Any help please The National Student Loan Survey collects da

Any help please.

The National Student Loan Survey collects data to esamine questions related so the amoum of money that borrowers owe. The survey selected a sammple of 1280 borrowers who begas repayment on their k undergraduase study was $18,900 and the standard deviation was about $49,000. construct a 95% confidence interval for the true mean debt for The student bemowers. 5. ment on their loans between 4 and 6 moeths prior to the survey. The mean debt or

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 =    18900          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    49000          
n = sample size =    1280          
              
Thus,              
Margin of Error E =    2684.350281          
Lower bound =    16215.64972          
Upper bound =    21584.35028          
              
Thus, the confidence interval is              
              
(   16215.64972   ,   21584.35028   ) [ANSWER]