To compute the 95% confidence interval for the true proportion of Gastown residents living below the poverty line, we will use the formula for a confidence interval for a proportion. The formula is:

$\text{CI} = \hat{p} \pm Z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

Where:

• $\hat{p}$ is the sample proportion (the proportion of residents living below the poverty line in the sample),
• $Z_{\alpha/2}$ is the Z-score corresponding to the confidence level (for 95%, $Z_{\alpha/2} = 1.96$),
• $n$ is the sample size.

Step-by-step procedure:

1. Determine the sample proportion ($\hat{p}$): $\hat{p} = \frac{x}{n}$ where $x$ is the number of residents living below the poverty line in the sample and $n$ is the total sample size.

2. Calculate the margin of error (ME): $\text{ME} = Z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

3. Construct the confidence interval: $\text{CI} = \left( \hat{p} - \text{ME}, \hat{p} + \text{ME} \right)$

Once you provide the sample data (values for $x$ and $n$), we can plug those into the formula to get the final 95% confidence interval.

Let me know if you have that data or need any clarifications!