To calculate the margin of error (MOE) for the sample proportions with a 95% confidence level, we can follow these steps:

1. Calculate the sample proportion $\hat{p}$: $\hat{p} = \frac{\text{number of "yes" responses}}{\text{total number of responses}} = \frac{14,252}{14,252 + 115,336}$ $\hat{p} = \frac{14,252}{129,588} \approx 0.110$

2. Calculate the complement of the sample proportion $\hat{q}$: $\hat{q} = 1 - \hat{p} = 1 - 0.110 = 0.890$

3. Determine the sample size $n$: $n = 14,252 + 115,336 = 129,588$

4. Find the critical value $z$ for a 95% confidence level: The critical value $z$ for a 95% confidence level is approximately 1.96.

5. Calculate the standard error (SE) of the proportion: $SE = \sqrt{\frac{\hat{p} \cdot \hat{q}}{n}} = \sqrt{\frac{0.110 \cdot 0.890}{129,588}}$ $SE = \sqrt{\frac{0.0979}{129,588}} \approx \sqrt{7.55 \times 10^{-4}} \approx 0.0275$

6. Calculate the margin of error (MOE): $MOE = z \cdot SE = 1.96 \cdot 0.0275 \approx 0.0539$

Rounded to one decimal place, the margin of error (MOE) is: $MOE \approx 5.4\%$

Thus, the margin of error associated with the sample proportions at a 95% confidence level is $\boxed{5.4}$.

Would you like more details on any step, or have any questions?

Here are 5 relative questions expanding on this topic:

1. How do you interpret the margin of error in the context of this survey?
2. What would happen to the margin of error if the sample size increased?
3. Why is the critical value $z$ for a 95% confidence level 1.96?
4. How would the margin of error change if the confidence level was increased to 99%?
5. What assumptions are made when calculating the margin of error for proportions?

Tip: When dealing with proportions and margins of error, always ensure that your sample size is large enough to justify the normal approximation used in the calculations.