Question

If you want to be 99% confident of estimating the population proportion to within a sampling error of ±0.06, what sample size is needed?

Answer #1

Solution :

Given that,

= 1 - =0.5

margin of error = E = 0.06

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_{/2}
= Z_{0.005} = 2.576

sample size = n = (Z_{
/ 2} / E )^{2} *
* (1 -
)

= (2.576 / 0.06)^{2} * 0.5 * 0.5

= 460.8 = 461

sample size = 461

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