You work for a marketing firm that has a large client in the automobile industry. You have been asked to estimate the proportion of households in Chicago that have two or more vehicles. You have been assigned to gather a random sample that could be used to estimate this proportion to within a 0.04 margin of error at a $90 \%$ level of confidence.
a) With no prior research, what sample size should you gather in order to obtain a 0.04 margin of error? Round your answer up to the nearest whole number.
\[
n=\square \text { households }
\]

Step 1 ：We are given that the desired margin of error (E) is 0.04 and the confidence level is 90%. The Z-score corresponding to a 90% confidence level is 1.645. Since we don't have any prior research, we don't know the true proportion (p). In this case, we use p = 0.5, which maximizes the product p * (1-p) and thus gives us the largest possible sample size, ensuring that our sample will be large enough regardless of the true proportion.

Step 2 ：We use the formula for the sample size in a proportion estimation problem: \(n = \frac{Z^2 * p * (1-p)}{E^2}\), where n is the sample size, Z is the Z-score, p is the estimated proportion, and E is the desired margin of error.

Step 3 ：Substituting the given values into the formula, we get \(n = \frac{(1.645)^2 * 0.5 * (1-0.5)}{(0.04)^2}\).

Step 4 ：Solving the above expression, we find that n = 423.

Step 5 ：So, the sample size needed to obtain a 0.04 margin of error with no prior research and a 90% confidence level is \(\boxed{423}\) households.