Assume that a sample is used to estimate a population proportion p. Find the margin of error M.E. that corresponds to a sample of size 110 with $85.5 \%$ successes at a confidence level of $99.8 \%$.
\[
\text { M.E. }=
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\%
Report answer accurate to one decimal place (as a number of percentage points). Answer should be obtained without any preliminary rounding (however, the critical value may be rounded to 3 decimal places).
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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 \(\boxed{9.4\%}\).