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