Interpretation of linear regression interaction term plot
$begingroup$
I am interested in looking at the relationship between plant productivity, temperature change and plant biomass change.
I have run a linear model in R using the below equation for this with plant productivity as an interaction term.
Plant biomass change = temperature change * plant productivity
The interaction is statistically significant (p-value = <0.01). I have plotted the estimated coefficient for temperature change from model above against the plant biomass figures and wanted to check I'm interpreting it right!
I think (hope) the plot tells me how the effect of temperature on biomass change changes as productivity changes.
The graph shows me the temperature coefficient becomes less negative as biomass increases. I interpret this to mean that temperature has a negative effect on the biomass change in plants with lower productivity.
Is my interpretation correct?
r regression interaction
$endgroup$
add a comment |
$begingroup$
I am interested in looking at the relationship between plant productivity, temperature change and plant biomass change.
I have run a linear model in R using the below equation for this with plant productivity as an interaction term.
Plant biomass change = temperature change * plant productivity
The interaction is statistically significant (p-value = <0.01). I have plotted the estimated coefficient for temperature change from model above against the plant biomass figures and wanted to check I'm interpreting it right!
I think (hope) the plot tells me how the effect of temperature on biomass change changes as productivity changes.
The graph shows me the temperature coefficient becomes less negative as biomass increases. I interpret this to mean that temperature has a negative effect on the biomass change in plants with lower productivity.
Is my interpretation correct?
r regression interaction
$endgroup$
add a comment |
$begingroup$
I am interested in looking at the relationship between plant productivity, temperature change and plant biomass change.
I have run a linear model in R using the below equation for this with plant productivity as an interaction term.
Plant biomass change = temperature change * plant productivity
The interaction is statistically significant (p-value = <0.01). I have plotted the estimated coefficient for temperature change from model above against the plant biomass figures and wanted to check I'm interpreting it right!
I think (hope) the plot tells me how the effect of temperature on biomass change changes as productivity changes.
The graph shows me the temperature coefficient becomes less negative as biomass increases. I interpret this to mean that temperature has a negative effect on the biomass change in plants with lower productivity.
Is my interpretation correct?
r regression interaction
$endgroup$
I am interested in looking at the relationship between plant productivity, temperature change and plant biomass change.
I have run a linear model in R using the below equation for this with plant productivity as an interaction term.
Plant biomass change = temperature change * plant productivity
The interaction is statistically significant (p-value = <0.01). I have plotted the estimated coefficient for temperature change from model above against the plant biomass figures and wanted to check I'm interpreting it right!
I think (hope) the plot tells me how the effect of temperature on biomass change changes as productivity changes.
The graph shows me the temperature coefficient becomes less negative as biomass increases. I interpret this to mean that temperature has a negative effect on the biomass change in plants with lower productivity.
Is my interpretation correct?
r regression interaction
r regression interaction
asked 5 hours ago
Clare PClare P
161
161
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
That interpretation seems correct and that's an interesting way to graph it.
What I usually do is have one IV on the x axis, the DV on the y axis and several lines for the other IV. Then I make the same plot with the IVs switched. That might be more interpretable.
$endgroup$
add a comment |
$begingroup$
Not quite sure I follow your argument. If both predictor variables in your model are assumed continuous, then the model summary should report an estimated intercept (b0), an estimated coefficient for temperature change (b1), an estimated coefficient for plant productivity (b2) and an estimated coefficient for the interaction between temperature change and plant productivity (b3). The summary of the model output will report these values in the column titled Estimate - since I don't know what they are, I called them b0, b1, b2 and b3. Thus, the expected (or average) plant biomass change can be expressed as:
Expected plant biomass change = b0 + b1*(temperature change) + b2*(plant productivity) +
b3*(temperature change)*(plant productivity).
Because the model includes an interaction term, the effect of plant productivity on expected plant biomass change actually depends on temperature change. You can see this by re-arranging the above equation:
Expected plant biomass change = [b0 + b1*(temperature change)] + [b2 + b3*(temperature change)]*(plant productivity).
The intercept and slopes describing the relationship between plant productivity are given by:
Intercept: [b0 + b1*(temperature change)]
Slope: [b2 + b3*(temperature change)]
For example, if the temperature change is zero degrees (Celsius?), then:
Expected plant biomass change = b0 + (b2 + b3)*(plant productivity).
