What if my linear regression data contains several co-mingled linear relationships?
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Let's say I am studying how daffodils respond to various soil conditions. I have collected data on the pH of the soil versus the mature height of the daffodil. I'm expecting a linear relationship, so I go about running a linear regression.
However, I didn't realize when I started my study that the population actually contains two varieties of daffodil, each of which responds very differently to soil pH. So the graph contains two distinct linear relationships:
I can eyeball it and separate it manually, of course. But I wonder if there is a more rigorous approach.
Questions:
Is there a statistical test to determine whether a data set would be better fit by a single line or by N lines?
How would I run a linear regression to fit the N lines? In other words, how do I disentangle the co-mingled data?
I can think of some combinatorial approaches, but they seem computationally expensive.
regression linear-model dataset
asked 2 hours ago
SlowMagicSlowMagic
1111
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SlowMagic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
Let's say I am studying how daffodils respond to various soil conditions. I have collected data on the pH of the soil versus the mature height of the daffodil. I'm expecting a linear relationship, so I go about running a linear regression.
However, I didn't realize when I started my study that the population actually contains two varieties of daffodil, each of which responds very differently to soil pH. So the graph contains two distinct linear relationships:
I can eyeball it and separate it manually, of course. But I wonder if there is a more rigorous approach.
Questions:
Is there a statistical test to determine whether a data set would be better fit by a single line or by N lines?
How would I run a linear regression to fit the N lines? In other words, how do I disentangle the co-mingled data?
I can think of some combinatorial approaches, but they seem computationally expensive.
regression linear-model dataset
asked 2 hours ago
SlowMagicSlowMagic
1111
New contributor
SlowMagic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
related stats.stackexchange.com/questions/245902/…
$endgroup$
– rep_ho
2 hours ago
1
$begingroup$
do you know which dafodil is which varaiety? If so, then you can just include that information into your model
$endgroup$
– rep_ho
2 hours ago
1
$begingroup$
This seems a classic case of statistical interaction, as in @Demetri Pananos's answer.
$endgroup$
– rolando2
1 hour ago
add a comment |
$begingroup$
Let's say I am studying how daffodils respond to various soil conditions. I have collected data on the pH of the soil versus the mature height of the daffodil. I'm expecting a linear relationship, so I go about running a linear regression.
However, I didn't realize when I started my study that the population actually contains two varieties of daffodil, each of which responds very differently to soil pH. So the graph contains two distinct linear relationships:
I can eyeball it and separate it manually, of course. But I wonder if there is a more rigorous approach.
Questions:
Is there a statistical test to determine whether a data set would be better fit by a single line or by N lines?
How would I run a linear regression to fit the N lines? In other words, how do I disentangle the co-mingled data?
I can think of some combinatorial approaches, but they seem computationally expensive.
regression linear-model dataset
asked 2 hours ago
SlowMagicSlowMagic
1111
New contributor
SlowMagic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let's say I am studying how daffodils respond to various soil conditions. I have collected data on the pH of the soil versus the mature height of the daffodil. I'm expecting a linear relationship, so I go about running a linear regression.
However, I didn't realize when I started my study that the population actually contains two varieties of daffodil, each of which responds very differently to soil pH. So the graph contains two distinct linear relationships:
I can eyeball it and separate it manually, of course. But I wonder if there is a more rigorous approach.
Questions:
Is there a statistical test to determine whether a data set would be better fit by a single line or by N lines?
How would I run a linear regression to fit the N lines? In other words, how do I disentangle the co-mingled data?
