For multiple sclerosis patients we wish to estimate the mean

For multiple sclerosis patients we wish to estimate the mean age at which the disease was first
diagnosed. We want a 95 percent confidence interval that is 10 years wide. If the population
variance is 90, how large should our sample be?

Solution

As

Margin of error = width/2 = 10/2 = 5

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 =    9.486832981  
E = margin of error =    5  
      
Thus,      
      
n =    13.82925175  
      
Rounding up,      
      
n =    14   [ANSWER]