Question

Confidence is 99 percent. Please help.

Problem 7-15 A job in an insurance office involves telephone conversations with policyholders. The office manager estimates that about half of the employee's time is spent on the telephone. How many observations are needed in a work sampling study to estimate that time percentage to within 7 percent and have a confidence of 990 percent? (Round z values to 2 decimal places. Do not round other intermediate calculations. Round your final answer to the nearest whole number.) ints Number of observations

Answer 1

For confidence of 99% , Z = 2.58 (See standard normal distribution table)

Margin of Error formula (E),

For 0 ≤ p ≤ 1, p*(1 - p) is maximum when p = 0.5

Hence, Number of observations (N) = Z-score^2 * 0. 5 * 0.5 / (margin of error)^2

= 2.58^2* 0.5 * 0.5 / 7% * 7% = 340 observations