Given a normal population with a standard deviation 162 A ra

Given a normal population with a standard deviation 1.62. A random sample of 60 observations was taken, and its mean was calculated to be 5.41

Calculate a 99%confidence interval for : show work

if we had wanted 99% confidence interval that had a maximum error of 0.50, what should our sample size have been?

Solution

a)

Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    5.41          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    1.62          
n = sample size =    60          
              
Thus,              
Margin of Error E =    0.538711776          
Lower bound =    4.871288224          
Upper bound =    5.948711776          
              
Thus, the confidence interval is              
              
(   4.871288224   ,   5.948711776   ) [ANSWER]

b)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.005  
      
Using a table/technology,      
      
z(alpha/2) =    2.575829304  
      
Also,      
      
s = sample standard deviation =    1.62  
E = margin of error =    0.5  
      
Thus,      
      
n =    69.65049056  
      
Rounding up,      
      
n =    70   [ANSWER]