Question

Starting salaries of 110 college graduates who have taken a statistics course have a mean of $42,647. Suppose the distribution of this population is approximately normal and has a standard deviation of $10,972.
Using a 75% confidence level, find both of the following:
(NOTE: Do not use commas or dollar signs in your answers.)

(a) The margin of error:  

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

Answer 1

Solution :

Given that,

= 42647

= 10972

n = 110

At 75% confidence level the z is ,

  = 1 - 75% = 1 - 0.75 = 0.25

/ 2 = 0.25 / 2 = 0.125

Z/2 = Z0.125 = 1.15

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

= 1.15* ( 10972/ 110 )

= 1203.06

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

- E < < + E

42647 - 1203.06 < < 42647 + 1203.06

41443.94 < < 43850.06

(41443.94 ,43850.06)