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 0.025 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 0.025 margin of error?
b) Your firm has decided that your plan is too expensive, and they wish to reduce the sample size required. You conduct a small preliminary sample, and you obtain a sample proportion of $\widehat{p}=0.225$. Using this new information. what sample size should you gather in order to obtain a 0.025 margin of error?
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Step 1 ：Given that the margin of error (E) is 0.025 and the Z-score for a 90% confidence level is 1.645 (Z).

Step 2 ：For part a), since we have no prior research, we can use the formula for the sample size needed for a proportion, which is \(n = \frac{{Z^2 * p * (1-p)}}{{E^2}}\), where p is the estimated proportion. Since we don't have an estimated proportion, we can use p = 0.5, which will give us the maximum sample size needed.

Step 3 ：Substituting the given values into the formula, we get \(n = \frac{{(1.645)^2 * 0.5 * (1-0.5)}}{{(0.025)^2}}\).

Step 4 ：Calculating the above expression, we find that the sample size needed for part a) is \(\boxed{1083}\).

Step 5 ：For part b), we can use the same formula, but this time we have an estimated proportion from a preliminary sample, which is p = 0.225.

Step 6 ：Substituting the given values into the formula, we get \(n = \frac{{(1.645)^2 * 0.225 * (1-0.225)}}{{(0.025)^2}}\).

Step 7 ：Calculating the above expression, we find that the sample size needed for part b) is \(\boxed{755}\).