unit 08 contingency tables

7/30/2019 Unit 08 Contingency Tables

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Biostatistics 200

Unit 8: Inference for ContingencyTables

P & G Sections 15.1, 15.3

Section 6.5

19 November 2012

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The chi-squared test in contingency tables, P&G Chapter

15, pp 342 - 349

Odds and Odds Ratios, P&G Chapter 6, pp 144 - 149,section 15.3

Confidence intervals for an OR, P&G, pp 352 - 357

Summary

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PROGRESS THIS UNIT

The chi-squared test in contingency tables, P&G Chapter15, pp 342 - 349

Odds and Odds Ratios, P&G Chapter 6, pp 144 - 149,

section 15.3


Summary

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OUTLINE FOR THIS SECTION

Reformulating the two-sample binomial problem as acontingency table.

Contingency tables and testing for associationsbetween categorical variables (22 and r c tables)

Main method will be the Pearson chi-squared (2)test, but we will also discuss Fishers exact test.

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ANALYZING BINARY DATA FROM TWO

POPULATIONS IN A TABLE

Let x1 and x2 be the observed number of successes for twobinomial variables, with numbers of trials n1 and n2.

The two sample proportions would be

p1 =x1

n1p2 =

x2n2

Inference for the difference p1 p2 can be based onp1 p2.

The text covers that approach in section 14.6. Instead, we

will use two-way tables.5 / 6 1


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TWO-WAY (CONTINGENCY) TABLESBinomial outcomes in two samples can be organized in atable

Table of Outcome by GroupOutcome Group

Sample 1 Sample 2 TotalSuccess a = x1 b = x2 a + bFailure c d c + dTotal a + c = n1 b + d = n2 n1 + n2 = n

With this notation,

p1 =a

a + c

p2 =b

b + d

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THE ORGANIZATION OF TABLES

Typically, the group variable is used for the columns,and the outcome variable (success or failure) for therows.

The group variable is sometimes called theexplanatory variable.

The outcome variable is sometimes called theresponse variable.

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THE ORGANIZATION OF TABLES. . .

Epidemiologists often use case-control designs,where cases are sampled according to outcome(typically, disease or no disease) and exposure ismeasured based on retrospective records.

In a case-control analysis, Stata uses theexposure status as the column label because it isthe explanatory variable.

Epi 202 uses exposure as the column variable. Epi 500 uses the exposure status as the row

variable.

I use the Epi 202/Stata convention.

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HARVARD UNDERGRADUATE ATTITUDES

TOWARD ABORTION

In 2007, I asked my undergraduate class Should awoman be able to obtain a legal abortion for any reason ifshe chooses not to have the child?

Answer by Sex of StudentOutcome Group

Female Male TotalYes 73 37 110

No 22 18 40Total 95 55 150

Sex is group variable, Yes/No is outcome.

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THE SURVEY ON ATTITUDES TOWARD

ABORTION . . .

Equivalent ways of asking the research question in myclass survey

Are female and male students equally likely toanswer yes to the abortion question?

Is response to the abortion question independent ofsex of the student?

Is there any association between response and sex ofthe student?

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ATTITUDES TOWARD ABORTION

In this setting, the null hypothesis is that response and

sex of the student (i.e, row and column variables) areindependent.

The alternative is that they are not independent.

Equivalently, let

pF = probability that a randomly chosen femalestudent answers yes

pM = probability that a randomly chosen malestudent answers yes

Then the hypotheses are H0 : pF = pM vs HA : pF = pM.

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TESTING THE NULL HYPOTHESIS OF

INDEPENDENCE

The (Pearson) Chi-Square Test: basic idea

1. If the row and column variables are independent(null hypothesis is true) what do we expect to see?

2. How do expected values in the table compare towhat has been observed?

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COMPUTING EXPECTED CELL COUNTS

UNDER THE NULL HYPOTHESIS OF

INDEPENDENCEConsider the 95 female respondents in the 2 2 table

If sex and response were independent, whatpercentage of the female students would be expectedto respond yes?110/150 = 0.73333 = 73% Why?

