Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. a. When designing a study to determine this population proportion, what is the minimum number you would need to survey to be $95 \%$ confident that the population proportion is estimated to within 0.03 ? b. If it were later determined that it was important to be more than $9 \%$ confident and a new survey was commissioned, how would that affect the minimum number of people you would need to survey? A. Increase B. Decrease C. No change
Solution
Step 1 ：Define the constants for the problem: the z-score for a 95% confidence level (Z) is 1.96, the most conservative estimate of the population proportion (p) is 0.5, and the desired margin of error (E) is 0.03.
Step 2 ：Use the formula for minimum sample size in a proportion estimation: \(n = \lceil \frac{{Z^2 \cdot p \cdot (1-p)}}{{E^2}} \rceil\).
Step 3 ：Substitute the values into the formula: \(n = \lceil \frac{{1.96^2 \cdot 0.5 \cdot (1-0.5)}}{{0.03^2}} \rceil\).
Step 4 ：Solve the equation to find the minimum sample size, which is \(\boxed{1068}\).
Step 5 ：For the second part of the question, if it were later determined that it was important to be more than 9% confident, the minimum number of people you would need to survey would \(\boxed{Increase}\). This is because a higher confidence level requires a larger sample size to maintain the same margin of error.
