Question

A person recently read that 82% of cat owners are women. How large a sample should the person take it she wishes to be 90% confident that her proportion is within 0.03 of the true population proportion?

Answer 1

Solution :

Given that,

= 0.82

1 - = 1 - 0.82 = 0.18

margin of error = E = 3% = 0.03

At 90% confidence level

= 1 - 0.90 = 0.10

/2 =0.05

Z/2 = 1.645 ( Using z table )

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

= (1.645 / 0.03)2 * 0.82 * 0.18

=443.78

Sample size = 444