Question

Starting salaries of 120 college graduates who have taken a statistics course have a mean of $44,449. Suppose the distribution of this population is approximately normal and has a standard deviation of $9,768.
Using a 99% confidence level, find both of the following:
(NOTE: Do not use commas nor dollar signs in your answers.)

(a) The margin of error E:  

(b) The confidence interval for the mean :   

Answer 1

Solution :

sample size = n = 120

Degrees of freedom = df = n - 1 = 119

t /2,df = 2.618

a)

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

= 2.618 * (9768 / 120)

Margin of error = E = 2335

b)

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

- E < < + E

44449 - 2335 < < 44449 + 2335

42114 < < 46784

The confidence interval for the mean is from 42114 to 46784