Question

An opinion poll reported that 36% believe a certain governor broke campaign financing laws in his election campaign. How many people would you have to survey to be 99% confident that you can estimate to within 1% the fraction of people who believe the governor broke campaign financing laws? Round your answer to whole number.

Answer 1

Answer:

The number of people would have participate in opinion poll

n =  15,267

Step-by-step explanation:

Step(i):-

Given data an opinion poll reported that 36% believe a certain governor broke campaign financing laws in his election campaign.

Given proportion = 36% =0.36

Given the margin of error = 1% =0.01

Level of significance = 99% 0 r 0.01

Z₀.₀₁ = 2.576

step(ii):-

The margin of error of Population proportion is determined by

$M.E = \frac{Z_{0.01} \sqrt{p(1-p)} }{\sqrt{n} }$

$0.01 = \frac{2.576 \sqrt{0.36(1-0.36)} }{\sqrt{n} }$

$\sqrt{n} = \frac{2.576 \sqrt{0.36(1-0.36)} }{0.01 }$

√n =  123.648

Squaring on both sides , we get

n =  15,267

final answer:-

The number of people would have participate in opinion poll

n =  15,267