Question

uppose you want to estimate the proportion of traditional college students on your campus who own their own car. From research on other college​ campuses, you believe the proportion will be near 25​%. What sample size is needed if you wish to be 95​% confident that your estimate is within 0.02 of the true​ proportion?

Answer 1

Answer:

1,269 students

Step-by-step explanation:

Z-score for a 90% confidence interval (z) = 1.645

The proportion of students who own their own car (p) = 0.25

Standard error = 0.02  

The standard error of a proportion is given by:

$SE = z*\sqrt{\frac{p*(1-p)}{n} }$

Applying the given values, the sample size 'n' needed is:

$0.02 = 1.645*\sqrt{\frac{0.25*(1-0.25)}{n} }\\n=0.25*0.75*(\frac{1.645}{0.02})^2 \\n=1,268.45\ students$

Rounding up to the next whole student, the sample size needed is 1,269 students.