Categorical data II

The association between two categorical

variables

/ 1388دکتر سید ابراهیم جباری فر تاریخ : 2010

دانشیار دانشگاه علوم پزشکی اصفهان بخش دندانپزشکی جامعه نگر

By the end of the session you should be able to

Construct 2-way table to examine association between two

categorical variatles

Conduct Chi-squared test to assess the association between

two categorical variables

Conduct Chi-square test for trend when the exposure is

ordered categorical

Example

Random sample of individuals were questioned about their

occupatim and their BP was measured. Based on SBP and DBP

measures they were classified as hypertensive or non

hypertensive. Among 300 people in non-manual jobs, there

were 72 hypertensive individuals. Among 240 people in

manual jobs, there were 96 hypertensive indwiduals.

Constructing 2- way table

As a first step we need to organize our data in a formal way

we construct 2-way table

Hypertension

Total No Yes

240 144 96 Manual

300 228 72 Non- manual

540 372 168 Total

What does it mean when we speak about an

association between two categorical variables?

What does it mean when we speak about an association

between two categorical variables ?

It means that knowing the value of one variable tells us

sanething about the value of the other variable .

Two variables are therefore saidto be associated if the

distribution of one variable varies according to the value of

the other varialde .

What does it mean when we speak about an association

between two categorical variables ?

In our example, the two variables, occupation and hypertension,

are associated if the distribution of hypertension varies between

occupatioml groups .

And, if distribution of hypertension is same in both occupational

groups, we can say that there is no association between

hypertension and occupational category - because knowing a

occupational category of indi\idual wil! not tell us anything

about hypertension.

What does it mean when we speak about an association between

two categorical variables ?

•Having constructed a two-way table, the next step is to look

whether the distribution of one variable differs according to the

value of the other variable .

We need to calculate either row or column percentages .

Often, one variable can be regarded as the response variable, while

the other is the explanatory variable, and this should help to decide

what percentages are shown

•If the columns represent the explanatory variable, then column

percentages are more appropriate, and vice versa.

Constructing 2-way table

As a second step we calculate proportion of hypertensive

individuals among manual workers. non-manual workers and in

the whole sample

Hypertension

Total No Yes

240 144(60.0%) 96(40.0%) Manual

300 228(76.0%) 72(24.0%) Non-manual

450 372(68.9%) 168(31.1%) Total

The numbers in the four categories in the 2 way

table in the previous slide all called

OBSERVED NUMBERS I

The data seem to suggest sane association between

hypertension and occLPation )40% of manual workers with

hypertension compared to 24% of non-manual workers with

hypertension(

The calculation and examination of such percentages is an

essential step in the analysis of a two-way table, and should

always be done before starting formal significance tests.

Significance test the association

Although it seemsthat there is an association in the table, the

question is whether this may be attributable to samping variability

Each of the percentages i1 the table is subject to sampling error,

and we need to assess whether the differences between them may

be due to chance

This is done by conducting a sig1ificance test

The null hypothesis is "there is no association between the two

variables'

Expected numbers

The significance test is Chi-squared test

This test canpares the observed numbers in each of four

categories of contingency table with the numbers to be

expected if there was no difference in proportion of

hypertensive indviduals in two occupational groups

Hypertension

Total No Yes

240 74.64 Manual

300 Non-manual

450 372(68.9%) 168(31.1%) Total

From the table above, the overall proportion of hypertensive

individuals is 168/372 )31.1 %(.

