Question

A recent survey asked 1244 adults how many years of education they had. The mean was 14.03 years with a standard deviation of 3.31 years

(a) Construct a 90% confidence interval for the mean number of years of education.

(b) An earlier study suggests the mean was 13.63 in 2001. Does your confidence interval support the claim that the mean remains approximately the same? Why?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 14.03

Population standard deviation =    =3.31

Sample size = n =1244

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 * (3.31 /  1244 )

E= 0.1544
At 90% confidence interval estimate of the population mean
is,

- E < < + E

14.03- 0.1544 <   < 14.03 + 0.1544

13.8756 <   < 14.1844

( 13.8756 , 14.1844 )

(B)yes because any sample mean can accept the confidence interval