f73db3 categorical data analysis workbook contents page preface aims summary...

F73DB3 CATEGORICAL DATA ANALYSIS

Workbook

Contents pagePreface

AimsSummaryContent/structure/syllabusplus other information

Background – computing (R)

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Examples

Single classifications (1-13)

Two-way classifications (14-27)

Three-way classifications (28-32)

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Example 1 Eye colours

Colour A B C D

Frequency observed 89 66 60 85

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Example 2 Prussian cavalry deaths

(a)Numbers killed in each unit in each year - frequency table

Number killed

0 1 2 3 4 5

Total

Frequency observed

144 91 32 11 2 0 280

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(b) Numbers killed in each unit in each year – raw data

0 0 1 0 0 2 0 0 0 0 . . . . . . . . . . . . . . . . . . . . . . . 00 0 2 0 1 0 1 2 0 1 . . . . . . . . . . . . . . . . . . . . . . . .0…..…..3 0 0 1 0 0 2 1 0 0 1 0 0 1 0 0 1 1 2 0 1 0 1 1

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(c) Total numbers killed each year

1875 ’76 ’77 ’78 ’79 ’80 ’81 ’82 ’83 ’84 ’85 ’86 ’87 ’88 ’89 ’90 ’91 ’92 ’93 ‘94

3 5 7 9 10 18 6 14 11 9 5 11 15 6 11 17 12 15 8 4

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Example 4 Political views

1 2 3 4 5 6 7 (very L) (centre) (very R)

Don’tKnow

Total

46 179 196 559 232 150 35 93 1490

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Example 7 Vehicle repair visits

Number of visits 0 1 2 3 4 5 6 Total

Frequency observed

295 190 53 5 5 2 0 550

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Example 15 Patients in clinical trial

Drug Placebo Total

Side-effects 15 4 19

No side-effects 35 46 81

Total 50 50 100

§1 INTRODUCTION

Data are counts/frequencies (not measurements)

Categories (explanatory variable)

Distribution in the cells (response)

Frequency distribution

Single classifications

Two-way classifications

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B: Cause of death

Cancer Other

A: Smoking status

Smoker 30 20

Not smoker 15 35

Illustration 1.1

Data may arise as

Bernoulli/binomial data (2 outcomes)

Multinomial data (more than 2 outcomes)

Poisson data

[+ Negative binomial data – the version with

range x = 0,1,2, …]

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§2 POISSON PROCESS AND

ASSOCIATED DISTRIBUTIONS

hwu2.1 Bernoulli trials and related distributions

Number of successes – binomial distribution

[Time before kth success – negative binomial distributionTime to first success – geometric distribution]

Conditional distribution of success times

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2.2 Poisson process and related distributions

time

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Poisson process with rate λ

Number of events in a time interval of length t, Nt , has a Poisson distribution with mean t

...,2,1,0,

! n

n

tenNP

nt

t

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Poisson process with rate λ

Inter-event time, T, has an exponential distribution with parameter (mean 1/)

, 0tf t e t

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given n events in time (0,t)

how many in time (0,s) (s < t)?

Conditional distribution of number of events

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given n events in time (0,t)

how many in time (0,s) (s < t)?

Conditional distribution of number of events

Answer Ns|Nt = n ~ B(n,s/t)

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Splitting into subprocesses

time

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0 10 20 30 40 50

02

04

06

08

01

00

t

N

Realisation of a Poisson process

# events

time

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X ~ Pn(), Y ~ Pn() X,Y independentthen we know X + Y ~ Pn( +)

Given X + Y = n, what is distribution of X?

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X ~ Pn(), Y ~ Pn() X,Y independentthen we know X + Y ~ Pn(+)

Given X + Y = n, what is distribution of X?

AnswerX|X+Y=n ~ B(n,p) where p = /( +)

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2.3 Inference for the Poisson distribution

ˆ /N r ˆ ˆ, . .E s e

r

ˆ , /N r

Ni , i = 1, 2, …, r, i. i. d. Pn(λ), N=ΣNi

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CI for

rz

ˆ

ˆ .

