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Math Help - Z-tables, Backwards and Interpolation: I just don't get it.

  1. #1
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    Exclamation Z-tables, Backwards and Interpolation: I just don't get it.

    I swear, every time I think I get it, I end up being completely wrong. Basically, I'm still confused as to how Z_{0.01}=2.326, or Z_{0.025}=1.96, or this:



    How did z_{a/2} become 2.575? I understand that
    1-a=0.99
    a=0.01
    a/2=0.005

    ...but not how that transforms into 2.575.
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  2. #2
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    Quote Originally Posted by trackies View Post
    I swear, every time I think I get it, I end up being completely wrong. Basically, I'm still confused as to how Z_{0.01}=2.326, or Z_{0.025}=1.96, or this:



    How did z_{a/2} become 2.575? I understand that
    1-a=0.99
    a=0.01
    a/2=0.005

    ...but not how that transforms into 2.575.
    Pr(z > z*) = 0.005 => Pr(z < z*) = 1 - 0.005 = 0.995.

    So find 0.995 and go backwards to the coreesponding value of z:

    So z* = 2.575 (using tables).
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  3. #3
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    My z table only goes down to 3.0 on the side, and 0.09 along the top, so I can't find .995.
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  4. #4
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    Quote Originally Posted by trackies View Post
    My z table only goes down to 3.0 on the side, and 0.09 along the top, so I can't find .995.
    No, those are the z-values. You are looking for the z value corresponding to 0.995 .... Look for 0.995 in the big array of numbers corresponding to probabilities ..... Then look at what value of z gives it .....
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  5. #5
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    Ah, no, that's what I mean. The largest number of the z-table given to us is 0.4990, which corresponds to the z-value 3.09. That's why 0.995 isn't on this table..
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  6. #6
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    Quote Originally Posted by trackies View Post
    Ah, no, that's what I mean. The largest number of the z-table given to us is 0.4990, which corresponds to the z-value 3.09. That's why 0.995 isn't on this table..
    There are (at least) two types of normal table one gives the area under the curve from -infty to z, and the other the area from 0 to z (z>0). You want the former, but you can get the numbers by using the table you have got but subtracting 0.5 of off the p value (this is because the area from -infty to 0 is exactly 1/2).

    So in this case you want to use p=0.995-0.5=0.495

    RonL
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