A random sample of 27 people employed by the Florida state authority established

Question

A random sample of 27 people employed by the Florida state authority established they earned an average wage (including benefits) of $57.00 per hour. The sample standard deviation was $5.81 per hour. (Use z Distribution Table.) What is the best estimate of the population mean? Develop a 99% 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 $1.00 at 99% confidence? (Round up your answer to the next whole number.)

Explanation / Answer

a)

It is the sample mean,

best estimate = 57 [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 =    0.005          
X = sample mean =    57          
z(alpha/2) = critical z for the confidence interval =    2.58          
s = sample standard deviation =    5.81          
n = sample size =    27          
              
Thus,              
Margin of Error E =    2.884788355          
Lower bound =    54.11521164          
Upper bound =    59.88478836          
              
Thus, the confidence interval is              
              
(   54.11521164   ,   59.88478836   ) [ANSWER]

ALTERNATIVE: If you use z = 2.575 for 99% confidence:

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 =    57          
z(alpha/2) = critical z for the confidence interval =    2.575          
s = sample standard deviation =    5.81          
n = sample size =    27          
              
Thus,              
Margin of Error E =    2.87919768          
Lower bound =    54.12080232          
Upper bound =    59.87919768          
              
Thus, the confidence interval is              
              
(   54.12080232   ,   59.87919768   ) [ANSWER]

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c)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.005  
      
Using a table/technology,      
      
z(alpha/2) =    2.58  
      
Also,      
      
s = sample standard deviation =    5.81  
E = margin of error =    1  
      
Thus,      
      
n =    224.694104  
      
Rounding up,      
      
n =    225   [ANSWER]

ALTERNATIVE: If you use z = 2.575 for 99% confidence:

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) =    2.575  
      
Also,      
      
s = sample standard deviation =    5.81  
E = margin of error =    1  
      
Thus,      
      
n =    223.8240406  
      
Rounding up,      
      
n =    224   [ANSWER]

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