Question

In 2012 the General Social Survey asked 840 adults how many years of education they had. The sample mean was 8.11 years with a standard deviation of 8.91 years.

An 80% interval for the mean number of years of education. Round the answers to two decimal places.

An 80% confidence interval for the mean number of years of education is _< μ <_.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 8.11

sample standard deviation = s = 8.91

sample size = n = 840

Degrees of freedom = df = n - 1 = 840 - 1 = 839

At 80% confidence level

= 1 - 80%

=1 - 0.80 =0.20

/2 = 0.10

t/2,df = t_{0.10, 839} = 1.283

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

= 1.283 * ( 8.91 / 840)

Margin of error = E = 0.39

The 80% confidence interval estimate of the population mean is,

- E + E

8.11 - 0.39 8.11 + 0.39

( 7.72 8.50 )