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BIOSTAT 3

BIOSTAT 3. Three tradition views of probabilities: Classical approach: make certain assumptions (such as equally likely, independence) about situation. [roll die P(2) = 1/6 = 0.167]

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BIOSTAT 3

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  1. BIOSTAT 3 • Three tradition views of probabilities: • Classical approach: make certain assumptions (such as equally likely, independence) about situation. [roll die P(2) = 1/6 = 0.167] • Relative frequency: assigning probabilities based on experimentation or historical data. [light 100 firecrackers, 80 pop, P(firecracker pops) = 80/100 = 0.8] • Subjective [Bayesian] approach: Assigning probabilities based on the assignor’s judgment. [Based on my past teaching experience I believe that P(a student in this class passes) = 0.98] As a student, would you change this probability if you made a 15 on the midterm exam?

  2. BIOSTAT 3 • Given k possible (mutually exclusive and collectively exhaustive) outcomes to an experiment {O1, O2, …, Ok}, the probabilitiesassigned to the outcomes must satisfy these requirements: • The probability of any outcome is between 0 and 1 • i.e. 0 ≤ P(Oi) ≤ 1 for each i, and • The sum of the probabilities of all the outcomes equals 1 • i.e. P(O1) + P(O2) + … + P(Ok) = 1

  3. BIOSTAT 3 • There are several types of combinations and relationships between events: • Complement of an event [everything other than that event] • Intersection of two events [event A and event B] or [A*B] • Union of two events [event A or event B] or [A+B] • Conditional probability P(A/B): read the probability of A given B has occurred.

  4. BIOSTAT 3 • Conditional Probability • Joint Probability (Multiplication rule) • Additional rule

  5. BIOSTAT 3 • Independent events: the probability of one event is not affected by the occurrence of the other event. • Two events A and B are said to be independent if • P(A|B) = P(A) • and • P(B|A) = P(B) • P(you have a flat tire going home/radio quits working)

  6. BIOSTAT 3 • A and B are mutually exclusive if the occurrence of one event makes the other one impossible. This means that • P(A and B) = P(A * B) = 0 • The addition rule for mutually exclusive events is • P(A or B) = P(A) + P(B) • Only if A and B are Mutually Exclusive.

  7. BIOSTAT 3 • Example: Investigation of the relationship of family history and the age at onset of bipolar disorder. Study included 318 subjects who had bipolar disorder resulting in the following data:

  8. BIOSTAT 3 • Calculate the following probabilities • P(E) = 141/318 = .4434 • P(A) = 63/318 = .1981 • P(A/E) = 28/141 = .1986

  9. BIOSTAT 3 • P(E*A) = 28/318 = .0881 or P(E*A) = P(E) * P(A/E) = (.4434)*(.1986) = .0881 • P(E/A) = 28/63 = .4444 or P(E/A) = P(E*A)/P(A) = (.0881)*(.1981) = .4447 • Why are these different??? • NOTE: Keep in mind that all of these probabilities are based on the fact that the subject does in fact have (or will have) bipolar disorder. HW PROBLEM: 3.4.1

  10. Breaking News: New test for early detection of cancer has been developed. Let C = event that patient has cancer Cc = event that patient does not have cancer + = event that the test indicates a patient has cancer • = event that the test indicates that patient does not have cancer Clinical trials indicate that the test is accurate 95% of the time in detecting cancer for those patients who actually have cancer: P(+/C) = .95 but unfortunately will give a “+” 8% of the time for those patients who are known not to have cancer: P(+/ Cc) = .08 • It has also been estimated that approximately 10% of the population have cancer and don’t know it yet: P(C) = .10 • You take the test and receive a “+” test results. Should you be worried? P(C/+) = ?????

  11. BIOSTAT 3 • What we know.P(+/C) = .95 P(+/ Cc) = .08 P(C) = .10From these probabilities we can findP(-/C) = .05 P(-/ Cc) = .92 P(Cc) = .90 • P(C/+) = ???

  12. BIOSTAT 3 The probabilities P(C) and P(CC) are called prior probabilities because they are determined prior to the decision about taking the cancer test. The conditional probability P(C | “test results”) is called a posterior probability (or revised probability), because the prior probability is revised after the results of the cancer test is known.

  13. BIOSTAT 3 • P(+/C) = .95 P(+/ Cc) = .08 P(C) = .10 • P(-/C) = .05 P(-/ Cc) = .92 P(Cc) = .90 • Calculate • P(C*+) = P(C)*P(+/C) = (.1)*(.95) = .095 • P(C*-) = P(C)*P(-/C) = (.1)*(.05) = .005 • P(Cc*+) = P(Cc)*P(+/Cc) = (.9)*(.08) = .072 • P(Cc*-) = P(Cc)*P(-/Cc) = (.9)*(.92) = .828

  14. BIOSTAT 3 • P(C/+) = (.095)/(.167) = .5689 • P(C/-) = (.005)/(.833) = .0060 • P(Cc/+) = (.072)/(.167) = .4311 • P(Cc/-) = (.828)/(.833) = .9940

  15. BIOSTAT 3 • HW Problems: • Work Example 3.5.1 in text to verify you can get the correct answer. • Work Exercise 3.5.1 Note: Sensitivity = P(+/C) Specificity = P(-/Cc) Predictive Value Positive = P(C/+) Predictive Value Negative = P(Cc/-)

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