Respuesta :
We can estimate the sample size needed for a certain confidence level, admisible error and population proportion using the Cochran's formula:
[tex]n_0=\frac{Z^2p(1-p)}{e^2}[/tex]We don't have an estimation for p, the population proportion, but we will take the worst case scenario (therefore, the sample size will cover if the proportion is different), that is p=0.5.
This value of p will give us the maximum value for p(1-p) and therefore a sample size that will cover all the possible values of p.
The value of Z depends on the level of confidence, that is 99.9%. Then, the value of Z can be looked up in the standard normal distribution:
[tex]1-\alpha=0.999\longrightarrow Z=3.291[/tex]As the true proportion has to be within +/- 2.5% of the estimated proportion. the error e is 2.5%, so its value is:
[tex]e=\frac{2.5}{100}=0.025[/tex]Then, we can calculate the sample size using Z=3.291, p=0.5 and e=0.025:
[tex]\begin{gathered} n_0=\frac{Z^2p(1-p)}{e^2} \\ n_0=\frac{(3.291)^2\cdot0.5\cdot0.5}{0.025^2} \\ n_0\approx\frac{10.83\cdot0.25}{0.000625} \\ n_0\approx\frac{2.71}{0.000625} \\ n_0\approx4332.27 \\ n_0=4333 \end{gathered}[/tex]Answer: n=4,333.
The sample size has to be at least 4,333 to have a 99.9% confidence that the estimation will be within +/- 2.5% of the true population.