As Peter suggested, you can choose several representative values for temperature change and then plot the corresponding lines obtained by substituting those representative values in the expressions of the above Intercept and Slope. Those lines would describe how the expected plant biomass change varies as a function of plant productivity.
To decide which representative values of temperature change to consider, you can plot the distribution of temperature changes observed in your study. If that distribution looks approximately normal, you can choose the average temperature change (m), as well as m - sd and m + sd, say, where sd is the standard deviation of that distribution. If the distribution is unimodal but skewed, you could replace m with the median and sd with the interquartile range of the distribution.
Plotting lines with different intercepts and slopes would allow you to see how the effect of plant productivity on expected plant biomass change depends on particular, representative values of temperature change. It's possible that some slopes will be positive, while others will be negative. In that case, you can note that the effect changes direction, etc.
Addendum:
If I understand @gung correctly, I think what you did was to re-express the first equation I wrote like so:
Expected plant biomass change = [b0 + b2*(plant productivity)] +
[b1 + b3*(plant productivity)]*(temperature change)
and then plot b1 + b3*(plant productivity) versus productivity to see how the rate of change in expected plant biomass change varies as a function of plant productivity. What is not clear to me though is how you computed the confidence band around b1 + b3*(plant productivity)? Did you compute the standard error (SE) of b1 + b3*(plant productivity) and then computed pointwise confidence bands via the formula b1 + b3*(plant productivity) +/- 1.96 SE? (The SE should take into account the correlation between b1 and b3). Or perhaps you used a critical value from a t-distribution instead of 1.96, with degrees of freedom given by the residual degrees of freedom?
$endgroup$
1
$begingroup$
I interpret the OP's plot as follows: Each point on that line is the slope on the simple effect of temperature at the level of productivity specified on the x-axis.
$endgroup$
– gung♦
2 hours ago
1
$begingroup$
@gung: Does my Addendum to the above capture what you think the OP's plot is showing? The OP stated that the coefficient plotted becomes more negative as biomass increases - but it should become more negative as plant productivity increases? That statement threw me off regarding what was actually being plotted.
$endgroup$
– Isabella Ghement
1 hour ago
1
$begingroup$
Yes. Confidence intervals for the slope of a simple effect can be computed at any point by the square root of the sum of the variances and 2*Cov(b1, b3), but I don't think that would address the simultaneity.
$endgroup$
– gung♦
1 hour ago
$begingroup$
Thanks so much, @gung! 😊
$endgroup$
– Isabella Ghement
1 hour ago
1
$begingroup$
You make a good point about "as biomass increases". My guess is that the OP misspoke, or is confused about the nature of the interaction.
$endgroup$
– gung♦
1 hour ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
That interpretation seems correct and that's an interesting way to graph it.
What I usually do is have one IV on the x axis, the DV on the y axis and several lines for the other IV. Then I make the same plot with the IVs switched. That might be more interpretable.
$endgroup$
add a comment |
$begingroup$
That interpretation seems correct and that's an interesting way to graph it.
What I usually do is have one IV on the x axis, the DV on the y axis and several lines for the other IV. Then I make the same plot with the IVs switched. That might be more interpretable.
$endgroup$
add a comment |
$begingroup$
That interpretation seems correct and that's an interesting way to graph it.
What I usually do is have one IV on the x axis, the DV on the y axis and several lines for the other IV. Then I make the same plot with the IVs switched. That might be more interpretable.
$endgroup$
That interpretation seems correct and that's an interesting way to graph it.
What I usually do is have one IV on the x axis, the DV on the y axis and several lines for the other IV. Then I make the same plot with the IVs switched. That might be more interpretable.
answered 4 hours ago
Peter Flom♦Peter Flom
76k11107209
76k11107209
add a comment |
add a comment |
$begingroup$
Not quite sure I follow your argument. If both predictor variables in your model are assumed continuous, then the model summary should report an estimated intercept (b0), an estimated coefficient for temperature change (b1), an estimated coefficient for plant productivity (b2) and an estimated coefficient for the interaction between temperature change and plant productivity (b3). The summary of the model output will report these values in the column titled Estimate - since I don't know what they are, I called them b0, b1, b2 and b3. Thus, the expected (or average) plant biomass change can be expressed as:
Expected plant biomass change = b0 + b1*(temperature change) + b2*(plant productivity) +
b3*(temperature change)*(plant productivity).
