Question

Hi, can you help me answer this question please thank you

Answer 1

We can calculate the margin of error, ME, using the following formula:

$ME=z\cdot\sqrt{\frac{p\cdot(1-p)}{n}}$

where, p is the proportion, n the sample size, and z is the z-score,

for this example:

z = 1.96 because the confidence level is 95%

p = 28/147

n = 147

let's replace and solve

$ME=1.96\cdot\sqrt[]{\frac{\frac{28}{147}\cdot(1-\frac{28}{147})}{147}}$

solving this, we obtain

$ME=0.06347$

which is the same as:

ME = 6.3%