Biostatistics course Part 13 Effect measures in 2 x 2 tables

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Biostatistics coursePart 13Effect measures in 2
x 2 tables

• Dr. Sc. Nicolas Padilla Raygoza
• Department of Nursing and Obstetrics
• Division Health Sciences and Engineering
• University of Guanajuato
• Campus Celaya-Salvatierra

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Biosketch

• Medical Doctor by University Autonomous of
  Guadalajara.
• Pediatrician by the Mexican Council of
  Certification on Pediatrics.
• Postgraduate Diploma on Epidemiology, London
  School of Hygiene and Tropical Medicine,
  University of London.
• Master Sciences with aim in Epidemiology,
  Atlantic International University.
• Doctorate Sciences with aim in Epidemiology,
  Atlantic International University.
• Associated Professor B, Department of Nursing and
  Obstetrics, Division of Health Sciences and
  Engineering, University of Guanajuato, Campus
  Celaya Salvatierra, Mexico.
• padillawarm_at_gmail.com

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Competencies

• The reader will obtain Risk Ratio or Odds Ratio
  from a 2 x 2 table.
• He (she) will calculate 95 confidence interval
  from RR or OR.
• He (she) will identify potential confounders
  and/or interactions.
• He (she) will apply Mantel Haenzsel test for RR,
  OR and Chi-squared.

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Introduction

• In part 12 of the course, we tested the
  association between two categorical variables.
• Now, we review the methods used to measure the
  association.
• We will work with binary variables, so we will
  use 2 x 2 tables.

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Example

• A nurse in a poor area of Mexico, was informed
  that many area children attending the nursery
  were sick of respiratory infections.
• She designed a cohort study to investigate the
  problem.
• During the following years 1000 children were
  followed.
• The main research question was
• Attending nursery is associated with respiratory
  infection?

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Example
Respiratory infection Respiratory infection Total
Attending nursery Yes n No n
Yes 37 33.9 72 66.1 109
No 43 4.8 848 95.2 891
Total 80 8 920 92 1000
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Risk Ratio (RR)

• In health research, the term "risk" is used
  instead of proportion.
• For example
• The risk of infection among children attending
  day care was 33.9.
• Thus, the risk ratio is the ratio of two
  proportions.
• The risk of respiratory infection for those
  attending the nursery 37 / (37 72)
  37/109 0.339
• The risk of respiratory infection in children not
  attending day care is 43 / (43 848) 43/891
  0.048.
• The risk ratio (RR) is the ratio of these two
  risks.
• Risk ratio 0.339 / 0.048 7.06

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Risk Ratio (RR)

• In general, the risk ratio can be obtained with
  the following formula, where a, b, c and d are
  the frequencies in the 2 x 2 table.

Outcome Outcome Total
Exposure Yes No
Yes a b a b
No c d c d
Total a c b d N
Risk Ratio (a /ab) / (c/c d)
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Odds Ratio (OR)

• The Odds Ratio (OR) is the ratio of the chance
  (probability) of the results between those
  exposed and the chance of the outcome among
  non-exposed.
• The chance of infection among attendees of the
  nursery is 37 / 72 0,514
• The chance of infection among children not
  attending day care is 43 / 848 0,051
• The Odds Ratio of these two probabilities OR
  0,514 / 0,051 10.08
• In general, the Odds Ratio was found with the
  following formula
• OR ad / bc (a / c) / (b / d)

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Confidence intervals

• In the analysis of data from children attending
  day care or not, we have the option to use RR or
  OR, to measure the effect of attendance at the
  nursery.
• Each value is an estimate only, so these values
  should be reported with confidence intervals.
• An approximate confidence interval at 95 for the
  RR is found using the following formula
• Minimum value RR / EF
• Maximum value RR x EF

EF exp(1.96v(1/a) (1/ab) (1/c) (1/cd))
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Confidence intervals

• CI for the data of children who attend day care
  or not, is
• EF exp (1.96 v 1 / 37 - 1 / 109 1 / 43 -1/891
  1.48
• RR 7.06
• Minimum 7.06/1.48 4.77
• Maximum value 7.06 x 1.48 10.45
• 95 CI 4.77 to 10.45

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Confidence intervals

• An approximate confidence interval at 95 for the
  OR is found using the following formula
• Minimum value OR / EF
• Maximum value OR x EF

EF exp(1.96v(1/a) (1/b) (1/c) (1/d))
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Confidence intervals

• CI for the data of children who attend day care
  or not, is
• EF exp (1.96 v 1 / 37 1 / 72 1 / 43 1 /
  848 1.65
• OR 10.08
• Minimum value 10.08/1.65 6.11
• Maximum value 10.08 x 1.65 16.63
• 95 CI 6.11 to 16.63

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Which measure is best?

