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ad-work [718]
3 years ago
9

A recent poll of 80 randomly selected Californians showed that 38% ( = 0.38) believe they are doing all that they can to conserv

e 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 %.
Mathematics
2 answers:
Elanso [62]3 years ago
8 0
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%
MaRussiya [10]3 years ago
7 0

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%




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