A researcher wishes to estimate the proportion of college students who cheat on exams. A poll of 490 college students showed that 33% of them had, or intended to, cheat on examinations. Find the margin of error for the 95% confidence interval.

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To find the margin of error for the 95% confidence interval, we need to use the formula:

Margin of Error = Z * sqrt(p * (1-p) / n)

Where:
Z is the Z-score for the desired confidence level (95% confidence level corresponds to a Z-score of approximately 1.96)
p is the observed proportion (33% or 0.33 in this case)
n is the sample size (490 in this case)

Let's plug in the values into the formula:

Margin of Error = 1.96 * sqrt(0.33 * (1-0.33) / 490)

Calculating this, we get:

Margin of Error ≈ 0.0327

So, the margin of error for the 95% confidence interval is approximately 0.0327 or 3.27%.