Question

There is a margin of error on a food survey of 6%. What is the range of the number of students who prefer chicken strips to meatloaf if 68 of 97 students said they prefer chicken strips and there are 1070 students at the school?

A. 750 to 795 students

B. 705 to 795 students

C. 705 to 784 students

D. 750 to 784 students

Answer 1

I might be wrong but, I think it's C

Answer 2

Option C. 705 to 784 students

Explanation:

margin error = 6%

range of students = 68 to 97

n = 1070

The margin of error is given by the formula:

ME =
`z\sqrt{((p(1-p))/(n) }`

where z is the critical value

p = the probability of an event happening

n = sample size

Therefore, taking the confidence level of 5 % and extrapolating in the tables, the value will range somewhere between 705 to 784 students.