To solve this problem, we need to calculate the margin of error and confidence interval for the given survey data.

Given Information

1. Sample size, $n = 344$
2. Sample proportion, $p = 0.31$ (since 31% reported owning a gun primarily for protection)
3. Confidence level = 95%

Step 1: Find the Margin of Error

For a confidence level of 95%, the z-score is approximately 1.96.

The formula for the margin of error (ME) for a proportion is: $\text{Margin of Error} = z \cdot \sqrt{\frac{p(1 - p)}{n}}$

Substitute the values: $\text{Margin of Error} = 1.96 \cdot \sqrt{\frac{0.31 \times (1 - 0.31)}{344}}$

Calculate each part:

1. $1 - 0.31 = 0.69$
2. $0.31 \times 0.69 = 0.2139$
3. $\frac{0.2139}{344} \approx 0.0006212$
4. $\sqrt{0.0006212} \approx 0.02492$
5. $1.96 \times 0.02492 \approx 0.0488$

So, the margin of error, rounded to two decimal places, is approximately 0.05.

Step 2: Find the Confidence Interval

The confidence interval is calculated as: $\text{Confidence Interval} = p \pm \text{Margin of Error}$

With $p = 0.31$ and Margin of Error $\approx 0.05$:

1. Lower limit = $0.31 - 0.05 = 0.26$
2. Upper limit = $0.31 + 0.05 = 0.36$

Final Answer

• Margin of Error: 0.05
• Confidence Interval: 0.26 to 0.36

Would you like more details on this solution or have any further questions?

Related Questions

1. What would the margin of error be if the confidence level was 90% instead of 95%?
2. How does increasing the sample size affect the margin of error?
3. What would the confidence interval be if the sample proportion was 0.25 instead of 0.31?
4. How is the z-score affected by changes in the confidence level?
5. How does the standard error change when the proportion $p$ gets closer to 0.5?

Tip

For quick estimations, remember that higher confidence levels yield wider intervals, while larger sample sizes reduce the margin of error.