Question

The union for a particular industry has determined that the standard deviation of the daily wages of its workers is $19. A random sample of 70 workers in this industry has a mean daily wage of $117. Find a 90% 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.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $117

Population standard deviation =    = $19
Sample size = n =70

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z/2    * ( /n)

= 1.645 * ( 19 /  70 )

=3.7
At 90% confidence interval estimate of the population mean
is,

- E < < + E

117 -3.7 <   <117 + 3.7

113.3 <   < 120.7

( 113.3 ,120.7 )