Auniversity newspaper is conducting a survey to determine what fraction of students support a $200 per year increase in fees to pay for a new football stadium. how big of a sample is required to ensure the margin of error is smaller than 0.04 using 95% confidence level?

The formula to calculate sample size is  Z * √((P(1-P)/n)

Z is the Z-score for the confidence level, in this problem the confidence level is given as 95%, so Z = 1.96

P is the population, which is unknown, but we want it to be the largest value possible, so in the equation above, P is 0.5.

n is the sample size.

Now we have: 1.96* √((0.5 * 0.5)/n)

They want this sample size to be less than 0.04

So now we have:

1.96* √((0.5 * 0.5)/n) < 0.04

Multiply both sides by 1/1.96:

√((0.5 * 0.5)/n) < 0.04  *1/1.96

Simplify:

0.5 / √n < 4/196

Take the reciprocal of both sides:

√n / 0.5 < 196/4

Simplify:

√n/0.5 < 49

Multiply both sides by 0.5:

√n = 49*0.5

√n = 24.5

to solve for n, raise both sides to the power of 2:

n = 24.5^2

n = 600.25

The sample needs to be about 600 people.

the answer is 8.64

step-by-step explanation:

show us the triangle so we can answer it