Check if dependency between two variables is linearIs $R^2$ useful or dangerous?What is the relationship between R-squared and p-value in a regression?How to check my data set is following non-linear to linear regression?How to prove linearity assumption in regression analysis for a continuous dependent and nominal independent variable?what does linear regression actually mean?Linear regression with depended predictor variables
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Check if dependency between two variables is linear
Is $R^2$ useful or dangerous?What is the relationship between R-squared and p-value in a regression?How to check my data set is following non-linear to linear regression?How to prove linearity assumption in regression analysis for a continuous dependent and nominal independent variable?what does linear regression actually mean?Linear regression with depended predictor variables
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
There are a number of values for dependent variable (let's name it Y) and the same number of corresponding values for independent variable (let's name it X).
How can i check if dependency Y(X) is linear?
If my table would looks like (i.e. independent variable X will take only two values 0 or 1):
Is it possible in that case that dependency Y(X) is linear? Why?
r regression
asked 8 hours ago
DenisDenis
837 bronze badges
$endgroup$
add a comment |
$begingroup$
There are a number of values for dependent variable (let's name it Y) and the same number of corresponding values for independent variable (let's name it X).
How can i check if dependency Y(X) is linear?
If my table would looks like (i.e. independent variable X will take only two values 0 or 1):
Is it possible in that case that dependency Y(X) is linear? Why?
r regression
asked 8 hours ago
DenisDenis
837 bronze badges
$endgroup$
1
$begingroup$
You should say what you mean by a linear relationship. Ordinarily, one says that ordinary (Pearson) correlation measures the linear component of association. So $|r|$ sufficiently near $1$ might be a criterion for linearity. // In your first dataset relationship seems linear, but not in the second. // I can't easily convert your 'pictures' to numbers, so I didn't actually find correlations for either.
$endgroup$
– BruceET
8 hours ago
1
$begingroup$
@Bruce Only some people make that claim about $|r|.$ I think it's overblown and misleading. A different interpretation of "linearity" is that alternative non-linear models aren't worth the additional complexity. There are two standard, textbook approaches to this: add a quadratic term or bin the independent variable(s). Run an ANOVA on the nested model. If it's not significant, conclude you haven't detected any nonlinearity. These are often called "goodness of fit" tests (which you surely know, but many of our visitors do not).
$endgroup$
– whuber♦
7 hours ago
$begingroup$
Thanks for link.
$endgroup$
– BruceET
7 hours ago
$begingroup$
@BruceET In the protocol i'm going to apply was mentioned:Note that if observations are, instead, linearly related to gradients, we have to use a Canonical Correlation Analysis (CANCOR) or Redundancy Analysis (RDA) instead of a Canonical Correspondence Analysis (CCA).So i'm not sure which definition author exactly mean.
$endgroup$
– Denis
7 hours ago
$begingroup$
@ whuber Thanks a lot. Could you provide some tutorial for these two methods (add a quadratic term and bin the independent variable) you mentioned please (if possible inr). I'm not a statistician and it's not easy to figure out that. What i already understand now from your post i have to make somelinear models(perhaps withlmfunction) and then test them withANOVA. But it's not clear which models i could make with one independent variable. Thank you again.
$endgroup$
– Denis
4 hours ago
add a comment |
$begingroup$
There are a number of values for dependent variable (let's name it Y) and the same number of corresponding values for independent variable (let's name it X).
How can i check if dependency Y(X) is linear?
If my table would looks like (i.e. independent variable X will take only two values 0 or 1):
Is it possible in that case that dependency Y(X) is linear? Why?
r regression
asked 8 hours ago
DenisDenis
837 bronze badges
$endgroup$
There are a number of values for dependent variable (let's name it Y) and the same number of corresponding values for independent variable (let's name it X).
How can i check if dependency Y(X) is linear?