How many students do we expect to fall into the celldefined by female, yes?

(95)(0.73333) = 69.7 Why?

What was the observed cell count? 73observed - expected= 73 69.7 = 3.3

Now compute the expected values for all cells.

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A SIMPLE FORMULA FOR EXPECTED COUNTS

Possible to show that for a cell in row i, column j, theexpected cell count under the hypothesis of independenceis

row i total column j total

total count in tableThis formula applies also to tables with more than tworows or columns (example coming later)

Lets test the hypothesis of no association (independence)between the sex of the respondent and response.

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NULL AND ALTERNATIVE HYPOTHESES IN

THIS SETTING

Equivalent ways to state the relevant hypotheses in this2 2 table:

H0: Sex of respondent and response are independentvs. HA: Sex of respondent and response are notindependent.

H0 : pF = pM vs. HA : pF = pM, where pF and pM are

the probabilities of female and male yes response,respectively.

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2


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(PEARSON) CHI-SQUARE (2) TEST

The test statistic for the hypotheses on the previous slide

is

2 =

all cells

(obs - exp)2

exp

The statistic has a sampling distribution that isapproximately 2 with degrees of freedomdf= (r 1)(c 1) where r = # rows, c = # columns.

In a 2 2 table, df = 1.

Important values from the distribution are given in TableA.8 on page A-26.

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2


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A SIMILAR 2 TABLE, BUT WITH MORE

DETAIL

Table entry for p is thecritical value (2) with

probability p lying to itsright.

Probability p

( 2)*

TABLE F

2 distribution critical values

Tail probability p

df .25 .20 .15 .10 .05 .025 .02 .01 .005 .0025 .001 .0005

1 1.32 1.64 2.07 2.71 3.84 5.02 5.41 6.63 7.88 9.14 10.83 12.122 2.77 3.22 3.79 4.61 5.99 7.38 7.82 9.21 10.60 11.98 13.82 15.203 4.11 4.64 5.32 6.25 7.81 9.35 9.84 11.34 12.84 14.32 16.27 17.734 5.39 5.99 6.74 7.78 9.49 11.14 11.67 13.28 14.86 16.42 18.47 20.005 6.63 7.29 8.12 9.24 11.07 12.83 13.39 15.09 16.75 18.39 20.51 22.116 7.84 8.56 9.45 10.64 12.59 14.45 15.03 16.81 18.55 20.25 22.46 24.107 9.04 9.80 10.75 12.02 14.07 16.01 16.62 18.48 20.28 22.04 24.32 26.028 10.22 11.03 12.03 13.36 15.51 17.53 18.17 20.09 21.95 23.77 26.12 27.879 11.39 12.24 13.29 14.68 16.92 19.02 19.68 21.67 23.59 25.46 27.88 29.67

10 12.55 13.44 14.53 15.99 18.31 20.48 21.16 23.21 25.19 27.11 29.59 31.42

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EVER GROWING CATALOGUE OF

DISTRIBUTIONS

Binomial

Normal t

F

2

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C S ( 2) T


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CHI-SQUARE (2) TEST

If the rows and columns are independent, theobserveds and expecteds shouldnt be verydifferent, and 2 value will be small

If there is an association between rows and columns,the observeds may be far from the expecteds,

leading to a large 2 value So, reject the null hypothesis when 2 value is

sufficiently large (look only at upper tail of chi squaredistribution).

Like the F-test in ANOVA, the alternative isinherently two-sided, but the right tail area is notdoubled.

