Question

Surveys conducted in American high schools concluded that 90% of the students in a sample of 400 students had more than one active email account. What would the margin of error be for the population proportion?
A.) 1.5%
B.) 3%
C.) 4.5%
D.) 9%

Answer 1

The sample proportion has an approximately normal distribution, with the mean being the population mean and the standard deviation (i.e. the standard error of the sampling proportion) equal to 

SE = sqrt(p*q/n) where p is the population proportion q is 1-p, and n is the sample size. 

(Typically, for a confkdence interval, the sample proportion and its complement are substituted for p and q.) 

The margin of error will be Zc*SE, where Zc is the critical value of Z for the level of confidence required. The confidence interval, then, will be the sample proportion, plus or minus the margin of error. 

For a 90% confidence interval, Zc is 1.645. 

So the margin of error is 1.645 * sqrt(.9*.1/400), roughly .025. So your confidence interval would be around [.875, .925] 


The best and most correct answer among the choices provided by your question is the second choice or letter C which is 3%.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.

Answer 2

I would say 9% because that's the percentage left