We have the following observations drawn from a normally dis

We have the following observations drawn from a normally distributed population: 0.497 0.501 0.469 0.548 0.452 Construct the 99% confidence interval for the population mean. Hint: Round off probabilities and values to 5 decimal places. Transform the sample mean into random variable t.

Determine probabilities based on some Excel value lookups: * 0.995 0.975 0.950 0.900 0.995 df 4 5 4 5 3 T.INV(*,df) 4.60409 2.57058 2.13185 1.47588 5.84091

Solution

Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    0.4934          
t(alpha/2) = critical t for the confidence interval =    4.604094871          
s = sample standard deviation =    0.036610108          
n = sample size =    5          
df = n - 1 =    4          
Thus,              
              
Lower bound =    0.418019282          
Upper bound =    0.568780718          

Thus, the confidence interval is              
              
(   0.418019282   ,   0.568780718   ) [ANSWER]