Question

What sample size is needed to give a margin of error within + or - 6% in estimating a population proportion with 90% confidence? Use z-values rounded to three decimal places. Round your answer up to the nearest integer. Please show equation and work.

Answer 1

Solution :

Given that margin of error E = 0.06 , p is not given then p = 0.5

=> q = 1 - p = 0.5

=> for 90% confidence interval , Z = 1.645

=> Sample size n = p*q*(Z/E)^2

= 0.5*0.5*(1.645/0.06)^2

= 187.9184

= 188 (round to nearest integer)

Answer 2

What sample size is needed to give a margin of error within ${\displaystyle \mathrm{}}$ in estimating a population mean with  90%confidence, assuming a previous sample had ${\displaystyle \mathrm{}}$.

Round your answer up to the nearest integer.