Question

College tuition: a simple random sample of 35 colleges and universities in the united states has a mean tuition of $19,400 with a standard deviation of $10,700. Construct a 95% confidence interval for the mean tuition for all colleges and universities in the United States. round the answers to the nearest whole number.

A 95% confidence interval for the mean tuition for all colleges and universities is __________________< mean < _____________________

Answer 1

Solution :

Given that,

$\bar x$ = $19400

s = $10700

n = 35

Degrees of freedom = df = n - 1 = 35 - 1 = 34

At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t_{$\alpha$ /2,df} = t_0.025,34 = 2.032

Margin of error = E = t_$\alpha$/2 * (s /$\sqrt$n) = 2.032 * (10700/ $\sqrt$35) = 3675

A 95% confidence interval for the mean tuition for all colleges and universities is

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

19400 - 3675 < $\mu$ < 19400 + 3675

$15725 < $\mu$ < $23075