Step 1 :We are given that the sample size is 2323 and the number of successes (people who have donated blood) is 423. We can calculate the sample proportion $p$ as the number of successes divided by the sample size.
Step 2 :Let's denote the sample size as $n$ and the number of successes as $x$. So, $n = 2323$ and $x = 423$.
Step 3 :The sample proportion $p$ is calculated as $\frac{x}{n} = \frac{423}{2323} = 0.182$ (rounded to three decimal places).
Step 4 :To verify the requirements for constructing a confidence interval about $p$, we need to check the following conditions: The sample is a simple random sample and the sampling distribution is approximately normal. This condition is satisfied if both $np$ and $n(1-p)$ are greater than or equal to 10.
Step 5 :Let's calculate $np$ and $n(1-p)$: $np = 2323 \times 0.182 = 423$ and $n(1-p) = 2323 \times (1-0.182) = 1900$.
Step 6 :Both $np$ and $n(1-p)$ are greater than 10, so the sampling distribution is approximately normal. The sample is a simple random sample as stated in the problem.
Step 7 :\(\boxed{\text{Therefore, the requirements for constructing a confidence interval about } p \text{ are satisfied.}}\)