Part 1: Describe the sampling distribution of $\hat{p}$

To determine the shape of the sampling distribution, we need to check two conditions:

1. Condition 1: The sample size $n$ should be less than or equal to 5% of the population size $N$:

   $n \leq 0.05N$ Here, $n = 150$ and $N = 25,000$. Check if this condition holds:

   $150 \leq 0.05(25,000)$

   $150 \leq 1,250$

   This condition is satisfied.

2. Condition 2: The product $np(1-p)$ should be greater than or equal to 10:

   $np(1 - p) \geq 10$ Here, $n = 150$ and $p = 0.6$. Check if this condition holds:

   $150 \times 0.6 \times (1 - 0.6) = 150 \times 0.6 \times 0.4 = 36$

   Since 36 is greater than 10, this condition is also satisfied.

Thus, since both conditions are satisfied, the sampling distribution of $\hat{p}$ is approximately normal. The correct choice is:

A. Approximately normal because $n \leq 0.05N$ and $np(1-p) \geq 10$.

Part 2: Determine the mean of the sampling distribution of $\hat{p}$

The mean of the sampling distribution of $\hat{p}$ is equal to the population proportion $p$.

$\mu_{\hat{p}} = p = 0.6$

So, the mean is $0.6$.

Part 3: Determine the standard deviation of the sampling distribution of $\hat{p}$

The standard deviation of the sampling distribution of $\hat{p}$ is given by the formula:

$\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} \times \sqrt{\frac{N - n}{N - 1}}$

Where:

• $p = 0.6$
• $n = 150$
• $N = 25,000$

First, calculate the first part:

$\sigma_{\hat{p}} = \sqrt{\frac{0.6(1 - 0.6)}{150}} = \sqrt{\frac{0.6 \times 0.4}{150}} = \sqrt{\frac{0.24}{150}} = \sqrt{0.0016} = 0.04$

Now, account for the finite population correction factor:

$\text{Correction factor} = \sqrt{\frac{N - n}{N - 1}} = \sqrt{\frac{25,000 - 150}{25,000 - 1}} = \sqrt{\frac{24,850}{24,999}} \approx \sqrt{0.9940} \approx 0.9970$

Finally, multiply the two values:

$\sigma_{\hat{p}} \approx 0.04 \times 0.9970 = 0.03988$

Rounding to six decimal places:

$\sigma_{\hat{p}} \approx 0.039880$

Summary:

1. Part 1 Answer: A. Approximately normal.
2. Part 2 Answer: $\mu_{\hat{p}} = 0.6$
3. Part 3 Answer: $\sigma_{\hat{p}} \approx 0.039880$

Would you like further clarification on any part? Here are some related questions to explore:

1. What happens to the sampling distribution when the sample size increases?
2. How would the sampling distribution change if the population proportion $p$ was closer to 0 or 1?
3. Why do we apply a finite population correction factor in the standard deviation formula?
4. What is the impact of ignoring the finite population correction factor in this scenario?
5. How can the normal approximation be used in hypothesis testing with sample proportions?

Tip: Always check both conditions (sample size and proportion criteria) when determining if the sampling distribution of a sample proportion is approximately normal.