# Contingency Tables Case st571-1/06-tables-2.pdfStacked Bar Graph Frequency 0 10 20 30 40 50...

date post

10-Mar-2018Category

## Documents

view

214download

2

Embed Size (px)

### Transcript of Contingency Tables Case st571-1/06-tables-2.pdfStacked Bar Graph Frequency 0 10 20 30 40 50...

Contingency Tables

Bret Hanlon and Bret Larget

Department of StatisticsUniversity of WisconsinMadison

October 46, 2011

Contingency Tables 1 / 56

Case Study

Case Study

Example 9.3 beginning on page 213 of the text describes an experimentin which fish are placed in a large tank for a period of time and someare eaten by large birds of prey. The fish are categorized by their level ofparasitic infection, either uninfected, lightly infected, or highly infected. Itis to the parasites advantage to be in a fish that is eaten, as this providesan opportunity to infect the bird in the parasites next stage of life. Theobserved proportions of fish eaten are quite different among the categories.

Uninfected Lightly Infected Highly Infected Total

Eaten 1 10 37 48Not eaten 49 35 9 93

Total 50 45 46 141

The proportions of eaten fish are, respectively, 1/50 = 0.02, 10/45 = 0.222,and 37/46 = 0.804.

Contingency Tables Case Study Infected Fish and Predation 2 / 56

Stacked Bar Graph

Fre

quen

cy

0

10

20

30

40

50

Uninfected Lightly Infected Highly Infected

Eaten Not eaten

Contingency Tables Case Study Graphics 3 / 56

Graphing Tabled Counts

A stacked bar graph shows:I the sample sizes in each sample; andI the number of observations of each type within each sample.

This plot makes it easy to compare sample sizes among samples andcounts within samples, but the comparison of estimates of conditionalprobabilities among samples is less clear.

Contingency Tables Case Study Graphics 4 / 56

Mosaic Plot

Rel

ativ

e F

requ

ency

0.0

0.2

0.4

0.6

0.8

1.0

Uninfected Lightly Infected Highly Infected

Eaten Not eaten

Contingency Tables Case Study Graphics 5 / 56

Mosaic Plot

A mosaic plot replaces absolute frequencies (counts) with relativefrequencies within each sample.

This plot makes comparisons of estimated conditional probabilitiesvery clear.

The cost is that the sample size information is lost.

Contingency Tables Case Study Graphics 6 / 56

Estimating Differences between Proportions

In the setting of the experiment, we observe a difference between theproportions of eaten fish in the lightly and highly infected fish.

A point estimate of this difference is

37

46 10

45= 0.804 0.222 = 0.582

How can we quantify uncertainty in this estimate?

Contingency Tables Case Study Estimation 7 / 56

Confidence Intervals

A confidence interval for a difference in proportions p1 p2 is basedon the sampling distribution of the difference in sample proportions.

If the two samples are independent,

E(p1 p2) = p1 p2

Var(p1 p2) =p1(1 p1)

n1+

p2(1 p2)n2

If both samples are large enough (depending on how close theproportions are to 0 or 1), this sampling distribution is approximatelynormal.

Contingency Tables Case Study Estimation 8 / 56

Confidence Interval

95%Confidence Interval for p1 p2A 95% confidence interval for p1 p2 is

p1 p2 1.96SE(

p1 p2)< p1 p2 < p1 p2 + 1.96SE

(p1 p2

)where pi = xi/ni for i = 1, 2 and

SE(

p1 p2)

=

p1(1 p1)

n1+

p2(1 p2)n2

This formula will be more accurate for large n1 and n2.

A rough rule of thumb is that each sample should have at least fiveobservations of each type.

Maybe this method can be improved by adding fake observations likethe one sample case?

Contingency Tables Case Study Estimation 9 / 56

Application

For the infected fish case study, a confidence interval for thedifference in probabilities of being eaten between highly and lightlyinfected fish is

0.415 < phigh plight < 0.749

(Show calculations on the board.)

In the settings of the experiment, we are 95% confident that theprobability a highly infected fish is eaten is greater than thecorresponding probability for a lightly infected fish by an amountbetween 0.415 and 0.749.

Contingency Tables Case Study Estimation 10 / 56

Odds Ratios

Odds ratios are an alternative way to think about probabilities.

Definition

The odds in favor of an event with probability p are p/(1 p).The odds ratio in favor of an event between two groups is the odds infavor for the first group divided by the odds in favor for the secondgroup.

odds ratio =p1/(1 p1)p2/(1 p2)

Odds ratios are estimated by plugging in sample proportions.

