The National Student Loan Survey asked the student loan borrowers in their sampl
ID: 3338706 • Letter: T
Question
The National Student Loan Survey asked the student loan borrowers in their sample about attitudes toward debt. Below are some of the questions they asked, with the percent who responded in a particular way. Assume that the sample size is 1285 for all these questions. Compute a 95% confidence interval for each of the questions, and write a short report about what student loan borrowers think about their debt. (Round your answers to three decimal places.)
(a) "To what extent do you feel burdened by your student loan payments?" 56.3% said they felt burdened.
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(b) "If you could begin again, taking into account your current experience, what would you borrow?" 55.8% said they would borrow less.
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(c) "Since leaving school, my education loans have not caused me more financial hardship than I had anticipated at the time I took out the loans." 32.7% disagreed.
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(d) "Making loan payments is unpleasant but I know that the benefits of education loans are worth it." 57.1% agreed.
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(e) "I am satisfied that the education I invested in with my student loan(s) was worth the investment for career opportunities." 58.1% agreed.
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(f) "I am satisfied that the education I invested in with my student loan(s) was worth the investment for personal growth." 72% agreed.
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Explanation / Answer
Solution:
Confidence interval for population proportion: sample proportion ± z * SE
Where z is a multiplier
SE = [p*(1-p)/n]
n = 1285
z for 95% confidence level = 1.96
a) p = 56.3% = 0.563
Standard Error = [0.563*(1-0.563)/1285] = 0.000191463 = 0.0138
95% confidence interval = 0.563 ± 1.96*0.0138
95% confidence interval = ( 0.5359, 0.5900)
b) p = 55.8% = 0.558
SE = [0.558*(1-0.558)/1285] = 0.0138
95% confidence interval = 0.558 ± 1.96*0.0138
95% confidence interval = (0.5309, 0.5850)
c) p = 32.7% = 0.327
SE = [0.327*(1-0.327)/1285] = 0.01309
95% confidence interval = 0.327 ± 1.96*0.01309
95% confidence interval = (0.3013, 0.3527)
d) p = 57.1% = 0.571
SE = [0.571*(1-0.571)/1285] = 0.0138
95% confidence interval = 0.571 ± 1.96*0.0138
95% confidence interval = ( 0.5439, 0.5980)
e) p = 58.1% = 0.581
SE = [0.581*(1-0.581)/1285] = 0.0137
95% confidence interval = 0.81 ± 1.96*0.0137
95% confidence interval = (0.7831, 0.8368)
f) p = 72% = 0.72
SE = [0.72*(1-0.72)/1285] = 0.0125
95% confidence interval = 0.72 ± 1.96*0.0125
95% confidence interval = (0.6955, 0.7445).
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