The 95% confidence interval can be created by using the formula that is given below;$$\mathrm{CI}=\bar{x} \pm z_{\alpha/2}\frac{s}{\sqrt{n}}$$Here, 95% confidence interval is to be calculated.The sample proportion of buildings meeting accessibility requirements, p is equal to 0.46.The sample size, n is 100.We have, $100(1-p)=100(1-0.46)=54$.Thus, the standard error is:$$\begin{aligned}s &=\sqrt{\frac{p(1-p)}{n}} \\ &=\sqrt{\frac{0.46 \times 0.54}{100}} \\ &=0.050\end{aligned}$$The z-score that corresponds to a 95% confidence level, i.e., $\alpha = 0.05$ is:$$\begin{aligned} z_{\alpha/2} &= z_{0.025} \\ &=1.96 \end{aligned}$$Therefore, the 95% confidence interval is given as:$$\begin{aligned} \mathrm{CI} &=\bar{x} \pm z_{\alpha/2} \frac{s}{\sqrt{n}} \\ &=0.46 \pm 1.96 \frac{0.050}{\sqrt{100}} \\ &=0.46 \pm 0.01 \end{aligned}$$Hence, the 95% confidence interval is (0.45, 0.47).Now, as the district estimated that 50% of its buildings met accessibility requirements, and the confidence interval does not contain 0.50, which implies that there is evidence that the district's estimation was incorrect.Answer: No, because 46% is not close to 50%.