The internal auditing staff of a local manufacturing company performs a sample a
ID: 1886683 • Letter: T
Question
The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are delinquent (more than 90 days overdue). For this quarter, the auditing staff randomly selected 400 customer accounts and found that 80 of these accounts were delinquent. What is the 95% confidence interval for the proportion of all delinquent customer accounts at this manufacturing company?A. .1608 to .2392
B. .1992 to .2008
C. .1671 to .2329
D. .1485 to .2515
E. .1714 to .2286
Explanation / Answer
ANSWER: 95% CONFIDENCE INTERVAL = [ 0.1671, 0.2329] (16.7% , 23.3%) POPULATION PROPORTION, CONFIDENCE INTERVAL, NORMAL DISTRIBUTION 95% CONFIDENCE INTERVAL p +/- (z critical value) * SQRT[p * (1 - p)/n] p = POPULATION PROPORTION [0.20] (20%) z critical value [1.645] for 95% CONFIDENCE INTERVAL n = SAMPLE SIZE [400] 95% CONFIDENCE INTERVAL = 0.20+/- 1.645 * SQRT [ 0.20 * (1 - 0.20)/400] = [0.1671, 0.2329] (16.7% , 23.3%) CONCLUSION: 95% CONFIDENCE INTERVAL of true POPULATION PROPORTION = [0.1671, 0.2329] (16.7% , 23.3%) Internal auditing staff can claim to 95% confidence, sample will represent true mean not exceeding 20% within this interval.
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