Question

margin of error for proportions
score: 3/5 penalty: 1 off
question
a survey was given to a random sample of 135 residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. of those surveyed, 108 respondents said they were in favor of the plan. at the 95% confidence level, what is the margin of error for this survey expressed as a proportion to the nearest thousandth? (do not write ±).
answer attempt 1 out of 2

Explanation:

Step1: Calculate sample proportion $\hat{p}$

$\hat{p}=\frac{108}{135} = 0.8$

Step2: Determine $z$-value for 95% confidence level

For 95% confidence level, $z = 1.96$

Step3: Calculate margin of error formula for proportion

$E=z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
Substitute $\hat{p}=0.8$, $n = 135$ and $z = 1.96$ into the formula:
$E=1.96\sqrt{\frac{0.8\times(1 - 0.8)}{135}}$
$E=1.96\sqrt{\frac{0.8\times0.2}{135}}$
$E=1.96\sqrt{\frac{0.16}{135}}$
$E=1.96\sqrt{0.001185185}$
$E=1.96\times0.0344$
$E\approx0.067$

Answer:

$0.067$