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 99% 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

Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 80 / 400 = 0.200

1 - = 1 - 0.200 = 0.8

Z/2 = 2.576

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.576 * (((0.200 * 0.8) / 400)

Margin of error = E = 0.0515

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.200 - 0.0515 < p < 0.200 + 0.0515

0.1485 < p < 0.2515

The 99% confidence interval for the population proportion p is :
**0.1485 to 0.2515**

**option D.** is correct

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