Let's go through the problem step-by-step.

Problem Breakdown

1. Given Information:

   • Population proportion $p = 0.37$.
   • Sample size $n = 500$.

2. Objective:

   • To describe the sampling distribution of $\hat{p}$ (the sample proportion).
   • To find the mean and standard deviation of the sampling distribution of $\hat{p}$.

Part (a) - Describe the Sampling Distribution of $\hat{p}$

For the sampling distribution of $\hat{p}$ to be approximately normal, two conditions need to be met:

• Condition 1: The sample size $n$ should be less than or equal to 5% of the population size, which is represented by $n \leq 0.05N$.
• Condition 2: Both $np$ and $n(1 - p)$ must be at least 10.

Verification:

1. Condition 1: Since this is a survey of adults, the population $N$ (number of adults in the country) is likely large enough that $n = 500 \leq 0.05N$ holds true.

2. Condition 2:

   • Calculate $np$: $np = 500 \times 0.37 = 185$
   • Calculate $n(1 - p)$: $n(1 - p) = 500 \times (1 - 0.37) = 500 \times 0.63 = 315$

   Since both values (185 and 315) are greater than 10, this condition is also satisfied.

Therefore, the sampling distribution of $\hat{p}$ is approximately normal, which matches option A.

Part (b) - Mean of the Sampling Distribution of $\hat{p}$

The mean of the sampling distribution of $\hat{p}$, denoted by $\mu_{\hat{p}}$, is equal to the population proportion $p$: $\mu_{\hat{p}} = p = 0.37$

This answer is already provided as 0.37.

Part (c) - Standard Deviation of the Sampling Distribution of $\hat{p}$

The standard deviation of the sampling distribution of $\hat{p}$, denoted by $\sigma_{\hat{p}}$, is calculated by the formula: $\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}}$

Substitute the values: $\sigma_{\hat{p}} = \sqrt{\frac{0.37 \times 0.63}{500}} = \sqrt{\frac{0.2331}{500}}$

Calculate this value: $\sigma_{\hat{p}} \approx \sqrt{0.0004662} \approx 0.0216$

So, the standard deviation of $\hat{p}$ is approximately 0.022 when rounded to three decimal places.

Summary of Answers

• (a) Option A: Approximately normal because $n \leq 0.05N$ and $np \geq 10$ and $n(1 - p) \geq 10$.
• Mean (b): $0.37$
• Standard Deviation (c): $0.022$

Would you like more details or have any questions?

Related Questions:

1. How would the sampling distribution change if the sample size were smaller?
2. Why is the condition $n \leq 0.05N$ important for the sampling distribution?
3. What happens if $np$ or $n(1 - p)$ are less than 10?
4. How does increasing the sample size affect the standard deviation of $\hat{p}$?
5. What would the mean and standard deviation be if $p$ changed to 0.5?

Tip:

Always check if the sample size and population proportion meet normality conditions before assuming the sampling distribution is approximately normal.