Question

A news agency publishes results of a recent poll. It reports that candidate A leads candidate B by 10% because 45% of the poll participants supported Ms. A whereas only 35% supported Mr. B. What margin of error should be reported for each of the listed estimates, 10%, 35%, and 45%? Notice that 900 people participated in the poll, and the reported margins of error typically correspond to 95% confidence intervals.

Answer 1

We need to use the formula to find Margin of Error but through the sample proportion, that is

$E=z*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

We will use a 95% confidence interval, that is a z value of 1.96 (Search in a Normal distribution table)

A) For A our $\hat{p}$ (proportion) is equal to 0.45. So applying the formula,

$E=1.96*\sqrt{\frac{0.45(1-0.45)}{900}}$

$E=3.25\%$

B) We make the same of point A, but change our proportion to 0.35

$E=1.96*\sqrt{\frac{0.35(1-0.35)}{900}}$

$E=3.116\%$

c) We need to calculate the SE through proportion for 0.1, that is

$SE = \sqrt{(\frac{\hat{p_1}(1-\hat{p_1})}{n})+(\frac{\hat{p_2}(1-\hat{p_2})}{n})}$

Then our Error is given by,

$E=z*SE$

$E=1.96*\sqrt(0.45*\frac{0.55}{900}+0.35\frac{0.65}{900}$

$E=0.045$