212 people surveyed, 72% favored home team. Find margin of error and give interval that is likely to contain exact percent. HELP?

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212 people surveyed, 72% favored home team. Find margin of error and give interval that is likely to contain exact percent. HELP?

Answer:

using the formula for margin of error, ±(1/√n)
±(1/√212)
±0.06868 this is the margin of error
to find the interval, you subtract and add this number from your %
72- 0.06868= 71.9313
72+0.06868=72.0687
the interval is 71.9313<M<72.0687