confidence interval (need help)

jay1234

New member
Joined
Nov 7, 2006
Messages
11
I need help with a problem, here it goes:

A sample of 400 observations from a population produced a sample proportion of 0.63
Find a 95% confidence Interval for p

I have been dealing with Z and T intervals, but I can't figure this one out. Thanks for anyone that takes a go at this.
 
p=0.63, q=1-0.63

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

95% CI corresponds to a z-score of 1.96

\(\displaystyle \L\\E=1.96\sqrt{\frac{(0.63)(0.37)}{400}}=0.0473\)

0.63-0.0473=0.583

0.63+0.0473=0.677

\(\displaystyle \L\\0.583<p<0.677\)

With 95% confidence, you can say the proportion is between 58.3% and 67.7%
 
I assume you're in a stats class?. The z-score is in the table. E is the margin of error.
 
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