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Solution
A)
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 =    0.025          
 X = sample mean =    40.15          
 t(alpha/2) = critical t for the confidence interval =    2.093024054          
 s = sample standard deviation =    15.89198739          
 n = sample size =    20          
 df = n - 1 =    19          
 Thus,              
               
 Lower bound =    32.71232096          
 Upper bound =    47.58767904          
               
 Thus, the confidence interval is              
               
 (   32.71232096   ,   47.58767904   ) [ANSWER]
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b)
YES, becasue 42 is inside this interval.
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c)
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 =    0.025          
 X = sample mean =    40.15          
 t(alpha/2) = critical t for the confidence interval =    2.093024054          
 s = sample standard deviation =    15.89198739          
 n = sample size =    20          
 df = n - 1 =    19          
 Thus,              
               
 Lower bound =    32.71232096          
 Upper bound =    47.58767904          
               
 Thus, the confidence interval is              
               
 (   32.71232096   ,   47.58767904   ) [ANSWER]
(I actually did some parts in part a using Excel).
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d)
We need to assume that at the population from where this sample came from is approximately normally distributed.


