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 344 with 96 successes at a confidence level of 98%.

Answer 1

Margin of error = (2.33)[√(pq/n)]
...where p = 96/344, q = 1 - p, and n = 344

Note: 2.33 represents 98% confidence interval.

Convert all fractions to decimals.
Plug the values into the formula and calculate.

I let you take it from here.