I can not understand which is the reason why the fitlm using variables without zscoring returns pvalues equal to NaN, whereas this does not happen using zscored variables. Below you can find the code I used:
medians_var1_scored = nanzscore(medians_var1);
medians_var2_scored = nanzscore(medians_var2);

medians_var1_scored_tr = medians_var1_scored’;
medians_var2_scored_tr = medians_var2_scored’;

tbl=table(medians_var1_scored_tr,medians_var2_scored_tr,’VariableNames’, …
{‘var1′,’var2’});
%build your model
mdl=fitlm(tbl,’var1 ~ var2′,’RobustOpts’,’on’)
which gives this result:
mdl =

Linear regression model (robust fit):
var1 ~ 1 + var2

Estimated Coefficients:
Estimate SE tStat pValue
_________ ________ ________ _______

(Intercept) 0.028674 0.094595 0.30312 0.76234
var2 -0.072919 0.094998 -0.76758 0.44429

Number of observations: 118, Error degrees of freedom: 116
Root Mean Squared Error: 1.03
R-squared: 0.00584, Adjusted R-Squared: -0.00273
F-statistic vs. constant model: 0.681, p-value = 0.411
If instead I use the original variables, I do:
tbl=table(medians_var1′,medians_var2′,’VariableNames’, …
{‘var1′,’var2’});
%build your model
mdl=fitlm(tbl,’var1 ~ var2′,’RobustOpts’,’on’)
and obtain:
mdl =

Linear regression model (robust fit):
var1 ~ 1 + var2

Estimated Coefficients:
Estimate SE tStat pValue
__________ __________ ______ __________

(Intercept) 0 0 NaN NaN
var2 8.6386e-13 2.1087e-14 40.966 7.8796e-71

Number of observations: 118, Error degrees of freedom: 117
Root Mean Squared Error: 31.4
R-squared: 0.263, Adjusted R-Squared: 0.263
F-statistic vs. constant model: Inf, p-value = NaN
I can’t understand why this happens. You can find attached both the original var1 and var2 variables and the zscored ones. But If I plot both of them, I obtain the same plot (just rescaled). Hence, there shouldn’t be a problem in the nanzscore function which is "hand written".I can not understand which is the reason why the fitlm using variables without zscoring returns pvalues equal to NaN, whereas this does not happen using zscored variables. Below you can find the code I used:
medians_var1_scored = nanzscore(medians_var1);
medians_var2_scored = nanzscore(medians_var2);

medians_var1_scored_tr = medians_var1_scored’;
medians_var2_scored_tr = medians_var2_scored’;

tbl=table(medians_var1_scored_tr,medians_var2_scored_tr,’VariableNames’, …
{‘var1′,’var2’});
%build your model
mdl=fitlm(tbl,’var1 ~ var2′,’RobustOpts’,’on’)
which gives this result:
mdl =

Linear regression model (robust fit):
var1 ~ 1 + var2

Estimated Coefficients:
Estimate SE tStat pValue
_________ ________ ________ _______

(Intercept) 0.028674 0.094595 0.30312 0.76234
var2 -0.072919 0.094998 -0.76758 0.44429

Number of observations: 118, Error degrees of freedom: 116
Root Mean Squared Error: 1.03
R-squared: 0.00584, Adjusted R-Squared: -0.00273
F-statistic vs. constant model: 0.681, p-value = 0.411
If instead I use the original variables, I do:
tbl=table(medians_var1′,medians_var2′,’VariableNames’, …
{‘var1′,’var2’});
%build your model
mdl=fitlm(tbl,’var1 ~ var2′,’RobustOpts’,’on’)
and obtain:
mdl =

Linear regression model (robust fit):
var1 ~ 1 + var2

Estimated Coefficients:
Estimate SE tStat pValue
__________ __________ ______ __________

(Intercept) 0 0 NaN NaN
var2 8.6386e-13 2.1087e-14 40.966 7.8796e-71

Number of observations: 118, Error degrees of freedom: 117
Root Mean Squared Error: 31.4
R-squared: 0.263, Adjusted R-Squared: 0.263
F-statistic vs. constant model: Inf, p-value = NaN
I can’t understand why this happens. You can find attached both the original var1 and var2 variables and the zscored ones. But If I plot both of them, I obtain the same plot (just rescaled). Hence, there shouldn’t be a problem in the nanzscore function which is "hand written". I can not understand which is the reason why the fitlm using variables without zscoring returns pvalues equal to NaN, whereas this does not happen using zscored variables. Below you can find the code I used:
medians_var1_scored = nanzscore(medians_var1);
medians_var2_scored = nanzscore(medians_var2);

medians_var1_scored_tr = medians_var1_scored’;
medians_var2_scored_tr = medians_var2_scored’;

tbl=table(medians_var1_scored_tr,medians_var2_scored_tr,’VariableNames’, …
{‘var1′,’var2’});
%build your model
mdl=fitlm(tbl,’var1 ~ var2′,’RobustOpts’,’on’)
which gives this result:
mdl =

Linear regression model (robust fit):
var1 ~ 1 + var2

Estimated Coefficients:
Estimate SE tStat pValue
_________ ________ ________ _______

(Intercept) 0.028674 0.094595 0.30312 0.76234
var2 -0.072919 0.094998 -0.76758 0.44429

Number of observations: 118, Error degrees of freedom: 116
Root Mean Squared Error: 1.03
R-squared: 0.00584, Adjusted R-Squared: -0.00273
F-statistic vs. constant model: 0.681, p-value = 0.411
If instead I use the original variables, I do:
tbl=table(medians_var1′,medians_var2′,’VariableNames’, …
{‘var1′,’var2’});
%build your model
mdl=fitlm(tbl,’var1 ~ var2′,’RobustOpts’,’on’)
and obtain:
mdl =

Linear regression model (robust fit):
var1 ~ 1 + var2

Estimated Coefficients:
Estimate SE tStat pValue
__________ __________ ______ __________

(Intercept) 0 0 NaN NaN
var2 8.6386e-13 2.1087e-14 40.966 7.8796e-71

Number of observations: 118, Error degrees of freedom: 117
Root Mean Squared Error: 31.4
R-squared: 0.263, Adjusted R-Squared: 0.263
F-statistic vs. constant model: Inf, p-value = NaN
I can’t understand why this happens. You can find attached both the original var1 and var2 variables and the zscored ones. But If I plot both of them, I obtain the same plot (just rescaled). Hence, there shouldn’t be a problem in the nanzscore function which is "hand written". fitlm, zscore, pvalue MATLAB Answers — New Questions

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