You have developed a survey of job questions for NYC Bus Dri

You have developed a survey of job questions for NYC Bus Drivers. A random sample of 100 drivers has an average score of 10.6, with a standard deviation of 2.8. What is your estimate of the average job satisfaction score for the population of all NYC Bus Drivers. Use the 95% confidence level.

I\'m not sure what formula to use and I\'m not sure what all the variables represent.

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 =    0.025          
X = sample mean =    10.6          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    2.8          
n = sample size =    100          
              
Thus,              
Margin of Error E =    0.548789916          
Lower bound =    10.05121008          
Upper bound =    11.14878992          
              
Thus, the confidence interval is              
              
(   10.05121008   ,   11.14878992   ) [ANSWER]