If the null hypothesis were true, the expected number of manual

subjects with hypertension is 31.1 % of 240, which is 74.64

Hypertension

Total No Yes

240 74.64 Manual

300 Non-manual

450 372(68.9%) 168(31.1%) Total

Expected numbers in the other cells of the table can be

calculated similarly, using the general formula:

Row×Culumn totalExpected number=

Overall total

Hypertension OBSERVED

Total No Yes

240 144 96 Manual

300 228 72 Non-manual

450 372(68.9%) 168(31.1%) Total

Next step – compare observed and expected numbers

Hypertension EXPECTED

Total No Yes

240 165.36 74.64 Manual

300 206.64 93.36 Non-manual

450 372(68.9%) 168(31.1%) Total

Chi-squared test (X2test)

Calculate )O-E(/E for each cell and sum over all cells

In our example

X2 = [)96-74.64( 2/74.64 + )144-165.36(2 / 165.36 + )72-

93.36(2 / 93.36 + )228-206.64(2 / 206.64] = 15.97

/E]E)OX 22

If X2 value is large then )O-E( is, in general, large and data do not

support H0 = association

If X2 value is small then )O-E( is, in general, Small and data do

support

H0= no association

Large values of X2 suggest that the data are inconsistent with the

null hypothesis, and therefore that there is an associaion between

the two variables.

Obtaining p-value

Under Ho: X2 distribution

Obtaining p-value

The P-value is obtained by referring the calculated value of X2

to tables of the en-squared

distribution .

The P-value in this case corresp:mds to the value shown as a in

the tables .

The degrees of freedom are given by the formula:

d.f=(r-1)x(c-1)

R= number of rows , c=number of columns

Back to our example:

X2=15.97

Table 2x2

d.f.=1

and from the table

P<O.001

Relation with normal test for the difference between 2

proportions:

Pl-P2= 40%-24% = 16%

=4.01

Z=)P1-P2(/SE=16/4.01=3.99

For 2 × 2 table, X2 = Z2, i.e. 15.97 = 3.992

Pvalues for both tests are identical

3001

2401

*68.9*31.1P2)SE(P1

Quick formula

There is a quick formula calculating chi-square test for 2×2 table

If we use following notation:

Quicker formula for calculating chi-squared test is

Total

e b a

f d c

N h g Total

Quicker formula for calculating chi- squared test is

d.f=1

efghbc)(ad*N

X2

2

Continuity correction

The chi-squared test for 2×2 table can be improved by using

a continuity correction, called Yates' continuity correction

When we compare two p-oportions using the Z test )or the

X2 test(, we are making the assumption that the sampling

distributions of each of the proportions, and of the

difference between them, are approximately normal.

Continuity correction - cont.

In fact, the distribution et a proportion is not exactly

normal, because it is a dscrete variable, whereas the

normal distribution elates to continuous variables.

Continuity correction

For large enough samples, the binomial distribution

approximates very closely tothe normal distribution

This is why we use of the Z and x2 tests for the comparison of

proportions

However, the approximation can be improved by making a

continuity correction, which adjusts the value of z to allow for

the fact thatwe are modelling a discrete variable with a

continuous distribution

Yates’c ontinuity correction - formula

The effect of the continuity correction is to reduce the value

of x2; with smaller samples, the effect is more substantial -

which is exactly what we want given that the normal

distribution is better approximation of binomial distribution

when sample is larger

Exact test for 2×2 tables

The exact test to compare 2 proportions is needed when the

numbers in the 2 × 2 table are very small )the consensus: if the

total number of individuals

is less than 40 and one of the cells )or more( is lesslhan 5(

It is based on calcLJating exact probabilities of the observed

table and of more 'extreme' tables with the same row and

column totas

Formula and more details - Kirkwood and Sterne, p.169-171

Larger tables )r x c tables(

No continuity correction

Valid if less than 20% of expected numbers are under 5 and none

is less than 1

If low expected numbers- combine either rows or columns to

overcome this problem

How to calculate expected number in particular cell

Row total Column total

Expected number=

Overall total

Ordered exposures 1X2 test for trend

A special case = there are two rows but more than two columns

)or it can be two columns araJ more than two rows(

AND an ordered categorical variable is involved )as the

variable with more than 2 categories(

Education M1

Total University Secondary Primary

132 20(22.2%) 64(32%) 48(40%) Yes

278 70 136 72 No

410 90 200 120 Total

3 2 1 Score for trend test

Example

In this 2 x 3 table, the aim is to examine if there is any

association between educaton and MI.

MI is the outcome of interest

Education is the exposure of nterest

We want to compare the proportions of subjects with MI in

the three education groups.

