Hello Mr. Perret, How can we determine the starting values and equation nature? For example, I have 3 independent variables, and two of my independent variables affect the dependent variable linearly, but one of my independent variables affects the dependent variable logarithmically. (Should I do like this ==> A*x1+ B*x2+ C* ln(D*x3) ). I could not set the function and starting values.
Your model looks good but instead of the D in the ln(.) you could also include it simply as an absolute term because your last part transforms into C*ln(x3) + C*ln(D). This also means that it would be way easier for you to simply logarithmize x3 and run a linear regression using the logarithmized x3 instead of the original. This would also eliminate the need to find good starting values.
Hi Mr. Perret i have 3 independents variable all of them affect the defendent parameters non linearly. How can i use model expression and what has to be starting values. Can you help me? T
How you model it depends on your conception of the underlying model, i.e. what theory tells you about the type of relation. Similarly, the starting values should be your first rough estimates of what the final coefficients could be like, e.g. from a preceding analysis.
In principle yes, you just have to model it accordingly. I would however question whether a second degree polynomial could not provide the same results, while being more stable.
If, in the context of a nonlinear regression, you specify your model accordingly, then yes SPSS can do this. Even if you include a respective terms into a normal regression model it will work. Personally however, I would see very few reasons to use sixth degree polynomials which cannot be covered by quadratic terms which would make the results more stable as well.
I do not know any method to determine the values based on the problem alone. You could try to get first estimate either via a preliminary survey or via expert interviews focussing on the approximate effects to be measured.
If you deduce the idea of variance analysis or rather or variance decomposition you calculate the variance between groups and the variance within groups. Now imagine you leave the 1/n in front of the formula for the variance within groups. That's what gives you the sum of squares. This article deduces this mathematically: yuehhua.github.io/2020/02/10/one-way-anova-and-sum-of-squares/ The corrected (between) is what they call SS_reg in this article.
R Squared can take values between 0 and 1 with 0 signifying that the model explains absolutely nothing and 1 signifying that the model perfectly describes the dependent variable. Practically, if you work with survey data, a value of 0.25 can already be considered rather good.
Thank you =شُكرا جزيلا
Hello Mr. Perret,
How can we determine the starting values and equation nature? For example, I have 3 independent variables, and two of my independent variables affect the dependent variable linearly, but one of my independent variables affects the dependent variable logarithmically. (Should I do like this ==> A*x1+ B*x2+ C* ln(D*x3) ). I could not set the function and starting values.
Your model looks good but instead of the D in the ln(.) you could also include it simply as an absolute term because your last part transforms into C*ln(x3) + C*ln(D). This also means that it would be way easier for you to simply logarithmize x3 and run a linear regression using the logarithmized x3 instead of the original. This would also eliminate the need to find good starting values.
Hi Mr. Perret i have 3 independents variable all of them affect the defendent parameters non linearly. How can i use model expression and what has to be starting values. Can you help me? T
How you model it depends on your conception of the underlying model, i.e. what theory tells you about the type of relation.
Similarly, the starting values should be your first rough estimates of what the final coefficients could be like, e.g. from a preceding analysis.
Hey Mr. Perret, Im wondering if spss can analyze 6 degree polynomial regression?
In principle yes, you just have to model it accordingly. I would however question whether a second degree polynomial could not provide the same results, while being more stable.
If, in the context of a nonlinear regression, you specify your model accordingly, then yes SPSS can do this. Even if you include a respective terms into a normal regression model it will work. Personally however, I would see very few reasons to use sixth degree polynomials which cannot be covered by quadratic terms which would make the results more stable as well.
How to choose good starting value ,any method?
I do not know any method to determine the values based on the problem alone. You could try to get first estimate either via a preliminary survey or via expert interviews focussing on the approximate effects to be measured.
Can we change the confidence intervals like 90%, 99% etc in nonlinear regression in SPSS?
If I remember correctly, up to SPSS 26 it is not yet possible.
What does the corrected and uncorrected sum of squares mean in the ANOVA table?
If you deduce the idea of variance analysis or rather or variance decomposition you calculate the variance between groups and the variance within groups. Now imagine you leave the 1/n in front of the formula for the variance within groups. That's what gives you the sum of squares.
This article deduces this mathematically: yuehhua.github.io/2020/02/10/one-way-anova-and-sum-of-squares/
The corrected (between) is what they call SS_reg in this article.
what is the best value of rsquared
R Squared can take values between 0 and 1 with 0 signifying that the model explains absolutely nothing and 1 signifying that the model perfectly describes the dependent variable. Practically, if you work with survey data, a value of 0.25 can already be considered rather good.