In a survey of 1 280 student loan borrowers there were 1050

In a survey of 1 280 student loan borrowers, there were 1050 borrowers whose total debt was $10,000 or more. Of these, 192 left school without completing a degree. Consider the population to be borrowers whose total debt was $10,000 or more.

Give the left endpoint (lower boundary) of a 95% confidence interval for the proportion of borrowers who left school without completing a degree in this population.

Use the sample proportion p(hat)=X/n for estimation

Question1. For the problem in question 2 give the right endpoint (upper boundary) of a 95% confidence interval for the proportion.

Question2. The P-value of a test of a null hypothesis is

Solution

proportion of population who left school and have borrowd 10000 or more p = 192/1050 = 0.18286 = 18.29%

standard deviation = sqrt(p*(1-p)/n) = 0.011929 = 1.19%

margin of error = Z(critical ) * standard deviation

= 1.96 * 0.011929

= 0.02338 = 2.34%

lower limt = P - margin of error

= 18.29% - 2.34%

= 15.95%

upper limit = 18.29% + 2.34%

= 20.62%

2)

the probability, assuming the null hypothesis is true, that the test statistic will take a value at least as extreme as that actually observed.