Because the model includes an interaction term, the effect of plant productivity on expected plant biomass change actually depends on temperature change. You can see this by re-arranging the above equation:
Expected plant biomass change = [b0 + b1*(temperature change)] + [b2 + b3*(temperature change)]*(plant productivity).
The intercept and slopes describing the relationship between plant productivity are given by:
Intercept: [b0 + b1*(temperature change)]
Slope: [b2 + b3*(temperature change)]
For example, if the temperature change is zero degrees (Celsius?), then:
Expected plant biomass change = b0 + (b2 + b3)*(plant productivity).
As Peter suggested, you can choose several representative values for temperature change and then plot the corresponding lines obtained by substituting those representative values in the expressions of the above Intercept and Slope. Those lines would describe how the expected plant biomass change varies as a function of plant productivity.
To decide which representative values of temperature change to consider, you can plot the distribution of temperature changes observed in your study. If that distribution looks approximately normal, you can choose the average temperature change (m), as well as m - sd and m + sd, say, where sd is the standard deviation of that distribution. If the distribution is unimodal but skewed, you could replace m with the median and sd with the interquartile range of the distribution.
Plotting lines with different intercepts and slopes would allow you to see how the effect of plant productivity on expected plant biomass change depends on particular, representative values of temperature change. It's possible that some slopes will be positive, while others will be negative. In that case, you can note that the effect changes direction, etc.
Addendum:
If I understand @gung correctly, I think what you did was to re-express the first equation I wrote like so:
Expected plant biomass change = [b0 + b2*(plant productivity)] +
[b1 + b3*(plant productivity)]*(temperature change)
and then plot b1 + b3*(plant productivity) versus productivity to see how the rate of change in expected plant biomass change varies as a function of plant productivity. What is not clear to me though is how you computed the confidence band around b1 + b3*(plant productivity)? Did you compute the standard error (SE) of b1 + b3*(plant productivity) and then computed pointwise confidence bands via the formula b1 + b3*(plant productivity) +/- 1.96 SE? (The SE should take into account the correlation between b1 and b3). Or perhaps you used a critical value from a t-distribution instead of 1.96, with degrees of freedom given by the residual degrees of freedom?
$endgroup$
1
$begingroup$
I interpret the OP's plot as follows: Each point on that line is the slope on the simple effect of temperature at the level of productivity specified on the x-axis.
$endgroup$
– gung♦
2 hours ago
1
$begingroup$
@gung: Does my Addendum to the above capture what you think the OP's plot is showing? The OP stated that the coefficient plotted becomes more negative as biomass increases - but it should become more negative as plant productivity increases? That statement threw me off regarding what was actually being plotted.
$endgroup$
– Isabella Ghement
1 hour ago
1
$begingroup$
Yes. Confidence intervals for the slope of a simple effect can be computed at any point by the square root of the sum of the variances and 2*Cov(b1, b3), but I don't think that would address the simultaneity.
$endgroup$
– gung♦
1 hour ago
$begingroup$
Thanks so much, @gung! 😊
$endgroup$
– Isabella Ghement
1 hour ago
1
$begingroup$
You make a good point about "as biomass increases". My guess is that the OP misspoke, or is confused about the nature of the interaction.
$endgroup$
– gung♦
1 hour ago
add a comment |
$begingroup$
Not quite sure I follow your argument. If both predictor variables in your model are assumed continuous, then the model summary should report an estimated intercept (b0), an estimated coefficient for temperature change (b1), an estimated coefficient for plant productivity (b2) and an estimated coefficient for the interaction between temperature change and plant productivity (b3). The summary of the model output will report these values in the column titled Estimate - since I don't know what they are, I called them b0, b1, b2 and b3. Thus, the expected (or average) plant biomass change can be expressed as:
Expected plant biomass change = b0 + b1*(temperature change) + b2*(plant productivity) +
b3*(temperature change)*(plant productivity).
Because the model includes an interaction term, the effect of plant productivity on expected plant biomass change actually depends on temperature change. You can see this by re-arranging the above equation:
Expected plant biomass change = [b0 + b1*(temperature change)] + [b2 + b3*(temperature change)]*(plant productivity).
