Question

6.18 Is college worth it? Part II: Exercise 6.16 presents the results of a poll where 48% of 331 Americans who decide to not go to college do so because they cannot afford it.


(a) Calculate a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it, and interpret the interval in context.
lower bound: _________ (please round to four decimal places)
upper bound: _________(please round to four decimal places)

Interpret the confidence interval in context: Pick one

• We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval
• We can be 90% confident that our confidence interval contains the sample proportion of Americans who choose not to go to college because they cannot afford it
• 90% of Americans choose not to go to college because they cannot afford it

(b) Suppose we wanted the margin of error for the 90% confidence level to be about 1.5%. How large of a survey would you recommend?
A survey should include at least _______ people?

Please Show all your work!

Answer 1

a)

z value at 90% = 1.645

CI = p +/- z *sqrt(p*(1-p)/n)
= 0.48 +/- 1.645 *sqrt(0.48 *(1-0.48)/331)
= (0.4348 , 0.5252)

Lower Bound = 0.4348
Upper bound = 0.5252

We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

b)
ME = 0.015
z value at 90% = 1.645

p = 0.48

ME =z *sqrt(p*(1-p)/n)
0.015 = 1.645 *sqrt(0.48 *(1-0.48)/n)
n = (1.645/0.015)^2 *0.48 *(1-0.48)
n = 3002