The National Student Loan Survey asked the student loan borrowers in their sampl

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 1260 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?" 57% said they felt burdened.

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(b) "If you could begin again, taking into account your current experience, what would you borrow?" 55.3% 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." 32.2% disagreed.

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(d) "Making loan payments is unpleasant but I know that the benefits of education loans are worth it." 57.6% 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." 58.6% 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." 69.1% agreed.

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A matched pairs experiment compares the taste of instant versus fresh-brewed coffee. Each subject tastes two unmarked cups of coffee, one of each type, in random order and states which he or she prefers. Of the 40 subjects who participate in the study, 14 prefer the instant coffee. Let p be the probability that a randomly chosen subject prefers fresh-brewed coffee to instant coffee. (In practical terms, p is the proportion of the population who prefer fresh-brewed coffee.)

(a) Test the claim that a majority of people prefer the taste of fresh-brewed coffee. Report the large-sample z statistic. (Round your answer to two decimal places.)


Report its P-value. (Round your answer to four decimal places.)


(b) Draw a sketch of a standard Normal curve and mark the location of your z statistic. Shade the appropriate area that corresponds to the P-value.

(c) Is your result significant at the 5% level?

YesNo    

What is your practical conclusion?

The result is significant at the 5% level, so we reject H0 and conclude that a majority of people prefer fresh-brewed coffee.The result is significant at the 5% level, so we reject H0 and conclude that a majority of people prefer instant coffee.    The result is not significant at the 5% level, so we do not reject H0 and conclude that a majority of people prefer instant coffee.The result is not significant at the 5% level, so we reject H0 and conclude that a majority of people prefer instant coffee.The result is not significant at the 5% level, so we do not reject H0 and conclude that a majority of people prefer fresh-brewed coffee.

Explanation / Answer

Solution:

Given = 0.05, |Z(0.025)|=1.96 (from standard normal table)

(a) The 95% CI is

p ± Z*sqrt(p*(1-p)/n)

--> 0.570 ± 1.96*sqrt(0.570*(1-0.570)/1260)

--> (0.5426635, 0.5973364)

(b) The 95% CI is

p ± Z*sqrt(p*(1-p)/n)

--> 0.553 ± 1.96*sqrt(0.553*(1-0.553)/1260)

--> (0.525547, 0.580452)

(c) The 95% CI is

p ± Z*sqrt(p*(1-p)/n)

--> 0.322 ± 1.96*sqrt(0.322*(1-0.322)/1260)

--> (0.2962003, 0.3477996)

(d) The 95% CI is

p ± Z*sqrt(p*(1-p)/n)

--> 0.576 ± 1.96*sqrt(0.576*(1-0.576)/1260)

--> (0.5487124, 0.603287)

(e) The 95% CI is

p ± Z*sqrt(p*(1-p)/n)

--> 0.586 ± 1.96*sqrt(0.586*(1-0.586)/1260)

--> (0.558803, 0.613196)

(f) The 95% CI is

p ± Z*sqrt(p*(1-p)/n)

--> 0.691 ± 1.96*sqrt(0.691*(1-0.691)/1260)

--> (0.665485, 0.716514)

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