The intercept and slopes describing the relationship between plant productivity are given by:
Intercept: [b0 + b1*(temperature change)]
Slope: [b2 + b3*(temperature change)]
For example, if the temperature change is zero degrees (Celsius?), then:
Expected plant biomass change = b0 + (b2 + b3)*(plant productivity).
As Peter suggested, you can choose several representative values for temperature change and then plot the corresponding lines obtained by substituting those representative values in the expressions of the above Intercept and Slope. Those lines would describe how the expected plant biomass change varies as a function of plant productivity.
To decide which representative values of temperature change to consider, you can plot the distribution of temperature changes observed in your study. If that distribution looks approximately normal, you can choose the average temperature change (m), as well as m - sd and m + sd, say, where sd is the standard deviation of that distribution. If the distribution is unimodal but skewed, you could replace m with the median and sd with the interquartile range of the distribution.
Plotting lines with different intercepts and slopes would allow you to see how the effect of plant productivity on expected plant biomass change depends on particular, representative values of temperature change. It's possible that some slopes will be positive, while others will be negative. In that case, you can note that the effect changes direction, etc.
Addendum:
If I understand @gung correctly, I think what you did was to re-express the first equation I wrote like so:
Expected plant biomass change = [b0 + b2*(plant productivity)] +
[b1 + b3*(plant productivity)]*(temperature change)
and then plot b1 + b3*(plant productivity) versus productivity to see how the rate of change in expected plant biomass change varies as a function of plant productivity. What is not clear to me though is how you computed the confidence band around b1 + b3*(plant productivity)? Did you compute the standard error (SE) of b1 + b3*(plant productivity) and then computed pointwise confidence bands via the formula b1 + b3*(plant productivity) +/- 1.96 SE? (The SE should take into account the correlation between b1 and b3). Or perhaps you used a critical value from a t-distribution instead of 1.96, with degrees of freedom given by the residual degrees of freedom?
$endgroup$
1
$begingroup$
I interpret the OP's plot as follows: Each point on that line is the slope on the simple effect of temperature at the level of productivity specified on the x-axis.
$endgroup$
– gung♦
2 hours ago
1
$begingroup$
@gung: Does my Addendum to the above capture what you think the OP's plot is showing? The OP stated that the coefficient plotted becomes more negative as biomass increases - but it should become more negative as plant productivity increases? That statement threw me off regarding what was actually being plotted.
$endgroup$
– Isabella Ghement
1 hour ago
1
$begingroup$
Yes. Confidence intervals for the slope of a simple effect can be computed at any point by the square root of the sum of the variances and 2*Cov(b1, b3), but I don't think that would address the simultaneity.
$endgroup$
– gung♦
1 hour ago
$begingroup$
Thanks so much, @gung! 😊
$endgroup$
– Isabella Ghement
1 hour ago
1
$begingroup$
You make a good point about "as biomass increases". My guess is that the OP misspoke, or is confused about the nature of the interaction.
$endgroup$
– gung♦
1 hour ago
add a comment |
$begingroup$
Not quite sure I follow your argument. If both predictor variables in your model are assumed continuous, then the model summary should report an estimated intercept (b0), an estimated coefficient for temperature change (b1), an estimated coefficient for plant productivity (b2) and an estimated coefficient for the interaction between temperature change and plant productivity (b3). The summary of the model output will report these values in the column titled Estimate - since I don't know what they are, I called them b0, b1, b2 and b3. Thus, the expected (or average) plant biomass change can be expressed as:
Expected plant biomass change = b0 + b1*(temperature change) + b2*(plant productivity) +
b3*(temperature change)*(plant productivity).
Because the model includes an interaction term, the effect of plant productivity on expected plant biomass change actually depends on temperature change. You can see this by re-arranging the above equation:
Expected plant biomass change = [b0 + b1*(temperature change)] + [b2 + b3*(temperature change)]*(plant productivity).
The intercept and slopes describing the relationship between plant productivity are given by:
Intercept: [b0 + b1*(temperature change)]
Slope: [b2 + b3*(temperature change)]
For example, if the temperature change is zero degrees (Celsius?), then:
Expected plant biomass change = b0 + (b2 + b3)*(plant productivity).
As Peter suggested, you can choose several representative values for temperature change and then plot the corresponding lines obtained by substituting those representative values in the expressions of the above Intercept and Slope. Those lines would describe how the expected plant biomass change varies as a function of plant productivity.
