In a time use study 40 randomly selected managers were found

In a time use study 40 randomly selected managers were found to spend a mean time of 2.4 hours per day on paperwork. The standard deviation of the 40 scores was 1.30 hours. Construct an 85% confidence interval for the mean time spent on paperwork by all managers.

Solution

Note that              
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 =    0.075          
X = sample mean =    2.4          
z(alpha/2) = critical z for the confidence interval =    1.439531471          
s = sample standard deviation =    1.3          
n = sample size =    40          
              
Thus,              
Margin of Error E =    0.295892884          
Lower bound =    2.104107116          
Upper bound =    2.695892884          
              
Thus, the confidence interval is              
              
(   2.104107116   ,   2.695892884   ) [ANSWER]