Can someone help me with this problem step by step please Th

Can someone help me with this problem step by step please. Thanks so much guys.

6. The cost of health care s the subject of many studies that use statistical methods. One such study estimated that the average length of service for home health care among people over the age of 65 who use this type of service is 96.0 days with a standard error of 5.1 days. Assuming that the degrees of freedom are large, calculate a 90% confidence interval for the true mean length of service.

Solution

As the degrees of freedom is large, we can simply use the z distribution.

Note that              
              
Lower Bound = X - z(alpha/2) * se              
Upper Bound = X + z(alpha/2) * se              
              
where              
              
alpha/2 = (1 - confidence level)/2 =    0.05          
              
Thus,              
X = sample mean =    96          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
se = standard error =    5.1          
              
              
Thus,              
              
Lower bound =    87.6112465          
Upper bound =    104.3887535          
              
Thus, the confidence interval is              
              
(   87.6112465   ,   104.3887535   ) [ANSWER]