Biostat 200 Lecture 2 1. Today Discussion of data cleaning Probability 2

Biostat 200 Lecture 2

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Today

• Discussion of data cleaning• Probability

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Data cleaning

• Data cleaning is always necessary with a new data set

• Assume your data set has errors and your job is to find them

• The first step is to use tables and summary statistics and graphs to identify outliers and anomalies

• Outliers are defined as extreme values• We do NOT automatically remove outliers !!!

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Outliers – what do we do?• First consider if the value is physically possible• Example: Our original data set had a person who

was 3’4” tall . Yes, that is physically possible but fairly unusual.

• Look at the other variables for clues. We found (last year) age=3.

• For this one, we remove the entire observation from the analysis data set because of ineligibility

• We document this, and retain a copy of the original data set

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Outliers – what do we do?• If age had been =20, we might have asked the

interviewer about this value.• Another example – there were a few other

strange heights: 5’12”, 5’20”, 5’41” ... • Probably typos? Check original source document.• You can prevent some of this by programming

your data entry programs not to accept out of range values.

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Outliers – what do we do?• We also had 2 observations with weight=25,

30 pounds...• If we can’t explain but we are pretty sure that

these values are not reasonable, we might exclude these values (but not the whole observation unless we suspect poor data throughout!)

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Outliers – what do we do?• What about these high values?

0.0

5.1

.15

.2F

ract

ion

0 50 100 150 200Number of minutes typically exercise

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Outliers – what do we do?• What about outliers that seem reasonable?• May have large influence on some analyses• Be aware of them, do not exclude them.• Think about more robust analyses. E.g. which

measures of central tendency might you use?

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Data management strategies• Keep a .do file for all your recodes

• At the beginning of the .do file read in the original raw data• At the end of the file save the data to another filename

• Use comments, set off by ***s, to remind yourself why you are making these recodes

• Make .do files for your analyses• I often keep these separate from my recodes files

• Make a generic .do file to create value labels that you might use across data sets– label define 0 “Male” 1 “Female”– label define 0 “Negative” 1 “Positive” 2 “Indeterminate”

• Use the command include *.do to include the value label .do file in your recode .do file

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• use "H:\Biostat200\colddata_2011.dta", clear• summ age, detail

** children were not eligible for the study ** • drop if age<18

• include "H:\Biostat200\label defines.do"

• label values educ educl• label values sex sexl

• save "H:\Biostat200\colddata_2011_v2.dta"

Example .do file for recoding and labeling variable levels

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Basic probability

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Basic probability• Probability is the foundation of statistical inference

– Statistical inference is what is needed to make statements about the characteristics of the population from which a sample was drawn

– p-values and confidence intervals tell us how our sample might relate to the population

• Many of the entities we use daily are probabilities – e.g. the probability of breast cancer given they are BRCA1/2 positive

Population

Sample

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Basic probability

• Event – Result of an experiment or observation– Occurs or does not occur– Denoted by uppercase letters e.g. A,B, X– We will apply probability to events – i.e. we will

want to know the probability that an event occurs

– E.g. a disease occurrence, an extreme laboratory value

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Basic probability• Frequentist definition of probability If an

experiment is repeated n times under essentially identical conditions, and if the event A occurs m times, then as n grows large, the ratio m/n approaches a fixed limit that is the probability of A

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Basic probability• Probability of an event – relative frequency

of its occurrence in a large number of trials repeated under the same conditions– E.g. Probability of picking a red ball out of a bag

of red and black balls– Always lies between 0 and 1 (inclusive)– Denoted P(A) or P(X)

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Basic probability• Complement of an event, Ā or AC (read Not A or A

complement)– E.g. the event that the person does not have malaria– P(A)= 1-P(Ā)

• In epidemiology, we often write E for exposed and Ē for not exposed

• Ω is the universe, all the possible outcomes of an event• P(Ω) = P(A) + P(Ā) = 1

A

A

Ā

Ω

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Complement example

• Probability that someone has extremely drug resistant (XDR TB) versus they do not

• P(XDR TB+) + P(XDR TB-) = 1

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Basic probability• The intersection of 2 events is written A ∩ B• The intersection is when both A and B occur

– E.g. The event that a person has both malaria and pulmonary tuberculosis

– The probability that both occur is written P(A ∩ B)

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Basic probability• The union of 2 events is written A U B• The union is if either A or B or both occur

