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

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:

=  

On substituting the values, we get

=  

=  

=  

=  

i.e.,

=  

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