Question

Helen is reading a news article that 75% ± 12% of students from Acsenda find a well-paid job after they finish their studies. What sample size is used in this study?

Answer 1

Given p=75%

or p =0.75

margin of error = 12% or 0.12

Now Standard error = sqrt {p(1-p) / n} where n is sample size

Mostly we take 95% confidence

so margin of error for 95%confidence = Z0.05* S.E.

= 1.96* sqrt {p(1-p) / n}

=1.96* sqrt{0.75*0.25/n}

now we have

1.96*0.4330127 /sqrt( n) =0.12

n = { 0.8487049/0.12}2

n= 50

when confidence interval is 99%

n= { 2.58*0.4330127/0.12}2

n= 87