Confidence Intervals & Margins of Error without Standard Dev

crossingtheDE

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Feb 18, 2008
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Okay, so this is the question I was given.

"A May 2005 Gallup Poll found that 38% of a random sample of 1012 adults said that they believe in ghosts. Find the margin of error for this poll if we want 90% confidence in our estimate of the number of Americans who believe in ghosts."

The formula I have for margin of error is m = z*(st. dev. / n^.5), so I'm really lost because I don't see any standard deviation given in the problem. Any help or suggestions would be amazing. :)
 
Confidence intervals for proportions do not include the SD.

Use \(\displaystyle E=z\cdot\sqrt{\frac{pq}{n}}\)

Where p=.38, q=1-p, z=1.645, n=1012
 
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