SOLUTION: In a newsstory distributed by the local news paper, it reports that a substantial fraction of mortgage loans that go into default within the rst year of the mortgage were approved

Question 1124393: In a newsstory distributed by the local news paper, it reports that a substantial fraction of mortgage loans that go into default within the rst year of the mortgage were approved on the basis of falsied applications. For instance, loan applications often exaggerate their income or fail to declare degrees. Suppose that a random sample of 1000 mortgage loans that were defaulted within therst year reveals that 410 of these loans were approved on the basis of falsied applications. A 90% condence interval for proportion of all the rst year defaults that are approved on the basis of falsied applications is given by:
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
410/1000 = 0.41 decimal equivalent
:
our proportion is 0.41
:
since we do not know the population's standard deviation, we use the standard error and the student t-statistic
:
standard error = square root(0.41 * (1-0.41) / 36) = 0.082
:
alpha(a) = 1 - 90/100 = 0.10
:
critical probability(p*) = 1 - a/2 = 0.95
:
degrees of freedom(df) = 36 -1 = 35
:
the t-statistic having 35 degrees of freedom and a cumulative probability of 0.95 is 1.69
:
Margin of Error(ME) = 1.69 * 0.082 = 0.1386
:
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90% confidence interval is 0.41 + or - 0.1386, which is 0.2714 to 0.5486
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