I can think of some combinatorial approaches, but they seem computationally expensive.
regression linear-model dataset
regression linear-model dataset
asked 2 hours ago
SlowMagicSlowMagic
1111
New contributor
SlowMagic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
SlowMagicSlowMagic
1111
New contributor
SlowMagic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
SlowMagicSlowMagic
1111
New contributor
SlowMagic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
SlowMagicSlowMagic
1111
asked 2 hours ago
SlowMagicSlowMagic
1111
1111
New contributor
SlowMagic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
SlowMagic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
SlowMagic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
related stats.stackexchange.com/questions/245902/…
$endgroup$
– rep_ho
2 hours ago
1
$begingroup$
do you know which dafodil is which varaiety? If so, then you can just include that information into your model
$endgroup$
– rep_ho
2 hours ago
1
$begingroup$
This seems a classic case of statistical interaction, as in @Demetri Pananos's answer.
$endgroup$
– rolando2
1 hour ago
add a comment |
$begingroup$
related stats.stackexchange.com/questions/245902/…
$endgroup$
– rep_ho
2 hours ago
1
$begingroup$
do you know which dafodil is which varaiety? If so, then you can just include that information into your model
$endgroup$
– rep_ho
2 hours ago
1
$begingroup$
This seems a classic case of statistical interaction, as in @Demetri Pananos's answer.
$endgroup$
– rolando2
1 hour ago
$begingroup$
related stats.stackexchange.com/questions/245902/…
$endgroup$
– rep_ho
2 hours ago
$begingroup$
related stats.stackexchange.com/questions/245902/…
$endgroup$
– rep_ho
2 hours ago
1
1
$begingroup$
do you know which dafodil is which varaiety? If so, then you can just include that information into your model
$endgroup$
– rep_ho
2 hours ago
$begingroup$
do you know which dafodil is which varaiety? If so, then you can just include that information into your model
$endgroup$
– rep_ho
2 hours ago
1
1
$begingroup$
This seems a classic case of statistical interaction, as in @Demetri Pananos's answer.
$endgroup$
– rolando2
1 hour ago
$begingroup$
This seems a classic case of statistical interaction, as in @Demetri Pananos's answer.
$endgroup$
– rolando2
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you know which sample comes from which variety of daffodil, you can estimate an interaction between variety and soil PH.
Your model will look like
$$ y = beta_0 + beta_1 text{variety} + beta_2text{PH} + beta_3text{variety}cdottext{PH} $$
Here is an example in R. I've generated some data that looks like this:
Clearly two different lines, and the lines correspond to two species. Here is how to estimate the lines using linear regression.
library(tidyverse)
#Simulate the data
N = 1000
ph = runif(N,5,8)
species = rbinom(N,1,0.5)
y = model.matrix(~ph*species)%*% c(20,1,20,-3) + rnorm(N, 0, 0.5)
y = as.numeric(y)
d = data_frame(ph = ph, species = species, y = y)
#Estimate the model
model = lm(y~species*ph, data = d)
summary(model)
And the result is
> summary(model)
Call:
lm(formula = y ~ species * ph, data = d)
Residuals:
Min 1Q Median 3Q Max
-1.61884 -0.31976 -0.00226 0.33521 1.46428
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 19.85850 0.17484 113.58 <2e-16 ***
species 20.31363 0.24626 82.49 <2e-16 ***
ph 1.01599 0.02671 38.04 <2e-16 ***
species:ph -3.03174 0.03756 -80.72 <2e-16 ***
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.4997 on 996 degrees of freedom
Multiple R-squared: 0.8844, Adjusted R-squared: 0.8841
F-statistic: 2541 on 3 and 996 DF, p-value: < 2.2e-16
For species labeled 0, the line is approximately
$$ y = 19 + 1cdot text{PH}$$
For species labeled 1, the line is approximately
$$ y = 40 - 2 cdot text{PH} $$
answered 2 hours ago
Demetri PananosDemetri Pananos
1,069317
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If you know which sample comes from which variety of daffodil, you can estimate an interaction between variety and soil PH.
