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 0.035 margin of error at a 98\% level of confidence.
a) With no prior research, what sample size should you gather in order to obtain a 0.035 margin of error? Round your answer up to the nearest whole number.
$$n=◻\text{ households }$$
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 $\hat{p}=0.18$. Using this new information. What sample size should you gather in order to obtain a 0.035 margin of erron? Round your answer up to the nearest whole number.
$$n=◻\text{ households }$$

Step 1 ：The problem is asking for the sample size needed to estimate the proportion of households in Chicago that have two or more vehicles with a margin of error of 0.035 at a 98% confidence level.

Step 2 ：For part a), we don't have any prior information about the proportion, so we'll use the most conservative estimate, which is 0.5. This is because the variance of a binomial distribution (which is what we're dealing with when we're talking about proportions) is highest when p = 0.5.

Step 3 ：For part b), we have a preliminary estimate of the proportion, which is 0.18. We'll use this to calculate the required sample size.

Step 4 ：The formula for the sample size needed to estimate a proportion with a given margin of error at a certain confidence level is: $n=\frac{Z^2\ast p\ast (1-p)}{E^2}$ where: - Z is the z-score corresponding to the desired confidence level (for a 98% confidence level, the z-score is approximately 2.33) - p is the estimated proportion - E is the desired margin of error

Step 5 ：Substituting the values into the formula, we get: $n=\frac{(2.33{)}^2\ast 0.5\ast (1-0.5)}{(0.035{)}^2}$ which simplifies to $n=1105$ for part a).

Step 6 ：For part b), substituting the values into the formula, we get: $n=\frac{(2.33{)}^2\ast 0.18\ast (1-0.18)}{(0.035{)}^2}$ which simplifies to $n=653$.

Step 7 ：Final Answer: For part a), the sample size needed is $\boxed{{\displaystyle 1105}}$ households. For part b), the sample size needed is $\boxed{{\displaystyle 653}}$ households.