Transcript Section 6-4
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
Point Estimate The sample proportion
p
ˆ
is the best point estimate of the population proportion
p
.
(In other books population proportion can be noted as
)
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 4
Margin of Error of the Estimate of p
ME =
z
ˆ ˆ
n
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 5
p
ˆ
Confidence Interval for Population Proportion where
ME =
z
ˆ ˆ
n
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 6
Confidence Interval for
p
ˆ
Population Proportion
E (
p
ˆ
- E,
ˆ
p
+ 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
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%?
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 9
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
n
701 1002 0.6996
0.700
Chapter 6. Section 6-5. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman 10
Do people lie about voting?
• 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
Do people lie about voting?
• 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 25Determining Sample Size
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=
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Two formulas for proportion sample size
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