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Math Help - calculating confidence

  1. #1
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    calculating confidence

    I have this problem in my homework and I can't seem to arrive at the correct answer

    In a random sample of 500 college students at WVU, 72% indicated they preferred to study while listening to music or some kind of background noise while others indicated they liked to study in a quiet setting. What is the 99% confidence interval for music/background noise studying?



    a. 72 3.44

    b. 72 14.14

    c. 72 .72

    d. 72 9.83

    e. None of the above

    Can anyone teach me how to solve these kinds of questions?
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  2. #2
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: calculating confidence

    Quote Originally Posted by DocNougat View Post
    I have this problem in my homework and I can't seem to arrive at the correct answer
    In a random sample of 500 college students at WVU, 72% indicated they preferred to study while listening to music or some kind of background noise while others indicated they liked to study in a quiet setting. What is the 99% confidence interval for music/background noise studying?
    a. 72 3.44
    b. 72 14.14
    c. 72 .72
    d. 72 9.83
    e. None of the above
    Approx normal dist to binomial dist:

    p=.72
    q=1-p

    approx SD = \sigma \text{:=}\sqrt{p q} = 0.449

    99% confidence interval of mean = 0.72\pm 2.58\sqrt{p q} = 0.72\pm 1.15842
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  3. #3
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    Re: calculating confidence

    So it would be none of the above?
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  4. #4
    Senior Member MaxJasper's Avatar
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    Re: calculating confidence

    (b) would be good too!
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  5. #5
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    Re: calculating confidence

    Also how do you get the 2.58?
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  6. #6
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: calculating confidence

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  7. #7
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    Re: calculating confidence

    Thanks, this makes more sense now. Any advice on the other question I posted?
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  8. #8
    Newbie ButterflyM's Avatar
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    Re: calculating confidence

    Quote Originally Posted by MaxJasper View Post
    Approx normal dist to binomial dist:

    p=.72
    q=1-p

    approx SD = \sigma \text{:=}\sqrt{p q} = 0.449

    99% confidence interval of mean = 0.72\pm 2.58\sqrt{p q} = 0.72\pm 1.15842
    Doesn't make sense, values are too large and go over 100%! Did you perhaps forget the 1/N in the square root? 0.72\pm 2.58\sqrt{p q/N} where N=500?
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