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= How do I calculate and interpret conditional probabilities? = | Gigerenzer (2002) suggests a way to obtain conditional probabilities using frequencies in a decision tree. Cortina and Dunlap (1997) give an example evaluating the detection rate of a test (positive/negative result) to detect schizophrenia (disorder). To do this one fixes the following: The base rate of schizophrenia in adults (2%) The test will correctly identify schizophrenia (give a positive result) on 95% of people with schizophrenia The test will correctly identify normal individuals (give a negative result) on 97% of normal people. Despite this we can show the [attachment:bayes.doc test is unreliable]. This is a more intuitive way of illustrating the equivalent Bayesian equation: $$\mbox{P(No disorder|+ result) = }\frac{\mbox{P(No disorder) * P(+ result | No disorder)}}{\mbox{P(No disorder) * P(+ result | No disorder) + P(Disorder) * P(- result | Disorder)}}$$ A talk with subtitles further illustrating aspects of conditional probabilities given by Ted Donnelly (Oxford), a geneticist, is available for viewing [http://blog.ted.com/2006/11/statistician_pe.php here.] * [attachment:bayes2.doc:More on Bayes theorem:Illustration of priors and likelihoods] __References__ |
How do I calculate and interpret conditional probabilities?
Gigerenzer (2002) suggests a way to obtain conditional probabilities using frequencies in a decision tree.
Cortina and Dunlap (1997) give an example evaluating the detection rate of a test (positive/negative result) to detect schizophrenia (disorder).
To do this one fixes the following:
The base rate of schizophrenia in adults (2%)
The test will correctly identify schizophrenia (give a positive result) on 95% of people with schizophrenia
The test will correctly identify normal individuals (give a negative result) on 97% of normal people.
Despite this we can show the [attachment:bayes.doc test is unreliable].
This is a more intuitive way of illustrating the equivalent Bayesian equation:
$$\mbox{P(No disorder|+ result) = }\frac{\mbox{P(No disorder) * P(+ result | No disorder)}}{\mbox{P(No disorder) * P(+ result | No disorder) + P(Disorder) * P(- result | Disorder)}}$$
A talk with subtitles further illustrating aspects of conditional probabilities given by Ted Donnelly (Oxford), a geneticist, is available for viewing [http://blog.ted.com/2006/11/statistician_pe.php here.]
- [attachment:bayes2.doc:More on Bayes theorem:Illustration of priors and likelihoods]
References
Cortina JM, Dunlap WP (1997) On the logic and purpose of significance testing Psychological methods 2(2) 161-172.
Gigerenzer G. (2002) Reckoning with risk: learning to live with uncertainty. London: Penguin.