Step 1 :Given a sample size (n) of 110, a success rate (p) of 85.5% or 0.855, and a z-score (Z) of 2.807 corresponding to a confidence level of 99.8%, we are to find the margin of error (M.E.).
Step 2 :The formula for the margin of error for a proportion is given by \(M.E. = Z * \sqrt{\frac{p(1-p)}{n}}\).
Step 3 :Substituting the given values into the formula, we get \(M.E. = 2.807 * \sqrt{\frac{0.855(1-0.855)}{110}}\).
Step 4 :Solving the above expression gives a margin of error (M.E.) of approximately 9.4%.
Step 5 :Thus, the margin of error that corresponds to a sample of size 110 with 85.5% successes at a confidence level of 99.8% is \(\boxed{9.4\%}\).