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Confidence Interval for a Population Proportion - Introduction to Statistics I | STAT 1000, Study notes of Statistics

Material Type: Notes; Professor: Pozdnyakov; Class: Introduction to Statistics I; Subject: Statistics; University: University of Connecticut; Term: Unknown 1989;

Typology: Study notes

Pre 2010

Uploaded on 09/17/2009

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Download Confidence Interval for a Population Proportion - Introduction to Statistics I | STAT 1000 and more Study notes Statistics in PDF only on Docsity! 11. Confidence Interval for a Population Proportion Sampling Distribution of p̂  n Xp ˆ (where n is number of trials and X is the total number of successes) is an unbiased consistent estimator of p  X has the binomial distribution ),( pnb  ppE )ˆ( and n pp pVar )1( )ˆ(    for large n (when )ˆ1(ˆ ppn  is greater than 5) by the Central Limit Theorem npp pp /)1( ˆ   is approximately normal )1,0(N Large-Sample Confidence Interval for p nppzp /)ˆ1(ˆˆ 2   Example 1 A retail lumberyard routinely inspects incoming shipments of lumber from suppliers. For select grade 8-foot 2-by-4 pine shipments, the lumberyard supervisor chooses one gross (144 boards) randomly from a shipment of several tens of thousands of boards. In the sample, 18 boards are not salable as select grade. Calculate a 95% CI for the proportion in the entire shipment that is not salable as select grade. Solution The sample proportion p̂ is equal to 18/144=.125. Since 575.15875.125.144)ˆ1(ˆ  ppn the confidence interval is given by 054.125. 144 875.125. 96.1125./)ˆ1(ˆˆ 2    nppzp  Small-Sample Confidence Interval for p If the sample size is small (or p is close to 0 or 1) statisticians used adjusted confidence interval 4 )~1(~~ 2    n pp zp  where )4( )2(~    n Xp 1 Sample Size Determination The required sample size to produce an interval estimator Wp  with 1 confidence is 2 2 )1(             W ppz n  Since the value of p is unknown, it can be estimated by using the sample proportion from a prior sample, or we can use the conservative choice of 5.p in which case 2 2 2          W z n  Example 2 A manufacturer of boxes of candy is concerned about the proportion of imperfect boxes – those containing cracked, broken, or otherwise unappetizing candies. How large a sample is needed to get 95% CI for this proportion with a width no greater than .02? Solution Since the value of p is unknown, we will use the conservative substitution: 9604 01.2 96.1 2 2 2 2                  W z n  Exercises p. 319: 5.41—5.43, p. 320: 5.46—5.50; p. 326: 5.63—5.64 References: 1. Chase and Bown, General Statistics 2. Hildebrand and Ott, Statistical Thinking for Managers 3. Keller and Warrack, Statistics for Management and Economics 4. McClave, Benson, and Sincich, A First Course In Business Statistics Exercises 1. The Aid to Families with Dependent Children (AFDC) program has an overall error rate of 4% in determining eligibility. The state of California uses sampling to monitor its counties to see whether they exceed the 4% error rate, which can result in economic sanctions. In one county, 9 cases out of 150 were found to be in error. (a) Find a 95% confidence interval for the error rate for the county (proportion of all cases in error). 2
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