Question

a) Find n for a 90% confidence interval for p with bound on the error of estimation (margin of error) = 0.038 using an estimate of p = 0.7. (Round up to the nearest whole number.) NOTE: If you have the 98% or the 99% confidence interval for this question, PLEASE READ THE INSTRUCTIONS at the top of the assignment.


b) Find the conservatively large value for n using the same confidence interval and bound on the error of estimation (margin of error) mentioned in part a with no estimate of p. (Round up to the nearest whole number.)

Answer 1

* Sample size = Z² $\alpha$ /2 * p (1 - p) / E²

a)

Sample size = 1.6449² * 0.7 * 0.3 / 0.038²

= 393.49

n = 394 (Rounded up to nearest integer)

b)

When prior estimate for proportion is not specified then p = 0.50

Sample size = 1.6449² * 0.5 * 0.5 / 0.038²

= 468.43

n = 469 (Rounded up to nearest integer)