I am very new to Statistic and Smartpls, hoping someone here can help me out.
I am building a model that is observing the relationship of the variable towards a variable that has only one indicator.
I read from this post viewtopic.php?f=5&t=3821
that this can be the variable observed and the indicator is equal to construct.
so that single indicator indicate the actual usage of a employee using company social media, which is the end goal of this thesis, trying to find out what variable influences most.
For that single indictor,
I have a path coefficients of 1.0 but when I do boostrapping it shows me 0.0.
since it is the variable observed , could it be that the indicator has 0 t value is ok ? (because it is equal to the construct?)
I tried to figure out if this is ok, since most of my variables are significant to that variable, the only thing is that indicator has 0 t value.
I have attached my results, could someone give me some feedback? thank you very much in advance!
Single indicator for the variable observed
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Single indicator for the variable observed
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- cringle
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Re: Single indicator for the variable observed
You can ignore the outer relationship of the single indicator construct. The latent variable and the indicator are identical. That's why the outer relation ship equals 1.0.
Best
Christian
Best
Christian
Prof. Dr. Christian M. Ringle, Hamburg University of Technology (TUHH), SmartPLS
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- Literature on PLS-SEM: https://www.smartpls.com/documentation
- Google Scholar: https://scholar.google.de/citations?use ... AAAJ&hl=de
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Re: Single indicator for the variable observed
I am getting Nan value instead of 1.000 for single indicator construct which is a categorical variable. I humbly request to kindly explain what could be the reason for this?
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Re: Single indicator for the variable observed
Single indicator construct have a fixed weight and loading of 1. As Christian has said previously: You can ignore the outer relationship of the single indicator construct. The latent variable and the indicator are identical.
Thus, there is no variance and therefore also no t or p-values to calculate. The t and p-value are just not defined and therefore reporting them is also not meaningful. Thus, you cannot get rid of the NaN.
We have updated this in the recent version of SmartPLS to better reflect this case.
Thus, there is no variance and therefore also no t or p-values to calculate. The t and p-value are just not defined and therefore reporting them is also not meaningful. Thus, you cannot get rid of the NaN.
We have updated this in the recent version of SmartPLS to better reflect this case.
Dr. Jan-Michael Becker, BI Norwegian Business School, 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