Question

1. The Dean of engineering is interested in estimating the proportion of incoming students at her university that plan to study engineering. During the university’s orientation, the office staff managed to survey 120 students during the lunch break. Of those surveyed, 25 students said they were interested in studying engineering.
   1. Compute a 95% confidence interval that estimates the proportion of incoming students that plan to study engineering.
   2. Using the information provided, what sample size would be needed for the width of the interval to be within 0.05?
   3. If the office staff had not been able to survey the students over lunch, what sample size would be needed in order to calculate the 95% confidence interval with an error of 0.05?

Answer 1

Ans:

a)

sample proportion=25/120=0.208

95% confidence interval for p

=0.208+/-1.96*sqrt(0.208*(1-0.208)/120)

=0.208+/-0.073

=(0.136, 0.281)

b)

Sample size required,n=1.96^2*0.208*(1-0.208)/0.05^2=254

c)

When no prior estimate is available,then consider p=0.5

Sample size required,n=1.96^2*0.5*(1-0.5)/0.05^2=384