Your model will look like
$$ y = beta_0 + beta_1 text{variety} + beta_2text{PH} + beta_3text{variety}cdottext{PH} $$
Here is an example in R. I've generated some data that looks like this:
Clearly two different lines, and the lines correspond to two species. Here is how to estimate the lines using linear regression.
library(tidyverse)
#Simulate the data
N = 1000
ph = runif(N,5,8)
species = rbinom(N,1,0.5)
y = model.matrix(~ph*species)%*% c(20,1,20,-3) + rnorm(N, 0, 0.5)
y = as.numeric(y)
d = data_frame(ph = ph, species = species, y = y)
#Estimate the model
model = lm(y~species*ph, data = d)
summary(model)
And the result is
> summary(model)
Call:
lm(formula = y ~ species * ph, data = d)
Residuals:
Min 1Q Median 3Q Max
-1.61884 -0.31976 -0.00226 0.33521 1.46428
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 19.85850 0.17484 113.58 <2e-16 ***
species 20.31363 0.24626 82.49 <2e-16 ***
ph 1.01599 0.02671 38.04 <2e-16 ***
species:ph -3.03174 0.03756 -80.72 <2e-16 ***
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.4997 on 996 degrees of freedom
Multiple R-squared: 0.8844, Adjusted R-squared: 0.8841
F-statistic: 2541 on 3 and 996 DF, p-value: < 2.2e-16
For species labeled 0, the line is approximately
$$ y = 19 + 1cdot text{PH}$$
For species labeled 1, the line is approximately
$$ y = 40 - 2 cdot text{PH} $$
answered 2 hours ago
Demetri PananosDemetri Pananos
1,069317
$endgroup$
add a comment |
$begingroup$
If you know which sample comes from which variety of daffodil, you can estimate an interaction between variety and soil PH.
Your model will look like
$$ y = beta_0 + beta_1 text{variety} + beta_2text{PH} + beta_3text{variety}cdottext{PH} $$
Here is an example in R. I've generated some data that looks like this:
Clearly two different lines, and the lines correspond to two species. Here is how to estimate the lines using linear regression.
library(tidyverse)
#Simulate the data
N = 1000
ph = runif(N,5,8)
species = rbinom(N,1,0.5)
y = model.matrix(~ph*species)%*% c(20,1,20,-3) + rnorm(N, 0, 0.5)
y = as.numeric(y)
d = data_frame(ph = ph, species = species, y = y)
#Estimate the model
model = lm(y~species*ph, data = d)
summary(model)
And the result is
> summary(model)
Call:
lm(formula = y ~ species * ph, data = d)
Residuals:
Min 1Q Median 3Q Max
-1.61884 -0.31976 -0.00226 0.33521 1.46428
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 19.85850 0.17484 113.58 <2e-16 ***
species 20.31363 0.24626 82.49 <2e-16 ***
ph 1.01599 0.02671 38.04 <2e-16 ***
species:ph -3.03174 0.03756 -80.72 <2e-16 ***
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.4997 on 996 degrees of freedom
Multiple R-squared: 0.8844, Adjusted R-squared: 0.8841
F-statistic: 2541 on 3 and 996 DF, p-value: < 2.2e-16
For species labeled 0, the line is approximately
$$ y = 19 + 1cdot text{PH}$$
For species labeled 1, the line is approximately
$$ y = 40 - 2 cdot text{PH} $$
answered 2 hours ago
Demetri PananosDemetri Pananos
1,069317
$endgroup$
add a comment |
$begingroup$
If you know which sample comes from which variety of daffodil, you can estimate an interaction between variety and soil PH.
