Question

A certain Journal reported that automobile crashes cost Kenya $162 billion annually. The average cost per person for crashes in Nairobi area was reported to be $1599. Suppose this average cost was based on a sample of 50 persons who had been involved in car crashes and that the population standard deviation is ? = $600. What is the margin of error for a 92% confidence level? What sample size would you recommend if the study required a margin of error of $170 with the same level of confidence?

Answer 1

= 1 - 0.92 = 0.08

From Z table,

Z/2 = 1.7507

Margin of error = Z/2 * / sqrt(n)

= 1.7507 * 600 / sqrt(50)

= 148.55

Sample size = (Z/2 * / E)²

= (1.7507 * 600 / 170)²

= 38.18

Sample size = 39 (Rounded up to nearest integer)