A survey of several 10 to 11 year olds recorded the followin

A survey of several 10 to 11 year olds recorded the following amounts spent on a trip to the mail: $18.41, $18.46. $17.83 Construct the 90% confidence interval for the average amount spent by 10 to 11 year olds on a trip to the mail. Assume the population is is approximately normal.

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       df = n - 1 =    2          

t(alpha/2) = critical t for the confidence interval =    9.924843201   [ANSWER, STEP 3]

*********************************      
  
X = sample mean =    18.23333333          
s = sample standard deviation =    0.350190424          
n = sample size =    3          
Thus,              
              
Lower bound =    16.22670337          
Upper bound =    20.2399633          
              
Thus, the confidence interval is              
              
(   16.22670337   ,   20.2399633   ) [ANSWER, CONFIDENCE INTERVAL]