If my table would looks like (i.e. independent variable X will take only two values 0 or 1):
Is it possible in that case that dependency Y(X) is linear? Why?
r regression
r regression
asked 8 hours ago
DenisDenis
837 bronze badges
asked 8 hours ago
DenisDenis
837 bronze badges
asked 8 hours ago
DenisDenis
837 bronze badges
asked 8 hours ago
DenisDenis
837 bronze badges
asked 8 hours ago
DenisDenis
837 bronze badges
837 bronze badges
1
$begingroup$
You should say what you mean by a linear relationship. Ordinarily, one says that ordinary (Pearson) correlation measures the linear component of association. So $|r|$ sufficiently near $1$ might be a criterion for linearity. // In your first dataset relationship seems linear, but not in the second. // I can't easily convert your 'pictures' to numbers, so I didn't actually find correlations for either.
$endgroup$
– BruceET
8 hours ago
1
$begingroup$
@Bruce Only some people make that claim about $|r|.$ I think it's overblown and misleading. A different interpretation of "linearity" is that alternative non-linear models aren't worth the additional complexity. There are two standard, textbook approaches to this: add a quadratic term or bin the independent variable(s). Run an ANOVA on the nested model. If it's not significant, conclude you haven't detected any nonlinearity. These are often called "goodness of fit" tests (which you surely know, but many of our visitors do not).
$endgroup$
– whuber♦
7 hours ago
$begingroup$
Thanks for link.
$endgroup$
– BruceET
7 hours ago
$begingroup$
@BruceET In the protocol i'm going to apply was mentioned:Note that if observations are, instead, linearly related to gradients, we have to use a Canonical Correlation Analysis (CANCOR) or Redundancy Analysis (RDA) instead of a Canonical Correspondence Analysis (CCA).So i'm not sure which definition author exactly mean.
$endgroup$
– Denis
7 hours ago
$begingroup$
@ whuber Thanks a lot. Could you provide some tutorial for these two methods (add a quadratic term and bin the independent variable) you mentioned please (if possible inr). I'm not a statistician and it's not easy to figure out that. What i already understand now from your post i have to make somelinear models(perhaps withlmfunction) and then test them withANOVA. But it's not clear which models i could make with one independent variable. Thank you again.
$endgroup$
– Denis
4 hours ago
add a comment |
1
$begingroup$
You should say what you mean by a linear relationship. Ordinarily, one says that ordinary (Pearson) correlation measures the linear component of association. So $|r|$ sufficiently near $1$ might be a criterion for linearity. // In your first dataset relationship seems linear, but not in the second. // I can't easily convert your 'pictures' to numbers, so I didn't actually find correlations for either.
$endgroup$
– BruceET
8 hours ago
1
$begingroup$
@Bruce Only some people make that claim about $|r|.$ I think it's overblown and misleading. A different interpretation of "linearity" is that alternative non-linear models aren't worth the additional complexity. There are two standard, textbook approaches to this: add a quadratic term or bin the independent variable(s). Run an ANOVA on the nested model. If it's not significant, conclude you haven't detected any nonlinearity. These are often called "goodness of fit" tests (which you surely know, but many of our visitors do not).
$endgroup$
– whuber♦
7 hours ago
$begingroup$
Thanks for link.
$endgroup$
– BruceET
7 hours ago
$begingroup$
@BruceET In the protocol i'm going to apply was mentioned:Note that if observations are, instead, linearly related to gradients, we have to use a Canonical Correlation Analysis (CANCOR) or Redundancy Analysis (RDA) instead of a Canonical Correspondence Analysis (CCA).So i'm not sure which definition author exactly mean.
$endgroup$
– Denis
7 hours ago
$begingroup$
@ whuber Thanks a lot. Could you provide some tutorial for these two methods (add a quadratic term and bin the independent variable) you mentioned please (if possible inr). I'm not a statistician and it's not easy to figure out that. What i already understand now from your post i have to make somelinear models(perhaps withlmfunction) and then test them withANOVA. But it's not clear which models i could make with one independent variable. Thank you again.
$endgroup$
– Denis
4 hours ago
1
1
$begingroup$
You should say what you mean by a linear relationship. Ordinarily, one says that ordinary (Pearson) correlation measures the linear component of association. So $|r|$ sufficiently near $1$ might be a criterion for linearity. // In your first dataset relationship seems linear, but not in the second. // I can't easily convert your 'pictures' to numbers, so I didn't actually find correlations for either.
