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Multiple Linear Regression - Hypothesis Testing

  1. Sep 26, 2013 #1
    1. The problem statement, all variables and given/known data
    I'm looking through some example problems that my professor posted and this bit doesn't make sense

    2up4bqf.png

    How do you come up with the values underlined?


    2. Relevant equations



    3. The attempt at a solution

    Upon researching it, I find that you should use α/2 for both of these values. So i'm not sure what's going on here
     
  2. jcsd
  3. Sep 26, 2013 #2

    statdad

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    You can't calculate the p-value exactly here, the best you can do is find a bound for it. The number of degrees of freedom is fixed at 5, so you look in the lower tail of the appropriate t-distribution, for 5 degrees of freedom, until you find two tabled values that bracket your calculated value. They just happen to be for 10% and 25%. This tells you that whatever the real p-value is, it is not smaller than 10%, so you don't reject.

    If your calculated value had fallen as
    [tex]
    -t_{0.025} < t < -t_{0.05}
    [/tex]

    you would know the p-value is smaller than 5%, so less than [itex] \alpha = 0.05 [/itex], so you would reject.
     
  4. Sep 26, 2013 #3

    Ray Vickson

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    Or, you could use a package such as Maple's 'stats' facility to get a precise result:
    stats[statevalf,cdf,studentst[5]](-1.1326);
    0.1543773607 <----- output
    So, the p value is about 0.154.
     
  5. Sep 26, 2013 #4
    Ok, thanks. That makes sense. I normally just try to stay away from tables and use my ti-89 instead.


    For another problem, same concept:

    I'm testing B3 = 0 vs b3 =/= 0 at 5% level of significance. I found a test statistic of -1.516. tistat.tcdf(-∞, -1.516, 12) = .0777. Since it's two tailed I multiply this by 2: 2(.0777) = .1554. Since .1554 > .05 I do not reject the null hypothesis. Correct? The null hypothesis is more likely than 5%

    I guess I could have also done 1 - tistat.tcdf(-1.516, 1.516, 12)
     
  6. Sep 27, 2013 #5

    statdad

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    I didn't check your ti determined p-value, but your reasoning is correct (especially since the one-sided p-value would already be larger than 5%).
     
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