A survey was conducted in 30 companies to study the percenta

A survey was conducted in 30 companies to study the percentage of engineering employees who contributed to some kind of community service. The survey gathered the following data: 76 81 77 82 80 85 60 80 79 82 70 88 85 80 79 83 75 87 78 80 84 72 75 90 84 82 77 75 86 80 Construct a 90% confidence interval for the mean participation rate for engineering employees

Solution

Note that              
              
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 =    0.05          
X = sample mean =    79.73333333          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    5.959248583          
n = sample size =    30          
              
Thus,              
              
Lower bound =    77.94372443          
Upper bound =    81.52294224          
              
Thus, the confidence interval is              
              
(   77.94372443   ,   81.52294224   ) [ANSWER]