#### Answer

p is between 0.92-0.1027=0.8173 and 0.92+0.1027=1.0227. The thing that is unusual that the upper bound exceeds 1, which is not possible for p. Modification is needed.

#### Work Step by Step

The best point estimate is equal to the proportion of the sample (x) divided by the sample size: $\hat{p}=\frac{x}{n}=\frac{44}{48}=0.92.$
b)$E=z_{\frac{\alpha}{2}}\cdot \sqrt{\frac{\hat{p}\cdot (1-\hat{p})}{n}}=2.575\cdot \sqrt{\frac{0.92\cdot (1-0.92)}{48}}=0.1027.$
Hence, the confidence interval: E is between $\hat{p}-E$ and $\hat{p}+E$, hence p is between 0.92-0.1027=0.8173 and 0.92+0.1027=1.0227. The thing that is unusual that the upper bound exceeds 1, which is not possible for p. Hence, modification is needed.