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2.4 Dispersion and LR tests for Poisson dataHomogeneity hypothesis

H0: the Ni s are i. i. d. Pn()

(for some unknown)

2

2 211

r

ii

r

N MX

M

Dispersion statistic

(M = sample mean)

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Likelihood ratio statistic

22

12 log /i i r

Y N N M

form for calculation – see p18 ◄◄

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§3 SINGLE CLASSIFICATIONS

Binary classifications

(a) N1 , N2 independent Poisson, with Ni ~ Pn(i)

or

(b) fixed sample size, N1 + N2 = n, with N1 ~ B(n,p1) where p1 = 1/(1 + 2)

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Qualitative categories

(a) N1 , N2, … , Nr independent Poisson, with

Ni ~ Pn(λi) or

(b) fixed sample size n, with joint multinomial distribution Mn(n;p)

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Testing goodness of fitH0: pi = i , i = 1,2, …, r

2

2

1

2observed frequency expected frequency

expected frequency

ri i

i i

all cells

N MX

M

This is the (Pearson) chi-square statistic

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2O E

E


expected frequency

The statistic often appears as

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21

21

It is distributed (approximately)

or when parameters have been estimated

in order to fit the model and calculate

the expected freqencies

r

r k k

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An alternative statistic is the LR statistic

2

1

2 21 1

2 log /

or

r

i i ii

r r k

Y N N M

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Sparse data/small expected frequencies

ensure mi 1 for all cells, and

mi 5 for at least about 80% of the cells

if not - combine adjacent cells sensibly

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Goodness-of-fit tests for frequency distributions

- very well-known application of the


expected frequencyall cells

statistic (see Illustration 3.4 p 22/23)

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Residuals (standardised)

0,1

1 /i i i i

i

i i i i

N n N Mr N

n M n M n

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Residuals (standardised)

0,1

1 /i i i i

i

i i i i

N n N Mr N

n M n M n

0,1i ii

i

N Mr N

M

simpler version

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Number of papers per author 1 2 3 4 5 6 7 8 9 10 11Number of authors 1062 263 120 50 22 7 6 2 0 1 1

MAJOR ILLUSTRATION 1Publish and be modelled

, 1, 2, 3,...!

x

P X x xx

Model

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MAJOR ILLUSTRATION 2Birds in hedges

Hedge type i A B C D E F GHedge length (m) li 2320 2460 2455 2805 2335 2645 2099Number of pairs ni 14 16 14 26 15 40 71

Model Ni ~ Pn(ili)

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Example 14 Numbers of mice bearing tumours in treated and control groups

Treated Control Total

Tumours 4 5 9

No tumours 12 74 86

Total 16 79 95

§4 TWO-WAY CLASSIFICATIONS

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Example 15 Patients in clinical trial

Drug Placebo Total



Total 50 50 100

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Patients in clinical trial – take 2

Drug Placebo Total



Total 50 50 100

4.1 Factors and responses

F × R tables

R × F , R × R

(F × F ?)

Qualitative, ordered, quantitative

Analysis the same - interpretation may be

different

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A two-way table is often called a

“contingency table”

(especially in R R case).

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Exposed Not exposed Total

Disease n11 n12 n1●

No disease n21 n22 n2●

Total n●1 n●2 n●● = n

Notation (2 2 case, easily extended)

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Three possibilities

One overall sample, each subject classified according to 2 attributes- this is R × R

Retrospective study

Prospective study (use of treated and control groups; drug and placebo etc)

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(a) R × R case

(a1) Nij ~ Pn(ij) , independent

or, with fixed table total

(a2) Condition on n = nij :

N|n ~ Mn(n ; p)

where N = {Nij} , p = {pij}.

4.2 Distribution theory and tests for r × s tables

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(b) F × R case

Condition on the observed marginal

totals n•j = nij for the s categories

of F ( condition on n and n•1)

s independent multinomials

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Usual hypotheses

(a1) Nij ~ Pn(ij) , independent

H0: variables/responses are independent

ij = i• •j / •• = ki•

(a2) Multinomial data (table total fixed)

H0: variables/responses are independent

P(row i and column j) = P(row i)P(column j)

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(b) Condition on n and n•j (fixed column totals)

Nij ~ Bi(n•j , pij) j = 1,2, …, s ; independent

H0: response is homogeneous (pij = pi• for all j)

i.e. response has the same distribution for all levels of the factor

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2

2

??

