An accounting firm annually monitors a certain mailing service\'s performance. O

Question

An accounting firm annually monitors a certain mailing service's performance. One parameter of interest is the percentage of mail delivered on time. In a sample of 322,000 items mailed between Dec. 10 and Mar. 3-the most difficult delivery season due to bad weather and holidays-the accounting firm determined that 256,400 items were delivered on time. Use this information to make a statement about the likelihood of an item being delivered on time by that mailing service. Assuming a confidence level of 95%, the likelihood of an item being delivered on time (when it was mailed between Dec. 10 and Mar. 3) is in the interval (Round to four decimal places as needed.)

Explanation / Answer

p = 256400/322000 = 0.7963

n = 322000

Z for 95% confidence interval = Z0.975 = 1.96

Confidence interval = (p + Z0.975 * sqrt(p*(1-p)/n)

= (0.7963 + 1.96 * sqrt(0.7963 * (1-0.7963)/322000)

= (0.7963 + 0.0007)

= (0.7956, 0.7970)

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