Sampling Distributions for Counts and Proportions

Sampling Distributions for Counts and Proportions

Example: Suppose we ask 1000 Transy students if they aretravelling out of the state for Spring Break, and 774 of them say “yes” and 226 say “no”

Let the number of people who say “yes” be the random variable X.

The random variable X is a count of the occurrences ofsome outcome in a fixed number of observations.

The number of observations is usually denoted by n

The sample proportion = X/np

p = 774/1000 = .774

The Binomial Setting

1) There are a fixed number n of observations

2) The n observations are independent

3) Each observation falls into just one of two categories, which, for convenience, we call “success” and “failure”.

4) The probability of a success, call it p, is the same for each observation

The Binomial Setting

Example: Toss a fair coin 10 times, and count the amount ofheads which will appear.

The random variable, X, is the number of heads that appear.

The probability of a success is p = 0.5

This experiment follows the four requirements for the Binomial Setting, so this is a Binomial Distribution.

Binomial Distributions

The distribution of the count X of successes in the binomialsetting is called the binomial distribution with parametersn and p.

The parameter n is the number of observations

The parameter p is is the probability of a success on anyone observation.

The possible values of X are the whole numbers from 0to n.

We say that X is B(n,p)

Binomial DistributionsIn the previous example, our distribution is B(10, 0.5)

New example: Draw a card from a deck and determine if itis a spade or not. Return the card to the deck, shuffle, andrepeat fourteen more times.

Are we in the binomial setting?

This is a B(15, 0.25)

Q: Is there ever a situation where we are “close enough” to use the binomial distribution?

Binomial DistributionsExample: Imagine we have a big ol’ bunch of transistors, sayroughly 10,000. We will say that 1000 of them are bad.

Pull an SRS of size 10 without replacement.

Let X be the amount of transistors which are bad in the SRS.

Q: Is this the binomial setting?

A: No. The first transistor has a 1000/10,000 = 0.1 chance ofbeing bad. What about the second?

The second has either a 1000/9,999 = 0.10001 or 999/9,999 chance = 0.09991 chance of being bad

We can consider this a B(10,0.1)

Sampling Distribution of a Count

When the population is much larger than the sample, thecount X of successes in an SRS of size n has approximatelythe B(n,p) distribution.

Note: We will use the binomial sampling distribution forcounts when the population is at least 10 times as largeas the sample.

Q: How can we find these binomial probabilities?

A: Tables

Binomial Probability Tables(Table C, page T-7 to T-10)

Consider the previous transistor problem using the B(10,.1)distribution.

p

n K 0.10

10 0 .34871 .38742 .19373 .05744 .01125 .00156 .00017 .00008 .00009 .000010 .0000

Q: What is the probability that wewill pull an SRS from this distributionwith exactly one bad transistor?

A: P(X = 1) =0.3874



p

n K 0.10

10 0 .34871 .38742 .19373 .05744 .01125 .00156 .00017 .00008 .00009 .000010 .0000

Q: What is the probability that wewill pull an SRS from this distributionwith at most one bad transistor?

A: P(X 1) =P(X=0) + P(X=1) =

0.3487 + 0.3874 = 0.7361



p

n K 0.10

10 0 .34871 .38742 .19373 .05744 .01125 .00156 .00017 .00008 .00009 .000010 .0000

Q: What is the probability that wewill pull an SRS from this distributionwith an even number of bad transistors?

A: P(X=2 or X=4 or X=6 or X=8 or X=10)

= P(X=2) + P(X=4) + P(X+6) + P(X=8) +P(X= 10)

= .3487 + .1937 + .0112 + .0001 + .0000

= .5537

Cautions about the Binomial TablesDo you notice anything about Table C?

The values of p are all 0.5 or smaller.

You need to adjust your count so you are using probabilitiesthat are at most 0.5

Example: Suppose I shoot free throws at a 92% rate ofsuccess. What is the probability I make 10 out of twelve?

This is a B(12, 0.92) distribution.

p

n K 0.08

12 0 .36771 .38372 .18353 .05324 .01045 .00146 .0001789101112

Cautions about the Binomial TablesExample: Suppose I shoot free throws at a 92% rate of success. What is the probability I make ten out of twelve?


