1. **State the problem:** We have a poll with a sample size $n=2030$ and a reported margin of error (MOE) of 2 percentage points (0.02) at a 95% confidence level. We want to determine which statement about the margin of error and sample size is true. 2. **Recall the margin of error formula for a proportion:** $$\text{MOE} = z^* \times \sqrt{\frac{p(1-p)}{n}}$$ where $z^*$ is the critical value for 95% confidence (approximately 1.96), $p$ is the sample proportion, and $n$ is the sample size. 3. **Calculate the margin of error using the given data:** Given $p=0.26$, $n=2030$, and $z^*=1.96$, $$\text{MOE} = 1.96 \times \sqrt{\frac{0.26 \times (1-0.26)}{2030}} = 1.96 \times \sqrt{\frac{0.26 \times 0.74}{2030}}$$ Calculate inside the square root: $$\frac{0.26 \times 0.74}{2030} = \frac{0.1924}{2030} \approx 9.48 \times 10^{-5}$$ Square root: $$\sqrt{9.48 \times 10^{-5}} \approx 0.00974$$ Multiply by $z^*$: $$1.96 \times 0.00974 \approx 0.0191$$ 4. **Interpretation:** The calculated margin of error is approximately 0.0191 (or 1.91 percentage points), which is very close to the reported 2 percentage points. 5. **Conclusion:** Since the calculated MOE is consistent with the reported MOE, the statement "The reported margin of error is consistent with the sample size." is TRUE. **Final answer:** The reported margin of error is consistent with the sample size.