Question

question
a survey was given to a random sample of 1000 residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. of those surveyed, 44% 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

Explanation:

Step1: Identify the formula for margin of error

For a proportion in a large - sample survey, the margin of error $E$ at a 95% confidence level is given by $E = 1.96\sqrt{\frac{p(1 - p)}{n}}$, where $p$ is the sample proportion and $n$ is the sample size.

Step2: Determine the values of $p$ and $n$

We are given that $p=0.44$ (since 44% = 0.44) and $n = 1000$.

Step3: Substitute the values into the formula

$E=1.96\sqrt{\frac{0.44\times(1 - 0.44)}{1000}}=1.96\sqrt{\frac{0.44\times0.56}{1000}}$.
First, calculate $0.44\times0.56 = 0.2464$. Then $\frac{0.2464}{1000}=0.0002464$.
$\sqrt{0.0002464}\approx0.0157$.
$1.96\times0.0157 = 0.0306$.

Step4: Convert to percentage

To convert the decimal to a percentage, multiply by 100. So $E = 3.1\%$.

Answer:

3.1%