– E.g. The event that a person has either malaria or tuberculosis or both

– P(A U B) = P(A) + P(B) – P(A ∩ B)– The probability of A or B is the sum of their individual

probabilities minus the probability of their intersection

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Basic probability• Two events are mutually exclusive if they cannot

occur together– In English: for mutually exclusive events, the

probability of A or B occurring is the sum of their individual probabilities; both cannot occur together so P(A ∩ B) = 0

– In probability lexicon: P(A U B) = P(A) + P(B) - P(A ∩ B) = P(A) + P(B)

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Basic probability• Two events are mutually exclusive if they

cannot occur together– This is true for complements– E.g.

• Being pregnant and not pregnant • You cannot be both

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Basic probability• If A and B are mutually exclusive,

P(A U B) = P(A) + P(B)• This is the additive rule of probability• E.g.

P(HCV genotype 1) in the US = .7P(HCV genotype 2) in the US = .15

P(HCV genotype 3,4,6) = .15 P(HCV genotype 1 or 2) = .85

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Basic probability• The additive rule of probability can be applied

to three or more mutually exclusive events• If none of the events can occur together, thenP(A1 U A2 U … U An ) = P(A1) + P(A2) + … P(An)

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Probability summary• Complement: P(A)= 1-P(Ā)• Union: Prob A or B or both = P(A U B)

P(A U B) =P(A) + P(B) – P(A ∩ B)

• Intersection: Prob A and B = P(A ∩ B)

• For mutually exclusive events: P(A ∩ B)=0P(A U B) = P(A) + P(B) additive rule

• So A and Ā are mutually exclusive

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Basic probability example• A = the event that an individual is exposed to

high levels of carbon monoxide• B = the event that an individual is exposed to

high levels of nitrogen dioxide– What is the event A ∩ B called? What is that in

this example?– What is the event A U B called? What is it in this

example?– What is the complement of A?– Are A and B mutually exclusive?

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Basic probability example– A ∩ B is the intersection of A and B. It is the

event that the person is exposed to both gases.– A U B is the union of A and B. It is the event that

the person is exposed to one or the other or both.– Ac is the event that the person is not exposed to

carbon monoxide.– Are A and B mutually exclusive? Can they both

occur? Yes. So NOT mutually exclusive.

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Conditional probability• The probability that an event B will occur given

that event A has occurred– Notation: P(B|A)– Read: the probability of B given A

• Example: Probability of a person becoming infected with malaria given that he/she uses a bed net at night

• Event A is using a bed net• Event B is becoming infected with malaria

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Conditional probability• Multiplicative rule of probability

P(A ∩ B) = P(A) P(B|A)So P(B|A) = P(A ∩ B) / P(A)

• Example: P(becoming infected with malaria | use a bed net)Answer: P( Becoming infected and using a bed net ) /

P(using a bed net)= number of people who become infected with

malaria who use a bed net / number of people who use a bed net

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Probability example1992 U.S. birth statistics• Probability that mother’s age was ≤24 = 0.003 + 0.124 + 0.263 = 0.390 (What probability rule?)

• Given that a mother is under age 30, what is the probability that she is under age 20?P( Mother’s age<20 | Mother’s age<30 ) = P ( Mother’s age<20 and <30 ) / P(Mother’s age <30) = ( 0.003 + 0.124 ) / ( 0.003 + 0.124 + 0.263 + 0.290 ) = 0.127 / 0.68 = 0.187

Age of mother Probability

<15 0.003

15-19 0.124

20-24 0.263

25-29 0.290

30-34 0.220

35-39 0.085

40-44 0.014

45-49 0.001

Total 1.000

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Examples of conditional probabilities

• Relative risk is the ratio of 2 conditional probabilities

P(disease | exposed) / P(disease | not exposed)

• Odds also include conditional probabilities P(disease | exposed) / (1- P(disease | exposed))

P(disease | not exposed) / (1- P(disease | not exposed))

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Independence

• If the occurrence of B does not depend on A, – then P(B|A) = P(B)– Example: Probability of becoming infected with

malaria given that you wear a blue shirt = probability of becoming infected with malaria

– Then the multiplicative rule is P(A ∩ B) = P(A) P(B)– Example: coin tosses – the probability of a heads on

the 2nd throw is independent of the outcome on the first throw

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Independence

Note that independence ≠ mutual exclusivity!– Mutual exclusivity

• 2 events cannot both occur• P(A ∩ B) =0

– Independence • 2 events do not depend on each other• P(B|A)=P(B)• P(A ∩ B) = P(A) P(B)

