Question

A sociologist is interested in surveying workers in computer-related jobs to estimate the proportion of such workers who have changed the jobs within the past year.

(a) In the absence of preliminary data, how large a sample must be taken to ensure that a 99% confidence interval will specify the proportion to within ±0.05?

(b) Is the sample size obtained in (a) large enough to ensure that a 95% confidence interval will specify the proportion to within ±0.05? Explain.

(c) Is the sample size obtained in (a) large enough to ensure that a 99% confidence interval will specify the proportion to within ±0.1? Explain.

Answer 1

Solution,

=  1 - =   0.5

a) margin of error = E = 0.05

Z/2 = Z0.005  = 2.576

sample size = n = (Z / 2 / E )2 * * (1 - )

= (2.576 / 0.05)2 * 0.5 * 0.5

= 663.57

sample size = n = 664

b) margin of error = E = 0.05

Z/2 = Z0.025  = 1.96

sample size = n = (Z / 2 / E )2 * * (1 - )

= (1.96 / 0.05)2 * 0.5 * 0.5

= 384.16

sample size = n = 385

c) margin of error = E = 0.1

Z/2 = Z0.005  = 2.576

sample size = n = (Z / 2 / E )2 * * (1 - )

= (2.576 / 0.1)2 * 0.5 * 0.5

= 165.89

sample size = n = 166