ij ij

ij

N m

m

where mij = ni• n•j /n as before

Tests of H0

The χ2 (Pearson) statistic:

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2

2

1 1

ij ij

r sij

N m

m

where mij = ni• n•j /n as before

Tests of H0

The χ2 (Pearson) statistic:

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OR: test based on the LR statistic Y2

Illustration: tonsils data – see p27

In R

Pearson/X2 : read data in using “matrix”then use “chisq.test”

LR Y2 : calculate it directly (or get it fromthe results of fitting a “log-linear model”- see later)

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Statistical tests

(a) Using Pearson’s χ2

4.3 The 2 2

table

Drug Placebo Total



Total 50 50 100

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2

2

1

ij ij

ij

N m

m

where mij = ni• n•j /n

row total × column total. .

grand totali e

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Yates (continuity) correction

Subtract 0.5 from |O – E| before squaring it

Performing the test in R

n.pat=matrix(c(15,35,4,46),2,2)chisq.test(n.pat)

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(b) Using deviance/LR statistic Y2

(c) Comparing binomial probabilities

(d) Fisher’s exact test

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Drug Placebo Total

Side-effects 15 4 N

19


Total 50 50 100

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Under a random allocation

50 50

4 15 50!50!19!81!( 4) 0.0039

100 4!46!15!35!100!

19

P N

:!

product of marginal factorialsNote probability

n product of cell factorials

one-sided P-value = P(N 4) = 0.0047

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In the 2 2 table, the H0 : independence

condition is equivalent to 1122 = 1221

Let λ = log(1122 /1221)

Then we have H0: λ = 0

λ is the “log odds ratio”

4.4 Log odds, combining and collapsing tables,interactions

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The “λ = 0” hypothesis is often called the “no association” hypothesis.

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The odds ratio is

1122 /1221

11 1 21 111 22 11 21

12 21 12 22 12 2 22 2

/ / //

/ / / /

n n n nn n n n

n n n n n n n n

1

2

( / )

odds on for column

odds on for column

odds ratio observed sample version

Sample equivalent is

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The odds ratio (or log odds ratio) provides a measure of association for the factors in the table.

no association odds ratio = 1 log odds ratio = 0

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Don’t combine heterogeneous

tables!

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Interaction

An interaction exists between two factorswhen the effect of one factor is different at different levels of another factor.

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45 50 55 60

0.0

00

0.0

02

0.0

04

0.0

06

0.0

08

0.0

10

0.0

12

age

d.r

ate

hwu§5 INTRODUCTION TO GENERALISED LINEAR MODELS (GLMs)

Normal linear model

Y|x ~ N with E[Y|x]= + xor E[Y|x]= 0 + 1x1 + 2x2 + … + rxr = x

i.e. E[Y|x] = (x) = x

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We are explaining (x) using a linear predictor (a linear function of the explanatory data)

Generalised linear model

Now we set g((x)) = x for some function g

We explain g((x)) using a linear function of the explanatory data, where g is called the link function

hwue.g. modelling a Poisson mean we use alog link g() = log

We use a linear predictor to explain log

rather than itself : the model is

Y|x ~ Pn with mean λx

with log λx = + x

or log λx = x

This is a log-linear model

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An example is a trend model in which we uselogi = + i

Another example is a cyclic model in which we use

logi =0 + 1 cosθi + 2 sinθi

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§6 MODELS FOR SINGLE CLASSIFICATIONS

6.1 Single classifications - trend models

Data: numbers in r categories

Model: Ni , i = 1, 2, …, r,

independent Pn(λi)

hwuBasic case

H0: λi’s equal v H1: λi’s follow a trend

Let Xj be category of observation j

P(Xj = i) = 1/r

XTest based on

see Illustration 6.1

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A more general model

Ni independent Pn(λi) with

ii e

log i i Log-linear model

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It is a linear regression model for logλi

and a non-linear regression model for λi .

It is a generalised linear model.

Here the link between the parameter we are

estimating and the linear estimator is the

log function - it is a “log link”.

