A local company wants to evaluate their quality of service b

A local company wants to evaluate their quality of service by surveying their customers. Their budget limits the number of surveys to 95. What is their maximum error of the estimated mean quality for a 81% level of confidence and an estimated standard deviation of 5?

Solution

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
      
              
where              
alpha/2 = (1 - confidence level)/2 =    0.095          
z(alpha/2) = critical z for the confidence interval =    1.310579112          
s = sample standard deviation =    5          
n = sample size =    95          
              
Thus,              

Margin of Error E =    0.672312899   [ANSWER]