Answer :
Step-by-step explanation:
To estimate the proportion of all Americans who are in favor of gun control legislation with a 95% confidence interval, we can use the formula for calculating the confidence interval for a proportion:
\[ \text{Confidence Interval} = \hat{p} \pm Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]
Where:
- \(\hat{p}\) is the sample proportion (proportion in favor of gun control legislation).
- \(Z\) is the Z-score corresponding to the desired confidence level (95% confidence level corresponds to \(Z = 1.96\) for large samples).
- \(n\) is the sample size.
Given:
- Sample size (\(n\)) = 2500
- Number of respondents in favor (\(\hat{p}\)) = 1600/2500 = 0.64 (64%)
Let's calculate the confidence interval:
\[ \text{Confidence Interval} = 0.64 \pm 1.96 \times \sqrt{\frac{0.64(1-0.64)}{2500}} \]
\[ = 0.64 \pm 1.96 \times \sqrt{\frac{0.64 \times 0.36}{2500}} \]
\[ = 0.64 \pm 1.96 \times \sqrt{\frac{0.2304}{2500}} \]
\[ = 0.64 \pm 1.96 \times 0.0152 \]
\[ = 0.64 \pm 0.0298 \]
Now, we can find the confidence interval:
Lower Limit = \(0.64 - 0.0298\)
Upper Limit = \(0.64 + 0.0298\)
Lower Limit ≈ \(0.6102\)
Upper Limit ≈ \(0.6698\)
So, the 95% confidence interval for the proportion of all Americans who are in favor of gun control legislation is approximately \(0.6102\) to \(0.6698\), or 61.02% to 66.98%.