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Fitting in R Example 13: stressful events data

>n=c(15,11, …, 1, 4) >r=length(n)>i=1:r

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>n=c(15,11, …, 1, 4) response vector>r=length(n)>i=1:r explanatory vector

model>stress=glm(n~i,family=poisson)

hwu>summary(stress)

Call:glm(formula = n ~ i, family = poisson) model being fittedDeviance Residuals: Min 1Q Median 3Q Max -1.9886 -0.9631 0.1737 0.5131 2.0362 summary information on the residualsCoefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 2.80316 0.14816 18.920 < 2e-16 ***i -0.08377 0.01680 -4.986 6.15e-07 *** information on the fitted parameters

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Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 (Dispersion parameter for poisson family taken to be 1)

Null deviance: 50.843 on 17 degrees of freedom Residual deviance: 24.570 on 16 degrees of freedom deviances (Y2 statistics)

AIC: 95.825Number of Fisher Scoring iterations: 4

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Fitted mean is

ˆ exp 2.80316 0.08377i i

e.g. for date 6, i = 6 and fitted mean is exp(2.30054) = 9.980

hwuFitted model

5 10 15

0

5

10

15

Date

log-linear trend model for stress data

num

be

r

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Test of H0: no trend the null fit, all fitted values equal (to theobserved mean) Y2 = 50.84 (~ 2 on 17df)

The trend model fitted values exp(2.80316-0.08377i)Y2 = 24.57 (~ 2 on 16df)

Crude 95% CI for slope is -0.084 ± 2(0.0168) i.e. -0.084 ± 0.034

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The lower the value of the residual deviance, the better in general is the fit of the model.

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Basic residuals

5 10 15

-6-4

-20

24

i

ba

sicr

esi

ds

hwu6.2 Taking into account a deterministic denominator – using an “offset” for the “exposure”

Model: Nx ~ Pn(λx) where

E[Nx] = λx = Exbθx

logλx = logEx + c + dx

See the Gompertz model example (p 40, data in Example 26)

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We include a term “offset(logE)” in the formulafor the linear predictor: in R

model = glm(n.deaths ~ age + offset(log(exposure)), family = poisson)

Fitted value is the estimate of the expected response per unit of exposure (i.e. per unit of the offset E)

hwu§7 LOGISTIC REGRESSION

• for modelling proportions

• we have a binary response for each item and a quantitative explanatory variable

for example: dependence of the proportion of insects killed in a chamber on the concentration of a chemical present – we want to predict the proportion killed from the concentration

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for example: dependence of the proportion of

women who smoke - on age

metal bars on test which fail - on pressure applied policies which give rise to claims – on sum insured

Model: # successes at value xi of explanatory variable: Ni ~ bi(ni , πi)

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We use a glm – we do not predict πi directly; we predict a function of πi called the logit of πi.

The logit function is given by:

logit( ) log1

It is the “log odds”.

See Illustration 7.1 p 43: proportion v dose

*

*

*

*

*

*

*

*

*

*

3.8 4.0 4.2 4.4 4.6 4.8

0.2

0.4

0.6

0.8

1.0

dose

pro

p

logit(proportion) v dose

*

*

*

*

*

*

*

*

*

3.8 4.0 4.2 4.4 4.6 4.8

-2-1

01

23

dose

log

itpro

p

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log1

ii

i

a bx

This leads to the “logistic regression” model

[ c.f. log linear model Ni ~ Poisson(λi) with log λi = a + bxi ]

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We are using a logit link

log1

g

We use a linear predictor to explain

rather than itself

log1

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The method based on the use of this model is called logistic regression

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Data: explanatory # successes group observed variable value size proportion

x1 n11 n1 n11/n1

x2 n21 n2 n21/n2

……. xs ns1 ns ns1/ns

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In R we declare the proportion of successes asthe response and include the group sizes as a set of weights

drug.mod1 = glm(propdead ~ dose, weights = groupsize, family = binomial)

explanatory vector is dose note the family declaration

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RHS of model can be extended if required to include additional explanatory variables and factors

e.g. mod3 = glm(mat3 ~ age+socialclass+gender)

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drug.mod – see output p44

Coefficients very highly significant (***) Null deviance 298 on 9df Residual deviance 17.2 on 8df

