To find the minimum sample size required for estimating a proportion with a specified level of confidence and margin of error, we can use the formula for sample size in proportion problems:

$n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right)$

where:

• $n$ = required sample size
• $Z$ = Z-score corresponding to the desired confidence level
• $p$ = estimated proportion from prior studies
• $E$ = margin of error

Given values:

• Confidence level = 95% ⟹ $Z = 1.96$ (Z-score for 95% confidence level)
• Estimated proportion $p = 0.273$ (or 27.3% from previous study)
• Margin of error $E = 0.05$ (or 5%)

Step 1: Substitute the values into the formula

$n = \frac{(1.96)^2 \cdot 0.273 \cdot (1 - 0.273)}{(0.05)^2}$

Step 2: Calculate each component

1. Calculate $Z^2$: $Z^2 = (1.96)^2 = 3.8416$

2. Calculate $p \cdot (1 - p)$: $p \cdot (1 - p) = 0.273 \cdot (1 - 0.273) = 0.273 \cdot 0.727 = 0.198231$

3. Calculate $E^2$: $E^2 = (0.05)^2 = 0.0025$

Step 3: Plug these values back into the formula

$n = \frac{3.8416 \cdot 0.198231}{0.0025}$

$n = \frac{0.761275776}{0.0025}$

$n \approx 304.51$

Since the sample size $n$ must be a whole number, we round up to the next integer.

Final Answer:

The minimum sample size required is: $n = 305$

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Would you like a more detailed breakdown or have any questions about these calculations?

Follow-up Questions

1. How would the sample size change if we wanted a smaller margin of error, such as 3%?
2. What would be the impact on the sample size if the confidence level were increased to 99%?
3. Why do we round up when calculating sample size in this context?
4. If no prior study was available, what value of $p$ should be used to estimate the sample size?
5. How does the sample size formula differ when estimating a mean instead of a proportion?

Tip

When estimating proportions and unsure of the proportion $p$, using $p = 0.5$ maximizes the required sample size, ensuring a conservative estimate.