To decide which representative values of temperature change to consider, you can plot the distribution of temperature changes observed in your study. If that distribution looks approximately normal, you can choose the average temperature change (m), as well as m - sd and m + sd, say, where sd is the standard deviation of that distribution. If the distribution is unimodal but skewed, you could replace m with the median and sd with the interquartile range of the distribution.
Plotting lines with different intercepts and slopes would allow you to see how the effect of plant productivity on expected plant biomass change depends on particular, representative values of temperature change. It's possible that some slopes will be positive, while others will be negative. In that case, you can note that the effect changes direction, etc.
Addendum:
If I understand @gung correctly, I think what you did was to re-express the first equation I wrote like so:
Expected plant biomass change = [b0 + b2*(plant productivity)] +
[b1 + b3*(plant productivity)]*(temperature change)
and then plot b1 + b3*(plant productivity) versus productivity to see how the rate of change in expected plant biomass change varies as a function of plant productivity. What is not clear to me though is how you computed the confidence band around b1 + b3*(plant productivity)? Did you compute the standard error (SE) of b1 + b3*(plant productivity) and then computed pointwise confidence bands via the formula b1 + b3*(plant productivity) +/- 1.96 SE? (The SE should take into account the correlation between b1 and b3). Or perhaps you used a critical value from a t-distribution instead of 1.96, with degrees of freedom given by the residual degrees of freedom?
$endgroup$
Not quite sure I follow your argument. If both predictor variables in your model are assumed continuous, then the model summary should report an estimated intercept (b0), an estimated coefficient for temperature change (b1), an estimated coefficient for plant productivity (b2) and an estimated coefficient for the interaction between temperature change and plant productivity (b3). The summary of the model output will report these values in the column titled Estimate - since I don't know what they are, I called them b0, b1, b2 and b3. Thus, the expected (or average) plant biomass change can be expressed as:
Expected plant biomass change = b0 + b1*(temperature change) + b2*(plant productivity) +
b3*(temperature change)*(plant productivity).
Because the model includes an interaction term, the effect of plant productivity on expected plant biomass change actually depends on temperature change. You can see this by re-arranging the above equation:
Expected plant biomass change = [b0 + b1*(temperature change)] + [b2 + b3*(temperature change)]*(plant productivity).
The intercept and slopes describing the relationship between plant productivity are given by:
Intercept: [b0 + b1*(temperature change)]
Slope: [b2 + b3*(temperature change)]
For example, if the temperature change is zero degrees (Celsius?), then:
Expected plant biomass change = b0 + (b2 + b3)*(plant productivity).
As Peter suggested, you can choose several representative values for temperature change and then plot the corresponding lines obtained by substituting those representative values in the expressions of the above Intercept and Slope. Those lines would describe how the expected plant biomass change varies as a function of plant productivity.
To decide which representative values of temperature change to consider, you can plot the distribution of temperature changes observed in your study. If that distribution looks approximately normal, you can choose the average temperature change (m), as well as m - sd and m + sd, say, where sd is the standard deviation of that distribution. If the distribution is unimodal but skewed, you could replace m with the median and sd with the interquartile range of the distribution.
Plotting lines with different intercepts and slopes would allow you to see how the effect of plant productivity on expected plant biomass change depends on particular, representative values of temperature change. It's possible that some slopes will be positive, while others will be negative. In that case, you can note that the effect changes direction, etc.
Addendum:
If I understand @gung correctly, I think what you did was to re-express the first equation I wrote like so:
Expected plant biomass change = [b0 + b2*(plant productivity)] +
[b1 + b3*(plant productivity)]*(temperature change)
and then plot b1 + b3*(plant productivity) versus productivity to see how the rate of change in expected plant biomass change varies as a function of plant productivity. What is not clear to me though is how you computed the confidence band around b1 + b3*(plant productivity)? Did you compute the standard error (SE) of b1 + b3*(plant productivity) and then computed pointwise confidence bands via the formula b1 + b3*(plant productivity) +/- 1.96 SE? (The SE should take into account the correlation between b1 and b3). Or perhaps you used a critical value from a t-distribution instead of 1.96, with degrees of freedom given by the residual degrees of freedom?
edited 1 hour ago
answered 2 hours ago
Isabella GhementIsabella Ghement
7,151320
7,151320
1
$begingroup$
I interpret the OP's plot as follows: Each point on that line is the slope on the simple effect of temperature at the level of productivity specified on the x-axis.
