Ordinal Logistic Regression or Proportional Odds Logistic Regression with R 

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Thanks and welcome!

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You're so welcome!

You're welcome!

You are welcome!

Thanks for comments!

Yes, it should work for such data.

Will try

Thanks, I've added it to my list.

You are welcome!

One way could be to use p-value.

Seeing this today. