Conditional Probability And the odds ratio and risk ratio as conditional probability.
Risk Ratio and Odds Ratiopioneer.chula.ac.th/~stosak/biostatlab/nonparametricstat_2.pdfΒ Β· Risk and...
Transcript of Risk Ratio and Odds Ratiopioneer.chula.ac.th/~stosak/biostatlab/nonparametricstat_2.pdfΒ Β· Risk and...
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Risk Ratio and Odds Ratio
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Risk and Odds
β’ Risk is a probability as calculated from
ππππππππππ‘π¦ =πππ ππ’π‘ππππ
π΄πΏπΏ πππ π ππππ ππ’π‘πππππ
β’ Odds is opposed to probability, and is calculated from
ππππ =πππ ππ’π‘ππππ
πππ πππ»πΈπ ππ’π‘πππππ β’ Both are measures of likelihood but differ in
β’ Denominator
β’ 0 β€ Risk (or probability) β€ 1 whereas 0 β€ Odds β€ β
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Calculation Examples
β’ The probability and the odds of flipping a coin and getting a headβ’ Outcome: H
β’ Other outcome: T
β’ All possible outcome: H, T
β’ ππππππππππ‘π¦ =πππ ππ’π‘ππππ
π΄πΏπΏ πππ π ππππ ππ’π‘πππππ =
1
2= 0.5
β’ ππππ =πππ ππ’π‘ππππ
πππ πππ»πΈπ ππ’π‘πππππ =
1
1= 1: 1
β’ The probability and the odds of flipping a coin twice and getting two headsβ’ Outcome: HH
β’ Other outcome: HT, TH, TT
β’ All possible outcome: HH, HT, TH, TT
β’ ππππππππππ‘π¦ =πππ ππ’π‘ππππ
π΄πΏπΏ πππ π ππππ ππ’π‘πππππ =
1
4= 0.25
β’ ππππ =πππ ππ’π‘ππππ
πππ πππ»πΈπ ππ’π‘πππππ =
1
3= 1: 3
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Another Exampleβ’ The probability and the odds of having 2 short-hair kittens in a litter with 5 kittens
β’ Outcome: 2 short-hair kittens (S S)
β’ Other outcome: 3 long-hair kittens (L L L)
β’ All possible outcome: S S L L L
β’ ππππππππππ‘π¦ =πππ ππ’π‘ππππ
π΄πΏπΏ πππ π ππππ ππ’π‘πππππ =
2
5= 0.4
β’ ππππ =πππ ππ’π‘ππππ
πππ πππ»πΈπ ππ’π‘πππππ =
2
3= 2: 3 β 0.67
β’ The probability and the odds of having 4 short-hair kittens in a litter with 5 kittensβ’ Outcome: 4 short-hair kittens (S S S S)
β’ Other outcome: 1 long-hair kittens (L)
β’ All possible outcome: S S S S L
β’ ππππππππππ‘π¦ =πππ ππ’π‘ππππ
π΄πΏπΏ πππ π ππππ ππ’π‘πππππ =
4
5= 0.8
β’ ππππ =πππ ππ’π‘ππππ
πππ πππ»πΈπ ππ’π‘πππππ =
4
1= 4: 1 = 4
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Graphical Representations of Risk and Odds
Odds =Risk =
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Epidemiology
β’ Wiki definition: βEpidemiology is the study and analysis about the distribution and determinants of health and diseases in defined populationβ
β’ Many types of epidemiological studiesβ’ Randomized control study
β’ Cohort study
β’ Case-control study
https://en.wikipedia.org/wiki/Epidemiologyhttps://youtu.be/sdFYHSxq_qo
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Epidemiological studies: Some assessmentsβ’ Experimental study or Observational study
β’ If a researcher assigns mixture of participants to groups (i.e. randomization), itβs the experimental study
β’ If the researcher does not assign participants to any groups, but let participantsβ characteristics determined which group they should fall in, itβs an observational study
β’ Directionalityβ’ Forward study β the exposure is known, then follow up to see what outcomes
occurred
β’ Backward study β the outcomes are occurred, then exposure is determined
β’ Timingβ’ Prospective β the study starts before the outcome occurred
β’ Retrospective β the study starts after the outcomes occurred
https://youtu.be/sdFYHSxq_qo
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Epidemiology Study Types: Cohort study
β’ An observational study; forward directionality; prospective timing, or can be a retrospective timing
β’ Conceptually,β’ Start with a population of disease-free individuals
β’ Identify individuals that are exposed to a risk factor(s) and those that are NOT exposed to the same risk factor(s)
β’ then follow up both groups over time to find out the risk of specific outcomes (e.g. diseases) occurring in each individual
β’ Relative Risk is used to determined association between the exposure and the outcomes
β’ Hypotheses (tentative!!!)β’ Ho: Proportions of the outcome in exposed and unexposed groups are equal
β’ H1: Proportions of the outcome in exposed and unexposed groups are not equal
https://youtu.be/sGfIKmKMRdg
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Cohort study
time
Exposuree.g. smoking
smoke
not smoke
outcomee.g. lung cancer
Lung cancer
No Lung cancer
versus
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Breast cancer and Hormone replacement therapy in the million-women study
OutcomeExposure
Breast cancer
No breast cancer
Used HRT 1934 140,936
Never used HRT 2894 389,863
The Lancet 362: 419-427
Question: what is the risk of using HRT on breast cancer occurrence in women?
OutcomeExposure
Breast cancer
No breast cancer
Used HRT a b
Never used HRT c d
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Relative Risk (RR) and confident intervals
πππππ‘π = ππ(π π )Β±π§ΰ΅π π
π + π+
ΰ΅π ππ + π
πΆπΌ = ππππ€ππ πππππ‘ , ππ’ππππ πππππ‘
OutcomeExposure
Breast cancer
No breast cancer
Used HRT a b
Never used HRT c d
π ππππ‘ππ£π π ππ π =
ππ + ππ
π + π
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Interpretation of RR
β’ If RR = 1 or CI includes 1, there is no risk for the outcome to the exposed group nor the unexposed group
β’ If RR is more than 1 and CI does not includes 1, the relative risk of the outcome in the exposed group was increased by 1 β π π Γ 100% relative to the unexposed groupβ’ Or the risk of the outcome has RR times more likely to occur in
the exposed group than in the unexposed group
β’ If RR is less than 1 and CI does not includes 1 in its range, the relation risk of the outcome in the exposed group was reduced by 1 β π π Γ 100% relative to the unexposed group
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Breast cancer and Hormone replacement therapy in the million-women study
OutcomeExposure
Breast cancer
No breast cancer
Used HRT 1934 140,936
Never used HRT 2894 389,863
πππππ‘ π = ln 1.824 β 1.96ΰ΅140936 1934
1394 + 140936+
ΰ΅389863 28942894 + 389863
πππππ‘ π = ln 1.824 + 1.96ΰ΅140936 1934
1394 + 140936+
ΰ΅389863 28942894 + 389863
π ππ ππ»π π =1934
1934 + 140936= 0.0135
π ππ πππ π»π π =2894
2894 + 389863= 0.0074
π ππππ‘ππ£π π ππ π =π ππ ππ»π π
π ππ πππ π»π π=0.0135
0.0074= 1.824
πππππ‘ π = 0.5573πππππ‘ π = 0.6120πΆπΌ = π .5573, π .6120
πΆπΌ = 1.746,1.958
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Relative Risk (RR) and confident intervals
β’ RR=1.824 with CI=(1.746,1.958) RR is significantly different from 1, reject Ho then accept H1 stating that proportions of breast cancer in both groups are not equal
β’ Interpretation: Relative risk of breast cancer in women who used HRT is increased by |1-1.824|100%=82.4% relative to women who did not use HRT.
β’ Or the risk of breast cancer is 1.8 times more likely to occur in women who used HRT than in the women who did not use HRT.
OutcomeExposure
Breast cancer
No breast cancer
Used HRT 1934 140,936
Never used HRT 2894 389,863
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Cardiovascular diseases among users of estrogen with progestin as compared to nonusers
β’ π π = Ξ€.000295 .001414 = 0.208 with CI=(0.103,0.419) RR is significantly different from 1, reject Ho ten accept H1 stating that proportions of breast cancer in both groups are not equal
β’ Interpretation: Relative risk of major coronary disease is reduced by|1-0.208|100% = 79.2% in users of estrogen with progestin relative to users who has not used any hormone
β’ Or the risk of major coronary disease is 0.2 times less likely to occur in user who of estrogen with progestin than in users who do not use any hormone.
OutcomeExposure
Major coronary disease
No disease Risk
Estrogen with progestin 8 27,153 ππ + πππππ
= π. ππππππ
Not Used 431 304,313 ππππππ + ππππππ
= π. ππππππ
Adapted from NEJM 1996: 335-453
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Epidemiology Study Types: Case-Control studyβ’ An observational study; backward directionality; retrospective
timing
β’ Conceptually,β’ Start with the case, i.e. a group of individuals having the outcomes
(e.g. disease), and the control, i.e. a group of individuals not having the outcomes,
β’ then look back in time in both groups to find out what exposure(s) in both case and control that lead to specific outcomes or diseases
β’ Odds ratio is used to determined association
β’ Hypotheses (tentative!!!)β’ Ho: Proportions of the exposure in case and control are equal
β’ H1: Proportions of the exposure in case and control are not equal
https://youtu.be/sGfIKmKMRdg
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Case-Control study
time
Exposuree.g. smoking
Lung cancer
No Lung cancer
outcomee.g. lung cancer
Smoking
Not smoking
versus
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Hay fever and eczema in 11 years old children
Case is children with hay fever and control is children without hay fever. Exposure is the children has experienced eczema or not. What is the odds of children having hay fever will develop eczema compared to children without hay fever?
Hay feverEczema
Yes (case)
No (control)
Yes 141 420
No 928 13525
BMJ. 2000 May 27; 320(7247): 1468.
Hay feverEczema
Yes (case)
No (control)
Yes a b
No c d
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Odds ratio (OR) and confident intervals
πππππ‘π = ππ(ππ )Β±π§1
π+1
π+1
π+1
π
πΆπΌ = ππππ€ππ πππππ‘ , ππ’ππππ πππππ‘
Hey feverEczema
Yes (case)
No (control)
Yes a b
No c d
ππππ π ππ‘ππ =Ξ€π π
ΰ΅π π
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Interpretation of ORβ’ If OR = 1 or CI includes 1, the odds are equal for the case
group and the control group to experience the exposures
β’ If OR is more than 1 and CI does not includes 1, the odds of the exposure in the case group was higher relative to the control group
β’ If OR is less than 1 and CI does not includes 1, the odd of the exposure in the case group was lower relative to the control groupβ’ Normally, there should be switched the case and the control (NOT
shuffling data!) so that OR is greater than 1
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Odds ratio (OR) and confident intervals
πππππ‘ π = ππ(4.893)β1.961
141+
1
420+
1
928+
1
13525
πππππ‘ π = ππ(4.893)+1.961
141+
1
420+
1
928+
1
13525
Hey feverEczema
Yes (case)
No (control)
Yes 141 420
No 928 13525
ππππ π ππ‘ππ =ΰ΅141 928
ΰ΅420 13525
= 4.893
πππππ‘ π = 1.386πππππ‘ π = 1.790
πΆπΌ = π1.386, π1.790
πΆπΌ = 3.998,5.998
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Hay fever and eczema in 11 years old children
β’ OR = 4.893 with CI=(3.998,5.998) OR is significantly different from 1, reject Ho then accept H1 stating that proportions of eczema developed in the case and the control are not equal
β’ Interpretation: children having hay fever has the odds of 4.893 times to develop eczema compared to children without hay fever
Hay feverEczema
Yes (case) No (control)
Yes 141 420
No 928 13525
BMJ. 2000 May 27; 320(7247): 1468.
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Leukemia and parental smoking in pregnancy
β’ Case is patients with leukemia and control is patients without leukemia. Exposure is whether or not there is parental smoking during pregnancy.
β’ What is the odd of patients with leukemia (the case) to have been exposed to parental smoking in pregnancy?
LeukemiaSmoking
Yes (case) No (control)
Yes 87 147
No 201 508
https://youtu.be/Sec4fewyUig
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Leukemia and parental smoking in pregnancy
β’ ππ = Ξ€0.433 0.289 = 1.496 with CI=(1.096,2.042) OR is significantly different from 1, reject Ho then accept H1 stating that proportions of exposing to parental smoking in pregnancy in case and control are not equal
β’ Interpretation: In patients with leukemia (case group), the odds is 1.496 times to have been exposed to parental smoking in pregnancy
LeukemiaSmoking
Yes (case) No (control)
Yes 87 147
No 201 508
Odds ΰ΅ππ πππ = π. πππ ΰ΅πππ
πππ = π. πππ
https://youtu.be/Sec4fewyUig
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Choosing a test [after a thought!]β’ If you want to know whether or not the observation deviates from the theory choose the test for goodness of fitβ’ If you have only 2 outcomes, use binomial test; else, use 2 test
β’ If observations is less than 1000, you may find exact probability [you are using a computer, arenβt you?]; else, asymptotic probability will suffice it
β’ However, you want to find association between 2 nominal variables, a) You may choose 2 test or Fisherβs exact test for a test of independence if what you
really want to know is whether one or more categories in variable A affect one or more categories in variable B; or
b) You may choose 2 test for a test of homogeneity if you just want to know whether proportions of one category in variable A are equal in 2 or more groups (=variable B); or
c) You may choose to find relative risk if you want to know causality of the outcome (=one of two categories in variable A) in 2 different exposed groups from the cohort study; or
d) You may choose to find odds ratio if you want to know odds of the exposure (=one of two categories in variable A) in the case and control groups from the case-control study
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Strength of association by crosstab 2 test
β’ For 2x2 tables, i.e. 2 binary nominal variables
β’ Phi that is defined as π =π2
π
β’ Example: if 2 = 9.375 and n=150, then π =9.375
150= 0.25
β’ For table larger than 2x2 tables
β’ Cramerβs V that is defined as π =π2
π min(πβ1,πβ1)
β’ Example: if 2 = 126.105, n=566, r=4 rows and c=3 columns, so r-1=4-1=3 and
c-1=3-1=2, then π =126.105
566 2= 0.1114 = 0.334
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Interpretation of Phi and Cramerβs V
β’ Reminder: Phi and Cramerβs V are the measures of association between two nominal variables, i.e. how strong the association is β’ Both Phi and Cramerβs V do not identify the pattern nor direction
β’ To assess the pattern of association, interpret the column percentages in the bivariate table
β’ Here the guideline
Measure of association Strength of association
Between 0.00 and 0.10 Weak
Between 0.11 and 0.30 Moderate
Greater than 0.30 Strong