Standard chi-squared test

X2=7.45

Table 2×3-d.f=2

P=0.024

So there is statistical evidence of an association between the two

variables.

However, when the column variable is an ordered categorical

variable, a more

sensitive procedure for detecting an association is to use a x2 test-

for-trend

X2 test – for trend

The null hypothesis is the same as for an ordinary 2 × c X2 test,

that there is no association between the two variables. But the

test-for-trend is particularly sensitive to an upward or downward

trend in proportions across the columns of the table

The test is conducted by assigning the numerical scores 1, 2, 3 ...

to the columns of the table. We then calculate the mean scores X1

and X2 observed in those with MI and in those without MI.

• Mean score among those withoutMI

X1=[48*1+64*2+20*3]/132=236/132=1.79

Mean score among those without MI

X2= [72*1 + 136* 2+ 70* 3]/278= 554/278= 1.99

If there is no difference between the three proportions, two mean

scores will be equal.

If there is an upward )or downward( trend in the proportions, the

mean score in the MJ group will be greater )or less( than that in the

no MI group.

We now need to calculate stancard deviation s.

Let's assume that we have a sampe of 41 0 points made up

of 120 ones, 200 twos and 90 threes. From this sample we

find the standard deviation of the whole sample in the

usual way.

S=

Mean is overall mean of ones, twos and threes

Example

mean = (120 *1 + 200 *2 +90 *3)/410= 1.93

And

S=

S=0.71

4091.93)(3*901.93)(2*2001.93)(1*120 222

Chi-squared test for trend

obtained by comparing the mean scores for the 2 groups:

d.f.=1

p = 0.006

Where n, and n2 are total numbers of individuals with and without

MI

You can see that square root of the formula is formula for Z test for

comparing means of 2 samples

Z is square root of X2

7.6

n1

n1

*s

)xx(X

21

21

You can see that p value is smaller than with ordinary chi-

squared test

There is even stronger evidence of an association between MI

and education

The alternative formula for chi-squared test for trend is

described by Kirkwood cnd Sterne, p.173 176

Chi- squared tests in STATA

Let's use data from practical 10 )last session(

We try to evaluate whether there is an association between current

smoking and age

We have age grouped into 4 groups )30-39, 40-49, 50-59, 60 69(

Smoking )variable smok( was coded 1 =current smokers, 2=non-

smokers

Firstly, we recode smoking

Recode smok 2=0 )we need the outcome to be coded as 0.1(

Let’s check proportion of smokers in each age category

Tab smok agegroup col

30-39,40-49,50-59,60-69

Total 60 50 40 30 1=Yes 0=no

1.67565.69

49178.81

49072.38

35756.31

33754.71

0

87534.31

13221.19

18727.62

27743.69

27945.29

1

2.550100.00

623100.00

677100.00

634100.00

616100.00

Total

Now , standard chi- squared test

Tab smok agegroup , col chi

30-39,40-49,50-59,60-69

Total 60 50 40 30 1=Yes 0=no

1.67565.69

49178.81

49072.38

35756.31

33754.71

0

87534.31

13221.19

18727.62

27743.69

27945.29

1

2.550100.00

623100.00

677100.00

634100.00

616100.00

Total

Pearson chi2)3(=118.7458 pr=0.000

Degrees of freedom chi-squared test value P<0.001

Chi – squared test for trend

COMMAND OUTCOME EXPOSURE

Tabodds smok ageroup

interval 95% conf odds controls cases agegroup

0.970220.907750.451660.32580

0.706440.663220.322460.22184

0.827890.775910.381630.26884

337357490491

279277187132

30405060

Test of homogeneity (equal odds): chi2(3)=118.70

pr>chi2=0.0000

Score test for trend of odds: chi2(1)=108.83

pr>chi2=0.0000

This is part that is interesting at themomenti

What have we learnt in the session?

1. Construct 2-way table to examine association

between two categorical variatles

2. Conduct Chi-squared test to assess the association

between two categorical variables

3. Conduct Chi-squared test for trend for ordered

exposures