A random sample of 60 workers in this industry has a mean da

A random sample of 60 workers in this industry has a mean daily wage of $ 113.

Find a 95% confidence interval for the true mean daily wage of all union workers in the industry. Then complete the table below.

Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place.

(If necessary, consult a list of formulas.)

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The union for a particular industry has determined that the standard deviation of the daily wages of its workers is $20.

A random sample of 60 workers in this industry has a mean daily wage of $ 113.

Find a 95% confidence interval for the true mean daily wage of all union workers in the industry. Then complete the table below.

Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place.

(If necessary, consult a list of formulas.)

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 =    113          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    20          
n = sample size =    60          
              
Thus,              
              
Lower bound =    107.9393948   [ANSWER]      
Upper bound =    118.0606052   [ANSWER]      

              
Thus, the confidence interval is              
              
(   107.9393948   ,   118.0606052   )

A random sample of 60 workers in this industry has a mean daily wage of $ 113. Find a 95% confidence interval for the true mean daily wage of all union workers

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