Question

A dean at a large state university (student population exceeds 40,000) is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid.  When the dean repeated this study the next year, she used a sample of 256 students and obtained an interval estimate of 41.95% to 58.05%. What is the confidence level and how many students in the sample received financial aid? Show your calculation(s).

Answer 1

here sample proprotion phat=(upper limit+lower limit)/2=(0.5805+0.4195)/2=0.5

std error of proprotion se =sqrt(phat*(1-phat)/n))=sqrt(0.5*0.5/256)=0.03125

margin of error ME =(upper limit-lower limit)/2=(0.5805-0.4195)/2=0.0805

also as margin of error ME =critical z*std error

0.03125=z*0.0805

z=0.0805/0.03125 =2.576

this is critical value for 99% confidence level

hence  confidence level is 99%.

number of students in the sample received financial aid =nphat=256*0.5=128