A survey of 2323 adults in a certain large country aged 18 and older conducted by a reputable polling organization found that 423 have donated blood in the past two years.

Verify that the requirements for constructing a confidence interval about $p$ are satisfied.
The sample
a simple random sample, the value of
(Round to three decimal places as needed.)

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 ：\(\boxed{\text{Therefore, the requirements for constructing a confidence interval about } p \text{ are satisfied.}}\)