Question

A survey of 1282 student loan borrowers found that 440 had loans totaling more than $20,000 for their undergraduate education. Give a 98% confidence interval for the proportion of all student loan borrowers who have loans of $20,000 or more for their undergraduate education. (Give answers accurate to 3 decimal places.)

Answer 1

The 98% confidence interval of the proportion = (0.312, 0.374)

Explanation:

(Give answers accurate to 3 decimal places.)

The formula for Confidence Interval of Proportion is given as:

p ± z × √p(1 - p)/n

Where p = Proportion = x/n

x = 440

n = 1282

p = 440/1282 = 0.34321372854

Approximately = 0.343

z = z-score of 98 % confidence interval

= 2.326

Confidence Interval =

= 0.343 ± 2.326 × √0.343(1 - 0.343)/1282

= 0.343 ± 2.326 × √0.225351/1282

= 0.343 ± 2.326 × √0.00017578081

= 0.343 ± 2.326 × 0.01325823555

= 0.343 ± 0.03083865589

0.343 - 0.03083865589

= 0.31216134411

Approximately = 0.312

0.343 + 0.03083865589

= 0.37383865589

Approximately to = 0.374

Therefore, the 98% confidence interval of the proportion = (0.312, 0.374)