We can re-word this as follows:

Example: Suppose I shoot free throws at an 8% rate of failure. What is the probability that I will miss two out of twelve?


A: P(X=2) =.1835

Binomial Mean and Standard Deviation

Q: If a count X is B(n,p), what are the mean X and the

standard deviation X ?

A: If a count X has the B(n,p) distribution, then :

X = np

X = np (1-p)

Binomial Mean and Standard Deviation

Example: Refer back to the free throw example.

The rate of misses was a B(12, 0.08)

X = np= (12)(0.08) = 0.96

X = np (1-p)

(12)(0.08)(0.92)=

0.8832= 0.9398=

Sample ProportionsIn statistical sampling, we often want to estimate the proportion p of “successes” in a population.

Our estimator is the sample proportion of successes :

p =Count of successes in sample

Size of sample

X

n=

Notice the difference between count and proportion :

Deep Thoughts

• The count is an integer between 0 and n

• The proportion is a number between 0 and 1

• In the binomial setting, the count has a binomial distribution

• The proportion does not have a binomial distribution, however, we can restate the proportion as a count and proceed from there

p

Example: We are going to survey 500 Transy students tosee if they like the new GE system. Suppose that 85% ofall Transy students would answer “yes”. What is the probability that at least 80% would agree in the sample proportion ?

The count X has binomial distribution :B(500, 0.85).

The sample proportion is : = X/500p

We are trying to find :P( 0.80 ) p

Notice that 80% of 500 is : (0.80)(500) = 400

This means we need to find :P( X 400)

P( X 400) =P( X = 400) +P( X = 401)+ … + P( X = 500)

Mean and Standard Deviation of a Sample Proportion

Let be the sample proportion of successes in an SRSof size n drawn from a large population having populationproportion p of successes.

p

The mean and standard deviation of are the following :p

= pp

p

= p(1-p)

n

Mean and Standard Deviation of a Sample Proportion

Example: Refer back to the previous question about thenew GE system.

Q: What are the mean and standard deviation of the proportion of the survey respondents who like the new GE system?

A: = pp = 0.85

p

= p(1-p)

n=

(0.85)(0.15)

500

= 0.0159687

Deeper Thoughts

• Recall that the sampling distribution of a sample proportion is close to normal. (See section 3.4, p. 270)

• Both the count X and the sample proportion are approximately normal when we have a large sample.

• We can use the normal curve to solve some difficult binomial distribution problems !

Normal Approximations for Counts and Proportions

• Draw an SRS of size n from a large population having population proportion p of successes.

• Let X be the count of successes in the sample, and = X / n be the sample proportion of successes.p

• When n is large, the sampling distributions of these statistics are approximately normal.

• X is approximately N( np, np(1-p) )

• is approximately Np ( p, ) p(1-p)

n

Note: Use this when np 10 and n(1-p) 10

Example: We are going to survey 500 Transy students tosee if they like the new GE system. Suppose that 85% ofall Transy students would answer “yes”. What is the probability that at least 80% would agree in the sample proportion ?

The count X has binomial distribution :B(500, 0.85).

The sample proportion is : = X/500p

We are trying to find :P( 0.80 ) p

Notice that 80% of 500 is : (0.80)(500) = 400

This means we need to find :P( X 400)

P( X 400) =P( X = 400) +P( X = 401)+ … + P( X = 500)



In this example :

• X is approximately N( np, np(1-p) )

• X is approximately N( (500)(0.85), (500)(0.85)(0.15) )

• X is approximately N( 425, 8 )

We are trying to find : P( X 400)

Z = 400 - 425

8= -3.13 .9991 = 99.91%


In this example :

• is approximately Np ( p, ) p(1-p)

n

We are trying to find : P( 0.80 ) p

• is approximately N(0.85, 0.0159687) p

Z =0.80 - 0.850.0159687

= -3.13 .9991 = 99.91%

Homework2, 4, 6, 8, 12, 14, 22, 24

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Sampling Distributions for Counts and Proportions