Eric Grodsky Sociology 360 Spring 2001 1 Lecture 14: probability and the binomial distribution...

Eric Grodsky Sociology 360 Spring 2001 1

Lecture 14: probability and the binomial distribution

Review: bias and efficiency Review: probability rules More probability rules Estimating counts: the binomial distribution

• The binomial coefficient

• Parameters for the binomial distribution


Bias

Bias refers to the difference between the mean of the sampling distribution and the parameter of interest

xbias


Efficiency

Efficiency refers to the sampling variance of the estimator

Efficiency is relative-- estimators are more or less efficient than one another

is more efficient than if1x 2x

22

21 xxss


Efficiency

Efficiency is influenced by sample size• Given two estimates of from the same sample

design and the same population, the estimate based on the larger sample will be more efficient

ns

x


Review

The distribution of x and the sampling distribution are different things• A distribution of x has only one

• A distribution of has many

• The mean of x is , but the mean of is

x

x x

x x

x


Review

The central limit theorem is about shape• The shape of any sampling distribution approaches

normality as n increases The law of large numbers is about efficiency

• As n increases, x-bar tends to get closer to Know the difference between the two


Review of probability rules

0P(A) 1 P(S)=1 P(not A)=1-P(A)

• P(not A)+P(A)=1

• P(A)=1-P(not A) P(A) and P(not A) are disjoint


Review of probability rules

If events A and B are disjoint:• P(A or B)=P(A) + P(B)

• P(A and B)=0


P(A and B)

If two events, A and B, are independent:• P(A and B)=P(A)P(B)


P(Chip votes Bush, Lynne votes Bush)

Chip and Lynne are independent observations P(Bush)=0.50, P(Gore)=0.40 P(Bush and Bush)=P(Bush)P(Bush) P(Bush and Bush)=(0.5)(0.5)=0.25 P(Bush and Gore)=(0.5)(0.4)=0.20


Extending the multiplication rule

What is the probability that, given three independently chosen voters, the first two vote for Bush and the last votes for Gore?

P(Bush)=0.5, P(Gore)=0.4


Probability of contraceptive failure

Studies suggest that the failure rate for male condoms is 1.5%

What is the probability of having 0 breaks for 100 condoms used?

What is the probability of having at least one break for 100 condoms used?


Probability of HS dropout

In the 1996-97 school year, approximately 5% of 18-24 year olds enrolled in the 10th through 12th grades dropped out of high school.

What is the probability that two randomly chosen students from this population dropped out?


The general addition rule

For any two events A and B, the probability of A or B occurring is:• P(A or B)=P(A)+P(B)-P(A and B)

This rule holds regardless of whether or not A an B are disjoint


The general addition rule

Why isn’t P(A or B)=P(A) + P(B) when A and B are not disjoint?


Nancy Reagan hypothesis revisited

What is the probability that a randomly selected high school student has tried smoking cigarettes or has tried LSD?

P(smoked or used LSD)=P(smoked)+P(used LSD)-P(smoked and used LSD)


Smoked and used LSD

| Ever smoked Ever used | cigarettes LSD | no yes | Total-----------+----------------------+---------- no | 6049 8153 | 14202 | 38.76 52.24 | 91.00 -----------+----------------------+---------- yes | 65 1340 | 1405 | 0.42 8.59 | 9.00 -----------+----------------------+---------- Total | 6114 9493 | 15607 | 39.17 60.83 | 100.00


Education and income

What is the probability that a randomly chosen adult will have at least a college education or will be in the top income quartile?


Education and income

income | education (categorical)

quartile | < HS HS some coll college > college | Total

-----------+-------------------------------------------------------+----------

1 | 278 395 294 215 85 | 1267

| 5.49 7.80 5.81 4.25 1.68 | 25.03

-----------+-------------------------------------------------------+----------

2 | 242 402 329 226 75 | 1274

| 4.78 7.94 6.50 4.47 1.48 | 25.17

-----------+-------------------------------------------------------+----------

3 | 229 411 321 209 96 | 1266

| 4.52 8.12 6.34 4.13 1.90 | 25.01

-----------+-------------------------------------------------------+----------

4 | 227 347 338 232 110 | 1254

| 4.49 6.86 6.68 4.58 2.17 | 24.78

-----------+-------------------------------------------------------+----------

Total | 976 1555 1282 882 366 | 5061

| 19.28 30.73 25.33 17.43 7.23 | 100.00


Estimating frequencies

The binomial distribution can be useful for estimating counts or frequencies

The distribution may apply if the following conditions are met:• The measure x is dichotomous (success or failure)

• There are a fixed number n of observations

• The n observations are independent

• The probability of success is uniform across observations


Estimating frequencies

What sort of things does this apply to?• Number of students who matriculate at a

university

• Number of contraceptive failures

• Number of cars on which a fuel tank will rupture on a 25 MPH rear impact


The binomial distribution

To identify the binomial distribution for the frequency of some event x, two parameters are necessary:• The number of observation n

• The probability of success p The point is to estimate the probability of

some number of successes or failures


Probabilities for counts

For continuous distributions, the probability of x-bar taking a specific value is nil. We always talk about a range of values.• Given an infinite number of possible values…

For dichotomous measures, the outcome (expected frequency) is discrete.


Building on probability math

What is the probability that 2 out of 5 randomly chosen condoms fails?• We know how to estimate the probability that the

first and fifth condom fail, given P(failure)=0.015

• P(FSSSF)=P(F)*P(S)*P(S)*P(S)*P(F)

• P(FSSSF)=P(F)2*P(S)3= (0.015)2*(0.985)3=0.000215


Contraceptive failure

This gives the probability that a particular 2 of five condoms fails (3 successes)

P(k successes, n-k failures)=pk(1-p)n-k

Next, we need to know how many such combinations there are


How many patterns in 5 trials?

