A survey conducted by the American Automobile Association AA

A survey conducted by the American Automobile Association (AAA) showed that a family of four spends an average of $215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagra Falls resulted in a sample mean of $252.45 per day and a sample deviation of $74.50.

a.) develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagra Falls. (to 2 decimals)

Solution

Note that              
              
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 =    252.45          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    74.5          
n = sample size =    64          
              
Thus,              
              
Lower bound =    234.1978354          
Upper bound =    270.7021646          
              
Thus, the confidence interval is              
              
(   234.1978354   ,   270.7021646   ) [ANSWER]