E LEMENTARY S

TATISTICS

Section 6-5 Estimating a Population Proportion

M ARIO F . T RIOLA E IGHTH Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman E DITION 1

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 are both satisfied.

ˆ

5 and nq ˆ

5

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 2

Notation for Proportions

p =

population proportion

p

ˆ

= x n

sample proportion of

x

successes in a sample of size

n

(pronounced ‘p-hat’) ˆ

=

1 ˆ

p =

sample proportion of

x

failures in a sample size of

n

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 3

Definition

p

ˆ

p

)

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 4

ME =

z



ˆ ˆ

n

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

p

ˆ

ME =

z



ˆ ˆ

n

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

ˆ

ˆ

ˆ

+ E)

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 7

Round-Off Rule for Confidence Interval Estimates of p Round the confidence interval limits to three significant digits.

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 8

• 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%?

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 9

• 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

n

701 1002  0.6996

 0.700

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 10

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

E

z

 / 2

n

 1.96

(.70)(.30)  .0284

1002 .671

E p p

.728

E

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 11

• 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%

Because 61% does not fall

inside

the interval, we can conclude our survey results are

not

consistent with the actual voter turnout.

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 12

ME

z

n

(solve for

n

by algebra)

n

=

(

z

 )

2 ME 2 ˆ

p q

ˆ

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 13

Sample Size for Estimating Proportion p

n

=

(

z

)

2 E 2 ˆ

p q

ˆ When no estimate of p is known:

n

=

(

z

)

2 0.25

E 2

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 14

n

(

z

 )

2 ME 2

n

=

(

z

 )

2

(

0.25

)

ME 2

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 15

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.

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 16

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 ˆ p q E 2

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 17

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 ˆ p q E 2 = [1.645] 2 (0.169)(0.831) 0.04

2

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 18

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 ˆ p q E 2 = [1.645] 2 (0.169)(0.831) 0.04

= 237.51965

2 = 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.

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 19

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.

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 20

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

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 21

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 = (1.645) 2 (0.25) 0.04

2 = 422.81641 = 423 households

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 22

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 = (1.645) 2 (0.25) 0.04

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%.

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 23

Finding the Point Estimate and E from a Confidence Interval Point estimate of ˆ

p

= ˆ

p

: (upper confidence interval limit) + (lower confidence interval limit) 2 Margin of Error: E = (upper confidence interval limit) - (lower confidence interval limit) 2

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 24

Finding the Point Estimate and E from a Confidence Interval Given the confidence interval .214< p < .678

Find the point estimate of ˆ

p

: ˆ

p

= .678 + .214

=

.446

2 Margin of Error:

E

= .678 - .214

=

.232

2

Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 25