Question

A recent poll of 80 randomly selected Californians showed that 38% ( = 0.38) believe they are doing all that they can to conserve water.
The state government would like to know, within a 99% confidence level, the margin of error for this poll. (99% confidence level z*-score of 2.58)

Remember, the margin of error, E, can be determined using the formula E = z*.

To the nearest whole percent, the margin of error for the poll is %.

Answer 1

The margin of error given the proportion can be found using the formula

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

Where
$z*$ is the z-score of the confidence level
$p$ is the sample proportion
$n$ is the sample size

We have
$z*=2.58$
$p=0.38$
$n=80$

Plugging these values into the formula, we have:
$2.58 \sqrt{ \frac{0.38(1-0.38)}{80} } =0.14$

The result 0.14 as percentage is 14%

Margin error is 38% ⁺/₋ 14%

Answer 2

Answer:  14%

Step-by-step explanation:

We know that Margin error $M.E.=z\sqrt{\frac{p(1-p)}{n}}$

Given: Sample size n=80

sample proportion p=0.38

Confidence level z=2.58

Substitute the values in the formula, we get

Margin error $M.E.=2.58\sqrt{\frac{0.38(1-0.38)}{80}}$

$\Rightarrow\ M.E.=2.58\sqrt{\frac{0.62\ \cdot\ 0.38}{80}}\\\Rightarrow\ M.E.=2.58 \sqrt{0.002945}\\\Rightarrow\ M.E.=2.58\times0.054\\\Rightarrow\ M.E.=0.13932\\\Rightarrow\ M.E.=13.932\%\approx14\%$

Thus, to the nearest whole percent, the margin of error for the poll is 14%