Question 1169827
Alright, let's calculate the required sample size for the given margin of error and confidence level.

**Understanding the Problem**

We need to find the minimum sample size (n) that will ensure a margin of error of at most 0.01 with a 90% confidence level.

**Formula for Sample Size (Proportion Unknown)**

When the population proportion (p) is unknown, we use the following formula:

$$n = \left(\frac{z_{\alpha/2}}{E}\right)^2 \cdot 0.25$$

Where:

* n is the sample size.
* z<sub>α/2</sub> is the z-critical value corresponding to the desired confidence level.
* E is the margin of error.
* 0.25 is used because it represents the maximum possible value of p(1-p), which occurs when p = 0.5.

**Steps**

1.  **Find the z-critical value (z<sub>α/2</sub>):**
    * For a 90% confidence level, α = 1 - 0.90 = 0.10.
    * α/2 = 0.10 / 2 = 0.05.
    * We need to find the z-value that corresponds to an area of 0.95 (0.90 + 0.05) in the cumulative standard normal distribution.
    * Using a z-table or calculator, the z-critical value for a 90% confidence level is approximately 1.645.

2.  **Plug the values into the formula:**
    * n = (1.645 / 0.01)^2 * 0.25
    * n = (164.5)^2 * 0.25
    * n = 27060.25 * 0.25
    * n = 6765.0625

3.  **Round up to the nearest whole number:**
    * Since we need a whole number for the sample size, and we want to ensure the margin of error is at most 0.01, we always round up.
    * n = 6766

**Final Answer**

The required sample size is 6766.