Suppose a simple random sample of size 

nequals=150150 

is obtained from a population whose size is 

Upper N equals 25 comma 000N=25,000 

and whose population proportion with a specified characteristic is 

p equals 0.6 .p=0.6.**** 

Complete parts ​(a)**** through​ (c)**** 

below**.** 

Question content area bottom

Part 1

​(a) Describe the sampling distribution of 

ModifyingAbove p with caretp.

Choose the phrase that best describes the shape of the sampling distribution below.

A.

Approximately normal because 

n less than or equals 0.05 Upper Nn≤0.05N 

and np left parenthesis 1 minus p right parenthesis greater than or equals 10.np(1−p)≥10.

Your answer is correct.

B.

Not normal because 

n less than or equals 0.05 Upper Nn≤0.05N 

and np left parenthesis 1 minus p right parenthesis less than 10.np(1−p)<10.

C.

Not normal because 

n less than or equals 0.05 Upper Nn≤0.05N 

and np left parenthesis 1 minus p right parenthesis greater than or equals 10.np(1−p)≥10.

D.

Approximately normal because 

n less than or equals 0.05 Upper Nn≤0.05N 

and np left parenthesis 1 minus p right parenthesis less than 10.np(1−p)<10.

Part 2

Determine the mean of the sampling distribution of 

ModifyingAbove p with caretp.

mu Subscript ModifyingAbove p with caret Baseline equalsμp=0.60.6 

​(Round to one decimal place as​ needed.)

Part 3

Determine the standard deviation of the sampling distribution of 

ModifyingAbove p with caretp.

sigma Subscript ModifyingAbove p with caretσpequals=enter your response here 

​(Round to six decimal places as​ needed.)

### 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.

Suppose a simple random sample of size n = 150 is obtained from a population whose size is N = 25,000 and whose population proportion with a specified characteristic is p = 0.6. Complete parts (a) through (c) below. (a) Describe the sampling distribution of p̂. (b) Determine the mean of the sampling distribution of p̂. (c) Determine the standard deviation of the sampling distribution of p̂.

Explore the sampling distribution of the proportion with a random sample size of n = 150 from a population of 25,000, where p = 0.6. This detailed solution covers calculating the mean and standard deviation of the sampling distribution using key statistical formulas.

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