Redundancy Analysis  Negative path coefficient

 PLS Senior User
 Posts: 21
 Joined: Fri Sep 18, 2015 6:40 pm
 Real name and title: Ms Priyanka
Redundancy Analysis  Negative path coefficient
Hi there,
For the formative measurement model, I conducted redundancy analysis. So I made the formative construct the IV and the reflective version of the same construct as a DV.
I was getting a positive path coefficient when my sample was 140. When I ran the data again with a sample of 200, my redundancy analysis for MSP gave a negative path coefficient.
Why is this happening? I don't understand. I searched through the forum. But I still don't understand what is happening.
Thanks in advance!
Regards,
Priyanka
For the formative measurement model, I conducted redundancy analysis. So I made the formative construct the IV and the reflective version of the same construct as a DV.
I was getting a positive path coefficient when my sample was 140. When I ran the data again with a sample of 200, my redundancy analysis for MSP gave a negative path coefficient.
Why is this happening? I don't understand. I searched through the forum. But I still don't understand what is happening.
Thanks in advance!
Regards,
Priyanka
 Attachments

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Re: Redundancy Analysis  Negative path coefficient
Hi
I have same question.
I have a negative path coefficient with a high tvalue, I wan to know this hypothesis is supported or not?
can be any mediation between Dis > Satisfaction > creativity ?
the effect size of distraction to satisfaction is 0.470 which is high...
I really appreciate if someone can help me in this regard
I have same question.
I have a negative path coefficient with a high tvalue, I wan to know this hypothesis is supported or not?
can be any mediation between Dis > Satisfaction > creativity ?
the effect size of distraction to satisfaction is 0.470 which is high...
I really appreciate if someone can help me in this regard
 Attachments

 88  Csopy.jpg (35.79 KiB) Viewed 8414 times

 PLS Senior User
 Posts: 21
 Joined: Fri Sep 18, 2015 6:40 pm
 Real name and title: Ms Priyanka
Re: Redundancy Analysis  Negative path coefficient
Hi Sanaz,
I wasn't looking at the structural model. My question was related to the measurement model.
In your case, I can't comment on the mediation.
From what I understand, your question is regarding the structural model.
Your path coefficient shows a negative value. Meaning: Distraction and satisfaction share a negative relationship. The path coefficent indicates the strength of the relationship.
While the tvalue (done through bootstrapping) is used to test whether the relationship being tested is significant.
Since you are testing mediation, you should be doing a 2 tail test.
Depending on the sig level you decided (I usually pick 0.05). You will run a bootstrap (which Im sure you have done to get these results)
Whether it is a two tail or one tail matters because should you decide your significance level is 0.05.
You will type in 0.05 in bootstrapping for a two tail, but for one tail I believe it is a diff value (but your sig level is still 0.05).
Once you have done the bootstrapping correctly. I usually pick 5000 subsamples. Then I get t critical value on excel = TINV(0.05, 5000) =1.960438593
And if my t value is higher than the t critical value. The the relationship is significant.
In your case, based on my understanding, your hypothesis is supported.
Any PLS Professionals, please free to correct me.
I am still learning. But that's what I understand.
Hope it helps,
Priyanka
I wasn't looking at the structural model. My question was related to the measurement model.
In your case, I can't comment on the mediation.
From what I understand, your question is regarding the structural model.
Your path coefficient shows a negative value. Meaning: Distraction and satisfaction share a negative relationship. The path coefficent indicates the strength of the relationship.
While the tvalue (done through bootstrapping) is used to test whether the relationship being tested is significant.
Since you are testing mediation, you should be doing a 2 tail test.
Depending on the sig level you decided (I usually pick 0.05). You will run a bootstrap (which Im sure you have done to get these results)
Whether it is a two tail or one tail matters because should you decide your significance level is 0.05.
You will type in 0.05 in bootstrapping for a two tail, but for one tail I believe it is a diff value (but your sig level is still 0.05).
Once you have done the bootstrapping correctly. I usually pick 5000 subsamples. Then I get t critical value on excel = TINV(0.05, 5000) =1.960438593
And if my t value is higher than the t critical value. The the relationship is significant.
In your case, based on my understanding, your hypothesis is supported.
Any PLS Professionals, please free to correct me.
I am still learning. But that's what I understand.
Hope it helps,
Priyanka
Re: Redundancy Analysis  Negative path coefficient
Than you so much for your replay and information
 Hengkov
 PLS SuperExpert
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 Joined: Sun Apr 24, 2011 10:13 am
 Real name and title: Hengky Latan
 Location: AMQ, Indonesia
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Re: Redundancy Analysis  Negative path coefficient
Hi,
Path coefficient is positive or negative depends on the hypothesis that you wake up and the relationship between variables. If you expect positive, but the result is negative (with large tstatistics), your hypothesis is rejected.
For the case of the addition of a sample of 140 to 200 changes the sign on the path coefficients, the data that you add to contain high variability.
Best,
Path coefficient is positive or negative depends on the hypothesis that you wake up and the relationship between variables. If you expect positive, but the result is negative (with large tstatistics), your hypothesis is rejected.
For the case of the addition of a sample of 140 to 200 changes the sign on the path coefficients, the data that you add to contain high variability.
Best,