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Law of Total Probability• The law of total probability:

P(B) = P(B ∩ A) + P(B ∩ Ā) P(B) = P(B|A)P(A) + P(B|Ā)P(Ā)

More generally P(B) = P(B ∩ A1) + P(B ∩ A2) + … + P(B ∩ An)

if P(A1 U A2 U … U An ) = 1

P(B) = P(B|A1)P(A1) + P(B|A2)P(A2) + … + P(B|An)P(An)

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Law of Total Probability• Helpful when you cannot directly calculate

a probability• Example:

– Suppose you know the TB prevalence in different areas and the population size in those areas, and you want to know the worldwide TB prevalence

– P(TB+) = P(TB+| live in lower income country)*P(live in lower income country) + P(TB+| live in upper income country)*P(live in upper income country)

– Weighted average of the 2 TB rates

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Diagnostic tests

• Diagnostic tests of disease are rarely perfect– True positives – the test is positive given the person has the

disease • The probability of this is P(T+|D+) = Sensitivity

– False positives – the test is positive although the person does not have the disease

– True negatives – the test is negative given the person does not have the disease

• The probability of this is P(T-|D-) = Specificity

– False negatives – the test is negative even though the person has the disease

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Diagnostic tests

• Sensitivity = P(T+|D+) = P(T+∩D+)/P(D+) = TP/(TP+FN)

• Specificity = P(T-|D-) = P(T-∩D-)/P(D-) = TN/(FP+TN)

TRUTH

D+ D-

Test T+ TP FP

T- FN TN

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Diagnostic tests

• Diagnostic test characteristics (sensitivity and specificity) are based on experiments in which the test is compared to a “gold standard”

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Diagnostic test validation example

• New biological markers of alcohol consumption are being developed. Phosphatidylethanol (PEth) is a metabolite of alcohol that is formed only in the presence of alcohol.

• We examined 77 HIV positives in Mbarara, Uganda. We followed them for 21 days and did daily breathalyzers and drinking surveys. If the breathalyzer result was ever >0 and/or the participant reported drinking, we considered this any alcohol consumption.

• We drew blood at the end of the 21-days to test for PEth.

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Diagnostic test example

• Number of positive PEth tests among those with any alcohol consumption in the prior 21 days >=10 ng/ml Sensitivity = 45/51 = 88.2%

• Number of negative PEth tests among the abstainers = Specificity

= 23/26 = 88.5%

“TRUTH”

Alc+ Alc-

Peth

Test

+ 45 3

- 6 23

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Diagnostic tests• The level of the cutoff for a diagnostic test can be set

to– Maximize sensitivity -- this will decrease specificity!

• This might be ideal if a follow up confirmatory test is easy and you want to be sure not to miss any positives

– Maximize specificity -- this will decrease sensitivity!• This might be necessary if there are grave ramifications of a false

positive test

• Receiver-operator curves illustrate this tension– The ROC curve plots the sensitivity versus the 1-specificity

for a test at every possible test cutoff

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ROC of PEth to detect alcohol consumption in persons with HIV in Mbarara, Uganda

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Application of laws of probability to diagnostic tests

• Suppose you have a panel of diagnostic tests and each give false positive results 2% of the time (98% specificity)

• If you test your patient with one of the tests and they do not have the disease, there is a 2% chance you’ll get a false positive result

• There is a 98% chance you will get the correct negative result.

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• If you give the patient 2 tests, what is the chance of at least 1 false positive?

• Possible results are:• You could get Neg Neg.

P(Neg test 1 ∩ Neg test 2) = 0.98*0.98=.9604• You could get Neg Pos

P (Neg test 1 ∩ Pos test 2) = 0.98*0.02=.0196• You could get Pos Neg

P (Pos test 1 ∩ Neg test 2) = 0.02*0.98=.0196• You could get Pos Pos

P (Pos test 1 ∩ Pos test 2) = 0.02*0.02=.000443


• All 4 of these possibilities add to 1.9604 + .0196 + .0196 + .0004 = 1

• P(1 or more test is pos) = (Neg test 1 ∩ Pos test 2) + (Pos test 1 ∩ Neg test 2) + P(Pos test 1 ∩ Pos test 2) = .0196 + .0196

+ .0004 =.0396

An easier way:P(1 or more test is pos) = 1-P(both tests are neg)

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• P(both tests are neg) = (Neg test 1 ∩ Neg test 2) =.98*.98

• So P(1 or more test is neg) = 1-.98*.98 = 0.0396• In general, P(At least one false positive)

= 1-P(no false positives occur over all tests)

= 1-P(test specificity)# of tests

Here = 1- 0.982

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• What is the probability of at least one false positive if 5 tests were run?

