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 _____%.
Accepted Solution
A:
The margin of error given the proportion can be found using the formula
[tex]z* \sqrt{ \frac{p(1-p)}{n} } [/tex]
Where [tex]z*[/tex] is the z-score of the confidence level [tex]p[/tex] is the sample proportion [tex]n[/tex] is the sample size
We have [tex]z*=2.58[/tex] [tex]p=0.38[/tex] [tex]n=80[/tex]
Plugging these values into the formula, we have: [tex]2.58 \sqrt{ \frac{0.38(1-0.38)}{80} } =0.14[/tex]