What is the maximum sample size needed to estimate the proportion of adults in US who vote regularly and want to be 95% sure that the error of our estimate will not exceed 0.06, if we have no information about the true proportion? (Answer should be rounded up to nearest integer)
solution :
Given that,
= 0.5
1 - = 1 - 0.5 = 0.5
margin of error = E = 0.06
At 95% confidence level the z is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2 = Z0.025 = 1.96
Sample size = n = (Z/2 / E)2 * * (1 - )
= (1.96* 0.06)2 * 0.5 * 0.5
= 266.77
Sample size = 267 (rounded)
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