Question

margin of error for proportions
score: 2/5 penalty: 1 off
question
a survey was given to a random sample of 115 residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. of those surveyed, 20% of the people 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 percentage to the nearest tenth? (do not write ±).
answer attempt 1 out of 2
% submit answer

Explanation:

Step1: Identify the formula

The formula for the margin of error for a proportion is $E = z\sqrt{\frac{p(1 - p)}{n}}$, where $z$ is the z - score, $p$ is the sample proportion, and $n$ is the sample size. For a 95% confidence level, $z=1.96$, $p = 0.20$, and $n = 115$.

Step2: Calculate the value inside the square - root

First, calculate $p(1 - p)$: $0.20\times(1 - 0.20)=0.20\times0.80 = 0.16$. Then, $\frac{p(1 - p)}{n}=\frac{0.16}{115}\approx0.0013913$.

Step3: Calculate the square - root value

$\sqrt{\frac{0.16}{115}}\approx\sqrt{0.0013913}\approx0.0373$.

Step4: Calculate the margin of error

$E=1.96\times0.0373\approx0.0731$.

Step5: Convert to percentage

To convert to a percentage, multiply by 100: $0.0731\times100 = 7.31\%$. Rounding to the nearest tenth, we get $7.3\%$.

Answer:

$7.3$