But … residual v fitted plotand … fitted v observed proportions plot

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-3 -2 -1 0 1 2

-2-1

01

23

Predicted values

Re

sid

ua

ls

glm(formula = num.mat ~ dose, family = binomial)

Residuals vs Fitted

10

5

1

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0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

prop

dru

g.m

od

1$

fit

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0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

prop

dru

g.m

od

2$

fit

model with aquadraticterm(dose^2)

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8.1 Log-linear models for two-way classifications

Nij ~ Pn(ij) , i= 1,2, …, r ; j = 1,2, …, s

H0: variables are independent

ij = i• •j / ••

§8 MODELS FOR TWO-WAY AND THREE-WAY CLASSIFICATIONS

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logij = logi• + log•j log••

row effect overall effect column effect

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We “explain” log ij in terms of additive effects: logij = + αi + βj

Fitted values are the expected frequencies

ˆˆˆ ˆexpij i j

Fitting process gives us the value of Y2 = -2logλ

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Nij ~ Pn(ij) , independent, with logij = + αi + βj

Declare the response vector (the cell frequencies) and the row/column codes as factorsthen use > name = glm(…)

Fitting a log-linear model

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Tonsils data (Example 16)

n.tonsils = c(19,497,29,560,24,269)rc = factor(c(1,2,1,2,1,2))cc = factor(c(1,1,2,2,3,3))

tonsils.mod1 = glm(n.tonsils ~ rc + cc, family=poisson)

Call:glm(formula = n.tonsils2 ~ rc + cc, family = poisson)Deviance Residuals: 1 2 3 4 5 6 -1.54915 0.34153 -0.24416 0.05645 2.11018 -0.53736 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 3.27998 0.12287 26.696 < 2e-16 ***rc2 2.91326 0.12094 24.087 < 2e-16 ***cc2 0.13232 0.06030 2.195 0.0282 * cc3 -0.56593 0.07315 -7.737 1.02e-14 ***---Null deviance: 1487.217 on 5 degrees of freedomResidual deviance: 7.321 on 2 degrees of freedom Y2 = - 2logλ

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The fit of the “independent attributes” model is not good

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> n.patients = c(15, 4, 35, 46)> rc = factor(c(1, 1, 2, 2)) > cc = factor(c(1, 2, 1, 2))

> pat.mod1 = glm(n.patients ~ rc + cc, family = poisson)

Patients data (Example 15)

Call:glm(formula = n.patients ~ rc + cc, family = poisson)Deviance Residuals: 1 2 3 4 1.6440 -2.0199 -0.8850 0.8457 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 2.251e+00 2.502e-01 8.996 < 2e-16 ***rc2 1.450e+00 2.549e-01 5.689 1.28e-08 ***cc2 2.184e-10 2.000e-01 1.09e-09 1 ---Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 (Dispersion parameter for poisson family taken to be 1)Null deviance: 49.6661 on 3 degrees of freedomResidual deviance: 8.2812 on 1 degrees of freedom AIC: 33.172

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fitted coefficients: coef(pat.mod1)(Intercept) rc2 cc2 2.251292e+00 1.450010e+00 2.183513e-10

fitted values: fitted(pat.mod1)1 2 3 4 9.5 9.5 40.5 40.5

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1ˆ j

2ˆ j

Estimates are

Predictors for cells 1,1 and 1,2 are 2.251292 :

exp(2.251292) = 9.5

exp(3.701302) = 40.5

Predictors for cells 2,1 and 2,2 are 2.251292 + 1.450010 = 3.701302 :

1 2 1 2ˆˆˆ 2.251292, 0, 1.450010, 0, 0

hwuResidual deviance: 8.2812 on 1 degree of freedom Y2 for testing the model i.e. for testing H0: response is homogeneous/ column distributions are the same/ no association between response and treatment

group

The lower the value of the residual deviance, the better in general is the fit of the model.Here the fit of the additive model is very poor(we have of course already concluded that there is an association – P-value about 1%).

