I have a problem with the calculation of the ANOVA. What is wrong with my code?
C_t = [30; 30; 30; 30; 30; 30; 30; 120; 120; 120; 120; 120; 120; 120; 120];
pick = [0; 0.026; 0.0550; 0.120 ; 0.320 ; 0.62 ; 1.2; 0 ;0.03; 0.066; 0.110 ; 0.220; 0.4; 0.740; 1.73];
tbl = table(C_t,pick,'VariableNames',{'C_t','pick'});
tbl.C_t = categorical(tbl.C_t);
mdl = fitlm(tbl,'pick')
tbl = anova(mdl,'summary'

 Akzeptierte Antwort

Jeff Miller
Jeff Miller am 6 Dez. 2022

0 Stimmen

Just change one line to this
mdl = fitlm(tbl,'pick~C_t')

4 Kommentare

Michael
Michael am 7 Dez. 2022
Thank you very much! Now it works. But I have one more question: why doesn't the function tbl = anova(mdl,'summary') work as in matlab help?
tbl = anova(mdl,'summary')
tbl=7×5 table
SumSq DF MeanSq F pValue
______ __ ______ _______ ________
Total 4757.8 99 48.059
Model 274.73 3 91.577 1.961 0.12501
. Linear 243.8 2 121.9 2.6103 0.078726
. Nonlinear 30.934 1 30.934 0.66242 0.41772
Residual 4483.1 96 46.699
. Lack of fit 1483.1 39 38.028 0.72253 0.85732
. Pure error 3000 57 52.632
I don't have (Liner, nonlineral, Lack of fit and pure error) and I care about these values
Jeff Miller
Jeff Miller am 7 Dez. 2022
You don't have linear, etc, because you have indicated that C_t is categorical. The missing terms are only calculated for numerical predictors.
But the example data you posted have only 2 values for C_t. It is not possible to distinguish linear versus nonlinear trends across only two C_t values.
Michael
Michael am 10 Dez. 2022
Ok, but how can i calculate leak of fit and purre error from this data?
Jeff Miller
Jeff Miller am 10 Dez. 2022
You can't. This is not a software problem, it is a limitation of your design. With only two values of X (30, 120), a straight line goes through the two (X, avg Y) pairs perfectly, so your data provide no information about whether there would be any deviation from a straight line fit ("lack of fit") if you had a third X value.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

James Water Bird
James Water Bird am 6 Dez. 2022

0 Stimmen

C_t = [30; 30; 30; 30; 30; 30; 30; 120; 120; 120; 120; 120; 120; 120; 120];
pick = [0; 0.026; 0.0550; 0.120 ; 0.320 ; 0.62 ; 1.2; 0 ;0.03; 0.066; 0.110 ; 0.220; 0.4; 0.740; 1.73];
% Use the table() function to create a table from the data
tbl = table(C_t,pick,'VariableNames',{'C_t','pick'});
% Convert the 'C_t' column to a categorical variable
tbl.C_t = categorical(tbl.C_t);
% Use the fitlm() function to fit a linear regression model to the data
mdl = fitlm(tbl,'pick')
% Use the anova() function to perform an analysis of variance on the model
tbl = anova(mdl,'summary')

1 Kommentar

Michael
Michael am 6 Dez. 2022
I understand that you have added descriptions, but this does not solve my problem.

Melden Sie sich an, um zu kommentieren.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by