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A S


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ABORTION ATTITUDES, IN STATA. tabi 73 37 \ 22 18, cchi2 chi2 expected

+--------------------+

| Key |

|--------------------|| frequency |

| expected frequency |

| chi2 contribution |

+--------------------+

| col

row | 1 2 | Total

-----------+----------------------+----------

1 | 73 37 | 110

| 69.7 40.3 | 110.0

| 0.2 0.3 | 0.4

-----------+----------------------+----------

2 | 22 18 | 40

| 25.3 14.7 | 40.0| 0.4 0.8 | 1.2

-----------+----------------------+----------

Total | 95 55 | 150

| 95.0 55.0 | 150.0

| 0.6 1.0 | 1.6

Pearson chi2(1) = 1.6311 Pr = 0.202

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U 2


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USING A TABLE OF THE 2 DISTRIBUTION

As with the t-distribution, tables in P&G yield only

approximations to p-values. Here are the first three rowsof the P&G table:

Area in the Upper Tail

df 0.10 0.05 0.025 0.01 0.001

1 2.706 3.841 5.024 6.635 10.828

2 4.605 5.991 7.378 9.210 13.816

3 6.251 7.815 9.348 11.345 16.266

........

From the table we see that p > 0.10

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EXAC O S O 2 2 A S


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EXACT METHODS FOR 22 TABLESFishers exact test is often used with small and moderatesample sizes. It is similar in spirit to using exact Binomial

calculations in a one sample problem.

Rule of thumb coming later for sample sizes whereFishers test should be used. Easy to get in Stata:

. tabi 73 37 \ 22 18, exact

| col

row | 1 2 | Total

-----------+----------------------+----------

1 | 73 37 | 110

2 | 22 18 | 40-----------+----------------------+----------

Total | 95 55 | 150

Fishers exact = 0.251

1-sided Fishers exact = 0.139

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IDEA BEHIND FISHERS EXACT TEST


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IDEA BEHIND FISHERS EXACT TEST

Condition on the observed row and column totals.

Given the row and column totals, the 4 cell counts aredetermined by any one of the cell counts. Upper leftcell is typically used.

For the sampling distribution: Use the conditionaldistribution of values for the upper left cell, given therow and column totals and under the hypothesis ofindependence.

A one-sided p-value is the probability of observing avalue as or more extreme as the cell count in the

upper left corner. Two-sided p-values even more complicated.

We will only consider two-sided tests in contingencytables.

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CONTINGENCY TABLES WITH MORE THAN 2


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CONTINGENCY TABLES WITH MORE THAN 2

ROWS OR 2 COLUMNS2 tests for tables with r > 2 rows and/or c > 2 columnspresent no difficulties.

Use the earlier formula for expected count for cell in rowi, col j under hypothesis of independence of rows andcolumns:

row i total column j total

total count in table

The chi-square statistic still has the form

2 =

all cells

(obs-exp)2exp

but has degrees of freedom df= (r 1)(c 1). 24/61


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Values in green are expected counts


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Values in green are expected counts.Certificate Status

Hospit. Confirmed Inaccurate Incorrect TotalAccurate No Change Recoded

Commun. 157 169.3 18 24.7 54 35.0 229Teaching 268 255.7 44 37.3 34 53.0 346

Total 425 62 88 575

2 = all cells

(obs-exp)2exp

= 21.52

df= (r 1)(c 1) = (1)(2) = 2

In Table A.8, 2df=2,0.001 = 13.82, p-value < 0.001, so wereject the null-hypothesis that accuracy in deathcertificates is independent of hospital type.

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STATA


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STATA. . .. tabi 157 18 54 \268 44 34, expected chi exact

....some output not shown

| col

row | 1 2 3 | Total

-------+---------------------------------+----------

1 | 157 18 54 | 229

| 169.3 24.7 35.0 | 229.0-------+---------------------------------+----------

2 | 268 44 34 | 346

| 255.7 37.3 53.0 | 346.0

-------+---------------------------------+----------

Total | 425 62 88 | 575| 425.0 62.0 88.0 | 575.0

Pearson chi2(2) = 21.5235 Pr = 0.000

Fishers exact = 0.000

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SOME COMMENTS


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SOME COMMENTS

The 2 test for r c tables does not take into account any

natural ordering of rows or columns that might bepresent in data.

The text mentions the Yates continuity correction (p 346)sometimes used in calculating the 2 statistic in smallsamples. Used far less often now; better to use Fishersexact test.

Fishers exact test can be used to assess associations in

general r c tables.Very common now for papers to report Fishers exact test,even in moderately large samples.

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COMMENTS


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COMMENTS. . .

