Question

Q 8: Assume that the population has a normal distribution. Estimate the population mean annual earning.

College students' annual earnings: 95% confidence; n=9,

sample mean = $21361, s=$865

Answer choices:

$20696 < u < $22026

$20796 < u < $21926

$19681 < u < $20128

$20584 < u < $21657

Answer 1

Solution :

degrees of freedom = n - 1 = 9 - 1 = 8

t$\alpha$/2,df = t_0.025,8 = 2.306

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 2.306 * ( 865/ $\sqrt$ 9)

Margin of error = E = 665

The 95% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

21361 - 665 < $\mu$ < 21361 + 665

( $20696 < $\mu$ < $22026 )