Contingency Tables Case Study Odds Ratios 11 / 56

Sampling Distribution of the Odds Ratio

We explore the sampling distribution of the odds ratio when n1 = 46,p1 = 0.8, n2 = 45, and p2 = 0.22 which are estimates from the casestudy.

We simulate 100,000 odds ratios from independent samples and graphthe results.

Contingency Tables Case Study Odds Ratios 12 / 56

Graph of Odds Ratio

odds

Per

cent

of T

otal

0

10

20

30

40

50

0 200 400 600

Contingency Tables Case Study Odds Ratios 13 / 56

Graph of Log of Odds Ratio

log(odds)

Per

cent

of T

otal

0

5

10

15

1 2 3 4 5 6

Contingency Tables Case Study Odds Ratios 14 / 56

Comparison

The sampling distribution of the odds ratio is very skewed to the right.

The sampling distribution of the log odds ratio is fairly symmetric andbell-shaped.

We will use the normal approximation for the log odds ratio and thentranslate back.

The standard error of the odds ratio can be estimated as

SE(

ln(odds ratio))

=

1

x1+

1

n1 x1+

1

x2+

1

n2 x2

Contingency Tables Case Study Odds Ratios 15 / 56

Confidence Interval for Odds Ratio

95%Confidence Interval forp1/(1 p1)p2/(1 p2)

A 95% confidence interval for the odds ratio is

exp(

ln OR 1.96SE) fisher.test(x, alternative = "greater")

Fisher's Exact Test for Count Data

data: xp-value < 2.2e-16alternative hypothesis: true odds ratio is greater than 195 percent confidence interval:35.49817 Infsample estimates:odds ratio108.3894

Contingency Tables Hypothesis Testing Fishers Exact Test 46 / 56

What you should know

You should know:

how to find a confidence interval for a difference in proportions;

how to find a confidence interval for an odds ratio;

how to test for independence in contingency tables using:I the 2 test of independence;I the G-test;I Fishers exact test

how to determine which tests are appropriate in which situations.

Contingency Tables Hypothesis Testing Fishers Exact Test 47 / 56

R Details

The R function chisq.test() can be used to automate calculationsfor the 2 test of independence.

The R function fisher.test() can be used to automate calculationsfor Fishers exact test.

There is no built-in function in R for the G-test, but the file gtest.Rcontains code for a function g.test() for this purpose. Source thiscode into R before use (see the file for details).

There is also a file mosaic.R with code for producing mosaic plots.

Contingency Tables Hypothesis Testing Fishers Exact Test 48 / 56

Appendix

The remaining slides contain details about the G-test that willnot be included for homework or exam.

Contingency Tables Appendix 49 / 56

Likelihood Ratio Tests

Definition

In a likelihood ratio test, the null hypothesis assumes a likelihood modelwith k0 free parameters which is a special case of the alternativehypothesis likelihood model with k1 free parameters. The two likelihoodmodels are maximized with likelihoods L0 and L1 respectively. The teststatistic is G = 2(ln L1 ln L0) which, for large enough samples, hasapproximately a 2(k1 k0) distribution when the null hypothesis is true.

Contingency Tables Appendix 50 / 56

The Multinomial Distribution

Definition

The multinomial distribution is a generalization of the binomial distributionwhere there is an independent sample of size n and each outcome is in oneof k categories with probabilities pi for the ith category (

ki=1 pi = 1).

The probability that there are xi outcomes of type i for i = 1, 2, . . . , k is(n

x1, . . . , xk

)(px11 p

xkk

)where (

n

x1, . . . , xk

)=

n!

x1! xk !is called a multinomial coefficient and x1 + + xk = n.

Contingency Tables Appendix 51 / 56

Likelihood for Multinomial Distribution

In the binomial distribution, we can rethink the parameters for fixed nwith probabilities p1 and p2 for the two categories with p1 + p2 = 1, sothere is only one free parameter. (If you know p1, you also know p2.)

The maximum likelihood estimates are p1 = x1/n andp2 = x2/n = 1 p1 = (n x1)/n.For more categories, the maximum likelihood estimates are pi = xi/nfor i = 1, . . . , k.

The maximum likelihood is then

L =

(n

x1, . . . , xk

)((x1n

)x1 (xkn

)xk)and the maximum log-likelihood is

ln L = ln

(n

x1, . . . , xk

)+

ki=1

xi ln(xi

n

)Contingency Tables Appendix 52 / 56

Contingency Tables

The observed outcomes {Oij} in a contingency table with r rows andc columns are jointly are modeled with a multinomial distributionwith parameters {pij} for i = 1, . . . , r and j = 1, . . . , c.There are rc probabilities.

Contingency Tables Appendix 53 / 56

Contingency Tables: Alternative Model

Under the alternative hypothesis

Recommended

*View more*