confidence interval (need help)

[jay1234]

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.

 

[galactus]

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%

 

[jay1234]

question:

How did you get the z score, and what does E stand for.

 

[galactus]

I assume you're in a stats class?. The z-score is in the table. E is the margin of error.