Inferences on Two-Way Contingency Tables

8/12/2019 Inferences on Two-Way Contingency Tables

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INFERENCES ON TWO-WAY

CONTINGENCY TABLES


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DIFFERENCE OF PROPORTIONS

Suppose denote the (conditional) probability

of success for row i. Then the difference of

proportions ( ) compares the success

probabilities in the two rows, i andj.

Note:1 1


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DIFFERENCE OF PROPORTIONS

estimates the true difference .

=

+

+

[Large Sample] %CI: [due to Walds]


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EXAMPLE # 1

The following table is from a report on the relationship between

aspirin use and myocardial infarction (heart attacks) by the

Physicians Health Study Research Group at Harvard MedicalSchool. The Physicians Health Study was a five-yearrandomized study testing whether regular intake of aspirin

reduces mortality from cardiovascular disease. Every other day,

the male physicians participating in the study took either one

aspirin tablet or a placebo. The study was blind thephysicians in the study did not know which type of pill theywere taking.


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EXAMPLE # 1


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EXAMPLE # 1

(a) Estimate the probability of suffering myocardial

infarction (MI) for both placebo and aspirin groups.

(b) Construct a 95% CI for the true difference of

probabilities of heart attack between male physicians who

took placebo and those who took aspirin. From this,

determine if aspirin is effective in diminishing the risk of

heart attack?


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RELATIVE RISK

For 2-by-2 tables, the relative risk(RR) is the ratio

= /

where it can be any non-negative number. RR = 1.0 iff

= .

/estimates the true ratio (RR) /.


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RELATIVE RISK

The importance of RR is due to the importance ofdifferences of a certain fixed size when proportions of

success (in all levels of ) are close to 0 or 1. That is, whilethe same difference was observed for (a) 0.010 and 0.001and (b) 0.410 and 0.401, (a) is more striking since thediscrepancy between the two proportions can be expressedas 10 times of the other. This goes to show that RR may

give better interpretative meaning for public healthimplications, than relying on the differences of proportions

alone (which may be misleading if i 0 or 1).


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RELATIVE RISK

The sampling distribution of RR is highly skewed

unless the sample sizes are quite large. Under which, anapproximate [large-sample due to Walds] 1

100%CI for the true log RRis given by:

/

+

+


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EXAMPLE # 2

Refer to the aspirin use and myocardial infarction (heartattacks) study by the PhysiciansHealth Study Research

Group at Harvard Medical School.(a) Estimate and interpret the RR of heart attackbetween male physicians who took placebo and thosewho took aspirin.

(b) Construct a 95% CI for the true RR of heart attackbetween male physicians who took placebo and thosewho took aspirin.


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ODDS RATIO

For a probability of success , the odds(of success)

are defined to be

= /( )

from which we can get

= /( )


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ODDS RATIO

For 2-by-2 tables, the odds ratio () is the ratio

=

=

/

/

where it can be any non-negative number.

Sample odds ratio () [through ML under multinomial

assumption, or independent binomial assumption]:

=


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ODDS RATIO

and independent = .

> . : higher success rate for row [Xlevel] 1

< . : higher success rate for row [Xlevel] 2

Values of farther from 1.0 in any direction represent strongerassociation between and.

is orientation invariant (unlike RR).

may be viewed as a cross-product ratio of joint probabilities ifinterdependence is desired.


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ODDS RATIO

The sampling distribution of is highly skewed unless

the sample sizes are quite large. Under which, anapproximate [large-sample] 1 100% CI for the

true log [which is symmetric about 0] is given by:


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ODDS RATIO

If some cell counts (nij) are 0, then can either be 0 or ,

or even undefined if both entries in a row or column are 0. To

adjust for this, an amended estimator is given by

=(. )( . )

( . )( . )

i.e., an adjustment of 0.5 was made on each cell count (also

applies for SE() for estimating a 1 100%CI).


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EXAMPLE # 3

Refer to the aspirin use and myocardial infarction (heartattacks) study by the PhysiciansHealth Study Research

Group at Harvard Medical School.(a) Estimate and interpret of heart attack betweenmale physicians who took placebo and those who tookaspirin.

(b) Construct a 95% CI for the true of heart attackbetween male physicians who took placebo and thosewho took aspirin.


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R E L A T I O N S H I P B E T W E E N

O D D S R A T I O A N D R E L A T I V E R I S K

=

Hence, whenever direct estimation of RR is not

possible, one can estimate instead, and use it to

approximate RR, as long asand .


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ODDS RATIO AND

CASE-CONTROL STUDIES

In most case-control studies, marginal distribution of the

response variable is usually fixed by the sampling design.

With this being retrospective, one can construct conditional

distributions for the explanatory variable, within levels of

the response outcome of interest. In this case, only can

be estimated due to its symmetric orientation (invariance).

Thus, for relatively rare successes [usually rare diseases],

RR is usually approximated by .


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TESTS OF IND EPENDENCE

Consider

:

For a sample of size with cell counts *nij, the values *ij = nijare expected frequencies, i.e. *(nij)under which is true.

To arrive at a decision, *nij is compared with *ij, such that for

is true, *nij ij must be small, i.e. larger differences provide

stringer evidences against .

Test statistics used to make such comparisons have large-sample

distributions.


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()

Mean:

Variance:

=

(,)


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PEARSON STATISTIC

=

score statistic

minimum at 0 if all nij = ij

p-value: -

* > for decent approximation


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LIKELIHOOD-RATIO STATISTIC

=

likelihood-ratio statistic [based on multinomial assumption]

minimum at 0 if all nij = ij

p-value:

-

* > for decent approximation


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In two-way tables, the null hypothesis of statistical independence

has the form

: = ++

: = ++

Note: *is estimated by the estimated expected frequencies

* =ni+n+j

n


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For testing independence in I x Jcontingency tables,

the and statistics are used, with both having

large-sample 2 distribution with degrees of

freedom = ( )( ).

converges in distribution more quickly than .