$endgroup$
– gung♦
2 hours ago
1
$begingroup$
@gung: Does my Addendum to the above capture what you think the OP's plot is showing? The OP stated that the coefficient plotted becomes more negative as biomass increases - but it should become more negative as plant productivity increases? That statement threw me off regarding what was actually being plotted.
$endgroup$
– Isabella Ghement
1 hour ago
1
$begingroup$
Yes. Confidence intervals for the slope of a simple effect can be computed at any point by the square root of the sum of the variances and 2*Cov(b1, b3), but I don't think that would address the simultaneity.
$endgroup$
– gung♦
1 hour ago
$begingroup$
Thanks so much, @gung! 😊
$endgroup$
– Isabella Ghement
1 hour ago
1
$begingroup$
You make a good point about "as biomass increases". My guess is that the OP misspoke, or is confused about the nature of the interaction.
$endgroup$
– gung♦
1 hour ago
add a comment |
1
$begingroup$
I interpret the OP's plot as follows: Each point on that line is the slope on the simple effect of temperature at the level of productivity specified on the x-axis.
$endgroup$
– gung♦
2 hours ago
1
$begingroup$
@gung: Does my Addendum to the above capture what you think the OP's plot is showing? The OP stated that the coefficient plotted becomes more negative as biomass increases - but it should become more negative as plant productivity increases? That statement threw me off regarding what was actually being plotted.
$endgroup$
– Isabella Ghement
1 hour ago
1
$begingroup$
Yes. Confidence intervals for the slope of a simple effect can be computed at any point by the square root of the sum of the variances and 2*Cov(b1, b3), but I don't think that would address the simultaneity.
$endgroup$
– gung♦
1 hour ago
$begingroup$
Thanks so much, @gung! 😊
$endgroup$
– Isabella Ghement
1 hour ago
1
$begingroup$
You make a good point about "as biomass increases". My guess is that the OP misspoke, or is confused about the nature of the interaction.
$endgroup$
– gung♦
1 hour ago
1
1
$begingroup$
I interpret the OP's plot as follows: Each point on that line is the slope on the simple effect of temperature at the level of productivity specified on the x-axis.
$endgroup$
– gung♦
2 hours ago
$begingroup$
I interpret the OP's plot as follows: Each point on that line is the slope on the simple effect of temperature at the level of productivity specified on the x-axis.
$endgroup$
– gung♦
2 hours ago
1
1
$begingroup$
@gung: Does my Addendum to the above capture what you think the OP's plot is showing? The OP stated that the coefficient plotted becomes more negative as biomass increases - but it should become more negative as plant productivity increases? That statement threw me off regarding what was actually being plotted.
$endgroup$
– Isabella Ghement
1 hour ago
$begingroup$
@gung: Does my Addendum to the above capture what you think the OP's plot is showing? The OP stated that the coefficient plotted becomes more negative as biomass increases - but it should become more negative as plant productivity increases? That statement threw me off regarding what was actually being plotted.
$endgroup$
– Isabella Ghement
1 hour ago
1
1
$begingroup$
Yes. Confidence intervals for the slope of a simple effect can be computed at any point by the square root of the sum of the variances and 2*Cov(b1, b3), but I don't think that would address the simultaneity.
$endgroup$
– gung♦
1 hour ago
$begingroup$
Yes. Confidence intervals for the slope of a simple effect can be computed at any point by the square root of the sum of the variances and 2*Cov(b1, b3), but I don't think that would address the simultaneity.
$endgroup$
– gung♦
1 hour ago
$begingroup$
Thanks so much, @gung! 😊
$endgroup$
– Isabella Ghement
1 hour ago
$begingroup$
Thanks so much, @gung! 😊
$endgroup$
– Isabella Ghement
1 hour ago
1
1
$begingroup$
You make a good point about "as biomass increases". My guess is that the OP misspoke, or is confused about the nature of the interaction.
$endgroup$
– gung♦
1 hour ago
$begingroup$
You make a good point about "as biomass increases". My guess is that the OP misspoke, or is confused about the nature of the interaction.
$endgroup$
– gung♦
1 hour ago
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