Question

The union for a particular industry has determined that the standard deviation of the daily wages of its workers is $16. A random sample of 60 workers in this industry has a mean daily wage of $111. Find a 99% 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 answer to one decimal place. What is the lower limit of the 99% confidence interval? What is the upper limit of the 99% confidence interval?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 111

Population standard deviation =    = 16

Sample size = n = 60

At 99% confidence level

= 1 - 99%  

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z_{0.005 =} 2.576

Margin of error = E = Z/2 * ( /n)

= 2.576 * ( 16 /  60 )

= 5.3

At 99% confidence interval estimate of the population mean is,

  ± E

111 ± 5.3   

(105.7, 116.3 )