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One month before the election, a poll of a large number of randomly selected voters showed 65% planning to vote for a certain candidate. The newspaper article reported a 90% conservative margin of error of 4 percentage points.

Use the above information to construct a 90% conservative confidence interval for the population proportion of all voters planning to vote for this candidate.

Can someone please explain why the answer is 0.65 +/- 0.04?

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The margin of error is usually defined as a particular way of expressing a confidence interval - see Wikipedia's definition, for example. So you are not really being asked to do anything other than re-express what the newspaper has already stated. By changing from 65% plus or minus 4 percentage points to 0.65 plus or minus 0.04 you have only changed it from percentages to proportions, by dividing by 100 (which obviously has no importance).

Looking again at your question I suspect it is actually exactly this relationship between a percentage and a proportion that is being tested here (as it is several months old I don't mind giving the answer away), not really the confidence interval issue.

Exactly what a "confidence interval" means is another matter (see @gung's comment). The usual explanation is that if you performed this exercise (sampling a large number of voters etc) many times and used the same technique each time to estimate your confidence interval, 90% of the time the confidence interval would include the true but unknown value.

BTW if only newspapers were as clear as this, stating that the margin of error is a certain number of "percentage points" - all too often they omit the crucial "points", which leaves a significant ambiguity in meaning.

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