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Recall:

The degrees of freedom is obtained by taking the differencebetween the number of parameters [cell counts] under the alternative

[for w/c there are IJ 1 non-redundant parameters] and null

[for w/c there are (I 1)+(J 1) non-redundant parameters]

hypotheses, i.e.,

= = ( )( )


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EXAMPLE # 4

The following table, from the 2000 General SocialSurvey, cross classifies gender and political party

identification. Subjects indicated whether they identifiedmore strongly with the Democratic or Republican partyor as Independents. This also contains estimated

expected frequencies for : Independence betweenGender and Political Party Identification.

Determine if a significant association between genderand political party identification exists or not.


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EXAMPLE # 4


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RESIDUALS FOR CEL LS

A cell-by-cell comparison of observed and estimated

frequencies help us better understand the nature of theevidence.

However, it is rather insufficient to simply rely on the

raw cell differences [due to the inherent

magnitude of the counts].


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STANDARDIZED RESIDUAL

+ +

follows a [large-sample] standard normal distribution under

: (as compared to 0) evidence towards lack of fit of

i.e., at a significance level , one expects 100% of the

standardized residuals to be beyond 2 (or 3, if many cells) by chance

alone under


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EXAMPLE # 5

Refer to the gender and political party identification

example. The following table shows the standardized

residuals for testing independence in the previous

example. Try to make sense of the computed standardized

residuals in relation with the observed global result for

testing independence between gender and political arty

identification.


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EXAMPLE # 5


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STANDARDIZED RESIDUALS

Notice that residuals for the females are the negative

of those of males. In general, the residuals in each

column must sum up to 0 as the observed counts and the

expected frequencies are constrained by the same row

and column totals. In particular, for 2 x J tables,

= ( )


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PARTITIONING

Recall: Let and be independent 2RVs w/ degrees of

freedom and 2, respectively. Then

= ~+

In essence, this enables one to separate/collapse rows or columns

of I x Jtables to several sub-tables, and obtain 2or 2statistics for

which the sum of each corresponding partitioned statistic is the

globalstatistics.


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PARTITIONING

Consider: For a test of independence in a 2 x J table, a

2

statistic can be broken down intoJ

1components: [1] thefirst two columns; [2] collapsing of the first two columns, then

compared with the 3rd column; [3] collapsing of the first three

columns, then compared with the 4th column, etc. until the Jth

column is considered. In particular, this is true for .


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PARTITIONING

While it might seem more natural to obtain statistic for each

2 x 2pairing, note that the sum of these individual statistics willnot total the global.[Issues due to non-independence]

has exact partitionings; does not (at least, algebraically).

Nevertheless, partitioning 2 is valid for both statisticsas long as

independence of partitions are met.


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SOME COMMENTS ON TESTS

These tests likewise require a very large sample size n

relative to IJ. Moreover, converges poorly as compared

to for very small sample sizes, i.e. for large IorJ,

still provides decent approximation even if some expected

frequencies are as small as 1.


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SOME COMMENTS ON TESTS

tests merely indicate the degree of evidence for an association;

they do not give anything about the strength and the nature of the

association.

Both and are orientation invariant, i.e. they do not change

values with reorderings of rows or columns. However, both are only

powerful when associations regarding nominal variables are of concern.

For ordinal, more powerful tests exist.


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FISHERS EXACT TEST

Recall:For 2 x 2 tables,independence = .

Consider the cell counts { }. A small-sample nullprobability distribution for the cell counts that does not

depend on unknown parameters results from considering the

set of tables having the same row and column total. Under this

condition, each * then have the hypergeometric

distribution.


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FISHERS EXACT TEST

It is sufficient to know alone to determine all other cell

counts. Under the null hypothesis of independence : = ,

is hypergeometricwith

=

Hence, thep-valueequals the sum of hypergeometric probabilitiesfor outcomes at least as favorable to as the observed outcome.


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EXAMPLE # 6

In his 1935 book, The Design of Experiments, Fisher described

the following experiment: When drinking tea, a colleague of

Fishers at Rothamsted Experiment Station near London

claimed she could distinguish whether milk or tea was added

to the cup first. To test her claim, Fisher designed an

experiment in which she tasted eight cups of tea. Four cups

had milk added first, and the other four had tea added first.


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EXAMPLE # 6

She was told there were four cups of each type and she

should try to select the four that had milk added first. The

cups were presented to her in random order. The following

table shows a potential result of the experiemtn. Perform a

test to check whether there is evidence of a positive

association between the true order of the pouring and her

guess. Compute for the exact p-value of the test.


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EXAMPLE # 6


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CONSERVATISM OF

FISHERS EXACT TEST

Being an exact test, the test is very conservative, i.e. the

actual error rate when the null hypothesis of independence istrue is much smaller than the intended one. This is essentially

true for one-sided alternatives. Hence, mid p-value is

preferred as an alternative to diminish the conservativeness.


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SMALL-SAMPLE

CONFIDENC E INTERVAL FOR

It is also possible to construct small-sample confidence

intervals for odds ratio. The procedure involved is a

generalization of Fishersexact test that tests an arbitrary value,

: = . Hence, a % CI would then

contain all values of for which the exact p-value of

: = is greater than 0.05. This can also be constructed

using mid p-value to preserve conservatism.


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SMALL-SAMPLE

CONFIDENC E INTERVAL FOR

For the tea-taste experiment, a 95% CI for can be

computed to be as follows:

Exact p-value: (0.21 , 626.17)

Mid p-value: (0.31 , 308.55)

Inferences on Two-Way Contingency Tables

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