Covariance and Autocorrelation |
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COVARIANCE example for bird damage in Sorghum Plots
You first need to collect data (score of 1 to 10, or 0 to 100, or
some other scale that can take qualitative data to a quantitative scale).
X person needs to know what 1 means and what 10 means. 10 would for me
mean significant damage. We need to “gauge damage” but using
individual people. If you want to add x-person, you can then compare
them, but both need to be first instructed as to what the agreed-upon-scale
meant.
Also, and unfortunately, the individual scoring the damage has to
have his/her values for all plots. ALL 72 plots where an independent
score was recorded by that person.
How can this data be used. First, if we plan on using this
as a covariate, we must first analyze our “bird-damage” data as a dependent
variable in the model. If for instance the independent effect of
treatment in the model was significant for “bird-damage” , then
“bird-damage” cannot be used as a covariate. If “bird damage”
is found to be independent of treatment, it can be used as a covariate to
remove this source of variability, and that allows the researcher to better
detect treatment differences using their dependent variable (e.g., grain
yield).
Covariance can be viewed as a linear regression adjustment,
within an analysis of variance model.
We cover all of this in the Experimental Methods class that will
be taught in the spring of 2017. For this data, I will work with those
interested to properly employ the covariate.
But, you have to remember, that “bird-damage” could actually be
influenced by treatment. Maybe the birds preferred the high N plots.
You just don’t know. Also, this bird-damage data may not be
normally distributed which throws and additional wrench into the mix, since
non-parametric statistics would then be in order.
For taking scores, or ranks, you just have to apply the “score” number to
the exact same area that would be harvested. If I am going to harvest
all of the center 2 rows, then the score has to be my “view” or my “rank” of
that very same area.
These areas (harvest and rating) have to be the same, and that makes sense.
If the score and harvested area are different, how could I match final
values up? Obviously you cant. So, give those 2 center rows that you will harvest, your “mental average” score.
EXAMPLE:
1 - no damage For me this is not appropriate because I seriously
doubt you have plots that have 100% damage (a total loss). Ten has to be
the plot with the most damage of all plots. A score of 1 has to be a
plot that has the least amount of damage.
You have to remember that you are entering a “score” for bird damage.
Nothing else. If a plot is going to yield zero-grain, but where the
reason for that damage score of 10 has nothing to do with “bird damage” then
that plot’s bird damage score has to be zero. |
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COVARIANCE Gwen data one; input loc $ rep trt score yld; cards; EFAW 1 1 1 4.172467842 EFAW 1 2 1 5.809007482 EFAW 1 3 1 5.325708055 EFAW 1 4 1 7.205499074 EFAW 1 5 1 7.736446123 EFAW 1 6 2 6.75440387 EFAW 1 7 1 7.66826357 EFAW 1 8 2 5.826040013 EFAW 1 9 1 7.16472762 EFAW 1 10 1 8.647987937 EFAW 1 11 1 7.947073382 EFAW 1 12 2 6.531951129 EFAW 2 1 1 3.302389924 EFAW 2 2 1 4.074529335 EFAW 2 3 2 6.578584877 EFAW 2 4 1 4.784524348 EFAW 2 5 1 1.276861579 EFAW 2 6 1 5.035148482 EFAW 2 7 1 4.999456874 EFAW 2 8 1 4.980246697 EFAW 2 9 1 3.344957309 EFAW 2 10 1 5.522799872 EFAW 2 11 1 5.740105067 EFAW 2 12 1 2.821315273 EFAW 3 1 1 3.988830645 EFAW 3 2 1 2.761136063 EFAW 3 3 2 4.821103296 EFAW 3 4 1 6.068641054 EFAW 3 5 2 5.703061566 EFAW 3 6 2 6.263577978 EFAW 3 7 1 5.915862371 EFAW 3 8 1 6.00198225 EFAW 3 9 1 4.243763648 EFAW 3 10 2 6.500921231 EFAW 3 11 1 7.179428377 EFAW 3 12 2 6.538540067 LCB 1 1 2 5.655882526 LCB 1 2 2 7.163848857 LCB 1 3 2 5.79625248 LCB 1 4 3 7.893927624 LCB 1 5 5 8.562844148 LCB 1 6 5 9.03617542 LCB 1 7 1 5.187611913 LCB 1 8 2 8.532728297 LCB 1 9 2 7.84532925 LCB 1 10 3 7.887392674 LCB 1 11 4 9.328681845 LCB 1 12 3 9.049988716 LCB 2 1 1 4.165793306 LCB 2 2 4 7.503853017 LCB 2 3 4 7.595765585 LCB 2 4 4 7.358587957 LCB 2 5 5 8.526402937 LCB 2 6 3 8.72137655 LCB 2 7 2 4.779970235 LCB 2 8 3 7.730226834 LCB 2 9 3 7.583431838 LCB 2 10 4 8.903242919 LCB 2 11 6 9.074334694 LCB 2 12 3 8.922225361 LCB 3 1 2 5.92788277 LCB 3 2 4 6.798394452 LCB 3 3 2 6.598045047 LCB 3 4 4 7.89878046 LCB 3 5 4 9.285757316 LCB 3 6 4 7.909769048 LCB 3 7 1 4.214283074 LCB 3 8 2 7.332422217 LCB 3 9 2 6.37976641 LCB 3 10 4 7.180812325 LCB 3 11 4 9.127999641 LCB 3 12 5 7.876696768 data two; set one; Proc sort; by loc; proc glm; by loc; class rep trt; model score yld = rep trt; means trt; run; proc glm; by loc; class rep trt; model yld = rep trt score; lsmeans trt; run; |
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data one;
Covariance (go to RCBD example, #222, use GN as a covariate) |
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Covariance using preplant
soil test P as the covariate: Rule: You have to run anova using preplant P as a dependent variable FIRST. Why? Because you have to establish that "treatment" was not significant when using your potential covariate as a dependent variable. Your potential covariate has to be independent of treatment and the only way to establish this is to use as a dependent variable in your model. Once you establish that your covariate is independent of treatment, you can legitimately use it.
data one; |
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