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Let's break down the problem step-by-step.

### Given Data:
- Population size (\(N\)) = 200
- Sample size (\(n\)) = 15
- Population mean (\(\mu\)) = 35.8
- Population standard deviation (\(\sigma\)) = 4.8

You are asked to calculate the **expected value** and **standard error** of the sampling distribution of means for two cases:
1. **With replacement** 
2. **Without replacement**

### (i) Sampling with replacement:
When sampling with replacement, the standard error of the sampling distribution of the sample mean is calculated as:

\[
SE_{\text{with replacement}} = \frac{\sigma}{\sqrt{n}}
\]

Where:
- \(\sigma\) is the population standard deviation
- \(n\) is the sample size

Thus, plugging in the values:

\[
SE_{\text{with replacement}} = \frac{4.8}{\sqrt{15}} \approx \frac{4.8}{3.87} \approx 1.24
\]

The **expected value** of the sample mean is always equal to the population mean, so:

\[
E(\bar{x}) = \mu = 35.8
\]

### (ii) Sampling without replacement:
When sampling without replacement, the standard error is adjusted using the finite population correction factor (FPC), which is:

\[
SE_{\text{without replacement}} = \frac{\sigma}{\sqrt{n}} \times \sqrt{\frac{N-n}{N-1}}
\]

Where:
- \(N\) is the population size
- \(n\) is the sample size

Plugging in the values:

\[
SE_{\text{without replacement}} = \frac{4.8}{\sqrt{15}} \times \sqrt{\frac{200-15}{200-1}} = 1.24 \times \sqrt{\frac{185}{199}} 
\]

\[
SE_{\text{without replacement}} \approx 1.24 \times \sqrt{0.9296} \approx 1.24 \times 0.964 \approx 1.20
\]

Again, the **expected value** of the sample mean remains the same as the population mean:

\[
E(\bar{x}) = \mu = 35.8
\]

### Final Answers:
1. **With replacement**:
   - Expected value = 35.8
   - Standard error = 1.24

2. **Without replacement**:
   - Expected value = 35.8
   - Standard error = 1.20

Would you like further details or have any questions?

### Related Questions:
1. How does increasing the sample size affect the standard error?
2. What is the finite population correction factor and when is it used?
3. Why does the expected value of the sample mean always equal the population mean?
4. How would the standard error change if the population size increased significantly?
5. What is the impact of a larger sample size on the reliability of the sample mean?

**Tip:** When the population size is much larger than the sample size, the finite population correction factor has a negligible effect, and the formulas for sampling with and without replacement yield similar results.

Obtain the expected value and standard error of the sampling distribution of means for samples of size 15, drawn from a population of 200 having mean = 35.8 and standard deviation = 4.8 when sampling is done (i) with replacement, and (ii) without replacement.

Learn how to calculate the expected value and standard error of a sampling distribution for samples of size 15 drawn from a population of 200 with a mean of 35.8 and a standard deviation of 4.8, with detailed steps for both sampling with and without replacement.

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