• Risk Ratios are calculated for cross-sectional
  and cohort studies.
• The formula for the 95 confidence interval for
  RR requires larger sample sizes than for OR.
• OR are calculated for case-control and
  cross-sectional studies.
• In case-control studies is not possible to
  calculate risks, and therefore can not calculate
  RR.
• There is an advantage in using OR.
• It is a consistent measure of effect, unlike RR.

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Example (Cont)

• Mexican children showed a strong association
  between exposure (attending nursery) and outcome
  (respiratory infection).
• However such an association may be confounded by
  other factor(s).
• For example, although children who attend day
  care, seem to have a 7 times higher risk of
  respiratory infection, the cause of the infection
  can also be something that is associated with
  children who go to daycare.
• In other words, to attend the nursery may be a
  marker of exposure that causes a respiratory
  infection.
• If this is true, we can say that the association
  between respiratory infections and assistance to
  the nursery, are confused.

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How identify a potential confounder?

• To evaluate a potential confounder, we should
  consider three aspects
• The exposure
• The outcome
• The confounder

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Example

• The nurse is interested in the association
  between day care attendance and presence of
  respiratory infection, but is aware that children
  might be exposed to other factors that cause
  respiratory infection.
• For example, overcrowding at home is a risk
  factor for respiratory infection.
• It is therefore a potential confounder of the
  association between attendance at day care and
  respiratory infections.

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Confounders

• For a variable has been a potential confounding,
  it should meet three conditions
• Must be
• an independent risk factor for the outcome of
  interest
• should be associated with the exposure of
  interest
• not be in the cause pathway between exposure and
  outcome.

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Confounders

• How do we check these conditions in the study of
  Mexican children?
• Condition 1 of confusion
• Risk factor for the outcome of interest
• Is there an association between overcrowding and
  respiratory infection?

Overcrowding in home RI Yes RI No Risk of RI
Yes 54 55 54/109 0.5
No 21 870 21/891 0.02
RR 25 95CI 15.72 a 39.75 X2 311.67 Pltlt0.05
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Confounders

• How do we check these conditions in the study of
  Mexican children?
• Condition 2 of confusion
• Association with exposure
• Is there an association between overcrowding and
  assistance to child care?

Overcrowding in home Attendance to nursery Yes Attendance to nursery No
Yes 43 66
No 35 856
X2 170.39 Pltlt0.05
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Confounders

• How do we check these conditions in the study of
  Mexican children?
• Condition 3 of confusion
• Is the potential confusion is the causal pathway?
  
• In this example, it is unlikely that child care
  assistance, is caused by overcrowding

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Do we have a confounder?

• In this study, overcrowding has satisfied the
  three conditions necessary for a confounding
  variable
• It is an independent risk factor for the outcome
  of interest. Overcrowding is associated with
  respiratory infection.
• It is associated with the exposure of interest.
  Overcrowding is associated with attendance at the
  nursery.
• It is not in the causal pathway. Overcrowding is
  unlikely to be the cause of attendance at nursery.

