Question

A random sample of elementary school children in New York state is to be selected to estimate the proportion p p who have received a medical examination during the past year. It is desired that the sample proportion be within 0.045 of the true proportion with a 98 98 % level of confidence (a) Assuming no prior information about p p is available, approximately how large of a sample size is needed?

If a planning study indicates that pp is around 0.20.2, approximately how large of a sample size is needed?

Answer 1

Solution:

Given that,

a ) = 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 0.045

At 98.98% confidence level the z is ,

= 1 - 98.98% = 1 - 0.98.98 = 0.0102

/ 2 = 0.0102/ 2 = 0.0051

Z_/2 = Z_0.0051 = 2.569

Sample size = n = ((Z _{/ 2}) / E)² * * (1 - )

= (2.569 / 0.045 )² * 0.5 * 0.5

= 814.80

= 815

n = sample size = 815

b ) = 0.20

1 - = 1 - 0.20 = 0.80

margin of error = E = 0.045

At 98.98% confidence level the z is ,

= 1 - 98.98% = 1 - 0.98.98 = 0.0102

/ 2 = 0.0102/ 2 = 0.0051

Z_/2 = Z_0.0051 = 2.569

Sample size = n = ((Z _{/ 2}) / E)² * * (1 - )

= (2.569 / 0.045 )² * 0.20 * 0.80

= 521.47

= 521

n = sample size = 521