Anonymous Person adding 1 sd to 0 (the centered mean) will move the new zero point down relative to the original data. This is somewhat confusing, but if you make three sorted columns of data: raw data, mean-centered data and adding 1 sd data, you will see that adding the SD moves the new zero point down 1 sd relative to the mean centered data. Subtracting 1sd will move the zero point up relative to the mean centered data. With the interaction term in the equation the relationship between the focal predictor and the dv is interpreted at the 0 value of the moderator.
@@JasonPopan Where could I find more information on this explanation? Not quite clear what you mean by the 'zero point' and relative to the mean centred data. Is it something like, as -1.62 = 0, everything relative to it is higher?!
So based on your video, if we were to intepret it, does that also mean: Having pets (1SD above quantity worth of pet) actually buffers the negative effect of Kids raising your blood pressure, compared to having pets (1SD below quantity worth of pet)?
Thanks for this comprehensive video! One question though, in my analysis only the interaction effect is significant, not the main effects. I guess that because with your data pets was significant and children was not, you did simple slopes for high and low pets. But I am not sure which one of my main effects I am supposed to choose to do my simple slopes analysis? Does it matter or should I even do both?
You can do both. Ideally, you have a clear moderator in mind based on previous theory. Also, there are different techniques for probing interactions that might give a more detailed analysis of the differences between the slopes. I believe one such method is called Johnson-neyman
Thanks for the video! My moderator is gender coded 0=boys 1=girls. How do I do this procidure to produce slopes for boys and girls instead of 1 SD +/-?
Multiply to create the interaction term. The slope of the focal predictor will be for the group coded 0. To see the slope for the other group recode that one to zero and the other to one and repeat the process
Hey, looks like your response times are still super impressive despite this coming out two years ago! I am probing a significant two-way interaction, but my "high" and "low" conditions do not change the coefficient value of the "focal" variable (or any of the variables). I've double checked that my variable construction is correct (they are definitely plus and minus 1 SD), and my interaction is significant at p< .001. Any ideas? Thank you!
Verify that you recomputed the interaction term after re-centering your variable and that you used the correct interaction term in the analysis. Also, you may want to look into the Process Macro, which automates most of this and has different options for probing interactions.
I followed your video, but I got a slightly different result from using PROCESS macro. Is it normal? .45, .36 from spss regression; .46,.35 from process conditional effects
@@JasonPopan thank you sir, I found the correct figure. If I want to compare two regression slope, and show their degree of freedom, do I need to double up their degree of freedom? Or just follow the D=n-k-1?
You can get the estimate of the focal predictors relationship with the dv by mean centering. You probably did this already in the analysis prior to the simple slopes analysis. See this: ruclips.net/video/gr8-8LD3bKo/видео.html
I am a freelancer and this came handy while I am running this procedure. Thank you!
Why do you subtract 1SD to get "high" and add 1SD to get "low" rather than vice versa?
Anonymous Person adding 1 sd to 0 (the centered mean) will move the new zero point down relative to the original data. This is somewhat confusing, but if you make three sorted columns of data: raw data, mean-centered data and adding 1 sd data, you will see that adding the SD moves the new zero point down 1 sd relative to the mean centered data. Subtracting 1sd will move the zero point up relative to the mean centered data. With the interaction term in the equation the relationship between the focal predictor and the dv is interpreted at the 0 value of the moderator.
@@JasonPopan Where could I find more information on this explanation? Not quite clear what you mean by the 'zero point' and relative to the mean centred data. Is it something like, as -1.62 = 0, everything relative to it is higher?!
So based on your video, if we were to intepret it, does that also mean: Having pets (1SD above quantity worth of pet) actually buffers the negative effect of Kids raising your blood pressure, compared to having pets (1SD below quantity worth of pet)?
依變項是血壓，自變項是孩子數，調節變項是寵物數。高寵物數(1SD)下，孩子數不能預測血壓。低寵物數(1SD)下，孩子數可以顯著預測血壓。
Thanks for this comprehensive video! One question though, in my analysis only the interaction effect is significant, not the main effects. I guess that because with your data pets was significant and children was not, you did simple slopes for high and low pets. But I am not sure which one of my main effects I am supposed to choose to do my simple slopes analysis? Does it matter or should I even do both?
You can do both. Ideally, you have a clear moderator in mind based on previous theory. Also, there are different techniques for probing interactions that might give a more detailed analysis of the differences between the slopes. I believe one such method is called Johnson-neyman
Thanks for the video! My moderator is gender coded 0=boys 1=girls. How do I do this procidure to produce slopes for boys and girls instead of 1 SD +/-?
Multiply to create the interaction term. The slope of the focal predictor will be for the group coded 0. To see the slope for the other group recode that one to zero and the other to one and repeat the process
Hey, looks like your response times are still super impressive despite this coming out two years ago! I am probing a significant two-way interaction, but my "high" and "low" conditions do not change the coefficient value of the "focal" variable (or any of the variables). I've double checked that my variable construction is correct (they are definitely plus and minus 1 SD), and my interaction is significant at p< .001. Any ideas? Thank you!
Verify that you recomputed the interaction term after re-centering your variable and that you used the correct interaction term in the analysis. Also, you may want to look into the Process Macro, which automates most of this and has different options for probing interactions.
@@JasonPopan YES! Recomputing the interaction term with the high level and then the low level did the trick. Thank you so much!
1:13 linear regression
I followed your video, but I got a slightly different result from using PROCESS macro. Is it normal? .45, .36 from spss regression; .46,.35 from process conditional effects
Look into whether you've specified mean centering of the predictors.
@@JasonPopan thank you sir, I found the correct figure. If I want to compare two regression slope, and show their degree of freedom, do I need to double up their degree of freedom? Or just follow the D=n-k-1?
@@111nyg the interaction term will have n-k-1 DF as do the other predictors.
Which value shows the effect of the mean-level of pets? I just get how I can find the slopes for high- and low-level.
You can get the estimate of the focal predictors relationship with the dv by mean centering. You probably did this already in the analysis prior to the simple slopes analysis. See this: ruclips.net/video/gr8-8LD3bKo/видео.html