4: Probability. Rarely is it possible to make assertions about populations* with complete certainty.

Transcription

1 4: Probability Rarely is it possible to make assertions about populations* with complete certainty. * Statistical populations may be real or hypothetical see Chap 1. 4: Probability 1 Definitions of Probability p. 4.1 P Probability / expected proportion of an event P Three ways to conceive probabilities < Logically /(no. of events) / (possible occurrences) 4 kings out of 52 cards; ˆ Pr(King) = 4 / 52 =.0769 < Empirically (experience) Observe 75% cure rate; ˆ Probabilty of cure =.75 Not reliable when n small < Subjectively Dr. says you have a 50% chance of recovery Definitions of Probability 2

2 Selected Properties of Probabilities p. 4.2 P Notation: < A / an event e.g, let A represent pet owning < Pr(A) / probability of A e.g., Pr(A) =.30 P Probabilities range from 0 to 1 < 0 # Pr(A) # 1 P Complements <} / the complement ( opposite ) of A e.g., } = not owning a pet < Law of complements: Pr(}) = 1! Pr(A) e.g., Pr(}) = 1!.30 =.70 Selected Properties of Probabilities 3 Properties (cont.) p. 4.2 P Random variable / quantity that varies depending on chance P Types of random variables < Discrete Whole number increments only e.g., number of asthma cases in a sample < Continuous Fractional values possible e.g., mean birthweight in a sample Properties (cont.) 4

3 Binomial Distributions p. 4.3 PApplies to some discrete random variables PBernoulli trial / random event characterized by success or failure < e.g., whether a coin turns up heads or tails < whether a person selected at random is a smoker < whether a particular treatment works PBinomial distribution / number of successes in n independent Bernoulli trials < number of heads in n flips < no. of smokers in sample of n < no of cures in n treatments Binomial Distributions 5 Portrait of Daniel Bernoulli 6

4 Another portrait of Dan 7 Binomial Family p. 3.4 PBinomial = a family w/ members identified by two parameters: < n / no. of Bernoulli trials < p / probability of success of trial PIllustrative example < X / no. of smokers in sample of 10; population prevalence =.15 < n = 10 < p =.15 PNotation: X~b(n,p) where < ~ / distributed as < b / binomial Binomial Family 8

5 Binomial Expectation and Variance p. 4.4 P :/ expected no. of successes <: = np [Formula 4.1] < Illustrative example: X~b(10,.15) : = (10)(1.5) = 1.5 PF² / variance of X <F² = npq where q = 1!p [Formula 4.2] < Illustrative example: X~b(10,.15) F² = (10)(.15)(1!.15) = Binomial Expectation and Variance 9 Choose Function n Ci = n! i!( n i)! p. 4.5 [Formula 4.3] P n C i / no. of ways to choose i items out of n < e.g., 3 C 2 = no. of ways to choose 2 items of 3 < Items A, B, and C < By logic:{a, B}, {A, C}, {B,C}; ˆ 3 C 2 = 3 PFactorial function: n! = n(n!1)(n!2)þ1 < e.g., 4! = (4)(3)(2)(1) = 24 < 1! = 1 and 0! = 1 (by definition) P 3 C 2 by formula: 3 C 2 = 3! = ( 2!)( 3 2)! ( 2 1)( 1) 6 = = 2 3 Choose Function 10

6 i Pr( X = i) = C p q Binomial Formula n i n i p. 4.5 [Formula 4.4] where X = binomial random variable i = given number of successes nc i / n Choose i (formula 4.3) n = number of trials (parameter) p = probability of success each trial (parameter) q = 1!p Binomial Formula 11 i Pr( X = i) = C p q n Illustrative Example i n i p. 4.6 PIllustrative example: n = 4 treatments p =.75 (probability of success each treatment) q = 1!.75 =.25 PWhat is probability of seeing 0 successes? (i = 0) Pr(X = 0)= 4C 4!0 4C 0 = (4!) / 4!) = (4@3@2@1) / 4@3@2@1) = = 1 (anything to the 0 power is 1).25 4!0 =.25 @.25 =.0039 ˆ Pr(X = 0) = =.0039 where n C i n! = i!( n i)! Illustrative Example 12

7 Illustrative Example (cont.) X~b(4,.75) Pr(X = 1) = 4 C 4!1 4C 1 = (4!) / (1!@3!) = (4@3@2@1) / 3@2@1) = = !1 =.25 =.0156 ˆ Pr(X=1) =.0469 i Pr( X = i) = C p q n i n i Illustrative Example (cont.) 13 Illustrative example (cont.) X~b(4,.75) p. 4.6 PPr(X = 2) = 4 C 4!2 = (6)(.5625)(.0625) =.2109 PPr(X = 3) = 4 C 4!3 = (4)(.4219)(.25) =.4219 PPr(X = 4) = 4 C 4!4 = (1)(.3164)(1) =.3164 Illustrative example (cont.) 14

