To determine the sample size required for a given confidence level and margin of error, we can use the following formula for sample size in estimating proportions:

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

Where:

• $n$ is the required sample size.
• $Z$ is the Z-value corresponding to the desired confidence level.
• $p$ is the estimated population proportion (from the current sample).
• $E$ is the desired margin of error.

Step 1: Gather the necessary values.

• Confidence level: 95%, so the Z-value for a 95% confidence interval is 1.96 (from the Z-distribution table).
• $p$, the estimated population proportion, is based on the 1571 respondents who answered "personal satisfaction" out of 3130, so:

$p = \frac{1571}{3130} \approx 0.5026$

• Margin of error $E = 2.9\%$, or 0.029.

Step 2: Plug values into the formula.

$n = \left( \frac{1.96^2 \cdot 0.5026 \cdot (1 - 0.5026)}{0.029^2} \right)$ $n = \left( \frac{3.8416 \cdot 0.5026 \cdot 0.4974}{0.000841} \right)$ $n = \left( \frac{0.9645}{0.000841} \right)$ $n \approx 1146.5$

Step 3: Round up to the nearest whole number.

Since the sample size must be a whole number, we round up: $n \approx 1147$

Thus, a sample size of at least 1147 respondents is needed to be 95% confident that the sample percentage is within 2.9% of the population percentage.

Would you like more details, or do you have any further questions?

Here are 5 related questions you might find interesting:

1. What is the Z-value for a 90% confidence interval, and how would it change the required sample size?
2. How would the sample size change if the margin of error was reduced to 1%?
3. What if the proportion $p$ was assumed to be 0.75 instead of 0.5026?
4. How does changing the confidence level from 95% to 99% affect the required sample size?
5. Why is it necessary to round up the calculated sample size?

Tip: The margin of error is inversely proportional to the square root of the sample size, meaning smaller margins of error require significantly larger sample sizes.