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:

• 1. 90% of Americans choose not to go to college because they cannot afford it
• 2. 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
• 3. 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) 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.

Answer 1

Solution :

Given that,

n = 331

Point estimate = sample proportion = = 48% = 0.48

1 - = 1 - 0.48 = 0.52

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 (((0.48 * 0.52) / 331 )

= 0.0452

A 90% confidence interval for population proportion p is ,

± E

= 0.48 ± 0.0452

= ( 0.4348, 0.5252 )

lower bound = 0.4348

upper bound = 0.5252

3. 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) margin of error = E = 1.5% = 0.015

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.645 / 0.015)² * 0.48 * 0.52

= 3001.88

sample size = n = 3002 people