Question

Starting salaries of 60 college graduates who have taken a statistics course have a mean of $44,600. Suppose the distribution of this population is approximately normal and has a standard deviation of $8,985.
Using an 81% confidence level, find both of the following:
(NOTE: Do not use commas nor dollar signs in your answers.)

(a) The margin of error:  

(b) The confidence interval for the mean μ: <μ<

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 44600

Population standard deviation =    = 8985

Sample size n =60

At 81% confidence level the z is ,

= 1 - 81% = 1 - 0.81 = 0.19

/ 2 = 0.095

Z/2 = Z0.095 = 1.31 ( Using z table )

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

= 1.31 * (8985 /  60 )

= 1519.5457
At 95% confidence interval estimate of the population mean
is,

- E < < + E

44600 - 1519.5457 <   <44600 + 1519.5457

43080.4543 <   < 46119.5457

( 43080.4543 , 46119.5457 )