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8.2 Two-way classifications - taking into account a deterministic denominator

See the grouse data (Illustration 8.3 p50, data in Example 25)

Model: Nij ~ Pn(λij) where

E[Nij] = λij = Eij exp( + αi + βj)

logE[Nij/Eij] = + αi + βj

i.e. logλij = logEij + + αi + βj

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We include a term “offset(logE)” in the formulafor the linear predictor

Fitted value is the estimate of the expected response per unit of exposure (i.e. per unit of the offset E)

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8.3 Log-linear models for three-way classifications

Each subject classified according to 3factors/variables with r,s,t levels respecitvely

Nijk ~ Pn(ijk) withlog ijk = + αi + βj + γk + (αβ)ij + (αγ)ik + (βγ)jk + (αβγ)ijk

r s t parameters

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Model with two factors and an

interaction (no longer additive) is

log ij = + αi + βj + (αβ)ij

Recall “interaction”

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Range of possible models/dependenciesFrom1 Complete independence

model formula: A + B + C

link: log ijk = + αi + βj + γk

notation: [A][B][C]df: rst – r – s – t + 2

8.4 Hierarchic log-linear models

Interpretation!

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…. through

2 One interaction (B and C say)model formula: A + B*C link: log ijk = + αi + βj + γk + (βγ)jk

notation: [A][BC]

df: rst – r – st + 1

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…. to

5 All possible interactionsmodel formula: A*B*C

notation: [ABC]

df: 0

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Model selection: by

backward elimination or forward selection

through the hierarchy of modelscontaining all 3 variables

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saturated [ABC]

[AB] [AC] [BC]

[AB] [AC] [AB] [BC] [AC][BC]

[AB] [C] [A] [BC] [AC] [B]

[A] [B] [C] independence

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Our models can include

mean (intercept)

+ factor effects

+ 2-way interactions

+ 3-way interaction

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Illustration 8.4 Models for lizards data (Example 29)

liz = array(c(32, 86, 11, 35, 61, 73, 41, 70), dim = c(2, 2, 2))n.liz = as.vector(liz)s = factor(c(1,1,1,1,2,2,2,2)) species d = factor(c(1, 1, 2, 2, 1, 1, 2, 2)) diameter of perchh = factor(c(1,2,1,2,1,2,1,2)) height of perch

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Forward selectionliz.mod1 = glm(n.liz ~ s + d + h, family = poisson) liz.mod2 = glm(n.liz ~ s*d + h, family = poisson) liz.mod3 = glm(n.liz ~ s + d*h, family = poisson)liz.mod4 = glm(n.liz ~ s*h + d, family = poisson)liz.mod5 = glm(n.liz ~ s*d + s*h, family = poisson)liz.mod6 = glm(n.liz ~ s*d + d*h, family = poisson)

hwuForward selectionliz.mod1 = glm(n.liz ~ s + d + h, family = poisson) 25.04 on 4df

liz.mod2 = glm(n.liz ~ s*d + h, family = poisson) † 12.43 on 3df

liz.mod5 = glm(n.liz ~ s*d + s*h, family = poisson)liz.mod6 = glm(n.liz ~ s*d + d*h, family = poisson)

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Forward selectionliz.mod1 = glm(n.liz ~ s + d + h, family = poisson)

liz.mod2 = glm(n.liz ~ s*d + h, family = poisson) †

liz.mod5 = glm(n.liz ~ s*d + s*h, family = poisson)† 2.03 on 2df

hwu> summary(liz.mod5)Call:glm(formula = n.liz ~ s * d + s * h, family = poisson)Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 3.4320 0.1601 21.436 < 2e-16 ***s2 0.5895 0.1970 2.992 0.002769 ** d2 -0.9420 0.1738 -5.420 5.97e-08 ***h2 1.0346 0.1775 5.827 5.63e-09 ***s2:d2 0.7537 0.2161 3.488 0.000486 ***s2:h2 -0.6967 0.2198 -3.170 0.001526 ** Null deviance: 98.5830 on 7 degrees of freedomResidual deviance: 2.0256 on 2 degrees of freedom

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*

*

*

*

*

*

*

*

20 40 60 80

20

40

60

80

n.liz

liz.m

od

5$

fit

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*

*

*

*

*

*

*

*

20 40 60 80

-0.1

0-0

.05

0.0

00

.05

0.1

0

liz.mod5$fit

liz.m

od

5$

res

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FIN

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Number of papers per author 1 2 3 4 5 6 7 8 9 10 11Number of authors 1062 263 120 50 22 7 6 2 0 1 1