1-0.985 = 0.096• What if the false positive proportion was .05?

1-0.955 = 0.226• What is the probability of at least one false

positive if 10 tests were run (where P(FP=0.02))? 1-0.9810 = 0.183

• What if the false positive proportion was .05?1-0.9510 = 0.401

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Bayes’ theorem for diagnostic tests• Suppose you know from diagnostic testing that

– The sensitivity of a new rapid HIV antibody test (P(T+|HIV+)) is 0.96

– The specificity P(T-|HIV-)) of the test is 0.99

• You want to know the probability that someone with a positive test using this test is truly infected with HIV – What is P(HIV+|T+) ?

• This is called the Positive Predictive Value (PPV) of the test

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Bayes’ theorem

• P(A|B)=P(B|A)P(A) / P(B)

• Proof:– By definition of conditional probability– P(A|B)=P(A∩B)/P(B)

• P(A∩B) = P(A|B)*P(B)– P(B|A)=P(A∩B)/P(A)

• P(A∩B) = P(B|A)P(A) so P(A|B)*P(B) = P(B|A)P(A)rearrange to get P(A|B)=P(B|A)*P(A) / P(B)

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By Bayes’ theorem:P(HIV+|T+) = P(T+|HIV+)*P(HIV+) / P(T+) using

P(A|B)=P(B|A)P(A) / P(B)

Probability of being truly infected with HIV (HIV+) if you have a positive test result

Bayes’ theorem for diagnostic tests

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Want to know P(HIV+|T+)Instead we know:

Sensitivity P(T+|HIV+) and Specificity P(T-|HIV-) and P(T-|HIV+) = 1-sensitivity (false negatives) and P(T+|HIV-) = 1-specificity (false positives)


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P(HIV+|T+) = P(T+|HIV+)*P(HIV+) / P(T+)

P(T+|HIV+) = 0.96 (sensitivity) P(HIV+) in sub-Saharan Africa is = 0.02P(T+) = the overall chances of having a positive test

P(T+) = P(T+|HIV+) P(HIV+) + P(T+|HIV-) P(HIV-) by the law of total probability

= 0.96*0.02 + 0.01*0.98 P(HIV+|T+) = 0.96*0.02/(0.0192+0.0098) = 0.662


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The prevalence of HIV was assumed to be 2%So before testing, the probability that a randomly

selected person is infected with HIV is .02This is the prior probability.

The probability that someone who tests positive has HIV is .662

This is the posterior probabilityIt incorporates the information gained by doing the test

In reality, HIV tests have much higher sensitivity than 96% … So the PPV is higher

Prior and posterior probability

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What is P(HIV+|T+) in a population in which the HIV prevalence is 0.004?

P(HIV+|T+) = P(T+|HIV+)*P(HIV+) / P(T+) P(T+|HIV+)=0.96 P(HIV+) is =0.004

P(T+) = P(T+|HIV+) P(HIV+) + P(T+|HIV-) P(HIV-) = 0.96*0.004 + 0.01*0.996

P(HIV+|T+) = 0.96*0.004/(0.00384+0.0096) = 0.278


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Bayes’ theoremBayes’ theorem allows you to use what you

know about the conditional probability of one event on another to help you understand the inverse

P(A1| B) = P(A1 ∩ B) / P(B)

= P( B | A1 ) P(A1) / P(B)

= P( B|A1 ) P(A1) / (P(B|A1)P(A1) + P(B|A2)P(A2) )

Remember P(B) = P(B|A1)P(A1) + P(B|A2)P(A2) by the law of total probability

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For next time

• Read Pagano and Gauvreau– Chapter 6 (Review of today’s material)– Chapter 7

• Bring your textbook to lecture next Tuesday

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Documents

Biostat 200 Lecture 2 1. Today Discussion of data cleaning Probability 2