Question

Salaries of 32 college graduates who took a statistics course in college have a​ mean,x overbarx​,of $ 65,300. Assuming a standard​ deviation,sigmaσ​,of​$13,299​,construct a 95​% confidence interval for estimating the population mean muμ.

​$nothingless than<muμless than<​$nothing

​(Round to the nearest integer as​ needed.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $65300

Population standard deviation =    = $13299

Sample size = n =32

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96   ( Using z table ( see the 0.025 value in standard normal (z) table corresponding z value is 1.96 )

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

= 1.96* (13299 /  32 )

E= 4607.8684
At 95% confidence interval estimate of the population mean
is,

- E < < + E

65300 - 4607.8684 <   < 65300 + 4607.8684

60692.1316 <   < 69907.8684

( 60692.1316 ,69907.8684)

( $60692 ,$69908)