To determine the sample size required for a 95% confidence interval with a given margin of error, we can use the following formula for sample size estimation:

$n = \left( \frac{Z_{\alpha/2} \cdot \sqrt{p^* (1 - p^*)}}{E} \right)^2$

Where:

• $Z_{\alpha/2}$ is the z-value corresponding to the desired confidence level (95% confidence corresponds to a z-value of 1.96),
• $p^*$ is the planning value for the population proportion (0.25 in this case),
• $E$ is the margin of error (0.1 in this case).

Now, let's calculate the sample size step-by-step.

Step 1: Plug the values into the formula

$n = \left( \frac{1.96 \cdot \sqrt{0.25 \cdot (1 - 0.25)}}{0.1} \right)^2$

Step 2: Simplify the equation

$n = \left( \frac{1.96 \cdot \sqrt{0.25 \cdot 0.75}}{0.1} \right)^2$ $n = \left( \frac{1.96 \cdot \sqrt{0.1875}}{0.1} \right)^2$ $n = \left( \frac{1.96 \cdot 0.433}{0.1} \right)^2$ $n = \left( \frac{0.848}{0.1} \right)^2$ $n = (8.48)^2$

Step 3: Calculate the sample size

$n = 71.91$

Since the sample size must be a whole number, round it up to the next whole number.

$n = 72$

Final Answer:

The required sample size is 72.

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Would you like a detailed explanation of the steps or have any other questions?

Here are five related questions to consider:

1. How does increasing the confidence level affect the sample size?
2. What happens to the required sample size if the margin of error decreases?
3. How does changing the planning value $p^*$ impact the sample size?
4. What is the role of the z-value in determining the sample size?
5. How would the sample size formula change for different confidence intervals?

Tip: Always round the calculated sample size up to ensure the margin of error is met.