Bayes Theorem. Bayes Theorem- Example. Evaluation of Medical Screening Procedure. Evaluation of Medical Screening Procedure

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Bayes’ Theorem

P(C | A) P(A) P(C | A) P(A) + P(C | B) P(B) = P(A | C) = P(E | B) P(B)

P(E | B) P(B) + P(E | A) P(A)

= P(B | E) = P(D | A) P(A) P(D | A) P(A) + P(D | B) P(B) = P(A | D) =

Bayes’ Theorem-

Example

Evaluation of Medical Screening Procedure

Cost of procedure is $1,000,000

Data regarding accuracy of the procedure is: Prob (+ test result | patient has diabetes) = .90 Prob (+ test result | patient has leukemia) = .95 Prob (+ test result | patient has neither) = .07

Bayes’ Theorem-

Example

Prob (Patient has diabetes) = .02 Prob (Patient has leukemia) = .01 Prob (Patient has neither) = .97 We also know that in the general population:

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Bayes’ Theorem-

Example

What probabilities do we really need to know?

Prob (patient has diabetes | + test result) = ??? Prob (patient has leukemia | + test result) = ??? Prob (patient has neither | + test result) = ???

How can we find these probabilities?

Bayes’ Theorem-

Example

Prob (patient has diabetes | + test result) =

P (+ test result | diabetes) P(diabetes) P (+ test result | diabetes) P(diabetes) + P (+ test result | leukemia) P(leukemia) + P (+ test result | neither) P(neither)

Bayes’ Theorem-

Example

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Bayes’ Theorem-

Example

Bayes’ Theorem-

Example

Bayes’ Theorem-

Example

To demonstrate that these probs. are correct, look at a random sample of 100,000 people

2000 1000 97,000 1800 950 6790 200 50 90,210 100,000 9,540 90,460 D L N 90% 95% 7%

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Bayes’ Theorem-

Example

Prob (+ test result | patient has diabetes) = .90 Prob (+ test result | patient has leukemia) = .95 Prob (+ test result | patient has neither) = .07

Prob (Patient has diabetes) = .02 Prob (Patient has leukemia) = .01 Prob (Patient has neither) = .97 Remember that:

and

Bayes’ Theorem-

Example

May also want to be concerned about false negatives:

Prob (Patient has diabetes | - ) = 200/90460 Prob (Patient has leukemia | - ) = 50/90460

Bayes’ Theorem-

Example

Cost of procedure is $1,000,000

Data regarding accuracy of the procedure is: Prob (+ test result | patient has diabetes) = .90 Prob (+ test result | patient has leukemia) = .95 Prob (+ test result | patient has neither) = .07

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Bayes’

Theorem-Practice problem

Product Manager Problem

There are two types of probabilities: Actual: P(superior product)

Survey results: P(Survey says superior product)

Bayes’

Theorem-Practice problem

Product manager Problem - p. 21

Find P(superior product|survey says not superior) Part a) First survey is negative

Part b) Second survey is negative

Find P(superior product|survey says not superior) Using results from (a) as starting point in (b) – use condit. prob. as new marginals

Quick Review of Discrete and

Continuous Distributions

Discrete Probability Distribution

Continuous Probability Distribution Random Variable

Discrete Continuous

Binomial

Normal (and Normal Approximation to Binomial)

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Random Variable

Variable that takes on a set of unique numerical values which represent every possible outcome of a random process

Discrete

Continuous

--Values are usually integers (finite number of values)

Values are real #s (infinite number of values within interval)

Probability Distribution

Discrete – Collection of probabilities that describe frequency with which random variable takes on each of its possible values

1. 0 <= P(x) <= 1 2. ΣP(x_i) = 1

i

Called probability mass function

Cumulative distribution

function

Probability that x takes on values <= X F(x) = P( x <= X) =

Σ

P(x_i)

xi<=X

1. 0 <= F(x) <= 1 2. F(a) <= F(b) if a < b

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Discrete Prob. Distribution

example

There are 20 students in a class, where the ages are distributed as follows:

Age 18 19 20 21 24 Frequency 5 6 4 4 1

Discrete Prob. Distribution

example

There are 20 students in a class, where the ages are distributed as follows:

Age 18 19 20 21 24 Frequency 5 6 4 4 1 P(x) .25 .30 .20 .20 .05 20 1.0 F(x) .25 .55 .75 .95 1.00

Discrete Prob. Distribution

example

Find

1. P(x = 19) = .30 2. P(x <= 21) = .95 3. P(19 < x <= 22) = .40 4. The mean age of students

in the class

µ= Σ x_iP(x_i) = 19.6

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Discrete Prob. Distribution

example

To find variance: σ2 ₌Σ _(x i-µ) 2 P(xi) i = E (x_i2 _{) -}µ2 Where E (x_i2 _{) =}Σ_x i2P(xi) (Easier to compute) i

Discrete Prob. Distribution

example

σ2 ₌Σ _(x i-µ) 2 P(xi) σ2 _{=.25(18-19.6 )}2 _{+ .3(19-19.6)}2_+.2(20-19.6)2 +.2(21-19.6)2_{+.05(24-19.6)}2 _{= 2.14} = E (x_i2 _{) -}µ2 _{= 386.3-384.16 = 2.14} E (x_i2 _{) =}Σ_x i2P(xi) = 182(.25) + 192(.3) + 202_{(.2) + 21}2_{(.2) + 24}2_(.05) σ2 1. 2.

