19 Verizon Wireless would like to estimate the proportion of

19. Verizon Wireless would like to estimate the proportion of households that use cell phones for their phone service without a land line. A random sample of 150 households was selected and 48 relied strictly on cell phones for their service. The margin of error for a 90% confidence interval for the proportion based on this sample is ________.

Solution

p=48/150=0.32

Given a=1-0.9=0.1, Z(0.05) = 1.645 (from standard normal table)

So the margin of error is

Z*sqrt(p*(1-p)/n)

=1.645*sqrt(0.32*(1-0.32)/150)

=0.06265413