Question 1174137
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Hi,

Normal Distribution:  σ = 25 and ME = 4 
Find the sample size necessary for a 99% confidence interval 

 ME ={{{( z*sigma/sqrt(n))}}}
Or
  n =  {{{ ((z*sigma)/ME)^2}}}

  n = {{{ ((2.576*25)/4)^2}}} = 16.1  0r  17 rounded up (to be on the safe side)

   17 the sample size necessary for a 99% confidence interval 

 = CI	z = value
90%	z =1.645
92%	z = 1.751
95%	z = 1.96
98%	z = 2.326
99%	z = 2.576

Wish You the Best in your Studies.
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