Your model will look like
$$ y = beta_0 + beta_1 text{variety} + beta_2text{PH} + beta_3text{variety}cdottext{PH} $$
Here is an example in R. I've generated some data that looks like this:
Clearly two different lines, and the lines correspond to two species. Here is how to estimate the lines using linear regression.
library(tidyverse)
#Simulate the data
N = 1000
ph = runif(N,5,8)
species = rbinom(N,1,0.5)
y = model.matrix(~ph*species)%*% c(20,1,20,-3) + rnorm(N, 0, 0.5)
y = as.numeric(y)
d = data_frame(ph = ph, species = species, y = y)
#Estimate the model
model = lm(y~species*ph, data = d)
summary(model)
And the result is
> summary(model)
Call:
lm(formula = y ~ species * ph, data = d)
Residuals:
Min 1Q Median 3Q Max
-1.61884 -0.31976 -0.00226 0.33521 1.46428
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 19.85850 0.17484 113.58 <2e-16 ***
species 20.31363 0.24626 82.49 <2e-16 ***
ph 1.01599 0.02671 38.04 <2e-16 ***
species:ph -3.03174 0.03756 -80.72 <2e-16 ***
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.4997 on 996 degrees of freedom
Multiple R-squared: 0.8844, Adjusted R-squared: 0.8841
F-statistic: 2541 on 3 and 996 DF, p-value: < 2.2e-16
For species labeled 0, the line is approximately
$$ y = 19 + 1cdot text{PH}$$
For species labeled 1, the line is approximately
$$ y = 40 - 2 cdot text{PH} $$
answered 2 hours ago
Demetri PananosDemetri Pananos
1,069317
$endgroup$
If you know which sample comes from which variety of daffodil, you can estimate an interaction between variety and soil PH.
Your model will look like
$$ y = beta_0 + beta_1 text{variety} + beta_2text{PH} + beta_3text{variety}cdottext{PH} $$
Here is an example in R. I've generated some data that looks like this:
Clearly two different lines, and the lines correspond to two species. Here is how to estimate the lines using linear regression.
library(tidyverse)
#Simulate the data
N = 1000
ph = runif(N,5,8)
species = rbinom(N,1,0.5)
y = model.matrix(~ph*species)%*% c(20,1,20,-3) + rnorm(N, 0, 0.5)
y = as.numeric(y)
d = data_frame(ph = ph, species = species, y = y)
#Estimate the model
model = lm(y~species*ph, data = d)
summary(model)
And the result is
> summary(model)
Call:
lm(formula = y ~ species * ph, data = d)
Residuals:
Min 1Q Median 3Q Max
-1.61884 -0.31976 -0.00226 0.33521 1.46428
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 19.85850 0.17484 113.58 <2e-16 ***
species 20.31363 0.24626 82.49 <2e-16 ***
ph 1.01599 0.02671 38.04 <2e-16 ***
species:ph -3.03174 0.03756 -80.72 <2e-16 ***
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.4997 on 996 degrees of freedom
Multiple R-squared: 0.8844, Adjusted R-squared: 0.8841
F-statistic: 2541 on 3 and 996 DF, p-value: < 2.2e-16
For species labeled 0, the line is approximately
$$ y = 19 + 1cdot text{PH}$$
For species labeled 1, the line is approximately
$$ y = 40 - 2 cdot text{PH} $$
answered 2 hours ago
Demetri PananosDemetri Pananos
1,069317
edited 1 hour ago
answered 2 hours ago
Demetri PananosDemetri Pananos
1,069317
answered 2 hours ago
Demetri PananosDemetri Pananos
1,069317
answered 2 hours ago
Demetri PananosDemetri Pananos
1,069317
1,069317
add a comment |
add a comment |
SlowMagic is a new contributor. Be nice, and check out our Code of Conduct.
SlowMagic is a new contributor. Be nice, and check out our Code of Conduct.
SlowMagic is a new contributor. Be nice, and check out our Code of Conduct.
SlowMagic is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
related stats.stackexchange.com/questions/245902/…
$endgroup$
– rep_ho
2 hours ago
1
$begingroup$
do you know which dafodil is which varaiety? If so, then you can just include that information into your model
$endgroup$
– rep_ho
2 hours ago
1
$begingroup$
This seems a classic case of statistical interaction, as in @Demetri Pananos's answer.
$endgroup$
– rolando2
1 hour ago