Question

What is the margin of sampling error if p=16% and n=400?Formula : ME=2 √p(1-p)/nOptions : 1- about 2% 2- 3% 3- 4% 4- 5%5- 6%

Answer 1

Given:

p = 16%

and, n = 400

We will find the margin of sampling error

The formula to find the margin of sampling error will be:

`ME=2\sqrt[]{(p(1-p))/(n)}`

substitute with (p) and (n):

`ME=2\sqrt[]{(0.16\cdot(1-0.16))/(400)}=0.03666\cdot100\%=3.666\%`

Rounding to the nearest whole percent

So, the answer will be 4%