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Danny09
Junior Member
USA
3 Posts |
 Posted - 12/01/2011 : 17:51:00
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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! |
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royhaas
Moderator
USA
3039 Posts |
Posted - 12/02/2011 : 08:09:32
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| The margin of error can be defined as one-half the width of the confidence interval. |
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Danny09
Junior Member
USA
3 Posts |
Posted - 12/02/2011 : 16:33:11
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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?
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Danny09
Junior Member
USA
3 Posts |
Posted - 12/02/2011 : 16:37:24
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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. |
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Quick Question on finding the Margin of Error
