Question

A survey of 1273 student loan borrowers found that 458 had loans totaling more than $20,000 for their undergraduate education. Give a 99% confidence interval for the proportion of all student loan borrowers who have loans of $20,000 or more for their undergraduate education. (Round your answers to three decimal places.)

Lower limit	_______
Upper limit	_______

Answer 1

Solution :

Given that,

n = 1273

x = 458

Point estimate = sample proportion = $\hat p$ = x / n = 458 / 1273 = 0.360

1 - $\hat p$ = 1 - 0.360 = 0.64

At 99% confidence level the z is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

Z_$\alpha$/2 = Z_0.005 = 2.576

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 2.576 * ($\sqrt$((0.360 * 0.64) / 1273)

= 0.035

A 95% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.360 - 0.035< p < 0.360 + 0.035

0.325 < p < 0.395

( 0.325 , 0.395 )

Lower limit	0.325
Upper limit	0.395