MAJOR ILLUSTRATION 1

, 1, 2, 3,...!

x

P X x xx

Model

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0.90 0.92 0.94 0.96 0.98 1.00

4.5

5.0

5.5

th

log

l2

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1 2 3 4 5 6 7 8 9 10 11+

02

00

40

06

00

80

01

00

0

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MAJOR ILLUSTRATION 2

Hedge type i A B C D E F GHedge length (m) li 2320 2460 2455 2805 2335 2645 2099Number of pairs ni 14 16 14 26 15 40 71

Model Ni ~ Pn(ili)

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1 2 3 4 5 6 7

01

02

03

04

05

0

type

de

nsi

ty

x xx

x

x

x

x

x x x

x

x

x

x

hwuCyclic models

DNOSAJJMAMFJ

60

50

40

30

Month

case

s

leukaemia data

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Model

Ni independent Pn(λi) with

0

0

exp cos

exp cos sin , 1, ...,

exp cos sin

i i

i i

i i

a b i r

c a b

Explanatory variable: the category/month i has been transformed into an angle i

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It is another example of a non-linear regression model for Poisson responses.

It is a generalised linear model.

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Fitting in R>n=c(40, 34, …, 33, 38) response vector>r=length(n)>i=1:r>th=2*pi*i/r explanatory vector

model>leuk=glm(n~cos(th) + sin(th),family=poisson)

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ˆ exp 3.73069 0.17177cos 0.11982sini i i

Fitted mean is

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J F M A M J J A S O N D

30

40

50

60

Month

case

s

cyclic model for leukaemia data

Fitted model

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Male Female

Cinema often 22 21

Not often 20 12

F73DB3 CDA Data from class

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Male Female

Cinema often 22 21 43

Not often 20 12 32

42 33 75

Male Female


Not often 20 12 32

42 33 75

P(often|male) = 22/42 = 0.524P(often|female) = 21/33 = 0.636significant difference (on these numbers)?is there an association between gender and cinema attendance?

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Null hypothesis H0: no association between gender and cinema attendance

Alternative: not H0

Under H0 we expect 42 43/75 = 24.08 in cell 1,1 etc.

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> matcinema=matrix(c(22,20,21,12),2,2)> chisq.test(matcinema)Pearson's Chi-squared test with Yates' continuity correctiondata: matcinema X-squared = 0.5522, df = 1, p-value = 0.4574

> chisq.test(matcinema)$expected [,1] [,2][1,] 24.08 18.92 [2,] 17.92 14.08

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> matcinema=matrix(c(22,20,21,12),2,2)> chisq.test(matcinema)Pearson's Chi-squared test with Yates' continuity correctiondata: matcinema X-squared = 0.5522, df = 1, p-value = 0.4574

> chisq.test(matcinema)$expected [,1] [,2] null hypothesis can stand [1,] 24.08 18.92 no association between gender[2,] 17.92 14.08 and cinema attendance

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Male Female


Not often 100 60 160

210 165

P(often|male) = 110/210 = 0.524P(often|female) = 105/60 = 0.636significant difference (on these numbers)?

more students, same proportions

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> matcinema2=matrix(c(110,100,105,60),2,2)> chisq.test(matcinema2)Pearson's Chi-squared test with Yates' continuity correctiondata: matcinema2

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> matcinema2=matrix(c(110,100,105,60),2,2)> chisq.test(matcinema2)Pearson's Chi-squared test with Yates' continuity correctiondata: matcinema2 X-squared = 4.3361, df = 1, p-value = 0.03731> chisq.test(matcinema2)$expected [,1] [,2][1,] 120.4 94.6[2,] 89.6 70.4

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> matcinema2=matrix(c(110,100,105,60),2,2)> chisq.test(matcinema2)Pearson's Chi-squared test with Yates' continuity correctiondata: matcinema2 X-squared = 4.3361, df = 1, p-value = 0.03731> chisq.test(matcinema2)$expected

[,1] [,2] null hypothesis is rejected[1,] 120.4 94.6 there IS an association between[2,] 89.6 70.4 gender and cinema attendance

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FIN

f73db3 categorical data analysis workbook contents page preface aims summary...

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