In: Statistics and Probability
When 415 college students are randomly selected and surveyed, it is found that 328 made online purchases during the last month. What sample size will produce a margin of error of 6% with 95% confidence level?
Number of Items of Interest, x = 328
Sample Size, n = 415
Sample Proportion , p̂ = x/n = 0.7904
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sample proportion , p̂ = 0.7904
sampling error , E = 0.06
Confidence Level , CL= 0.95
alpha = 1-CL = 0.05
Z value = Zα/2 = 1.960 [excel formula =normsinv(α/2)]
Sample Size,n = (Z / E)² * p̂ * (1-p̂) = ( 1.960 / 0.06 ) ² * 0.79 * ( 1 - 0.79 ) = 176.8
so,Sample Size required= 177