A research group conducted an extensive survey of 3130 wage and salaried workers on issues ranging from relationships with their bosses to household chores. The data were gathered through hour-long telephone interviews with a nationally representative sample. In response to the question, "What does success mean to you?" 1571 responded, "Personal satisfaction from doing a good job." Let p be the population proportion of all wage and salaried workers who would respond the same way to the stated question. How large a sample is needed if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 2.9% of the population percentage?

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.

A research group conducted an extensive survey of 3130 wage and salaried workers on issues ranging from relationships with their bosses to household chores. The data were gathered through hour-long telephone interviews with a nationally representative sample. In response to the question, 'What does success mean to you?' 1571 responded, 'Personal satisfaction from doing a good job.' Let p be the population proportion of all wage and salaried workers who would respond the same way to the stated question. How large a sample is needed if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 2.9% of the population percentage?

Determine the sample size required to estimate the population proportion of workers who define success as personal satisfaction with a 95% confidence level and a margin of error of 2.9%. This problem explores key statistical concepts like confidence intervals and proportion estimation using real survey data.

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