I am mechanically testing devices from multiple companies that have slightly different geometry. I want to know whether the geometry plays a role in the results (or if just the company and material drive the difference).

Linear regression shows variable X with correlation (p=0.3); general linear regression shows variable X as having no effect on the model (p>0.4). What is going on?

I am assuming that the more simple linear regression model is being confounded by the lack of predictors and outputting inaccurate results. Is this correct?

Is the lack of significance in a regression model enough to confidently say that variable x is (likely) not influencing results?

If this is a silly basic question, I apologize.

Thanks in advance.

General Linear Model (ANOVA Based)

Material is nested in Company for this model

Backward Elimination of Terms (alpha to remove = 0.1)
Candidate terms: Shaft Diameter, Thread Diameter, Company, Material(Company)

                   ----Step 1---    ----Step 2---   -----Step 3----

                    Coef    P       Coef    P       Coef    P
Constant             116            3758            843.5    
Shaft Diameter      -948    0.401   -993    0.377        
Thread Diameter     1036    0.482                
Company             -526    0.038   -617    0.000   -476.7  0.000
Material(Company)    242    0.175    472    0.079   284.8   0.000

Analysis of Variance
   Source         DF    Seq SS Contribution Adj SS  Adj MS  F-Value
   Company         2    7247546 56.04%      7059609  3529805  71.65
   Material        2    2630442 20.34%      2630442  1315221  26.70
   Error          62    3054346 23.62%      3054346  49264   
   Lack-of-Fit    46    2356103 18.22%      2356103  51220     1.17
   Pure Error     16    698243  5.40%       698243   43640   
   Total          66    12932334100.00%          
      Source            P-Value
      Company           0.0000000000000001
      Material(Company) 0.0000000043283494
       Lack-of-Fit      0.3763855668940961
        Pure Error   

      S            223.241      222.326     221.954
      R-sq          76.88%      76.69%      76.38%
      R-sq(adj)     74.57%      74.77%      74.86%
      R-sq(pred)    70.64%      70.95%      71.40%
      Mallows’ Cp   7.00        5.50        4.29
     α to remove = 0.1

Linear Regression

Regression Analysis: Maximum Force (N) versus Shaft Diameter
The regression equation is
Maximum Force (N) = - 5370 + 2091 Shaft Diameter
Model Summary
S       R-sq    R-sq(adj)
273.027 62.53%  61.96%
Analysis of Variance
 Source     DF       SS      MS     F       P
 Regression 1   8086979 8086979 108.49  0.0000000000000017
   Error    65  4845355 74544        
   Total    66  12932334             
  Fitted Line: Maximum Force (N) versus Shaft Diameter
  • 3
    $\begingroup$ What is "general linear regression"? Can you show your model results? $\endgroup$ – The Laconic Dec 10 '18 at 4:21
  • $\begingroup$ Of course! Edited to show model results. I'm using Minitab if that makes a difference $\endgroup$ – Ryan Dec 13 '18 at 4:20
  • 1
    $\begingroup$ My guess is that shaft diameter varies with material and/or company, which I think is what you’re getting at with your worry about confounding. Do different companies specialize in different geometries? Do geometries depend on the material? Domain knowledge (that I don’t have) would probably suggest an answer. $\endgroup$ – The Laconic Dec 13 '18 at 4:48
  • 1
    $\begingroup$ This cannot be answered without more context. $\endgroup$ – kjetil b halvorsen Mar 29 at 14:07

OLS regression and ANOVA are the same model. They give the same results, albeit the output is often formatted differently.

However, in your code, in the first example you use backward elimination, which is a bad idea as all the results can be shown to be wrong and in the second case you limit it to one variable, which is not one of the ones selected by backward elimination. So, of course the results are different.

There are several reasons why backward elimination might not select shaft diameter, including collinearity.

No, the lack of significance is not sufficient to say that X is not influencing results. You could run a model with and without X and see what happens.


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