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Danny09
Junior Member

USA
3 Posts

 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)nWhere p= the same proportion and n= the poulation totalSo 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!

royhaas
Moderator

USA
3059 Posts

 Posted - 12/02/2011 :  08:09:32 The margin of error can be defined as one-half the width of the confidence interval.

Danny09
Junior Member

USA
3 Posts

 Posted - 12/02/2011 :  16:33:11 quote:Originally posted by royhaasThe 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 = 0.0001511768Surely this cant be right...? Am I imputing the data into the formula wrong?

Danny09
Junior Member

USA
3 Posts

 Posted - 12/02/2011 :  16:37:24 Sorry for the double post, but for problem #2 where p=.78 and n=2,289I calculated the standard dev. as S = 2.462This seems somewhat more believable than what I got for #1.
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