$endgroup$
– BruceET
8 hours ago
$begingroup$
You should say what you mean by a linear relationship. Ordinarily, one says that ordinary (Pearson) correlation measures the linear component of association. So $|r|$ sufficiently near $1$ might be a criterion for linearity. // In your first dataset relationship seems linear, but not in the second. // I can't easily convert your 'pictures' to numbers, so I didn't actually find correlations for either.
$endgroup$
– BruceET
8 hours ago
1
1
$begingroup$
@Bruce Only some people make that claim about $|r|.$ I think it's overblown and misleading. A different interpretation of "linearity" is that alternative non-linear models aren't worth the additional complexity. There are two standard, textbook approaches to this: add a quadratic term or bin the independent variable(s). Run an ANOVA on the nested model. If it's not significant, conclude you haven't detected any nonlinearity. These are often called "goodness of fit" tests (which you surely know, but many of our visitors do not).
$endgroup$
– whuber♦
7 hours ago
$begingroup$
@Bruce Only some people make that claim about $|r|.$ I think it's overblown and misleading. A different interpretation of "linearity" is that alternative non-linear models aren't worth the additional complexity. There are two standard, textbook approaches to this: add a quadratic term or bin the independent variable(s). Run an ANOVA on the nested model. If it's not significant, conclude you haven't detected any nonlinearity. These are often called "goodness of fit" tests (which you surely know, but many of our visitors do not).
$endgroup$
– whuber♦
7 hours ago
$begingroup$
Thanks for link.
$endgroup$
– BruceET
7 hours ago
$begingroup$
Thanks for link.
$endgroup$
– BruceET
7 hours ago
$begingroup$
@BruceET In the protocol i'm going to apply was mentioned:
Note that if observations are, instead, linearly related to gradients, we have to use a Canonical Correlation Analysis (CANCOR) or Redundancy Analysis (RDA) instead of a Canonical Correspondence Analysis (CCA). So i'm not sure which definition author exactly mean.$endgroup$
– Denis
7 hours ago
$begingroup$
@BruceET In the protocol i'm going to apply was mentioned:
Note that if observations are, instead, linearly related to gradients, we have to use a Canonical Correlation Analysis (CANCOR) or Redundancy Analysis (RDA) instead of a Canonical Correspondence Analysis (CCA). So i'm not sure which definition author exactly mean.$endgroup$
– Denis
7 hours ago
$begingroup$
@ whuber Thanks a lot. Could you provide some tutorial for these two methods (add a quadratic term and bin the independent variable) you mentioned please (if possible in
r). I'm not a statistician and it's not easy to figure out that. What i already understand now from your post i have to make some linear models (perhaps with lm function) and then test them with ANOVA. But it's not clear which models i could make with one independent variable. Thank you again.$endgroup$
– Denis
4 hours ago
$begingroup$
@ whuber Thanks a lot. Could you provide some tutorial for these two methods (add a quadratic term and bin the independent variable) you mentioned please (if possible in
r). I'm not a statistician and it's not easy to figure out that. What i already understand now from your post i have to make some linear models (perhaps with lm function) and then test them with ANOVA. But it's not clear which models i could make with one independent variable. Thank you again.$endgroup$
– Denis
4 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Sometimes a simple scatterplot alone is sufficient to make this determination. If the relationship is linear, I would expect a scatterplot of the data to show the data somewhat evenly scattered about a straight line - here, this is not the case, as the shape of the data from the first table has obvious curvature:
As an simple example equation that would fit this curvature, here is a hyperbolic type equation "y = (a + (b * x)) / (c + x)" fitted to the data:
and here is a straight line "y = a + (b * x)" fitted to the data:
answered 7 hours ago
James PhillipsJames Phillips
8113 gold badges7 silver badges7 bronze badges
$endgroup$
$begingroup$
I'm wondering if it's possible in theory to have a linear dependency betweenXandYif X take only two possible values 0 or 1?