8 Probability Distribution for X~b(4,.75) p. 4.6 Pr(X = 0)=.0039 Pr(X = 1) =.0469 Pr(X = 2) =.2109 Pr(X = 3) = Pr(X = 4) = No. of Successes Probability Distribution for X~b(4,.75) 15 Area Under Bar = Probability p. 4.6 e.g., For X~b(4,, area corresponding to Pr(X=2): height width = = No. of Successes Area Under Bar = Probability 16

9 Cumulative Probability Function p. 4.7 P Cumulative probability = probability less than or equal to given value P Illustrative example: X~b(4,.75) < Pr(X # 0) = Pr(X = 0) =.0039 < Pr(X # 1) = Pr(X # 0) + Pr(X = 1) = =.0508 < Pr(X # 2) = Pr(X # 1) + Pr(X = 2) = =.2617 < Pr(X # 3) = Pr(X # 2) + Pr(X = 3) = =.6836 < Pr(X # 4) = Pr(X # 3) + Pr(X = 4) = = Cumulative Probability Function 17 Cumulative Probability = Left Tail X~b(4,.75) e.g., Pr(X # 2) = = No. of Successes Cumulative Probability = Left Tail 18

10 Normal Distribution (p. 4.8) P Applies to some continuous random variables P Recognized as smooth bell-shaped curve < cf., chunky binomial histogram functions! < Area under function [still] = probability P Curve properties < Total area under curve = 1 < Probability of any exact value = 0 < Pr(x 1 < X < x 2 ) = area under curve between points Normal Distribution 19 Area Under the Curve (pp ) PNormal curve shaped so points of inflection at ±1F (arrows) PLandmarks : ± F encompasses 68% area under curve : ± 2F encompasses 95% area under curve : ± 3F encompasses almost all area under curve inflection inflection 68% 95% :!2F :!F : :+F :+2F Area Under the Curve 20

11 Use Common Sense When Determining Areas! :!2F :!F : :+F :+2F Use Common Sense When Determining Areas! 21 Left Tail = Cumulative Probability Use Law of Complements to Determine Right-Tail Area.025 1!.025 =.975 :!2F :!F : :+F :+2F Left Tail = Cumulative Probability 22

12 Illustrative Example: Wechsler IQ Scores p. 4.9 PWechsler IQ scores normally distributed with : = 100, F = 15 PNotation: X~N(:, F) / X is distributed as a normal random variable with mean : and standard deviation F < e.g. X~N(: = 100, F = 15) Illustrative Example: Wechsler IQ Scores 23 Standard Normal (Z) Variable (p. 4.10) P Z / standard anormal random variable = normal random variable with mean 0 and standard deviation 1 Z~N(0,1) PPercentile / cumulative probability = left tail region percentile p is greater than or equal to p% of distribution e.g., 50 th percentile on Z = 0 PNotation: z p / p th percentile on Z e.g., z.5 = Standard Normal (Z) Variable 24

13 Don t Forget These Landmarks on Z (p. 4.10) P 2.5 th percentile: z.025 =!1.96.!2 P 97.5 th percentile: z.975 = % z.025 z.975 Don t Forget These Landmarks on Z 25 Other Points 6Use Z Tables p PZ table Appendix in Reader < Columns and rows determine Z value (x axis) < Body = cumulative probabilities (area under left tail) PDetermining probabilities from z < e.g., z = 1.00 < Left tail region ( p)=.8413 < Right tail region ( q) = 1!.8413 =.1587 < Draw curve & shade region! < Notation z.8413 = Other Points 6Use Z Tables 26

14 Determining Z Percentiles p PWhat is 95 th percentile on Z? PDraw curve and work backwards PUse closest value (or interpolate) Pz.9495 = 1.64 Pz.9505 = 1.65 PTherefore, z.95 between 1.64 and Determining Z Percentiles 27 PSubtract : & divide by F Z scores p x µ z = σ PDistance from mean in standard deviation units PExample: Wechsler IQ score of 115: X~N(100, 15) PWhen X = 115, z = (115!100) / 15 = 1 Formula Z scale X scale Z scores 28

15 Application of Normal Distribution p PDetailed instruction on Reader p PFour steps < Draw normal curve < Mark middle of curve with : < Turn value into z score < Place z score on curve and determine relevant probability with Z table Application of Normal Distribution 29