# What are the statistics for successfully paying off home loans in the USA from the time home loans were first?

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When home loans were first instituted. I know my Grandparents bought a house in Baltimore City, MD for about \$ 15,000 back in the 1960s and now its worth about in the 300ks.

I am not sure where I am going with this question other than that the practices within the mortgage industry need an overhaul so that home buyers that default on a loan

are not left in the dust and penniless while the mortgage lenders get the properties plus some extra cash from the down payments and monthly loan payments.

That is how I see it what about you?

In a news story distributed by the Washington Post, Lew Sichelman reports that a substantial fraction of mortgage loans that go into default within the first year of the mortgage were approved on the basis of falsified applications. For instance, loan applicants often exaggerate their income or fail to declare debts. Suppose that a random sample of 1,000 mortgage loans that were defaulted within the first year reveals that 410 of these loans were approved on the basis of falsified applications.
a. Find a point estimate of and 95% confidence interval for p, the proportion of all first year defaults that are approved on the basis of falsified applications.
b. Based on your interval, what is a reasonable estimate of the minimum percentage of first-year defaults that are approved on the basis of falsified applications?

Michelle J
April 29, 2011 at 9:05 pm

My parents paid 13,500.00 for thier house in 1972. They paid it off in 2002. I can remember how happy they were. I can’t imagine ever being able to actually pay off ours. The home prices have been driven so high, it seems like ends will never meet.

Merlyn
April 29, 2011 at 9:11 pm

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

in this question we have

pHat = 410/1000
n = 1000

the z-score for a 95% CI is 1.96

the 95% CI is

0.41 ± 1.96 * sqrt( 0.41 * (1-0.41) / 1000 )

(0.3795159 , 0.4404841)

b) about 38% is a good estimate for the minimum proportion of first – year defaults that were approved on the basis of falsified applications.

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