Question

An administrator at UF wants to estimate the proportion of students who would support an increase in the student activity fee. This increase would be used to fund a $450 million renovation of the campus football stadium. How many students would need to be selected in the sample if the administrator wants a margin of error of 5% for a 99% con

Answer 1

Answer:

The correct answer is "664".

Step-by-step explanation:

The given values are:

Margin of error,

E = 5%

i.e.,

   = 0.05

Level of confidence,

C = 99%

i.e.,

   = 0.99

Level of significance,

α = 0.01

From the z value table,

Critical z = 2.576

As we know,

The required sample size n will be:

=  $(\frac{z}{E} )^2\times p(1-p)$

On substituting the values, we get

=  $(\frac{2.576}{0.05} )^2\times 0.5(1-0.5)$

=  $(51.52)^2\times 0.25$

=  $2,654.31\times 0.25$

=  $663.5776$

i.e.,

=  $664$