A random sample of 27 people employed by the Florida state a

A random sample of 27 people employed by the Florida state authority established they earned an average wage (including benefits) of $63.00 per hour. The sample standard deviation was $6.11 per hour. (Use z Distribution Table.)

Develop a 95% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.)

How large a sample is needed to assess the population mean with an allowable error of $2.00 at 90% confidence? (Round up your answer to the next whole number.)

Solution

a) It is the sample mean, 63.00 [ANSWER]

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b)

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 =    0.025          
X = sample mean =    63          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    6.11          
n = sample size =    27          
              
Thus,              
Margin of Error E =    2.304662945          
Lower bound =    60.69533705          
Upper bound =    65.30466295          
              
Thus, the confidence interval is              
              
(   60.69533705   ,   65.30466295   ) [ANSWER]

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c)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.05  
      
Using a table/technology,      
      
z(alpha/2) =    1.644853627  
      
Also,      
      
s = sample standard deviation =    6.11  
E = margin of error =    2  
      
Thus,      
      
n =    25.2509047  
      
Rounding up,      
      
n =    26 [ANSWER]