Question

Starting salaries of 41 college graduates who have taken a statistics course have a mean of $44,753 and a sample standard deviation of $10,905.
Using 99% confidence, find both of the following:

A. The margin of error ?

B. The confidence interval for the mean ?:
? <?< ?

Answer 1

Solution :

Given that,

Point estimate = sample mean = =$44,753

sample standard deviation = s = $10,905

sample size = n = 41

Degrees of freedom = df = n - 1 =41-1=40

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df = t0.005,40 = 2.704

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

= 2.704* (10905 /41)

Margin of error = E = 4605.1

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

- E < < + E

44753 - 4605.1 < < 44753 + 4605.1

40147.9 < < 49358.1

(40147.9,49358.1) ,40147.9 < < 49358.1