The following rule of thumb is used for the validity for

the Pearson 2 test:

In 2 2 tables, each expected cell count (calculatedunder the hypothesis of independence) should be at

least 5. In tables with more than 4 cells (excluding the cells

with the row and column totals), the averageexpected count should be at least 5, and no expectedcount should be smaller than 1.

When these conditions do not hold, use Fishersexact test.

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COMMENTS


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COMMENTS. . .Alternate form of2 test for 2 2 tables.

If the table has entries

Outcome GroupSample 1 Sample 2 Total

Success a b a + b

Failure c d c + dTotal n1 n2 n

then the 2 test can be written

2 =n(ad bc)2

(a + c)(b + d)(a + b)(c + d)

df = 1

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PROGRESS THIS UNIT


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PROGRESS THIS UNIT



section 15.3


Summary

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INTRODUCTION


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INTRODUCTION

The 2

and Fishers exact test provide methods for testingthe null hypothesis of independence between row andcolumn variables.

But neither test provides an estimate of the nature of the

association when the hypothesis of independence isrejected.

We will use odds ratios for estimating associationbetween row and column variables.

To study odds ratios, we first need to study odds.

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USING ODDS IN BINOMIAL MODELS


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USING ODDS IN BINOMIAL MODELS

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BETTING IN A FAIR GAME


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BETTING IN A FAIR GAME

An American roulette wheel has 38 slots: 1,2,3,. . . ,36, 0, 00

If you place a $1 bet on 00 for a single spin of the wheel,

You have 1/38 chance of winning in a single spin

1 way to win, 37 ways to lose, or

The casino has 37 ways to win, 1 way to lose

The odds of winning for the house are 37 to 1 and are

1 to 37 for you

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BETTING IN ROULETTE


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BETTING IN ROULETTEFor the game to be fair,

Casino keeps your $1 if 00 does not come up

Casino pays $37 if 00 comes up, and you keep your$1 bet

IfXrepresents your winnings from a $1 bet and E(X)the average winnings in many such bets

E(X) = 1(37/38) + 37(1/38) = 0

Casinos stay in business

by paying out 35 to 1, the casinos insure that rouletteis not a fair game.

In this case

E(X) = 1(37/38) + 35(1/38) = (2/38) = 0.05335/61

MATHEMATICAL DEFINITION OF ODDS


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M M C N ON O O S

In roulette the 37:1 odds for the house is the same as

37/38

1/38

More generally, if the probability of an event A is p,

the odds of the event is p/(1 p)

Sometimes written as p : (1 p) (read as p "to" 1 p).

(1/3) : (2/3) odds is the same as 1:2 odds

Ifp is small (say p < 0.10) then (1 p) 1 and soodds p. The approximation improves as p approaches 0.

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ODDS VS. PROBABILITIES


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Probability Odds = p/(1 p) Odds0 0/1 = 0 0

1/100 = 0.01 1/99 = 0.0101 1 : 991/10 = 0.10 1/9 = 0.11 1 : 9

1/4 1/3 1 : 31/3 1/2 1 : 21/2 ( 1

2)/( 1

2)=1 1 : 1

2/3 (2/3)/(1/3)=2 2 : 13/4 3 3 : 1

1 1/0

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ODDS RATIO OR RELATIVE ODDS


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Suppose we have a disease (e.g., lung cancer denoted byD) and two groups (e.g., smokers denoted by E for

exposure, non-smokers denoted by Ec

)Odds Ratio (OR) or relative odds of disease comparingsmokers to non-smokers is

=Pr(D|E)

1 Pr(D|E) Pr(D|Ec)1 Pr(D|Ec)This is the odds of disease for smokers divided by the

odds of disease for non-smokers. OR > 1 implies smokers have higher probability of

disease

OR < 1 implies smokers have lower probability.

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FUNDAMENTAL RESULT FOR


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Why is this surprising?

Because when cases and controls are sampled andexposure is determined retrospectively, it is onlypossible to estimate Pr(E|D) or Pr(E|Dc), not Pr(D|E)

and Pr(D|Ec

).

