Question

a survey was given to a random sample of 1650 voters in the united states to ask about their preference for a presidential candidate. of those surveyed, 52% of the people said they preferred candidate a. 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 ±).

Explanation:

Step1: Identify the formula for margin of error for proportions

For a 95% confidence - level, the formula for the margin of error $E$ 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, the z - score $z=1.96$, $p = 0.52$ (52% expressed as a decimal), and $n = 1650$.

Step2: Calculate $(1 - p)$

$1-p=1 - 0.52=0.48$

Step3: Calculate $\frac{p(1 - p)}{n}$

$\frac{p(1 - p)}{n}=\frac{0.52\times0.48}{1650}=\frac{0.2496}{1650}\approx0.00015127$

Step4: Calculate $\sqrt{\frac{p(1 - p)}{n}}$

$\sqrt{\frac{p(1 - p)}{n}}=\sqrt{0.00015127}\approx0.0123$

Step5: Calculate the margin of error $E$

$E = 1.96\times0.0123\approx0.0241$

Step6: Convert the margin of error to a percentage

To convert to a percentage, multiply by 100: $E = 0.0241\times100 = 2.41\%\approx2.4\%$

Answer:

$2.4$