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Multiple Linear Regression - Hypothesis Testing

  • Thread starter Phox
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  • #1
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Homework Statement


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?


Homework Equations





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
 

Answers and Replies

  • #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.
 
  • #3
Ray Vickson
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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.
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.
 
  • #4
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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.
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)
 
  • #5
statdad
Homework Helper
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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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