Contingency tables
Data from a semiconductor factory (Faraway p.)
- sample of chip wafers classified as - whether a particle was found on the die that produced the wafer - whether the wafer was good (satisfactory) or bad (defective)
Representations in R
## Loading required package: kableExtra| None | Some | |
|---|---|---|
| Good | 320 | 14 |
| Bad | 80 | 36 |
| Good | Bad | particles |
|---|---|---|
| 320 | 80 | None |
| 14 | 36 | Some |
| count | quality | particles |
|---|---|---|
| 320 | Good | None |
| 80 | Bad | None |
| 14 | Good | Some |
| 36 | Bad | Some |
Types of Analysis
Classical
##
## Pearson's Chi-squared test with Yates' continuity correction
##
## data: cts
## X-squared = 60.124, df = 1, p-value = 8.907e-15##
## Fisher's Exact Test for Count Data
##
## data: cts
## p-value = 2.955e-13
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
## 5.090628 21.544071
## sample estimates:
## odds ratio
## 10.21331Logistic regression
##
## Call:
## glm(formula = cbind(Good, Bad) ~ particles, family = binomial,
## data = binomial.data)
##
## Deviance Residuals:
## [1] 0 0
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 1.3863 0.1250 11.090 < 2e-16 ***
## particlesSome -2.3308 0.3389 -6.878 6.07e-12 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 54.03 on 1 degrees of freedom
## Residual deviance: 0.00 on 0 degrees of freedom
## AIC: 14.161
##
## Number of Fisher Scoring iterations: 3Log-linear representation
llfit <- glm(count ~ quality * particles, family=poisson, data=count.data)
summary(llfit)##
## Call:
## glm(formula = count ~ quality * particles, family = poisson,
## data = count.data)
##
## Deviance Residuals:
## [1] 0 0 0 0
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 5.7683 0.0559 103.187 < 2e-16 ***
## qualityBad -1.3863 0.1250 -11.090 < 2e-16 ***
## particlesSome -3.1293 0.2730 -11.461 < 2e-16 ***
## qualityBad:particlesSome 2.3308 0.3389 6.878 6.07e-12 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for poisson family taken to be 1)
##
## Null deviance: 4.7410e+02 on 3 degrees of freedom
## Residual deviance: 4.4409e-14 on 0 degrees of freedom
## AIC: 31.744
##
## Number of Fisher Scoring iterations: 3llfit <- glm(count ~ quality + particles, family=poisson, data=count.data)
summary(llfit)##
## Call:
## glm(formula = count ~ quality + particles, family = poisson,
## data = count.data)
##
## Deviance Residuals:
## 1 2 3 4
## 1.324 -2.370 -4.350 5.266
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 5.6934 0.0572 99.535 <2e-16 ***
## qualityBad -1.0575 0.1078 -9.813 <2e-16 ***
## particlesSome -2.0794 0.1500 -13.863 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for poisson family taken to be 1)
##
## Null deviance: 474.10 on 3 degrees of freedom
## Residual deviance: 54.03 on 1 degrees of freedom
## AIC: 83.774
##
## Number of Fisher Scoring iterations: 5cat("Deviance based chi-squared (1 d.f.) test: P=", 1-pchisq(llfit$deviance,1))## Deviance based chi-squared (1 d.f.) test: P= 1.973977e-13Multinomial data (two variables)
Snee (1974) presents data on 592 students cross-classified by hair and eye color.
