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Presentation Transcript

Last Time

- Probability

The Classical Statistical Trail

Rates &

Proportions

Inferential

Statistics

Application

Descriptive Statistics

Discrete Random

Variables

Binomial

Probability

Discrete Probability Distributions; Moments

Problem 6.61

- A survey of middle aged men reveals that 28% of them are balding at the crown of their head. Moreover, it is known that such men have an 18% probability of suffering a heart attack in the next ten years. Men who are not balding in this way have an 11% probability of a heart attack. Find the probability that a middle aged man will suffer a heart attack in the next ten years.

P(HA/Bald and MA) = 0.18

P(HA/Not Bald and MA)

= 0.11

Bald

Not Bald

Middle Aged men

Probability of a heart attack in the next ten years

- P(HA) = P(HA and Bald and MA) + P(HA and Not Bald and MA)
- P(HA) = P(HA/Bald and MA)*P(BALD and MA) + P(HA/Not BALD and MA)* P(Not Bald and MA)
- P(HA) = 0.18*0.28 + 0.11*0.72 = 0.054 + .0792 = 0.1296

Random Variables

- There is a natural transition or easy segue from our discussion of probability and Bernoulli trials last time to random variables
- Define k to be the random variable # of heads in 1 flip, 2 flips or n flips of a coin
- We can find the probability that k=0, or k=n by brute force using probability trees. We can find the histogram for k, its central tendency and its dispersion

Outline

- Random Variables & Bernoulli Trials
- example: one flip of a coin
- expected value of the number of heads
- variance in the number of heads
- example: two flips of a coin
- a fair coin: frequency distribution of the number of heads
- one flip
- two flips

Outline (Cont.)

- Three flips of a fair coin, the number of combinations of the number of heads
- The binomial distribution
- frequency distributions for the binomial
- The expected value of a discrete random variable
- the variance of a discrete random variable

Concept

- Bernoulli Trial
- two outcomes, e.g. success or failure
- successive independent trials
- probability of success is the same in each trial
- Example: flipping a coin multiple times

Flipping a Coin Once

The random variable k is the number of heads

it is variable because k can equal one or zero

it is random because the value of k depends on

probabilities of occurrence, p and 1-p

Heads, k=1

Prob. = p

Prob. = 1-p

Tails, k=0

Flipping a coin once

- Expected value of the number of heads is the value of k weighted by the probability that value of k occurs
- E(k) = 1*p + 0*(1-p) = p
- variance of k is the value of k minus its expected value, squared, weighted by the probability that value of k occurs
- VAR(k) = (1-p)2 *p +(0-p)2 *(1-p) = VAR(k) = (1-p)*p[(1-p)+p] =(1-p)*p

Flipping a coin twice: 4 elementary outcomes

h, h; k=2

heads

h, h

Prob =p

heads

Prob=1-p

Prob =p

tails

h, t; k=1

h, t

Prob=p

heads

t, h; k=1

t, h

Prob =1-p

tails

Prob =1-p

t, t

tails

t, t; k=0

Flipping a Coin Twice

- Expected number of heads
- E(k)=2*p2 +1*p*(1-p) +1*(1-p)*p + 0*(1-p)2 E(k) = 2*p2 + p - p2 + p - p2 =2p
- so we might expect the expected value of k in n independent flips is n*p
- Variance in k
- VAR(k) = (2-2p)2 *p2 + 2*(1-2p)2 *p(1-p) + (0-2p)2 (1-p)2

Continuing with the variance in k

- VAR(k) = (2-2p)2 *p2 + 2*(1-2p)2 *p(1-p) + (0-2p)2 (1-p)2
- VAR(k) = 4(1-p)2 *p2 +2*(1 - 4p +4p2)*p*(1-p) + 4p2 *(1-p)2
- adding the first and last terms, 8p2 *(1-p)2 + 2*(1 - 4p +4p2)*p*(1-p)
- and expanding this last term, 2p(1-p) -8p2 *(1-p) + 8p3 *(1-p)
- VAR(k) = 8p2 *(1-p)2 + 2p(1-p) -8p2 *(1-p)(1-p)
- so VAR(k) = 2p(1-p) , or twice VAR(k) for 1 flip

Three Flips of a Fair Coin

- It is not so hard to see what the value of the number of heads, k, might be for three flips of a coin: zero, one ,two, three
- But one head can occur two ways, as can two heads
- Hence we need to consider the number of ways k can occur, I.e. the combinations of branching probabilities where order does not count

