an accounting firm annually monitors a certain mailing servi

an accounting firm annually monitors a certain mailing service\'s performance. one parameter of interest is the percentage of mail delivered on time. in a sample of 307,000 items mailed between Dec.10 and March 3--- the most difficult delivery season due to bad weather and holidays--- the accounting firm determined that 287,900 items were delivered on time. use this information to make a statement about the likelihood of an item being delivered on time by that mailing service.

Assuming a confidence level of95%, the likelihood of an item being delivered on time( where it mailed between Dec.10 and Mar.3) is in the interval ( , ) round to four decimal places

Solution

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.937785016          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.000435943          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
              
lower bound = p^ - z(alpha/2) * sp =   0.936930583          
upper bound = p^ + z(alpha/2) * sp =    0.938639449          
              
Thus, the confidence interval is              
              
(   0.936930583   ,   0.938639449   ) [ANSWER]