Question

4. A study in New England Journal of Medicine wanted to estimate the proportion of all needle injection-drug users, who test HIV-positive in New Haven, Connecticut by selecting a sample. How large should the sample size be so that the confidence interval for proportion has a margin of error within ±5% of the population proportion with 99% confidence level if a preliminary sample found that 67.5% of needles used by injection-drug users are HIV-positive?

Answer 1

Solution :

Given that,

= 67.5% =0.675

1 - = 1 - 0.675 = 0.325

margin of error = E = 5% = 0.05

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58    ( Using z table )

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

= (2.58 / 0.05)2 * 0.675 * 0.325

=584.099

Sample size = 584