june 10, 2008stat 111 - lecture 9 - proportions1 introduction to inference sampling distributions...

June 10, 2008 Stat 111 - Lecture 9 - Proportions 1

Introduction to Inference

Sampling Distributions

for Counts and Proportions

Statistics 111 - Lecture 9


Administrative Notes

• Homework 3 is due on Monday, June 15th – Covers chapters 1-5 in textbook

• Exam on Monday, June 15th • Review session on Thursday


Last Class• Focused on models for continuous data: using the

sample mean as our estimate of population mean

• Sampling Distributionof the Sample Mean• how does the sample mean change over different samples?

PopulationParameter:

Distributionof thesevalues?

Sample 1 of size n xSample 2 of size n xSample 3 of size n xSample 4 of size n xSample 5 of size n xSample 6 of size n x

.

. .


Today’s Class

• We will now focus on count data: categorical data that takes on only two different values

“Success” (Yi = 1) or “Failure” (Yi = 0)

• Goal is to estimate population proportion:

p = proportion of Yi = 1 in population


Examples

• Gender: our class has 83 women and 42 men • What is proportion of women in Penn student

population?

• Presidential Election: out of 2000 people sampled, 1150 will vote for McCain in upcoming election• What proportion of total population will vote for

McCain?

• Quality Control: Inspection of a sample of 100 microchips from a large shipment shows 10 failures• What is proportion of failures in all shipments?


Inference for Count Data

• Goal for count data is to estimate the population proportion p

• From a sample of size n, we can calculate two statistics:1. sample count Y2. sample proportion = Y/n

• Use sample proportion as our estimate of population proportionp

• Sampling Distributionof the Sample Proportion• how does sample proportion change over different samples?

PopulationParameter: p

Distributionof thesevalues?

Sample 1 of size n xSample 2 of size n xSample 3 of size n xSample 4 of size n xSample 5 of size n xSample 6 of size n x

.

. .


The Binomial Setting for Count Data

1. Fixed number n of observations (or trials)

2. Each observation is independent

3. Each observation falls into 1 of 2 categories:1. Success (Y = 1) or Failure (Y = 0)

4. Each observation has the same probability of success: p = P(Y = 1)


Binomial Distribution for Sample Count

• Sample count Y (number of Yi=1 in sample of size n) has a Binomial distribution

• The binomial distribution has two parameters:• number of trials n and population proportion p

P(X=k) = nCk * pk (1-p)(n-k)

• Binomial formula accounts for• number of success: pk

• number of failures : (1-p)n-k

• different orders of success/failures: nCk = n!/(k!(n-k)!)


Binomial Probability Histogram• Can make histogram out of these probabilities

• Can add up bars of histogram to get any probability we want: eg. P(Y < 4)

• Different values of n and p have different histograms, but Table C in book has probabilities for many values of n and p


Binomial Table


Example: Genetics• If a couple are both carriers of a certain

disease, then their children each have probability 0.25 of being born with disease

• Suppose that the couple has 4 children• P(none of their children have the disease)?

P(X=0) = 4!/(0!*4!) * .250 * (1-.25)4

• P(at least two children have the disease)?P(Y ≥ 2) = P(Y = 2) +P(Y = 3) +P(Y = 4)

= 0.2109 +0.0469 +0.0039 (from table)

= 0.2617


Example: Quality Control

• A worker inspects a sample of n=20 microchips from a large shipment

• The probability of a microchip being faulty is 10% (p = 0.10)

• What is the probability that there are less than three failures in the sample?

P(Y < 3) = P(Y = 0) + P(Y =1) + P(Y = 2)

= 0.1216 + 0.2702 + 0.2852 (from table)

= 0.677


Sample Proportions• Usually, we are more interested in a sample

proportion = Y/n instead of a sample count

P ( < k ) = P( Y < n*k)• Example: a worker inspects a sample of 20

microchips from a large shipment with probability of a microchip being faulty is 0.1

• What is the probability that our sample proportion of faulty chips is less than 0.05?

• P ( < .05 ) = P( Y < 1) = P(Y=0) = .1216

0.05 x 20


Mean and Variance of Binomial Counts

• If our sample count Y is a random variable with a Binomial distribution, what is the mean and variance of Y across all samples?• Useful since we only observe the value of Y for our

sample but what are the values in other samples?

• We can calculate the mean and variance of a Binomial distribution with parameters n and p:

μY = n*p

σ2 = n*p*(1-p)

σ = √ (n*p*(1-p))


Mean/Variance of Binomial Proportions

• Sample proportion is a linear transformation of the sample count ( = Y/n )

μ = 1/n * mean(Y) = 1/n * np = p

• Mean of sample proportion is true probability of success p

σ2 = 1/n2 Var(Y) = 1/n2 * n*p*(1-p) = p(1-p)/n

• Variance of sample proportion decreases as sample size n increases!


Variance over Long-Run• Lower variance with larger sample size means that

sample proportion will tend to be closer to population mean in larger samples

• Long-run behaviour of two different coin tossing runs. Much less likely to get unexpected events in larger samples


Binomial Probabilities in Large Samples• In large samples, it is often tedious to calculate

probabilities using the binomial distribution• Example: Gallup poll for presidential election• Bush has 49% of vote in population. What is the

probability that Bush gets a count over 550 in a sample of 1000 people?

P(Y > 550) = P(Y = 551) + P(Y = 552) + … + P(Y =1000)

= 450 terms to look up in the table!

• We can instead use the fact that for large samples, the Binomial distribution is closely approximated by the Normal distribution


Normal Approximation to Binomial• If count Y follows a binomial distribution with

parameters n and p, then Y approximately follows a Normal distribution with mean and variance:

μY = n*p

• This approximation is only good if n is “large enough”. • Rule of thumb for “large enough”:n·p≥ 10 and n(1-p) ≥ 10

• Also works for sample proportion: = Y/n follows a Normal distribution with mean and variance


Example: Quality Control• Sample of 100 microchips (with usual 10% of

microchips are faulty. What is the probability there are at least 17 bad chips in our sample?

• Using Binomial calculation/table is tedious. Instead use Normal approximation:

• Mean = n·p = 1000.10 = 10

• Var = n·p·(1-p) = 1000.100.90 = 9

= P(Z ≥ 2.33)

=1- P(Z ≤ 2.33)

= 0.01 (from table)


Example: Gallup Poll• Bush has 49% of vote in population• What is the probability that Bush gets sample

proportion over 0.51 in sample of size 1000? • Use normal distribution with

mean = p = 0.49 and variance p·(1-p)/n = 0.000245

= P(Z ≥1.27) =1- P(Z ≤1.27)

= 0.102


Why does Normal Approximation work?

• Central Limit Theorem: in large samples, the distribution of the sample mean is approx. Normal

• Well, our count data takes on two different values:“Success” (Yi = 1) or “Failure” (Yi = 0)

• The sample proportion is the same as the sample mean for count data!

• So, Central Limit Theorem works for sample proportions as well!


Next Class - Lecture 10

• Review session on Wednesday/Thursday– Show up with questions!

june 10, 2008stat 111 - lecture 9 - proportions1 introduction to inference sampling distributions...

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