Question

A researcher wishes to estimate the number of households with two computers. How large a sample is needed in order to be 99​% confident that the sample proportion will not differ from the true proportion by more than 3​%? A previous study indicates that the proportion of households with two computers is 24%

Answer 1

Solution :

Given that,

= 0.24

1 - = 1 - 0.24 = 0.76

margin of error = E = 0.03

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58 ( Using z table ( see the 0.005 value in standard normal (z) table corresponding z value is 2.58 )

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

= (2.58 / 0.03)2 * 0.24 * 0.76

=1349.0

Sample size = 1349