Looking for 2 failures and 3 successes

FSSSF SFSSF SSFFS

FSSFS SFSFS SSSFF

FSFSS SFFSS

FFSSS SSFSF There are ten such combinations



Now that we know the probability of 2 failures in a particular order and the number of combinations of two failures, we can answer the question

P(2 failures in 5 trials)=P(2 failures, 3 successes)*count(combinations with 2 failures, 3 successes)

P(2 failures in 5 trials)=.000215*10=.00215


Estimating the number of combinations

Finding this by hand for 2 successes and 3 failures is not too bad

This gets unwieldly for a larger number of trials

Fortunately, there is a formula for figuring out how many combinations there are


The binomial coefficient

You want to know how many combinations there are of k successes in n independent trials• This is like wanting to count the number of events

in the sample space of n trials for which there are k successes and n-k failures


The binomial coefficient

Where n is the number of trials and k is the number of successes (or failures)

)!(!

!

knk

n

k

n


What is n! ?

n! is “n factorial” n!=n*(n-1)*(n-2)*…*1 5!=5*4*3*2*1=120 0!=1 by definition



How many patterns of 3 failures out of 10 condoms are there?


Estimating probabilities with the binomial distribution

Same idea we have already talked about• Probability is number of combinations times

probability for one of those combinations

knk ppk

nkXP

)1()(


Binomial probabilities

What is the probability of one contraceptive failure in 12 condoms?

111211 )985.01(985.011

12)11(

XP


Aid for low income college

What is the probability that 11 of 13 people randomly selected support government aid for low income college students?

n=13, p=0.862

111311 )862.01(862.011

13)11(

XP


Characterizing the binomial distribution

Like other distributions, the binomial can be characterized by its mean and standard deviation

These are characteristics of the theoretical distribution for binomial counts

)1( pnp

np


The binomial distribution

What is the sampling distribution for an SRS of 10 individuals on the college aid question?

p=0.86, n=10 so =10(0.86)=8.6

097.1)14.0)(86.0(10


The binomial distribution (n=10)

Estimate of number supporting aid, using binomial distribution500 samples of 10 observations each

Fra

ctio

n

mu=8.621, sigma=1.0974 5 6 7 8 9 10

0

.2

.4

.6


The binomial distribution (n=10)

count | Freq. Percent Cum.

------------+-----------------------------------

3 | 1 0.20 0.20

5 | 2 0.40 0.60

6 | 20 4.00 4.60

7 | 53 10.60 15.20

8 | 132 26.40 41.60

9 | 172 34.40 76.00

10 | 120 24.00 100.00

------------+-----------------------------------

Total | 500 100.00


Binomial distribution (n=100)Estimate of number supporting aid, using binomial distribution500 samples of 100 observations each

Fra

ctio

n

mu=86.2, sigma=3.47075 80 85 90 95

0

.1

.2


Binomial distribution (n=100) 7* | 7 7* | 8888 7* | 99999999 8* | 00000000000000000 8* | 111111111111111111 8* | 2222222222222222222222 8* | 33333333333333333333333333333333333333333333333333 8* | 44444444444444444444444444444444444444444444444444444 8* | 555555555555555555555555555555555555555 8* | 6666666666666666666666666666666666666666666666666666666666 8* | 777777777777777777777777777777777777777777777777777777777777 8* | 8888888888888888888888888888888888888888888888888 8* | 99999999999999999999999999999999999999999999 9* | 00000000000000000000000000000000000 9* | 11111111111111111111 9* | 22222222222 9* | 333333 9* | 4444 9* | 5


The normal approximation

As n increases, the binomial distribution approximates the normal distribution

Can use the normal approximation (CLT) Formally, as n gets larger, the binomial

distribution is approximately N(np, )

Closer p is to 0.5, closer binomial is to normal

)1( pnp


The binomial and the normal

How close is close enough?• Rule of thumb from Moore is if np10 and

n(1-p)10, the binomial distribution approximates the normal distribution

This shows the importance of p close to 0.5 and of larger n



For n==30

p np n(1-p)

.1 3 27

.3 9 21

.5 15 15

.7 21 9

.9 27 3



For p==.25

n np n(1-p)

10 2.5 7.5

20 5 15

30 7.5 22.5

40 10 30

50 12.5 37.5


The binomial and CLT go to court

Hazelwood School District v. United States (433 U. S. 299. 1977)• A largely rural district in northern St. Louis county

• U. S. government alleges that Hazelwood is guilty of employment discrimination

• Of 405 teachers hired by Hazelwood in the 1972 and 1973 school years, 15 were black


The binomial and CLT go to court

Of all teachers in the St. Louis County employment market, 5.7% were black at that time

If you include the city of St. Louis, 15.5% were black

Under each scenario, what is the probability that by chance only 15 of 405 teachers hired would be black?


Homework due 4/4

Moore: 5.2,3,7,13,16,22,25,29

Project: Analysis plan. Due 3/29. Respond informally in writing to my comments, telling us what analyses you plan to include in your paper and why. Turn this in with your assignment 4. Should be no more than a page or two.

Eric Grodsky Sociology 360 Spring 2001 1 Lecture 14: probability and the binomial distribution...

Documents

Transcript of Eric Grodsky Sociology 360 Spring 2001 1 Lecture 14: probability and the binomial distribution...