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A recent poll of 80 randomly selected Californians showed that [tex]$38 \%(\hat{p}=0.38)$[/tex] believe they are doing all that they can to conserve water.

The state government would like to know, within a [tex]$99 \%$[/tex] confidence level, the margin of error for this poll. The [tex]$99 \%$[/tex] confidence level [tex]$z^\ \textless \ em\ \textgreater \ $[/tex]-score is 2.58.

Remember, the margin of error, [tex]$E$[/tex], can be determined using the formula [tex]$E=z^\ \textless \ /em\ \textgreater \ \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$[/tex].

To the nearest whole percent, the margin of error for the poll is [tex]$\square \%$[/tex].



Answer :

GinnyAnswer
To determine the margin of error for the given poll at a 99% confidence level, we will use the provided formula for margin of error:

[tex]\[ E = z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \][/tex]

Given values:
- \(\hat{p} = 0.38\) (proportion of Californians who believe they are doing all they can to conserve water)
- \(n = 80\) (sample size)
- \(z^* = 2.58\) (z-score for 99% confidence level)

Step-by-Step Solution:

1. Calculate the standard error:

[tex]\[ \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.38 \times (1 - 0.38)}{80}} \][/tex]

2. Simplify inside the square root:

[tex]\[ 0.38 \times (1 - 0.38) = 0.38 \times 0.62 = 0.2356 \][/tex]

[tex]\[ \frac{0.2356}{80} = 0.002945 \][/tex]

3. Find the square root:

[tex]\[ \sqrt{0.002945} \approx 0.05426 \][/tex]

4. Multiply by the z-score to find the margin of error:

[tex]\[ E = 2.58 \times 0.05426 \approx 0.14001 \][/tex]

5. Convert the margin of error to a percentage:

[tex]\[ 0.14001 \times 100 \approx 14.001 \][/tex]

6. Round to the nearest whole percent:

[tex]\[ 14.001 \approx 14 \][/tex]

Therefore, to the nearest whole percent, the margin of error for the poll is [tex]\(14\%\)[/tex].

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