Three Flips of a Coin

- There is only one way of getting three heads or of getting zero heads
- But there are three ways of getting two heads or getting one head
- One way of calculating the number of combinations is Cn(k) = n!/k!*(n-k)!
- Another way of calculating the number of combinations is Pascal’s triangle

The Probability of Getting k Heads

- The probability of getting k heads (along a given branch) in n trials is: pk *(1-p)n-k
- The number of branches with k heads in n trials is given by Cn(k)
- So the probability of k heads in n trials is Prob(k) = Cn(k) pk *(1-p)n-k
- This is the discrete binomial distribution where k can only take on discrete values of 0, 1, …k

Expected Value of a discrete random variable

- E(x) =
- the expected value of a discrete random variable is the weighted average of the observations where the weight is the frequency of that observation

Expected Value of the sum of random variables

- E(x + y) = E(x) + E(y)

Expected Number of Heads After Two Flips

- Flip One: kiI heads
- Flip Two: kjII heads
- Because of independence p(kiI and kjII) = p(kiI)*p(kjII)
- Expected number of heads after two flips: E(kiI + kjII) = (kiI + kjII) p(kiI)*p(kjII)
- E(kiI + kjII) = kiI p(kiI)* p(kjII) +

Cont.

- E(kiI + kjII) = kiI p(kiI)* p(kjII) + kjII *p(kjII) p(kiI)
- E(kiI + kjII) = E(kiI) + E(kjII) = p*1 + p*1 =2p
- So the mean after n flips is n*p

Variance of a discrete random variable

- VAR(xi) =
- the variance of a discrete random variable is the weighted sum of each observation minus its expected value, squared,where the weight is the frequency of that observation

Cont.

- VAR(xi) =
- VAR(xi) =
- VAR(xi) =
- So the variance equals the second moment minus the first moment squared

The variance of the sum of discrete random variables

- VAR[xi + yj] = E[xi + yj - E(xi + yj)]2
- VAR[xi + yj] = E[(xi - Exi) + (yj - Eyj)]2
- VAR[xi + yj] = E[(xi - Exi)2 + 2(xi - Exi) (yj - Eyj) + (yj - Eyj)2]
- VAR[xi + yj] = VAR[xi] + 2 COV[xi*yj] + VAR[yj]

The variance of the sum if x and y are independent

- COV [xi*yj] = E(xi - Exi) (yj - Eyj)
- COV [xi*yj]= (xi - Exi) (yj - Eyj)
- COV [xi*yj]= (xi - Exi) p[x(i)]* (yj - Eyj)* p[y(j)]
- COV [xi*yj] = 0

Variance of the number of heads after two flips

- Since we know the variance of the number of heads on the first flip is p*(1-p)
- and ditto for the variance in the number of heads for the second flip
- then the variance in the number of heads after two flips is the sum, 2p(1-p)
- and the variance after n flips is np(1-p)

Application

- Rates and Proportions

Field Poll

- The estimated proportion, from the sample, that will vote for Guliani is:
- where is 0.35 or 35%
- k is the number of “successes”, the number of likely voters sampled who are for Guliani, approximately 122
- n is the size of the sample, 348

Field Poll

- What is the expected proportion of voters Nov. 7 who will vote for Guliani?
- = E(k)/n = np/n = p, where from the binomial distribution, E(k) = np
- So if the sample is representative of voters and their preferences, 35% should vote for Guliani next February

Field Poll

- How much dispersion is in this estimate, i.e. as reported by the Field Poll, what is the sampling error?
- The sampling error is calculated as twice the standard deviation or square root of the variance in
- = VAR(k)/n2 = np(1-p)/n2 =p(1-p)/n
- and using 0.35 as an estimate of p,
- = 0.35*0.65/348 =0.000654

Field Poll

- So the sampling error should be 2*0.026 or 5.2%.
- The Field Poll reports a 95% confidence interval or about two standard errors , I.e 2*2.6% ~ 5.4%

Field Poll

- Is it possible that Guliani might get 50% of the vote or more? Not likely since the probabilty of Guliani reciving more then 40% of the vote is only 2.5%
- Based on a normal approximation to the binomial, the true proportion voting for Guliani should fall between 29.5% and 40.5% with probability of about 95%, unless sentiments change.

Lab Two

- The Binomial Distribution, Numbers & Plots
- Coin flips: one, two, …ten
- Die Throws: one, ten ,twenty
- The Normal Approximation to the Binomial

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