If you want to be 95% confident of estimating the population proportion to within a sampling error of ±0.06 and there is historical evidence that the population proportion is approximately .40, what sample size is needed?
Solution :
Given that,
= 0.40
1 - = 1 - 0.40 = 0.60
margin of error = E = % = 0.06
At 90% confidence level z
= 1 - 90%
= 1 - 0.90 =0.10
/2
= 0.05
Z/2
= Z0.05 = 1.645 ( Using z table ( see the 0.05 value in standard
normal (z) table corresponding z value is 1.645 )
Sample size = n = (Z/2 / E)2 * * (1 - )
= (1.645 / 0.06)2 * 0.40 * 0.6
=180.40
Sample size = 181
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