Question

A sample of 65 information system managers had an average hourly income of $42.75 and a standard deviation of $7.50.

a. When the 95% confidence interval has to be developed for the average hourly income of all system managers, its margin of error is

b. The 95% confidence interval for the average hourly income of all information system managers is

Answer 1

Solution :

Given that,

= 42.75

s = 7.50

n = 65

Degrees of freedom = df = n - 1 = 65 - 1 = 64

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,64 = 1.998

(a)

Margin of error = E = t_/2,df * (s /n)

= 1.998 * (7.50 / 65)

= 1.86

Margin of error = E = 1.86

(b)

The 95% confidence interval estimate of the population mean is,

- E < < + E

42.75 - 1.86 < < 42.75 + 1.86

40.89 < < 44.61

(40.89 , 44.61)