Why is this important?

Because even when exposure is estimated by

sampling from cases and controls, it is possible toestimate the correct OR.

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EXPLOITING THE SYMMETRY OF THE ODDS


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RATIO

Thus, OR can be estimated in two ways:

Prospective studies of exposed and unexposed, to seewho develops disease, as in a cohort study design

Retrospective studies of diseases vs. healthy subjects,to see who is exposed, as in a case-control studydesign

Both types of studies can estimate the OR of disease,comparing exposed to unexposed.

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EXPLOITING THE RARE DISEASE


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ASSUMPTION

When a disease D is rare in both exposed and unexposedgroups

1 Pr(D|E) and 1 Pr(D|Ec) are both close to 1.

In this case

OR Pr(D|E)

Pr(D|Ec),

which is called relative risk.

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A SIMPLE FORMULA FOR AN ODDS RATIO IN


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A CASE CONTROL STUDY

Exposed Unexposed Total

Disease a b a + bNo Disease c d c + d

Total a + c b + d n

OR = P(E|D)/(1 P(E|D))P(E|DC)/(1 P(E|DC))

=

(a/(a + b))

(b/(a + b))

(c/(c + d))

(d/(c + d))

=ad

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A SIMPLE FORMULA FOR AN ODDS RATIO IN


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A PROSPECTIVE STUDY

Exposed Unexposed Total

Disease a b a + bNo Disease c d c + d

Total a + c b + d n

OR = P(D|E)/(1 P(D|E))P(D|EC)/(1 P(D|EC))

=

(a/(a + c))

(c/(a + c))

(b/(b + d))

(d/(b + d))

=ad

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ELECTRONIC FETAL MONITORING (EFM),


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P&G PP 354 - 357Does EFM have an impact on Caesarean-section(C-section) delivery decisions?

Assume that in a sample of 5,824 births:

EFM ExposureCaesarean Delivery Yes No Total

Yes 358 229 587No 2,492 2,745 5,237

Total 2,850 2,974 5,824

This is a case-control study, sampled according to type ofdelivery. EFM is the exposure variable.

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THE EPIDEMIOLOGISTS APPROACH TO THIS


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PROBLEM

Even though this is a case-control study that sampledaccording to type of delivery, it is possible to estimatethe odds ratio

=Pr(C|E)

1 Pr(C|E) Pr(C|Ec)1 Pr(C|Ec) ,where C = (woman delivers by C-Section) and E =(EFM used during pre-natal care).

Invoke the rare disease assumption to estimate

Pr(C-section|EFM)

Pr(C-section|no EFM)

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C-SECTION AND EFM


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EFM ExposureCaesarean Delivery Yes No Total

Yes 358 229 587No 2,492 2,745 5,237

Total 2,850 2,974 5,824

Relative Risk =Pr(C-section|EFM)

Pr(C-section|no EFM)

Odds Ratio

= (358)(2745)(229)(2492)

= 1.72

Can we check the rare disease assumption with these

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INTEGRATING THE OR AND THE TEST


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The small pvalue from the table leads to rejection ofH0.

Data from the study suggests that the odds of a C-sectionin women with pre-natal EFM is 72% higher than inwomen without pre-natal EFM

If the rare disease assumption is justified,RR OR = 1.72, and study suggests women withpre-natal EFM are 72% more likely to have C-section.

Can also test H0 by examining a confidence interval for

ORNext section gives the formula for this confidenceinterval.

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PROGRESS THIS UNIT


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section 15.3


Summary

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STATA AGAIN. . .


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FORMULAS FOR ATTRIBUTABLE RISK


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In a case-control study,

Attr. frac. ex. =OR 1

OR

=1.722 1

1.722= 0.419

Attr. frac. pop = Attr. frac. ex. proportion exposed cases

= (0.419)(0.6099)

= 0.2557

Formulas for cohort studies are different.

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PROGRESS THIS UNIT


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The chi-squared test in contingency tables, P&G Chapter

15, pp 342 - 349


section 15.3


Summary

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unit 08 contingency tables

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