Question

Starting salaries of 64 college graduates who have taken a statistics course have a mean of $42,500 with a standard deviation of $6,800. Find an 90% confidence interval for ?μ. (NOTE: Do not use commas or dollar signs in your answers. Round each bound to three decimal places.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $42500

Population standard deviation =    = $6800
Sample size = n =64

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 * ( 6800/  64 )

= 1398.25
At 90% confidence interval estimate of the population mean
is,

- E < < + E

42500 -1398.25<   <42500 + 1398.25

41101.750<   < 43898.250
lower limit=41101.750

upper limit=43898.250