If you wish to estimate a population mean with a margin of e

If you wish to estimate a population mean with a margin of error of ME = .3 using a 95% confidence interval, and you know from prior sampling that variance is approximately equal to 7.2, how many observations would have to be included in your sample?

Solution

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.025  
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also,      
      
s = sample standard deviation = sqrt(7.2) =   2.683281573  
E = margin of error =    0.3  
      
Thus,      
      
n =    307.3167057  
      
Rounding up,      
      
n =    308   [ANSWER]