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Math Help - Z-value using Z-table or t-table.

  1. #1
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    Z-value using Z-table or t-table.

    Hi there. I don't really understand how the solution to the following came about:

    z_{a/2}=z_{0.01}=2.326
    P(Z>2.326)=0.01

    The answers after dictate: "To find where this value comes from, look up 0.49 backwards on the Z table, or look up t_{0.01} for \infty degrees of freedom the t table."

    Firstly, I don't understand why I'm supposed to look up 0.49 on the Z table and not 0.01? I'm also confused as to how to use the table 'backwards', especially if/when interpolation comes into it.

    Secondly, why does the degree of freedom = \infty?
    Last edited by trackies; June 10th 2008 at 07:50 PM. Reason: layout
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  2. #2
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    Quote Originally Posted by trackies View Post
    Hi there. I don't really understand how the solution to the following came about:

    z_{a/2}=z_{0.01}=2.326
    P(Z>2.326)=0.01

    The answers after dictate: "To find where this value comes from, look up 0.49 backwards on the Z table, or look up t_{0.01} for \infty degrees of freedom the t table."

    Firstly, I don't understand why I'm supposed to look up 0.49 on the Z table and not 0.01? I'm also confused as to how to use the table 'backwards', especially if/when interpolation comes into it.

    Secondly, why does the degree of freedom = \infty?
    If you use the t-statistic instead of the z-statistic, n = 250 => df = 249 which is effectively df = infinite as far as the values used in the calculation are concerned.

    You'd use 0.99 because the tables give you Pr(z < z*), not Pr(z > z*). So you want the value z* such that Pr(z < z*) = 0.99, which means going backwards from 0.99 in your tables.

    I'm not sure where the 0.49 comes from ......
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  3. #3
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    Ah, okay! I understand about the t-table now.

    However, I don't understand where you got 0.99 from.
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    Quote Originally Posted by trackies View Post
    Ah, okay! I understand about the t-table now.

    However, I don't understand where you got 0.99 from.
    You need to find the value of z* such that Pr(z > z*) = 0.01. But your tables don't give Pr(z > z*), they only give Pr(z < z*). So you need to consider Pr(z < z*) = 1 - 0.01 = 0.99.
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  5. #5
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    I double checked and it definitely says 0.49. Perhaps a typo? But in any case, that makes perfect sense now. Thank-you!
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  6. #6
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    Quote Originally Posted by trackies View Post
    I double checked and it definitely says 0.49. Perhaps a typo? [snip]
    I'd bet on it. Note that 0.49 is close to 0.5 and Pr(z < z*) = 0.5 => z* = 0 .....

    Quote Originally Posted by trackies View Post
    [snip]But in any case, that makes perfect sense now. Thank-you!
    You're welcome.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    If you use the t-statistic instead of the z-statistic, n = 250 => df = 249 which is effectively df = infinite as far as the values used in the calculation are concerned.

    You'd use 0.99 because the tables give you Pr(z < z*), not Pr(z > z*). So you want the value z* such that Pr(z < z*) = 0.99, which means going backwards from 0.99 in your tables.

    I'm not sure where the 0.49 comes from ......
    OK, I think post #6 of this thread: http://www.mathhelpforum.com/math-he...-dont-get.html

    explains why 0.49 rather than 0.99 is used .......
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