Question

suppose the true proportion of voters in the county who support a restaurant tax is 0.52. consider the sampling distribution for the proportion of supporters with sample size n = 167. what is the mean of this distribution? what is the standard error of this distribution? round answer to three decimal places.

Explanation:

Step1: Recall the mean formula for sampling distribution of proportion

The mean of the sampling - distribution of the sample proportion $\hat{p}$ is equal to the population proportion $p$. That is, $\mu_{\hat{p}}=p$.

Step2: Identify the population proportion

Given that $p = 0.52$, so the mean of the sampling distribution of the proportion of supporters $\mu_{\hat{p}}=0.52$.

Step3: Recall the standard - error formula for sampling distribution of proportion

The standard error of the sampling distribution of the sample proportion is given by $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}$, where $p$ is the population proportion and $n$ is the sample size.

Step4: Substitute the values of $p$ and $n$ into the formula

We have $p = 0.52$, $1-p=1 - 0.52 = 0.48$, and $n = 167$. Then $\sigma_{\hat{p}}=\sqrt{\frac{0.52\times0.48}{167}}$.
First, calculate $0.52\times0.48 = 0.2496$. Then $\frac{0.2496}{167}\approx0.00149461$. And $\sqrt{0.00149461}\approx0.039$.

Answer:

Mean: $0.52$
Standard Error: $0.039$