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 Quick Question on finding the Margin of Error

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T O P I C    R E V I E W
Danny09 Posted - 12/01/2011 : 17:51:00
Hello all! I have a few quick question regarding confidence intervals/margin of error.

1.) Suppose that 518 of the 1,493 respondents for a survey agree with a question. What is the margin of error for the 95% confidence interval?

2.) A study surveyed 2,829 randomly selected 13- to 19-year-olds. Of those teenagers, 78% had a television in their room. What is the upper bound on the 95% confidence interval for the proportion of all teens who have a TV set in their room?


I know in both problems you use the formula for finding the square root of a discrete probability model, or S=p(1 - p)n

Where p= the same proportion
and n= the poulation total

So for question two, p=0.78 and n=2,829

Could anyone help point me in the right direction as to where I go from here? Thanks very much in advance!
3   L A T E S T    R E P L I E S    (Newest First)
Danny09 Posted - 12/02/2011 : 16:37:24
Sorry for the double post, but for problem #2 where p=.78 and n=2,289

I calculated the standard dev. as S = 2.462

This seems somewhat more believable than what I got for #1.
Danny09 Posted - 12/02/2011 : 16:33:11
quote:
Originally posted by royhaas

The margin of error can be defined as one-half the width of the confidence interval.



Thanks for the reply! I seem to be stuck on calculating the confidence interval though...

For question #1, if the population proportion = 0.347 and N = 1494,

I calculated S = 0.0001511768

Surely this cant be right...? Am I imputing the data into the formula wrong?
royhaas Posted - 12/02/2011 : 08:09:32
The margin of error can be defined as one-half the width of the confidence interval.

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