1 Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison...

1Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman

MARIO F. TRIOLAMARIO F. TRIOLA EIGHTHEIGHTH

EDITIONEDITION

ELEMENTARY STATISTICS

Section 6-5 Estimating a Population Proportion


Assumptions 1. The sample is a simple random

sample.

2. The conditions for the binomial distribution are satisfied

3. The normal distribution can be used to approximate the distribution of sample proportions because np 5 and nq 5 are both satisfied.

ˆ ˆ


q = 1 - p = sample proportion of x failures in a sample size of n

p =ˆ xn sample proportion

ˆ

p = population proportion

(pronounced ‘p-hat’)

of x successes in a sample of size n

Notation for Proportions

ˆ


DefinitionPoint Estimate

The sample proportion p is the best point estimate of the population

proportion p.

(In other books population proportion can be noted as )

ˆ


Margin of Error of the Estimate of p

z

ME =

nˆ ˆp q


Confidence Interval for Population Proportion

p - ME < < + ME where

ˆ p ˆ p

z

ME =

nˆ ˆp q


Confidence Interval for Population Proportion

p - E < < + E

p = p + E

(p - E, p + E)

ˆ p ˆ p

ˆ

ˆ ˆ


Round-Off Rule for Confidence Interval Estimates of p

Round the confidence interval limits to

three significant digits.


Do people lie about voting?

• In a survey of 1002 people, 701 people said they voted in a recent election. Voting records showed that 61% of eligible voters actually did vote. Using these results, find the following about the people who “said” they voted:

• a) Find the point estimate• b) Find the 95% confidence interval estimate of

the proportion • c) Are the survey results are consistent with the

actual voter turnout of 61%?



• In a survey of 1002 people, 701 people said they voted in a recent election. Voting records showed that 61% of eligible voters actually did vote. Using these results, find the following about the people who said they voted:

• a) Find the point estimate

701ˆ 0.6996 0.700

1002

xp

n



• b) Find the 95% confidence interval estimate of the proportion

/ 2

ˆ ˆ (.70)(.30)1.96 .0284

1002

pqE z

n

ˆ ˆ

.7 .0284 .7 .0284

.671 .728

p E p p E

p

p



• c) Are the survey results consistent with the actual voter turnout of 61%?

We are 95% confident that the true proportion of the people who said they vote is in the interval 67.1% <p<72.8% .

Because 61% does not fall inside the interval, we can conclude our survey results are not consistent with the actual voter turnout.


( )2 ˆp q

Determining Sample Size

zME =

p qnˆ ˆ

(solve for n by algebra)

zn =

ˆME2


Sample Size for Estimating Proportion p

When an estimate of p is known: ˆ

ˆ( )2 p qn =

ˆE2

z

When no estimate of p is known:

( )2 0.25n =

E2z


ME2n =

z( )2(0.25)

Two formulas for proportion sample size

zn =

( )2 p q

ME2

ˆ ˆ


Example: We want to determine, with a margin of error of four percentage points, the current percentage of U.S. households using e-mail. Assuming that we want 90% confidence in our results, how many households must we survey? A 1997 study indicates 16.9% of U.S. households used e-mail.


n = [z/2 ]2 p q̂ˆ

E2



n = [z/2 ]2 p q

E2

= [1.645]2 (0.169)(0.831)

0.042


ˆˆ


= [1.645]2 (0.169)(0.831)

n = [z/2 ]2 p q

E2

= 237.51965= 238 households


To be 90% confident that our sample percentage is within four percentage points of the true percentage for all households, we should randomly select and survey 238 households.

0.042

ˆˆ


Example: We want to determine, with a margin of error of four percentage points, the current percentage of U.S. households using e-mail. Assuming that we want 90% confidence in our results, how many households must we survey? There is no prior information suggesting a possible value for the sample percentage.


n = [z/2 ]2 (0.25)

E

2



n = [z/2 ]2 (0.25)

E = (1.645)2 (0.25)

2

0.042


= 422.81641 = 423 households


n = [z/2 ]2 (0.25)

E = (1.645)2 (0.25)

2

= 422.81641 = 423 households

With no prior information, we need a larger sample to achieve the same results with 90% confidence and an error of no more than 4%.


0.042


Finding the Point Estimate and E from a Confidence Interval

Point estimate of p:

p = (upper confidence interval limit) + (lower confidence interval limit)

2

Margin of Error:

E = (upper confidence interval limit) - (lower confidence interval limit)

2

ˆˆ


Finding the Point Estimate and E from a Confidence Interval

Find the point estimate of p:

p = .678 + .214 = .446

2ˆ

ˆ

Margin of Error:

E = .678 - .214 = .232

2

Given the confidence interval .214< p < .678

1 Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison...

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