 PLS Junior User
 Posts: 1
 Joined: Mon Dec 14, 2015 10:22 am
 Real name and title: Richner Dominique
Re: Redundancy Analysis  Negative path coefficient
Hi, I also have a question concerning the redundancy analysis:
My pathcoefficents are below 0.8. However I have a lot of items and the previous study (from where I got the items) omitted several of those items. How do I decide which ones to omit? Is there no statistically measured critical value or something? (In the literature I only find that it has theoretically be decided, but I'm really having troubles in doing just so)
Best regards
Domae
My pathcoefficents are below 0.8. However I have a lot of items and the previous study (from where I got the items) omitted several of those items. How do I decide which ones to omit? Is there no statistically measured critical value or something? (In the literature I only find that it has theoretically be decided, but I'm really having troubles in doing just so)
Best regards
Domae

 SmartPLS Developer
 Posts: 1094
 Joined: Tue Mar 28, 2006 11:09 am
 Real name and title: Dr. JanMichael Becker
Re: Redundancy Analysis  Negative path coefficient
Rules for dropping indicators are not straightforward. First, theory plays an important role in defining the set of indicators. If an indicator is theoretically not important, then it should not be part of the formative measurement model. Second, empirical assessment of statistical significance can help determine indicators that are not related to the construct. Weights should ideally be significant, but as the number of formative indicators increases, the magnitude of the weights decreases. Hence, with many indicators there are always indicators with low weights that are not significant. Does this mean they are not important and should be dropped? No. They may still contribute to the construct. But remember the weights represent relative contribution. The loadings, however, represent absolute contribution (ignoring all the other indicators). Hence, if the loadings are small or not significant, then the indicator should be dropped. If weights are small and not significant, the indicator may still have a high loading (absolute contribution).
Nevertheless, dropping indicators will not help you increase your R² in the redundancy analysis. Dropping indicators should always reduce the R² as PLS will construct the formative LV as the one that best explains the dependent construct. If fewer indicators are available, the explained variance should always be smaller. However, that should not mean that you retain indicators that are not important for the formative construct.
Nevertheless, dropping indicators will not help you increase your R² in the redundancy analysis. Dropping indicators should always reduce the R² as PLS will construct the formative LV as the one that best explains the dependent construct. If fewer indicators are available, the explained variance should always be smaller. However, that should not mean that you retain indicators that are not important for the formative construct.
Dr. JanMichael Becker, University of Cologne, SmartPLS Developer
Researchgate: https://www.researchgate.net/profile/Jan_Michael_Becker
GoogleScholar: http://scholar.google.de/citations?user ... AAAJ&hl=de
Researchgate: https://www.researchgate.net/profile/Jan_Michael_Becker
GoogleScholar: http://scholar.google.de/citations?user ... AAAJ&hl=de

 PLS Senior User
 Posts: 21
 Joined: Fri Sep 18, 2015 6:40 pm
 Real name and title: Ms Priyanka
Re: Redundancy Analysis  Negative path coefficient
Hey all,
Firstly thanks for your insights Dr. JanMichael Becker!
I have one more question.
Why is there are a negative value in the MMSPR2 item on the left construct (which is the formative construct) after running PLS Algorithm? (please refer to image attached)
The item is measured on a 7 point likert scale. I recoded the item in SPSS (so the value 7 becomes 1, 6 becomes 2 and so on) prior to moving data to SMARTPLS. Because MMSPR2 was inverse of the other items in the questionnaire.
However it is negative after I run redundancy analysis.
Why?
What does a negative value mean?
Even when I use the original item without recoding it, MMSPR2 still comes negative. Should I be worried.
I really don't know what the negative value means. I would appreciate any insight from the community on this.
Many thanks,
Priyanka
Firstly thanks for your insights Dr. JanMichael Becker!
I have one more question.
Why is there are a negative value in the MMSPR2 item on the left construct (which is the formative construct) after running PLS Algorithm? (please refer to image attached)
The item is measured on a 7 point likert scale. I recoded the item in SPSS (so the value 7 becomes 1, 6 becomes 2 and so on) prior to moving data to SMARTPLS. Because MMSPR2 was inverse of the other items in the questionnaire.
However it is negative after I run redundancy analysis.
Why?
What does a negative value mean?
Even when I use the original item without recoding it, MMSPR2 still comes negative. Should I be worried.
I really don't know what the negative value means. I would appreciate any insight from the community on this.
Many thanks,
Priyanka
 Attachments

 Redundancy Analysis.png (8.66 KiB) Viewed 8269 times

 SmartPLS Developer
 Posts: 1094
 Joined: Tue Mar 28, 2006 11:09 am
 Real name and title: Dr. JanMichael Becker
Re: Redundancy Analysis  Negative path coefficient
The weight looks very small. Is it significant? If not, the sign does not convey any information. It might be a chance finding, especially, as recoding also leads to a negative weight (probably also very small / insignificant?).
Dr. JanMichael Becker, University of Cologne, SmartPLS Developer
Researchgate: https://www.researchgate.net/profile/Jan_Michael_Becker
GoogleScholar: http://scholar.google.de/citations?user ... AAAJ&hl=de
Researchgate: https://www.researchgate.net/profile/Jan_Michael_Becker
GoogleScholar: http://scholar.google.de/citations?user ... AAAJ&hl=de

 PLS Senior User
 Posts: 21
 Joined: Fri Sep 18, 2015 6:40 pm
 Real name and title: Ms Priyanka
Re: Redundancy Analysis  Negative path coefficient
Thank you for your input Dr. JanMichael Becker!
I decided to remove the negative item from the construct. After much discussion with my supervisors, I felt the item was open to being misunderstood by the respondents.
Many thanks,
Priyanka
I decided to remove the negative item from the construct. After much discussion with my supervisors, I felt the item was open to being misunderstood by the respondents.
Many thanks,
Priyanka