A simple random sample of 18 adults with incomes below the p

A simple random sample of 18 adults with incomes below the poverty level gave the following daily calcium intakes 886,633,943,847,934,841,1193,820,774,8834,1050,1058,1192,975,1313,872,1079,809 Determine the sample size required to have a margin of error of 50mg with a 95% confidence interval (show work)

Solution

Getting the standard deviation,

s = 1865.111202

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 =    1865.111202  
E = margin of error =    50  
      
Thus,      
      
n =    5345.220611  
      
Rounding up,      
      
n =    5346   [ANSWER]

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[Hi! One of the data points is 8834. Is that a typo? If yes, please resubmit this so we can continue helping you! Thanks!]