3 10 In a survey conducted to determine among other things t

3. [10] In a survey conducted to determine, among other things, the cost of vacations taken by single adults in a particular region, 144 individuals were randomly sampled. Each person was asked to assess the total cost of his or her most recent vacation. The average cost was $2,386 and the standard deviation was $400.

[5] a. Construct a 99% confidence interval for the average cost of a vacation trip for all single adults in the region.
[5] b. How large a sample would have to be taken to estimate the true average cost to within $60 of the unknown population average cost with 99% confidence?

Solution

a)

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 =    2386          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    400          
n = sample size =    144          
              
Thus,              
Margin of Error E =    85.86097678          
Lower bound =    2300.139023          
Upper bound =    2471.860977          
              
Thus, the confidence interval is              
              
(   2300.139023   ,   2471.860977   ) [ANSWER]

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b)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.005  
      
Using a table/technology,      
      
z(alpha/2) =    2.575829304  
      
Also,      
      
s = sample standard deviation =    400  
E = margin of error =    60  
      
Thus,      
      
n =    294.8842934  
      
Rounding up,      
      
n =    295   [ANSWER]