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Stratified tables

• Now, we know that the data must be additionaly
  analyzed for to have the effect of overcrowding.
• To adjust for confounder variable, we stratified
  the table 2 x 2 of interest.
• The table without stratify is called raw table.
• Can be divided into strata defined by the
  confounder variable.
• The sample is divided into two groups, each of
  them the status of overcrowding is the same.
• The two groups are
• Overcrowding and without overcrowding

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Stratified tables

• If we want to find childcare assistance is
  associated with respiratory infection when
  comparing children within the same category of
  overcrowding.
• The raw table for the relationship between
  respiratory infections and child care assistance

Respiratory infection Respiratory infection Total
Attendance to nursery Yes n No n
Yes 37 33.9 72 66.1 109
No 43 4.8 848 95.2 891
Total 80 8 920 92 1000
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Stratified tables

• Now, it is show stratified tables by overcrowding
  and without overcrowding

Overcrowding
Without overcrowding
Respiratory infection Yes Respiratory infection No Total
Nursery Yes 10 24 34
Nursery No 4 861 865
Total 14 885 899
Respiratory infection Yes Respiratory infection No Total
Nursery Yes 61 14 75
Nursery No 5 21 26
Total 66 35 101
RR 4.23 X232.88 p0.0000 95CI 1.91 a 9.37
RR 63.6 X2178.84 p0.0000 95CI 21.01 a 192.56
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Stratified tables

• Do you think that attendance at nursery is a risk
  factor for respiratory infections among children
  with overcrowding?
• Yes, children attending day care are 63 times
  more at risk of respiratory infection than those
  who do not attend nursery.
• The p value indicates a strong association
  between attendance at daycare and respiratory
  infection in the group without overcrowding.

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Stratified tables

• Do you think that attendance at nursery is a risk
  factor for respiratory infection in the group
  without overcrowding?
• Yes, children attending day care are more than 3
  times more at risk of respiratory infection than
  those not attending the nursery.
• The p value indicates a strong association
  between attendance at daycare and respiratory
  infection in this group.
• Within each stratum, the association between
  attendance at day care and respiratory infections
  is now independent of overcrowding at home.

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Comparison of results

• How to compare these results with those of the
  raw table?
• The raw table shows a strong relationship between
  attendance at day care and respiratory infection,
  RR is different in both tables stratified but
  remains a significant statistical association.

RR 95CI X2 P-value
Raw 7.06 4.77 a 10.45 111.88 lt0.05
Overcrowding 4.23 1.91 a 9.37 32.88 lt0.05
Without overcrowding 63.6 21.01 a 192.56 178.84 lt0.05
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Adjusted Risk Ratios

• Nurse do not want show data divided into strata,
  prefer a global estimate of the effect of
  attended to nursery in respiratory tract
  infection adjusted by overcrowding.
• This can be done by calculate RR using a Mantel
  Haenzsel method.
• First, look 2 x s table in each strata.

Exposure Disease Yes Diasease No Total
Yes ae be
No ce de
Total ne
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Risk Ratios from Mantel Haenzsel

• Adjusted RR (summarized), can be obtained with
• ? a (cd)/n
• RRMantel Haenzsel ---------------
• ? c (ab)/n
• This give us a average of RR initially estimate
  into each table more important each table with
  more sample size.

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Adjusted Risk Ratio

• We calculate overcrowding adjusted RR with Mantel
  Haenzsel formula

Overcrowding
Non-overcrowding
Respiratory infection Yes Respiratory infection No Total
Nursery Yes 61 14 75
Nursery No 5 21 26
Total 66 35 101
Respiratory infection Yes Respiratory infection No Total
Nursery Yes 10 24 34
Nursery No 4 861 865
Total 14 885 899
61 (5 21)/ 101 10 (4 861)/899 15.70
9.62 25.32 ---------------------------------
--------------- ----------------- -----------
6.56 5 (61 14)/101 4 (10 24)/899
3.71 0.15 3.86
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Adjusted Odds Ratio

• Adjusted OR is calculate in similar form that
  adjusted RR.
• ? ad/n
• RMMantel Haenzel -----------
• ? bc/n

Exposure Disease Yes Diasease No Total
Yes ae be
No ce de
Total ne
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Adjusted Odds Ratio

• In a cross-sectional study, on the use of
  quinfamide after a amoebic dysentery, it was
  reported how many are carriers of Entamoeba
  histolytic.

