According to the Institute for College Access and Success, 69% of students who g

Question

According to the Institute for College Access and Success, 69% of students who graduated from college in 2013 had student loan debt. You want to know if the proportion is the same in Wisconsin as in the US as a whole, so you survey a random sample of 100 people who graduated from Wisconsin colleges in 2013.

a. Assuming that the proportion of people with student loan debt is the same in Wisconsin as in the whole US, what is the probability that between 65 and 75% of your sample has student loan debt? Use a normal approximation for p^ hat.

b. Fill in the blank: If the proportion of people with student loan debt is the same in Wisconsin as in the whole US, then there is a 0.15 probability that the sample proportion with loan debt is less than____.

c. Fill in the blanks: If the proportion of people with student loan debt is the same in Wisconsin as in the whole US, then there is a 0.95 probability that the sample proportion with loan debt is above ____and below ____.

Explanation / Answer

a) standard error, s = sqrt(p*(1-p)/n) = sqrt(0.69*0.31/100) = 0.0462493

For probability that between 65% and 75% people has student loan debt,

P(0.65 <=p <= 0.75) = P(p<= 0.75) - P(p <= 65) = P(z <= (0.75 - 0.69)/0.0462493) - P(z <= (0.65 - 0.69)/0.0462493)

=P(z<=1.2973) - P(z<=-0.864878) = 0.9027 - 0.1936 = 0.7091

b) p value given is 0.15, since it specifies less than, the z score for left sided with p 0.15 is,

15.000% of a standard Gaussian distribution has values less than - 1.036

z = -1.036

z = (p - p0)/SE; -1.036 = (p-0.69)/0.0462493; p = 0.642

c) This is a 95% CI. zcrit = 1.96

lower limit = p0 - zcrit*SE = 0.69 - 1.96*0.0462493 = 0.599

upper limit = p0 + zcrit*SE = 0.69 + 1.96*0.0462493 = 0.781

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