Biostat 200 Lecture 2

Trimmed mean (chapter 3) In order to remove extreme values that might affect the mean, you can calculate a trimmed mean Remove the bottom 5% and the top 5% of values. Be careful not to remove too much data – sometimes the 5 th percentile is also the 10 th percentile... There is no easy way to do this in Stata (I went in did it by hand) Extra credit for doing this on Assignment 1

Grouped data (chapter 3) Sometimes you are given data in aggregate form The data consist of frequencies of each individual value or range of values For example:

Grouped mean The mean uses the midpoint of each group For the highest group, the use the midpoint between the cutpoint and the maximum Grouped Mean f i = the frequency in the i th group m i = the midpoint of the i th group = (25* * * *98) / 268 = cells/mm 3 (mean from original data was 296.9)

Grouped standard deviation The standard deviation = sqrt ( ( ) 2 *40 + ( ) 2 *72 + ( ) 2 *58 + ( ) 2 *98 ) / 267 ) = cells/mm 3 (SD from original data was 255.4)

Basic probability

Why – probability is the foundation of statistical inference – Methods needed to infer the characteristics of the population from which a sample was drawn Pagano and Gavreau, Chapter 6 Population Sample

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 Pagano and Gavreau, Chapter 6

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 Pagano and Gavreau, Chapter 6

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 child has malaria at time of study – Always lies between 0 and 1 (inclusive) – Denoted P(A) or P(X) Pagano and Gavreau, Chapter 6

Basic probability Complement of an event, Ā or A C (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 Pagano and Gavreau, Chapter 6 Ω

Complement example Probability that someone has extremely drug resistant (XDR TB) versus they do not P(XDR TB+) + P(XDR TB-) = 1

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) Pagano and Gavreau, Chapter 6

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 Pagano and Gavreau, Chapter 6

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) Pagano and Gavreau, Chapter 6

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 Pagano and Gavreau, Chapter 6

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 =.7 P(HCV genotype 2) in the US =.15 P(HCV genotype 3,4,6) =.15  P(HCV genotype 1 or 2) =.85 Pagano and Gavreau, Chapter 6

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, then P(A 1 U A 2 U … U A n ) = P(A 1 ) + P(A 2 ) + … P(A n ) Pagano and Gavreau, Chapter 6

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)=0 P(A U B) = P(A) + P(B) additive rule So A and Ā are mutually exclusive, Pagano and Gavreau, Chapter 6

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? Pagano and Gavreau, Chapter 6

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. – A c 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. Pagano and Gavreau, Chapter 6

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

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

Probability example 1992 U.S. birth statistics Probability that mother’s age was ≤24 = = (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) = ( ) / ( ) = / 0.68 = Age of motherProbability < Total1.000

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)) An odds ratio is the ratio of the two odds above

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 2 nd throw is independent of the outcome on the first throw Pagano and Gavreau, Chapter 6

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) Pagano and Gavreau, Chapter 6

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 ∩ A 1 ) + P(B ∩ A 2 ) + … + P(B ∩ A n ) if P(A 1 U A 2 U … U A n ) = 1 P(B) = P(B|A 1 )P(A 1 ) + P(B|A 2 )P(A 2 ) + … + P(B|A n )P(A n ) Pagano and Gavreau, Chapter 6

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 Pagano and Gavreau, Chapter 6

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 Pagano and Gavreau, Chapter 6

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+D+D+ D-D-D-D- Test T+T+T+T+TPFP T-T-T-T-FNTN Pagano and Gavreau, Chapter 6

Diagnostic tests Diagnostic test characteristics (sensitivity and specificity) are based on experiments in which the test is compared to a “gold standard” Pagano and Gavreau, Chapter 6

Diagnostic test validation example New biological markers of chronic alcohol consumption are being developed. Phosphatidylethanol (PEth) is a metabolite of alcohol that is formed only in the presence of alcohol. Researchers examined a group of alcoholics being admitted to inpatient alcohol detoxification (n=56) and a group of abstainers in a closed psychiatric ward (n=35). Hartmann, Addiction Biology, 2006 Pagano and Gavreau, Chapter 6

Diagnostic tests example Number of positive PEth tests among the alcoholics using a cutoff of 0.36 µmol/l = Sensitivity = 53/56 = 94.6% Number of negative PEth tests using a cutoff of 0.36 µmol/l among the abstainers = Specificity = 35/35 = 100% Hartmann, Addiiction Biology, 2006 “TRUTH” Alc+Alc- PethTest Pagano and Gavreau, Chapter 6

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 specificity for a test at every possible test cutoff Pagano and Gavreau, Chapter 6

Diagnostic tests example ROC of PEth to detect maximum breathalyzer result over 21 days ≥.1% g/l in Mbarara, Uganda Pagano and Gavreau, Chapter 6

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 Pagano and Gavreau, Chapter 6

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) Pagano and Gavreau, Chapter 6 Bayes’ theorem for diagnostic tests

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)  P(A|B)=P(B|A)*P(A) / P(B) Pagano and Gavreau, Chapter 6

By Bayes’ theorem: P(HIV+|T + ) = P(T + |HIV+)*P(HIV+) / P(T + ) P(T + |HIV+) = 0.96 (sensitivity) P(HIV+) in sub-Saharan Africa is = 0.02 P(T + ) = P(T+|HIV+) P(HIV+) + P(T+|HIV-) P(HIV-) by the law of total probability = 0.96* *0.98 P(HIV+|T + ) = 0.96*0.02/( ) = Pagano and Gavreau, Chapter 6 Bayes’ theorem for diagnostic tests

The prevalence of HIV was assumed to be 2% So before testing, the probability that a randomly selected person is infected with HIV is.02 This is the prior probability. The probability that someone who tests positive has HIV is.662 This is the posterior probability It incorporates the information gained by doing the test Pagano and Gavreau, Chapter 6 Prior and posterior probability

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* *0.996 P(HIV+|T + ) = 0.96*0.004/( ) = Pagano and Gavreau, Chapter 6 Bayes’ theorem for diagnostic tests

Bayes’ theorem Bayes’ theorem allows you to use what you know about the conditional probability of one event on another to help you understand the inverse P(A 1 | B) = P(A 1 ∩ B) / P(B) = P( B | A 1 ) P(A 1 ) / P(B) = P( B|A 1 ) P(A 1 ) / (P(B|A 1 )P(A 1 ) + P(B|A 2 )P(A 2 ) ) Remember P(B) = P(B|A 1 )P(A 1 ) + P(B|A 2 )P(A 2 ) Pagano and Gavreau, Chapter 6

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