$endgroup$
– Denis
4 hours ago
$begingroup$
In trivially simple cases yes, for example a data set with only two points such as [0, 5] and [1, 10]. A scatterplot of those two points will show that a straight line could trivially be made between them. However if you scatter plot the data in your second table, you should see that a straight line model is not going to fit that data at all - those data points do not lie on anything even close to a straight line.
$endgroup$
– James Phillips
7 mins ago
add a comment |
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$begingroup$
Sometimes a simple scatterplot alone is sufficient to make this determination. If the relationship is linear, I would expect a scatterplot of the data to show the data somewhat evenly scattered about a straight line - here, this is not the case, as the shape of the data from the first table has obvious curvature:
As an simple example equation that would fit this curvature, here is a hyperbolic type equation "y = (a + (b * x)) / (c + x)" fitted to the data:
and here is a straight line "y = a + (b * x)" fitted to the data:
answered 7 hours ago
James PhillipsJames Phillips
8113 gold badges7 silver badges7 bronze badges
$endgroup$
$begingroup$
I'm wondering if it's possible in theory to have a linear dependency betweenXandYif X take only two possible values 0 or 1?
$endgroup$
– Denis
4 hours ago
$begingroup$
In trivially simple cases yes, for example a data set with only two points such as [0, 5] and [1, 10]. A scatterplot of those two points will show that a straight line could trivially be made between them. However if you scatter plot the data in your second table, you should see that a straight line model is not going to fit that data at all - those data points do not lie on anything even close to a straight line.
$endgroup$
– James Phillips
7 mins ago
add a comment |
$begingroup$
Sometimes a simple scatterplot alone is sufficient to make this determination. If the relationship is linear, I would expect a scatterplot of the data to show the data somewhat evenly scattered about a straight line - here, this is not the case, as the shape of the data from the first table has obvious curvature:
As an simple example equation that would fit this curvature, here is a hyperbolic type equation "y = (a + (b * x)) / (c + x)" fitted to the data:
and here is a straight line "y = a + (b * x)" fitted to the data:
answered 7 hours ago
James PhillipsJames Phillips
8113 gold badges7 silver badges7 bronze badges
$endgroup$
$begingroup$
I'm wondering if it's possible in theory to have a linear dependency betweenXandYif X take only two possible values 0 or 1?
$endgroup$
– Denis
4 hours ago
$begingroup$
In trivially simple cases yes, for example a data set with only two points such as [0, 5] and [1, 10]. A scatterplot of those two points will show that a straight line could trivially be made between them. However if you scatter plot the data in your second table, you should see that a straight line model is not going to fit that data at all - those data points do not lie on anything even close to a straight line.
$endgroup$
– James Phillips
7 mins ago
add a comment |
$begingroup$
Sometimes a simple scatterplot alone is sufficient to make this determination. If the relationship is linear, I would expect a scatterplot of the data to show the data somewhat evenly scattered about a straight line - here, this is not the case, as the shape of the data from the first table has obvious curvature:
As an simple example equation that would fit this curvature, here is a hyperbolic type equation "y = (a + (b * x)) / (c + x)" fitted to the data:
and here is a straight line "y = a + (b * x)" fitted to the data:
answered 7 hours ago
James PhillipsJames Phillips
8113 gold badges7 silver badges7 bronze badges
$endgroup$
Sometimes a simple scatterplot alone is sufficient to make this determination. If the relationship is linear, I would expect a scatterplot of the data to show the data somewhat evenly scattered about a straight line - here, this is not the case, as the shape of the data from the first table has obvious curvature:
As an simple example equation that would fit this curvature, here is a hyperbolic type equation "y = (a + (b * x)) / (c + x)" fitted to the data:
and here is a straight line "y = a + (b * x)" fitted to the data:
answered 7 hours ago
James PhillipsJames Phillips
8113 gold badges7 silver badges7 bronze badges
answered 7 hours ago
James PhillipsJames Phillips
8113 gold badges7 silver badges7 bronze badges
answered 7 hours ago
James PhillipsJames Phillips
8113 gold badges7 silver badges7 bronze badges
answered 7 hours ago
James PhillipsJames Phillips
8113 gold badges7 silver badges7 bronze badges
8113 gold badges7 silver badges7 bronze badges
$begingroup$
I'm wondering if it's possible in theory to have a linear dependency betweenXandYif X take only two possible values 0 or 1?
