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Interpretation of linear regression interaction term plot


Nonsignificant interaction still causes main effect to flip?How does the interpretation of main effects in a Two-Way ANOVA change depending on whether the interaction effect is significant?How to determine the significance of an interaction?Interpretation of Coefficients in linear regression using 'fitlm'Why does sign of a main effect change in logistic regression when adding an interaction?Interpretation of interaction term with a logged dependent variableinterpretation of the main effect when interaction term is included and main effect changes signinterpretation of interaction term in regressionWhy and how does adding an interaction term affects the confidence interval of a main effect?Plotting correlation coefficient against regression coefficient






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2












$begingroup$


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!



enter image description here



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 plant productivity 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?










share|cite|improve this question











$endgroup$











  • $begingroup$
    The term a*b in a linear model (as fit by lm) usually only makes sense if at least one of a and b is categorical. Is this the case?
    $endgroup$
    – Hong Ooi
    Mar 9 at 6:16

















2












$begingroup$


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!



enter image description here



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 plant productivity 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?










share|cite|improve this question











$endgroup$











  • $begingroup$
    The term a*b in a linear model (as fit by lm) usually only makes sense if at least one of a and b is categorical. Is this the case?
    $endgroup$
    – Hong Ooi
    Mar 9 at 6:16













2












2








2





$begingroup$


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!



enter image description here



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 plant productivity 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?










share|cite|improve this question











$endgroup$




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!



enter image description here



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 plant productivity 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 9 at 12:09







Clare P

















asked Mar 8 at 23:58









Clare PClare P

185




185











  • $begingroup$
    The term a*b in a linear model (as fit by lm) usually only makes sense if at least one of a and b is categorical. Is this the case?
    $endgroup$
    – Hong Ooi
    Mar 9 at 6:16
















  • $begingroup$
    The term a*b in a linear model (as fit by lm) usually only makes sense if at least one of a and b is categorical. Is this the case?
    $endgroup$
    – Hong Ooi
    Mar 9 at 6:16















$begingroup$
The term a*b in a linear model (as fit by lm) usually only makes sense if at least one of a and b is categorical. Is this the case?
$endgroup$
– Hong Ooi
Mar 9 at 6:16




$begingroup$
The term a*b in a linear model (as fit by lm) usually only makes sense if at least one of a and b is categorical. Is this the case?
$endgroup$
– Hong Ooi
Mar 9 at 6:16










2 Answers
2






active

oldest

votes


















4












$begingroup$

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.






share|cite|improve this answer









$endgroup$




















    3












    $begingroup$

    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?






    share|cite|improve this answer











    $endgroup$








    • 2




      $begingroup$
      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
      Mar 9 at 3:06






    • 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
      Mar 9 at 3:28






    • 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
      Mar 9 at 3:39






    • 2




      $begingroup$
      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
      Mar 9 at 4:19






    • 1




      $begingroup$
      Apologies I mispoke! I plotted the graph using the interplot library in R which I believe calculates the confidence intervals using critical t-statistics.
      $endgroup$
      – Clare P
      Mar 9 at 12:35











    Your Answer





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    2 Answers
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    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$

















      4












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$















        4












        4








        4





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 9 at 0:28









        Peter FlomPeter Flom

        77.1k12109215




        77.1k12109215























            3












            $begingroup$

            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?






            share|cite|improve this answer











            $endgroup$








            • 2




              $begingroup$
              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
              Mar 9 at 3:06






            • 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
              Mar 9 at 3:28






            • 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
              Mar 9 at 3:39






            • 2




              $begingroup$
              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
              Mar 9 at 4:19






            • 1




              $begingroup$
              Apologies I mispoke! I plotted the graph using the interplot library in R which I believe calculates the confidence intervals using critical t-statistics.
              $endgroup$
              – Clare P
              Mar 9 at 12:35















            3












            $begingroup$

            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?






            share|cite|improve this answer











            $endgroup$








            • 2




              $begingroup$
              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
              Mar 9 at 3:06






            • 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
              Mar 9 at 3:28






            • 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
              Mar 9 at 3:39






            • 2




              $begingroup$
              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
              Mar 9 at 4:19






            • 1




              $begingroup$
              Apologies I mispoke! I plotted the graph using the interplot library in R which I believe calculates the confidence intervals using critical t-statistics.
              $endgroup$
              – Clare P
              Mar 9 at 12:35













            3












            3








            3





            $begingroup$

            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?






            share|cite|improve this answer











            $endgroup$



            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?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 9 at 3:30

























            answered Mar 9 at 3:00









            Isabella GhementIsabella Ghement

            7,856422




            7,856422







            • 2




              $begingroup$
              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
              Mar 9 at 3:06






            • 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
              Mar 9 at 3:28






            • 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
              Mar 9 at 3:39






            • 2




              $begingroup$
              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
              Mar 9 at 4:19






            • 1




              $begingroup$
              Apologies I mispoke! I plotted the graph using the interplot library in R which I believe calculates the confidence intervals using critical t-statistics.
              $endgroup$
              – Clare P
              Mar 9 at 12:35












            • 2




              $begingroup$
              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
              Mar 9 at 3:06






            • 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
              Mar 9 at 3:28






            • 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
              Mar 9 at 3:39






            • 2




              $begingroup$
              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
              Mar 9 at 4:19






            • 1




              $begingroup$
              Apologies I mispoke! I plotted the graph using the interplot library in R which I believe calculates the confidence intervals using critical t-statistics.
              $endgroup$
              – Clare P
              Mar 9 at 12:35







            2




            2




            $begingroup$
            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
            Mar 9 at 3:06




            $begingroup$
            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
            Mar 9 at 3:06




            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
            Mar 9 at 3:28




            $begingroup$
            @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
            Mar 9 at 3:28




            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
            Mar 9 at 3:39




            $begingroup$
            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
            Mar 9 at 3:39




            2




            2




            $begingroup$
            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
            Mar 9 at 4:19




            $begingroup$
            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
            Mar 9 at 4:19




            1




            1




            $begingroup$
            Apologies I mispoke! I plotted the graph using the interplot library in R which I believe calculates the confidence intervals using critical t-statistics.
            $endgroup$
            – Clare P
            Mar 9 at 12:35




            $begingroup$
            Apologies I mispoke! I plotted the graph using the interplot library in R which I believe calculates the confidence intervals using critical t-statistics.
            $endgroup$
            – Clare P
            Mar 9 at 12:35

















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