-Square Density, One Degree of Freedom

Density

5%5%

5.025.02

2424

Using Goodness of Fit to Choose Between Competing

Proabaility Models Men on base when a home run is hitMen on base when a home run is hit

2525

Men on base when a home run is hit

# 0 1 2 3 Sum

Observed 421 227 96 21 765

Fraction 0.550 0.298 0.125 0.027 1

2626

Conjecture

Distribution is binomialDistribution is binomial

2727

Average # of men on base# 0 1 2 3

fraction 0550 0.298 0.125 0.027

product 0 0.298 0.250 0.081

Sum of products = n*p = 0.298+0.250+0.081 = 0.63Sum of products = n*p = 0.298+0.250+0.081 = 0.63

21.03/63.0/ˆˆ npnp

2828

Using the binomialk=men on base, n=# of trials

P(k=0) = [3!/0!3!] (0.21)P(k=0) = [3!/0!3!] (0.21)00(0.79)(0.79)33 = 0.493 = 0.493 P(k=1) = [3!/1!2!] (0.21)P(k=1) = [3!/1!2!] (0.21)11(0.79)(0.79)22 = 0.393 = 0.393 P(k=2) = [3!/2!1!] (0.21)P(k=2) = [3!/2!1!] (0.21)22(0.79)(0.79)11 = 0.105 = 0.105 P(k=3) = [3!/3!0!] (0.21)P(k=3) = [3!/3!0!] (0.21)33(0.79)(0.79)00 = 0.009 = 0.009

2929

Goodness of Fit# 0 1 2 3 Sum

Observed 421 227 96 21 765

binomial 377.1 300.6 80.3 6.9 764.4

(Oj – Ej) 43.9 -73.6 15.7 14.1

(Oj–Ej)2/Ej 5.1 18.0 2.6 28.8 54.5

0.00

0.05

0.10

0.15

0.20

0.25

0 5 10 15 20

CHI

DE

NS

ITY

Chi Square, 3 degrees of freedomChi Square, 3 degrees of freedom

5%5%

7.817.81

3131

Conjecture: Poisson where np = 0.63

P(k=3) = 1- P(k=2)-P(k=1)-P(k=0)P(k=3) = 1- P(k=2)-P(k=1)-P(k=0) P(k=0) = eP(k=0) = e--k k /k! = e/k! = e-0.63 -0.63 (0.63)(0.63)00/0! = 0.5326/0! = 0.5326 P(k=1) = eP(k=1) = e--k k /k! = e/k! = e-0.63 -0.63 (0.63)(0.63)11/1! = 0.3355/1! = 0.3355 P(k=2) = eP(k=2) = e--k k /k! = e/k! = e-0.63 -0.63 (0.63)(0.63)22/2! = 0.1057/2! = 0.1057

3232

Goodness of Fit# 0 1 2 3 Sum

Observed 421 227 96 21 765

Poisson 407.4 256.7 80.9 20.0 765

(Oj–Ej)2/Ej 0.454 3.44 2.82 0.05 6.76