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Math Help - Solve for B

  1. #1
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    Solve for B

    This is driving me nuts!

    -Z < (XBar - B) / (B/sqrt(n)) < Z

    I want to solve this for B (and keep it in the center). I've burned through an eraser.....can anyone show me the steps?

    Thanks!
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  2. #2
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    Quote Originally Posted by B_Miner View Post
    This is driving me nuts!

    -Z < (XBar - B) / (B/sqrt(n)) < Z

    I want to solve this for B (and keep it in the center). I've burned through an eraser.....can anyone show me the steps?

    Thanks!
    Do you mean  -Z < \frac{(\bar{x} - B)} {\frac{B}{\sqrt{n}}} < Z ?

    We can split that problem into two inequalities:

    1) -Z < \frac{(\bar{x} - B)} {\frac{B}{\sqrt{n}}} and 2) \frac{(\bar{x} - B)} {\frac{B}{\sqrt{n}}} < Z

    For the first one, we can first multiply both sides of the inequality by \frac{B}{\sqrt{n}},
    and we arrive at \frac{-Z(B)}{\sqrt{n}} = \bar{x} - B.
    Then, add B to both sides:
    \frac{-Z(B)}{\sqrt{n}} + B = \bar{x}
    We can factor out B:
    B(\frac{-Z}{\sqrt{n}}+1) < \bar{x}
    and then divide both sides by \frac{-Z}{\sqrt{n}}+1, so the final inequality is:
    B < \frac{\bar{x}}{\frac{-z}{\sqrt{n}}+1}.
    It's the same process for the second inequality. Once you have both inequalities, you can mesh them together to form one big inequality with B in the center.
    Last edited by lisczz; April 5th 2009 at 06:22 PM. Reason: I hate LaTeX
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  3. #3
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    Quote Originally Posted by lisczz View Post
    Do you mean  -Z < \frac{(\bar{x} - B)} {\frac{B}{\sqrt{n}}} < Z ?

    We can split that problem into two inequalities:

    1) -Z < \frac{(\bar{x} - B)} {\frac{B}{\sqrt{n}}} and 2) \frac{(\bar{x} - B)} {\frac{B}{\sqrt{n}}} < Z

    For the first one, we can first multiply both sides of the inequality by \frac{B}{\sqrt{n}},
    and we arrive at \frac{-Z(B)}{\sqrt{n}} = \bar{x} - B.
    Then, add B to both sides:
    \frac{-Z(B)}{\sqrt{n}} + B = \bar{x}
    We can factor out B:
    B(\frac{-Z}{\sqrt{n}}+1) < \bar{x}
    and then divide both sides by \frac{-Z}{\sqrt{n}}+1, so the final inequality is:
    B < \frac{\bar{x}}{\frac{-z}{\sqrt{n}}+1}.
    It's the same process for the second inequality. Once you have both inequalities, you can mesh them together to form one big inequality with B in the center.
    Just a note that when getting the inequalities you should consider if it matters whether B is greater than 0 or less than zero ....
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  4. #4
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    B is >0 (this is using the CLT to estimate a confidence interval of the parameter from the exponential distribution (aka theta)

    Thanks so much for the detailed response lisczz!

    You guys are great for doing this!!
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  5. #5
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    I actually do have a question. This is what confuses me about inequalities. You have to assume that

    -Z / sqrt(N) +1 is >0 so that the sign of the inequality does not flip when you divide by it? Is that correct?
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