you are a pollster conducting a survey to est...
you are a pollster conducting a survey to estimate the proportion of voters who support a certain candidate in an upcoming election. you take a random sample of 500 voters and find that 280 of them support the candidate. estimate the population proportion of the voters who support the candidate using two standard errors. round your answer to the two decimal places. (2 points)\n□≤p≤□
Answer
# Explanation:
## Step1: Calculate sample proportion
Let $n = 500$ (sample size) and $x=280$ (number of successes in sample). The sample proportion $\hat{p}=\frac{x}{n}=\frac{280}{500}=0.56$.
## Step2: Calculate standard error
The formula for the standard error of a proportion is $SE=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.56$ and $n = 500$ into the formula:
\[
\begin{align*}
SE&=\sqrt{\frac{0.56\times(1 - 0.56)}{500}}\\
&=\sqrt{\frac{0.56\times0.44}{500}}\\
&=\sqrt{\frac{0.2464}{500}}\\
&=\sqrt{0.0004928}\\
&\approx0.0222
\end{align*}
\]
## Step3: Calculate confidence interval bounds
The margin of error $E = 2\times SE$ (since we want two - standard errors). So $E=2\times0.0222 = 0.0444$.
The lower bound of the confidence interval for the population proportion $p$ is $\hat{p}-E=0.56 - 0.0444=0.5156\approx0.52$.
The upper bound is $\hat{p}+E=0.56+ 0.0444=0.6044\approx0.60$.
# Answer:
$0.52\leq p\leq0.60$