Discrete Prob. Distribution

example

√

To find standard deviation: σ = σ2 _{= 2.14 = 1.46}_√

µ= Σ x_iP(x_i) = 19.6 = E (x_i2 _{) -}µ2 _{= 2.14}

σ2

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Binomial distribution

1) Only 2 possible outcomes, success or failure 2) There are n independent trials

3) Prob. of S or F remains constant

4) Random variable represents # of successes in n trials

Binomial distribution

r = # of successes n = # of trials p = probability of success P(x=r) = Cn_(p)r _(1-p)n-r Where Cn₌ n! r!(n-r)! r r µ= np σ2_{= np(1-p)}

Binomial distribution

Quiz Problem example:

r = # of successes n = # of trials = 9

p = probability of success = 0.4

P(x=3) = C9_(0.4)3 _{(1- 0.4)}9-3

3

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Binomial distribution

Use Table C from textbook - look under n=9 and prob. = 0.4 p=0.40 r 1 .9899 2 .9295 3 .7682 4 .5174 5 .2666 6 .0994 7 .0250 8 .0003 prob. of x=> 1 prob. of x=> 2 prob. of x=> 3 prob. of x=> 4 prob. of x=> 5 prob. of x=> 6 prob. of x=> 7 prob. of x=> 8

Binomial distribution

p=0.40 r 1 .9899 2 .9295 3 .7682 4 .5174 5 .2666 6 .0994 7 .0250 8 .0003 1. P(x=3) = P(x=>3) - P(x=>4) = .7682 - .5174 = .2508 2. P(x<5) = 1 - P(x=>5) = 1 -.2666 = .7334

Binomial distribution

p=0.40 r 1 .9899 2 .9295 3 .7682 4 .5174 5 .2666 6 .0994 7 .0250 8 .0003 3. P(3<x<8) = P(x=>4)-P(x=>8) = .5174 - .0003 = .5171 4. µ= np = 9(.4) = 3.6

5. Most likely = value with highest probability

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Normal Distribution

(Continuous)

P(a <= x <= b) = F(b) - F(a) for b>a

Infinite number of values for x within an interval; values within a range

Normal Distribution: Z = x -_σµ 0 1 2 3 4 -1 -2 -3 -4 N (µ, σ)

Normal Distribution

(Continuous)

Production process fills 46-oz. fruit juice cans Juice filling cans is N(46.5 oz., 0.2 oz.) a) State law requires that no more than 1 in 20 cans have less than 46 oz. Does this process meet the state law?

P(x<46) <= 1/20 = .05 ????

Normal Distribution

(Continuous) Normal Distribution: Z = x -_σµ 0 1 2 3 4 -1 -2 -3 -4 = 46 - 46.5 .2 = - 2.5 46.5 Look up -2.5 in Normal Table Z = 1 - .9938 = .0062 < .05

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Normal Distribution

(Continuous)

Production process fills 46-oz. fruit juice cans Juice filling cans is N(46.5 oz., 0.2 oz.)

b) P(46.3<=x<=46.6) = ??? c) P(x=>46.8) = ???

Normal Distribution

(Continuous)

b) P(46.3<=x<=46.6) = ??? Z = x -_σµ 46.3 - 46.5 .2 = -1.0 = Z = x -_σµ 46.6 - 46.5 .2 = .5 =

Normal Distribution

(Continuous)

b) P(46.3<=x<=46.6) = P(-1.0<=x<=.5) = .6915 - (1-.8413) = .5328 0 1 2 3 4 -1 -2 -3 -4

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Normal Distribution

(Continuous)

c) P(x=>46.8) = ??? Z = x -_σµ 46.8 - 46.5

.2 = 1.5

=

P(x=>1.5) = 1 - .9332 = .0668

Normal Approximation to the

Binomial

This applies when you have a Binomial setting but for very large n values. Normal Distribution: 0 1 2 3 4 -1 -2 -3 -4 N (µ, σ) Binomial Distribution: B (n, p ) Normal Approximation: µ= np σ2_{= np(1-p)} σ = np(1-p)

Normal Approximation to the

Binomial

Out of graduating Seniors, 80% have jobs by graduation. Take a sample of 200. a) What is the likelihood that at least 125

have jobs?

b) That between 170 and 185 have jobs? Binomial setting (S vs. F), but very large n

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Normal Approximation to the

Binomial

n = 200 p = .80 µ= np = 200 (.8) = 160 σ2_{= np(1-p) = 200 (.8) (.2)= 32} σ= np(1-p) = 32 = 5.66 a) P(x=>125) = P(Z=> -6.27) = 1.0 Z = x -_σµ 124.5 - 160 5.66 = - 6.27 =

Normal Approximation to the

Binomial

This applies when you have a Binomial setting but for very large n values. Normal Distribution: 0 1 2 3 4 -1 -2 -3 -4 N (µ, σ) Binomial Distribution: B (n, p ) Normal Approximation: µ= np σ2_{= np(1-p)} σ = np(1-p)

Normal Approximation to the

Binomial

n = 200 p = .80 µ= np = 200 (.8) = 160 σ2_{= np(1-p) = 200 (.8) (.2)= 32} σ= np(1-p) = 32 = 5.66 a) P(x=>125) = P(Z=> -6.27) = 1.0 Z = x -_σµ 124.5 - 160 5.66 = - 6.27 =

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Normal Approximation to the

Binomial

n = 200 p = .80 µ= np = 200 (.8) = 160 σ2_{= np(1-p) = 200 (.8) (.2) =32} σ= np(1-p) = 32 = 5.66 b) P(170<=x<=185) = P(1.68<=Z<=4.5) Z = 169.5 - 160_5.66 = 1.68 Z = 185.5 - 160_5.66 = 4.5

Normal Approximation to the

Binomial

b) P(170<=x<=185) = P(1.68<=Z<=4.5) = 1.0 - .95352 = .04648 0 1 2 3 4 -1 -2 -3 -4