a sample of 25 adult elephants had an average weight of 9500

a sample of 25 adult elephants had an average weight of 9,500 pounds, with a sample standard deviation of 150 pounds. Find the 99% confidence interval of the true mean. Assume that the variable is normally distributed

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 =    0.005          
X = sample mean =    9500          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    150          
n = sample size =    25          
              
Thus,              
Margin of Error E =    77.27487911          
Lower bound =    9422.725121          
Upper bound =    9577.274879          
              
Thus, the confidence interval is              
              
(   9422.725121   ,   9577.274879   ) [ANSWER]