Question

2. You're interested in the proportion of students in your school who favor the elimination of a local curfew on teenagers. Assume your school has 1,000 students. A. To generate a 95% confidence interval for the proportion of students who favor the elimination of a curfew, what's the minimum number of students you need if you want the margin of error to be 5%? (9 points)

Answer 1

Answer: The minimum number of students you need if you want the margin of error to be 5% IS 278.

Explanation:

Cochran’s Sample Size Formula gives the minimum number of students as $n= \frac{z^{2}pq }{e^{2} }$

Where:

   e is the desired level of precision (i.e. the margin of error),

   p is the (estimated) proportion of the population which has the attribute in question and q is 1 – p.

The z-value  for 95% confidence interval is found to be 1.96 in a Z table.

Assuming that half of the teenagers favor the elimination of a curfew: this gives us maximum variability. So p = 0.5 and q=0.5.

Then $n= (\frac{1.96^{2}*0.5*0.5 }{0.05^{2} } )$

$n= frac{0.9604}{0.0025} }$

$n= {384.16 }$

Rounding up, $n= \frac{385 }$

But considering that 1000 is a small population,  we can modify the sample size we calculated above formula by using this equation:

$s = \frac{n}{1 + \frac{n - 1}{N} }$

Where s is the adjusted sample size, n is the original sample size we calculated and N is the population size.

$s = \frac{385}{1 + \frac{385 - 1}{1000} }$

$s = \frac{385}{1 + \frac{384}{1000} }$

$s = 278$