Non-carrier Carrier Total
Quinfamide 100 54 154
Non quinfamide 15 72 87
Total 115 126 241
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Adjusted Odds Ratio

• We calculate adjusted OR by residence area, with
  the Mantel Haenzsel formula

Urban
Rural
Non-carrier Carrier Total
Quinfamide Yes 35 39 74
Quinfamide No 10 51 61
Total 45 90 135
Non-carrier Carrier Total
Quinfamide Yes 65 14 79
Quinfamide No 5 21 26
Total 70 35 105
(35 x 51 /135) (65 x 21/105) 13.2 13
26.2 ----------------------------------------
----------------- ---------- 7.4 (39 x 10 /
135) (14 x 5 /105) 2.89 0.67 3.56
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Mantel Haenzsel X2

• The nurse now knows that the association between
  respiratory infection and attend to nursery still
  is after adjusted by overcrowding, confounder
  variable.
• Now, she want to calculate a Chi squared test to
  significance of this association, adjusted by
  confounder.
• This can be do, calculating X2Mantel-Haenzsel
  test.

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Mantel Haenzsel X2

• To calculate adjusted Chi squared test for the
  confounder, we calculate Mantel Haenzsel Chi
  squared. Null hypothesis is that there is not
  association between attend to nursery and
  respiratory infection.
• Ho OR 1.

?ae-?E(ae)2 X2Mantel
Haenzsel -------------------
?V(ae)
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Mantel Haenzsel X2

• We should go, step by step, beginning with 2 x 2
  of each strata.

Exposure Disease Yes Disease No Total
Yes ae be
No ce de
Total ne
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Mantel Haenzsel X2

• Mantel Haenzsel Chi squared test is an average of
  individuals Chi squared of each table.
• To calculate Mantel Haenzsel Chi squared test, we
  need three values of each table
• ae number of ill and exposed
• E(ae) value expected of ae
• V(ae) variance (standard error squared) of ae,
• where,
• E(ae) total row x total column / grand total
  (ae be) x (ae ce)/ne  
• (ae be) x (ce de) x (ae
  ce) x (be de)
• V(ae)
  --------------------------------------------------
  ------
• 
  ne²(ne - 1)

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Example

• Overcrowding table
• a 61
• E(a) 75 x 66 / 101 49.01
• V(a) (75 x 66 x 26 x 35) / (101² x (101 - 1))
  4.42
• Non-overcrowding table
• a 10
• E(a) 34 x 14 / 899 0.53
• V(a) 34 x 14 x 865 x 885 / (899² x (899 - 1))
  0.50
• To obtain Mantel Haenzsel Chi squared test
  (adjusted Chi squared by overcrowding), we add
  these values from the two strata, using the
  formula

?ae-?E(ae)2 X2Mantel
Haenzsel -------------------
?V(ae)
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Example

• To obtain Mantel Haenzsel Chi squared test
  (Adjusted Chi squared test by overcrowding), we
  add these values, using the formula
• a
  E(a) V(a)
• Overcrowding 61
  49.01 4.42
• Non-overcrowding 10
  0.53 0.50
• Total 71
  49.54 4.92
•  
• X2Mantel-Haenzsel (71 49.54)²/4.92 93.60

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Confusion or not confusion

• How we decide if there is confusion?
• There are nor statistical tests to demonstrate
  confusion.
• We do calculate statistical tests and measure the
  effect raw and stratified tables.
• Then, we calculate summarized statistical test
  and we compare them with the raws, and we
  conclude if there is confusion or not.

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Confusion or not confusion

• If there is an important difference between raw
  and adjusted estimates, we say that the
  association of interest is confounding by another
  factor.
• We look the data of children that attend to
  nursery and respiratory infection.
• After adjust by overcrowding, RR diminish from
  7.06 to 6.56.

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Posibles effects from confusion

• Generally there are more than one confounder.
• They can have different effects
• The association in study, can be or not
  significative before of adjust for a confounder
  and not significative after.
• The association can be significative after adjust
  for a confounder but with a p-value less
  significative.
• Strata can show oposite results and in this case,
  it is better, show stratified results. This is
  interaction or effect modified.
• Confounder can hide an existing relationship.

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Bibliografía

• 1.- Last JM. A dictionary of epidemiology. New
  York, 4ª ed. Oxford University Press, 2001173.
• 2.- Kirkwood BR. Essentials of medical
  ststistics. Oxford, Blackwell Science, 1988 1-4.
• 3.- Altman DG. Practical statistics for medical
  research. Boca Ratón, Chapman Hall/ CRC 1991
  1-9.