$endgroup$
– Denis
4 hours ago
$begingroup$
In trivially simple cases yes, for example a data set with only two points such as [0, 5] and [1, 10]. A scatterplot of those two points will show that a straight line could trivially be made between them. However if you scatter plot the data in your second table, you should see that a straight line model is not going to fit that data at all - those data points do not lie on anything even close to a straight line.
$endgroup$
– James Phillips
7 mins ago
add a comment |
$begingroup$
I'm wondering if it's possible in theory to have a linear dependency betweenXandYif X take only two possible values 0 or 1?
$endgroup$
– Denis
4 hours ago
$begingroup$
In trivially simple cases yes, for example a data set with only two points such as [0, 5] and [1, 10]. A scatterplot of those two points will show that a straight line could trivially be made between them. However if you scatter plot the data in your second table, you should see that a straight line model is not going to fit that data at all - those data points do not lie on anything even close to a straight line.
$endgroup$
– James Phillips
7 mins ago
$begingroup$
I'm wondering if it's possible in theory to have a linear dependency between
X and Y if X take only two possible values 0 or 1?$endgroup$
– Denis
4 hours ago
$begingroup$
I'm wondering if it's possible in theory to have a linear dependency between
X and Y if X take only two possible values 0 or 1?$endgroup$
– Denis
4 hours ago
$begingroup$
In trivially simple cases yes, for example a data set with only two points such as [0, 5] and [1, 10]. A scatterplot of those two points will show that a straight line could trivially be made between them. However if you scatter plot the data in your second table, you should see that a straight line model is not going to fit that data at all - those data points do not lie on anything even close to a straight line.
$endgroup$
– James Phillips
7 mins ago
$begingroup$
In trivially simple cases yes, for example a data set with only two points such as [0, 5] and [1, 10]. A scatterplot of those two points will show that a straight line could trivially be made between them. However if you scatter plot the data in your second table, you should see that a straight line model is not going to fit that data at all - those data points do not lie on anything even close to a straight line.
$endgroup$
– James Phillips
7 mins ago
add a comment |
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1
$begingroup$
You should say what you mean by a linear relationship. Ordinarily, one says that ordinary (Pearson) correlation measures the linear component of association. So $|r|$ sufficiently near $1$ might be a criterion for linearity. // In your first dataset relationship seems linear, but not in the second. // I can't easily convert your 'pictures' to numbers, so I didn't actually find correlations for either.
$endgroup$
– BruceET
8 hours ago
1
$begingroup$
@Bruce Only some people make that claim about $|r|.$ I think it's overblown and misleading. A different interpretation of "linearity" is that alternative non-linear models aren't worth the additional complexity. There are two standard, textbook approaches to this: add a quadratic term or bin the independent variable(s). Run an ANOVA on the nested model. If it's not significant, conclude you haven't detected any nonlinearity. These are often called "goodness of fit" tests (which you surely know, but many of our visitors do not).
$endgroup$
– whuber♦
7 hours ago
$begingroup$
Thanks for link.
$endgroup$
– BruceET
7 hours ago
$begingroup$
@BruceET In the protocol i'm going to apply was mentioned:
Note that if observations are, instead, linearly related to gradients, we have to use a Canonical Correlation Analysis (CANCOR) or Redundancy Analysis (RDA) instead of a Canonical Correspondence Analysis (CCA).So i'm not sure which definition author exactly mean.$endgroup$
– Denis
7 hours ago
$begingroup$
@ whuber Thanks a lot. Could you provide some tutorial for these two methods (add a quadratic term and bin the independent variable) you mentioned please (if possible in
r). I'm not a statistician and it's not easy to figure out that. What i already understand now from your post i have to make somelinear models(perhaps withlmfunction) and then test them withANOVA. But it's not clear which models i could make with one independent variable. Thank you again.$endgroup$
– Denis
4 hours ago