Question

The CHE reported that 60% of those responding to a national survey of college freshmen were attending the college of their first choice. Suppose that 500 students responded to the survey the actual sample size was much larger.

a) Using the sample size of 500, calculate a 98% confidence interval for the proportion of college students who were attending their first choice of college.

b) The actual sample size of survey was much larger than 500. Would a confidence interval based on the actual sample size have been narrower or wider than the one computed in part a?

Answer 1

Solution:

a) Given:
n=500
p=60%
=0.60
(1-alpha)%= 98%
1-alpha= 0.98
alpha= 0.02
alpha/2=0.01
Z(alpha/2)=2.326 ..... From standard normal table.
Margin of error=Z(alpha/2) ×✓{[p(1-p)]/n}
= (2.326)×✓{[0.60(1-0.60)]/500}
=(2.326)×0.0224
=0.0521
98% confidence interval for the proportion of college students who were attending their first choice of college is given as,
p ± Margin of error=(0.60-0.0521 ,0.60+0.0521)
=(59.9479,60.0521)
98% confidence interval for the proportion of college students who were attending their first choice of college is (59.9479,60.0521)
b) As we observed that the confidence interval mainly depends upon the margin of error and margin of error is inversely proportion to the sample size so if the sample size is more than the margin of error would be less and therefore the confidence interval would be narrow.
Here,the actual sample size of survey was much larger than 500 so confidence interval based on the actual sample size have been narrower than the one computed in part( a)