problem with nls....
dear members,
On a closer inspection, I can see that the scatterplot of HF1 and HF5 is of the form y ~ -(1/x), while that of HF1 and HF6 is of the form y ~ (1/x). Is it possible that HF43nl is converging almost due to chance? I mean, for HF53nl, a straight line minimizes the RSS rather than for a curve like y ~ (1/x). Is it possible? If that is the case, should I model it linearly rather than nonlinearly? It is unsettling(this would always gives the wrong prediction given a predictor!). Or rather picewise nonlinear regression(for HF6 < 0 and HF6 > 0)?
very many thanks for your time and effort....
yours sincerely,
AKSHAY M KULKARNI
From: R-help <r-help-bounces at r-project.org> on behalf of akshay kulkarni <akshay_e4 at hotmail.com>
Sent: Thursday, March 21, 2019 5:26 PM
To: R help Mailing list
Subject: [R] problem with nls....
Sent: Thursday, March 21, 2019 5:26 PM
To: R help Mailing list
Subject: [R] problem with nls....
dear members,
I have the following nls call:
> HF53nl <- nls(HF1 ~ ((m/HF6) + 1),data = data.frame(HF6,HF1),start = list(m = 0.1))
> overview(HF53nl)
------
Formula: HF1 ~ ((m/HF6) + 1)
Parameters:
Estimate Std. Error t value Pr(>|t|)
m 2.147e-07 1.852e-06 0.116 0.908
Residual standard error: 0.03596 on 799 degrees of freedom
Number of iterations to convergence: 1
Achieved convergence tolerance: 1.246e-06
------
Residual sum of squares: 1.03
------
t-based confidence interval:
2.5% 97.5%
1 -3.420983e-06 3.850292e-06
------
Correlation matrix:
m
m 1
The scatter plot of HF6 and HF1 and the corresponding fitted line according to the above output of nls is attached(HF53nl). The fitted line is almost a straight line. But it should be a curve something of: y ~ 1/x. I think the very small value of m is making the curve a straight line.
But the fitted curve of the following call makes sense(attached: HF43nl):
> HF43nl <- nls(HF1 ~ ((k/HF5) + 1),data = data.frame(HF5,HF1),start = list(k = 0.1))
> overview(HF43nl)
------
Formula: HF1 ~ ((k/HF5) + 1)
Parameters:
Estimate Std. Error t value Pr(>|t|)
k -5.367e-04 5.076e-05 -10.57 <2e-16 ***
---
Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1
Residual standard error: 0.03368 on 799 degrees of freedom
Number of iterations to convergence: 1
Achieved convergence tolerance: 3.076e-07
------
Residual sum of squares: 0.906
------
t-based confidence interval:
2.5% 97.5%
1 -0.0006363717 -0.0004370954
------
Correlation matrix:
k
k 1
The queer thing is that the RSS for HF53nl and HF43nl is almost the same, which points to the purported validity of HF53nl. How is this possible? Can I go with the above estimates for the coefficient m of HF6 being equal to 2.147 * 10^(-7)? How do I make an nls call so that there is a better fit to HF1 and HF6.
NB: If you can't access the attached graphs, how do I send it to you otherwise? I can also give you HF1,HF6,HF5 if needed....
very many thanks for your time and effort....
yours sincerely,
AKSHAY M KULKARNI
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