3/15/11 Lecture 14 1

STOR 155 Introductory Statistics

(Chap 5)

Lecture 14: Sampling Distributions for

Counts and Proportions

The UNIVERSITY of NORTH CAROLINA

at CHAPEL HILL

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Statistic & Its Sampling Distribution

• A statistic is any numeric measure calculated from data. It is a random variable, its value varies from sample to sample. – count/proportion:

• Ex: number/proportion of free throws made by a Tar Heel player who shoots 20 free throws in a practice

– sample mean • Ex: average SAT score of a group of 10 students randomly

selected from STAT 155

• The probability distribution of a statistic is called its sampling distribution. – It depends on the population distribution, and the sample

size.

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Binomial Experiment

• n trials (with n fixed in advance).

• Each trial has two possible outcomes,

“success” (S) and “failure” (F).

• The probability of success, p, remains the

same from one trial to the next.

• The trials are independent, i.e. the outcome of

each trial does not affect outcomes of other

trials.

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Example

• The experiment: randomly draw n balls with replacement from an urn containing 10 red balls and 20 black balls.

• Let S represent {drawing a red ball} and F represent {drawing a black ball}.

• Then this is a binomial experiment with p =1/3.

• Q: Would it still be a binomial experiment if the balls were drawn without replacement? No!

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Binomial Distribution

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Do they follow binomial distributions (approximately) ?

• X = number of stocks on the NY stock exchange whose prices increase today

• X = number of games the Tar Heel will win next season

• A couple decides to have children until they have a girl.

X = number of boys the couple will have

Answer: NO in all 3 cases. Why ?

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Binomial Distribution

• If X ~ B(n, p), then X = np, 2X = np(1-p).

• P(X=x) depends on n and p, which can be

calculated using software or Table C (for some

n and p), or a Binomial Formula (page 329)

--- a simple argument given in class …

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•Binomial Table:

for n 20, and

certain values of p.

•Table C: Page T-6

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Credit Card Example

• Records show that 5% of the customers in a shoe store make their payments using a credit card.

• This morning 8 customers purchased shoes.

• Use the binomial table to answer the following questions.

1. Find the probability that exactly 6 customers did not use a credit card.

– X: number of customers who did not use a credit card. Then X ~ B(8, 0.95), which is not on the table.

– Y: number of customers who did use a credit card. Then Y ~ B(8, 0.05), which is on the table.

– P(X= 6) = P(Y = 2) = .0515.

2. What is the probability that at least 3 customers used a credit card? (See the board …)

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3. What is the expected number of customers

who used a credit card?

– Y = np = 8(.05) = 0.4.

4. What is the standard deviation of the number

of customers who used a credit card? – 2

Y = np (1 – p) = 8(. 05)(.95) = 0.38.

The standard deviation is

.62.038.0 Y

Credit Card Example (continued)

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Parking Example (bad impact ?)

• Sarah drives to work everyday, but does not own a parking permit. She decides to take her chances and risk getting a parking ticket each day. Suppose – A parking permit for a week (5 days) cost $ 30.

– A parking fine costs $ 50.

– The probability of getting a parking ticket each day is 0.1.

– Her chances of getting a ticket each day is independent of other days.

– She can get only 1 ticket per day.

• What is her probability of getting at least 1 parking ticket in one week (5

days)?

• What is the expected number of parking tickets that Sarah will get per week?

• Is she better off paying the parking permit in the long run?

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Sample Proportion

• If X ~ B(n, p), the sample proportion is defined as

• mean & variance of a sample proportion:

./)1( , ˆˆ nppp pp

.size sample

successes of #ˆ

n

Xp

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Example: Clinton's vote

• 43% of the population voted for Clinton in 1992.

• Suppose we survey a sample of size 2300 and see if they voted for Clinton or not in 1992.

• We are interested in the sampling distribution of the sample proportion , for samples of size 2300.

• What's the mean and variance of ?

p̂

p̂

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Count & Proportion of “Success”

• A Tar Heel basketball player is a 95% free throw

shooter.

• Suppose he will shoot 5 free throws during each

practice.

• X: number of free throws he makes in a practice.

• : proportion of free throws made during

practice.

• P(X=3) = P( =0.6). Why ?

p̂

p̂

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Normal Approximation for Counts and Proportions

• Let X ~ B(n, p) and

• If n is large, then

• Rule of Thumb: np 10, n(1 - p) 10.

./ˆ nXp

)./)1( ,( approx. is ˆ

))1( ,( approx. is

n-pppNp

-pnpnpNX

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Switches Inspection

• A quality engineer selects an SRS of 100 switches from a large shipment for detailed inspection.

• Unknown to the engineer, 10% of the switches in the shipment fail to meet the specifications.

• Software tells us that the actual probability that no more than 9 of the switches in the sample fail inspection is P(X 9) = 0.4513.

• What will the normal approximation say?

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Switches Inspection

• The normal approximation to the probability of

no more than 9 bad switches is the area to the

left of X = 9 under the normal curve.

• Using Table A, we have

• The approximation .3707 to the binomial

probability of .4513 is not very accurate. In this

case np = 10.

.3707.)33.()3

109

3

10()9(

ZP

XPXP

.3)9)(.1(.100)1( ,10)1)(.100( pnpnp XX

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Continuity Correction

• The normal approximation is more accurate if

we consider X=9 to extend from 8.5 to 9.5,

X = 10 to extend from 9.5 to 10.5, and so on.

• Example (Cont.):

.4325.)17.(

)3

105.9

3

10()5.9()9(

ZP

XPXPXP

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P(X 8) replaced by P(X < 8.5)

For large n the effects of the

continuity correction factor is

very small and will be omitted.

P(X 14) replaced by P(X > 13.5)

Continuity Correction

P(X < 8) = P(X<=7) replaced by P(X < 7.5)

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Coin Tossing Example

• Toss a fair coin 200 times, what is the probability that the total number of heads is between 90 and 110?

• X= the total number of heads

• X ~ B(200, 0.5). Want: P(90 X 110).

• X =200 × .5 = 100, X = (200× .5× .5)1/2 = 7.07.

• With continuity correction:

P(90 X 110) = P(X 110) - P(X 89)

P( Z (110 + .5 - 100)/7.07) - P (Z (89 + .5 - 100)/7.07)

= P (Z 1.48) - P (Z -1.48)= .8611.

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Normal Approximation for Sample Proportions

• Let X ~ B(n, p) and

• If n is large, then

• No continuity correction !

• Rule of Thumb: np 10, n(1 - p) 10.

./ˆ nXp

)./)1( ,( approx. is ˆ n-pppNp

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• The Laurier company’s brand has a market share of

30%. In a survey 1000 consumers were asked which

brand they prefer.

• What is the probability that more than 32% of the

respondents say they prefer the Laurier brand?

• Solution: The number of respondents who prefer

Laurier is binomial with n = 1000 and p = .30. Also, np

= 1000(.3) > 10, n(1-p) = 1000(1-.3) > 10.

.0838.)38.1(01449.

30.32.

)1(

ˆ)32.ˆ(

ZP

npp

ppPpP

Example

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Take Home Message

• Sampling distribution

• Binomial experiments

• Binomial distribution

• Binomial formula

• How to use Binomial Table

• Sample Proportion

• Normal approximation

• Continuity correction (only for counts, not for

proportions)