To find the standard error of the mean (SEM) for a sample, we use the formula:

$\text{SEM} = \frac{\sigma}{\sqrt{n}} \times \text{FPC}$

Where:

• $\sigma$ is the population standard deviation,
• $n$ is the sample size,
• FPC is the finite population correction factor given by $\sqrt{\frac{N - n}{N - 1}}$ when the population size $N$ is finite.

Given:

• Population standard deviation $\sigma = 1.56$
• Sample size $n$ (not specified, assuming it's the same for all cases)

a. Population size is infinite (N → ∞)

For an infinite population, the FPC is 1.

$\text{SEM}_{\infty} = \frac{1.56}{\sqrt{n}} \times 1 = \frac{1.56}{\sqrt{n}}$

b. Population size is N = 50,000

For $N = 50,000$:

$\text{FPC} = \sqrt{\frac{50000 - n}{50000 - 1}} \approx \sqrt{\frac{50000 - n}{49999}}$

$\text{SEM}_{50000} = \frac{1.56}{\sqrt{n}} \times \sqrt{\frac{50000 - n}{49999}}$

c. Population size is N = 5,000

For $N = 5,000$:

$\text{FPC} = \sqrt{\frac{5000 - n}{5000 - 1}} \approx \sqrt{\frac{5000 - n}{4999}}$

$\text{SEM}_{5000} = \frac{1.56}{\sqrt{n}} \times \sqrt{\frac{5000 - n}{4999}}$

d. Population size is N = 500

For $N = 500$:

$\text{FPC} = \sqrt{\frac{500 - n}{500 - 1}} \approx \sqrt{\frac{500 - n}{499}}$

$\text{SEM}_{500} = \frac{1.56}{\sqrt{n}} \times \sqrt{\frac{500 - n}{499}}$

Summary of SEM Values

To calculate specific values, we need the sample size $n$. If you provide $n$, I can compute the exact SEM for each case.

Do you have a specific sample size $n$ in mind, or would you like to assume a value?

Questions for Further Exploration

1. What is the sample size $n$ you would like to use?
2. How does the finite population correction factor impact the SEM?
3. Can you explain how SEM is different from standard deviation?
4. What are practical applications of SEM in research?
5. Would you like examples of calculating SEM in real-world scenarios?

Tip

Understanding the relationship between sample size and standard error can help in designing experiments and determining sample sizes needed for desired precision.