## count eye hair
## 1 5 green BLACK
## 2 29 green BROWN
## 3 14 green RED
## 16 7 brown BLONDCross-tabulation; Row percents
## eye
## hair green hazel blue brown
## BLACK 5( 4.6%) 15(13.9%) 20(18.5%) 68(63.0%)
## BROWN 29(10.1%) 54(18.9%) 84(29.4%) 119(41.6%)
## RED 14(19.7%) 14(19.7%) 17(23.9%) 26(36.6%)
## BLOND 16(12.6%) 10( 7.9%) 94(74.0%) 7( 5.5%)Classical tests
##
## Pearson's Chi-squared test
##
## data: ct
## X-squared = 138.29, df = 9, p-value < 2.2e-16Fisher’s exact test based on multivariate hypergeometric is theoretically possible
Test for independence based on log-linear model
##
## Call: glm(formula = count ~ hair + eye, family = poisson, data = haireye)
##
## Coefficients:
## (Intercept) hairBROWN hairRED hairBLOND eyehazel
## 2.4575 0.9739 -0.4195 0.1621 0.3737
## eyeblue eyebrown
## 1.2118 1.2347
##
## Degrees of Freedom: 15 Total (i.e. Null); 9 Residual
## Null Deviance: 453.3
## Residual Deviance: 146.4 AIC: 241The residual deviance 146.4 on 9 degrees of freedom provides a test of independence between hair and eye colour, \(Pr (\chi^2_{(9)} \ge 146.4)\)= 0
Test for independence based on multinomial logit model
## # weights: 20 (12 variable)
## initial value 820.686262
## iter 10 value 681.465249
## iter 20 value 676.828633
## iter 20 value 676.828632
## iter 20 value 676.828632
## final value 676.828632
## converged## # weights: 8 (3 variable)
## initial value 820.686262
## final value 750.050421
## converged## Likelihood ratio tests of Multinomial Models
##
## Response: cbind(green, hazel, blue, brown)
## Model Resid. df Resid. Dev Test Df LR stat. Pr(Chi)
## 1 1 9 1500.101
## 2 hair 0 1353.657 1 vs 2 9 146.4436 0Examining the margins
In Stuart (1955), data on the vision of a sample of women is presented. The left and right eye performance is graded into four categories:
## left
## right best second third worst
## best 1520 266 124 66
## second 234 1512 432 78
## third 117 362 1772 205
## worst 36 82 179 492- Large counts on the diagonal of cross-tabulation
- Suggests strong association left and right eyes
chisq.test(ct)##
## Pearson's Chi-squared test
##
## data: ct
## X-squared = 8096.9, df = 9, p-value < 2.2e-16llfit <- glm(count ~ right+left, family=poisson, eyegrade)
xtabs(llfit$residuals ~ right + left, eyegrade)## left
## right best second third worst
## best 2.0160139 -0.5470210 -0.8128423 -0.7030465
## second -0.5933194 1.2552569 -0.4288927 -0.6926120
## third -0.8132184 -0.5040207 1.1518322 -0.2579095
## worst -0.8211034 -0.6502802 -0.3233734 4.5439523- \(\chi^2\) test rejects independence
- cross-tabulation of residuals confirms structure of dependence
Real question of interest: is there a systematic difference between left and right eyes.
- suggests we examine marginal tables
## Right eyes## best second third worst
## [1,] 1976(26.4%) 2256(30.2%) 2456(32.8%) 789(10.6%)## Left eyes## best second third worst
## [1,] 1907(25.5%) 2222(29.7%) 2507(33.5%) 841(11.2%)Appear similar, but can apply formal test
- but hypothesis of marginal homogeneity can’t be represented in log-linear parameterization
- “ad hoc” Stuart-Maxwell test
library(nonpar)
#stuart.maxwell(ct)
cat("Error in stuart.maxwell(ct) : Data must be a 3x3 matrix\n")## Error in stuart.maxwell(ct) : Data must be a 3x3 matrixstewart.test(ct)## $chisq
## [1] 11.95657
##
## $df
## [1] 3
##
## $p.value
## [1] 0.007533425Other ad hoc tests can be derived from use of contrasts - Are right eyes better or worse than left eyes
rightBetter <- with(eyegrade,sum(count[as.integer(right)>as.integer(left)]))
leftBetter <- with(eyegrade,sum(count[as.integer(right)<as.integer(left)]))We observer that 1010 girls have right eye better than left eye and 1171 vice versa.
ctr.coefs <- with(eyegrade, sign(as.integer(right) - as.integer(left)))
xtabs(ctr.coefs ~ right + left, eyegrade)## left
## right best second third worst
## best 0 -1 -1 -1
## second 1 0 -1 -1
## third 1 1 0 -1
## worst 1 1 1 0cntr <- sum(ctr.coefs*eyegrade$count)
var.cntr <- t(matrix(ctr.coefs))%*%diag(eyegrade$count)%*%ctr.coefs
cat("P-value=", 1-pchisq(cntr^2/var.cntr,1))## P-value= 0.000565904