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Following up on this question, I would like to make an intuitive explanation of why i'm failing to reject $H_0$ ie that all varieties of wheat have same effect on yield. So, here is a boxplot:

enter image description here

And here is the model for it (without a global intercept):

lm=lm(yield~0+factor(variety)+clay+I(rainfall-mean(rainfall)))

Here is the summary for it:

Call:
lm(formula = yi ~ 0 + factor(va) + clay + I(ra - mean(ra)))

Residuals:
    Min      1Q  Median      3Q     Max 
-3.4685 -1.1002 -0.0688  1.3723  3.2639 

Coefficients:
                 Estimate Std. Error t value Pr(>|t|)    
factor(va)A      13.44778    0.63204  21.277  < 2e-16 ***
factor(va)B      12.57547    0.63234  19.887  < 2e-16 ***
factor(va)C      12.40505    0.60899  20.370  < 2e-16 ***
factor(va)D      12.59439    0.61368  20.523  < 2e-16 ***
clay             -1.97900    0.63081  -3.137  0.00351 ** 
I(ra - mean(ra))  0.05549    0.04209   1.318  0.19622    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.832 on 34 degrees of freedom
Multiple R-squared:  0.9811,    Adjusted R-squared:  0.9777 
F-statistic: 293.5 on 6 and 34 DF,  p-value: < 2.2e-16

My reasoning of why i'm failing to reject $H_0 : \mu_A=\mu_B=\mu_C=\mu_D$ So, by looking at the means and boxplot, they have the same yield averagely, and A has a bit higher average. But the fact that A has a slightly higher average doesn't justify the rejection of $H_0$.

Would this be right?

Analysis of Variance Table

Response: yi
                 Df  Sum Sq Mean Sq F value   Pr(>F)   
factor(va)        3   5.607  1.8689  0.5569 0.647109   
clay              1  30.875 30.8753  9.1996 0.004612 **
I(ra - mean(ra))  1   5.833  5.8326  1.7379 0.196218   
Residuals        34 114.109  3.3561 
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  • $\begingroup$ You’re getting $p<2.2\times 10^{-16}$, which is about as small of a number as R can give. $\endgroup$ – Dave May 26 at 11:06
  • $\begingroup$ @Dave this confirms my reasoning, right? $\endgroup$ – The Poor Jew May 26 at 11:08
  • $\begingroup$ No, we want small p-values in order to reject. Yours is about as small as it can be. I don’t totally follow your regression, however. Which value(s) do you see as your p-value(s)? $\endgroup$ – Dave May 26 at 11:11
  • 1
    $\begingroup$ Pairwise comparisons are a separate issue. You have strong evidence against your claim that the factor groups all have the same mean. $\endgroup$ – Dave May 26 at 11:39
  • 1
    $\begingroup$ Please stop doing this without an intercept. Basically everyone uses an intercept in ANOVA, and it makes it hard to communicate when you omit it. $\endgroup$ – Dave May 26 at 11:50

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