Can somebody please help me with this statistics question?
A survey of 1930 student loan borrowers found that 505 had loans totaling more than $20,000 for their undergraduate education. Give a 95% confidence interval for the proportion of all student loan borrowers who have loans of $20,000 or more for their undergraduate education.
95% CI = (______, ______)
For this survey there were 1730 borrowers whose total debt was $10,000 or more. Of these, 190 left school without completing a degree. Consider the population to be borrowers whose total debt was $10,000 or more. Find a 95% confidence interval for the proportion of borrowers who left school without completing a degree in this population.
95% CI = (______, ______)
Please please please explain if you can. I don’t know how to do this “type” of question and need guidance. thank you =)
large sample confidence intervals are used to find a region in which we are 100 (1-α)% confident the true value of the parameter is in the interval.
For large sample confidence intervals for the proportion in this situation you have:
pHat ± z * sqrt( (pHat * (1-pHat)) / n)
where pHat is the sample proportion
z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α
n is the sample size
the zscore for a 95% CI is 1.96
Question 1:
505 / 1930 ± 1.96 * sqrt( 505/1930 * ( 1 – 505/1930) / 1930 )
= 0.2616580 ± 0.01960980
= (0.2420482, 0.2812678)
Question 2:
190 / 1730 ± 1.96 * sqrt( 190 / 1730 * ( 1 – 190 / 1730) / 1730 )
= (0.09509247, 0.12456071)
