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

The correct answer is "664".

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:

=
`((z)/(E) )^2* p(1-p)`

On substituting the values, we get

=
`((2.576)/(0.05) )^2* 0.5(1-0.5)`

=
`(51.52)^2* 0.25`

=
`2,654.31* 0.25`

=
`663.5776`

i.e.,

=
`664`