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M ARIO F . T RIOLA

S TATISTICS. E LEMENTARY. Section 6-5 Estimating a Population Proportion. M ARIO F . T RIOLA. E IGHTH. E DITION. 1. The sample is a simple random sample. 2. The conditions for the binomial distribution are satisfied

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M ARIO F . T RIOLA

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  1. STATISTICS ELEMENTARY Section 6-5 Estimating a Population Proportion MARIO F. TRIOLA EIGHTH EDITION

  2. 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. Assumptions ˆ ˆ

  3. ˆ p= x n sample proportion Notation for Proportions p= population proportion of xsuccesses in a sample of size n (pronounced ‘p-hat’) q = 1 - p =sampleproportion of xfailures in a sample size of n ˆ ˆ

  4. The sample proportion pis the best point estimate of the population proportion p. (In other books population proportion can be noted as ) DefinitionPoint Estimate ˆ

  5. Margin of Error of the Estimate of p ˆ ˆ z p q ME =  n

  6. Confidence Interval for Population Proportion ˆ ˆ p p - ME < < + ME where p ˆ ˆ z p q ME =  n

  7. Confidence Interval for Population Proportion ˆ ˆ p p p - E < < + E p = p + E (p - E, p + E) ˆ ˆ ˆ

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

  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 • b) Find the 95% confidence interval estimate of the proportion • c) Are the survey results are consistent with the actual voter turnout of 61%?

  10. 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

  11. Do people lie about voting? • b) Find the 95% confidence interval estimate of the proportion

  12. 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% <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.

  13. ˆ ˆ p q z ME =  n z  Determining Sample Size (solve for n by algebra) ˆ ˆ ()2 p q n= ME2

  14. z ˆ ()2 p q  n= E2 z  Sample Size for Estimating Proportion p ˆ When an estimate of p is known: ˆ When no estimate of p is known: ()2 0.25 n= E2

  15. ()2 p q  n= ME2 z ()2  (0.25) n= Two formulas for proportion sample size z ˆ ˆ ME2

  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.

  17. n = [z/2 ]2p q 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. 2 E

  18. n = [z/2 ]2p q 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. 2 E = [1.645]2(0.169)(0.831) 0.042

  19. n = [z/2 ]2p q 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. 2 E 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. = [1.645]2(0.169)(0.831) 0.042 = 237.51965 = 238 households

  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.

  21. n = [z/2 ]2(0.25) 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. E 2

  22. n = [z/2 ]2(0.25) 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. E 2 = (1.645)2 (0.25) 0.042 = 422.81641 = 423 households

  23. n = [z/2 ]2(0.25) 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. E 2 = (1.645)2 (0.25) 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 = 422.81641 = 423 households

  24. Finding the Point Estimate and E from a Confidence Interval Point estimate of p: p =(upper confidence interval limit) + (lower confidence interval limit) 2 ˆ ˆ Margin of Error: E = (upper confidence interval limit) - (lower confidence interval limit) 2

  25. Finding the Point Estimate and E from a Confidence Interval Given the confidence interval .214< p < .678 Find the point estimate of p: p =.678 + .214= .446 2 ˆ ˆ Margin of Error: E =.678 - .214= .232 2

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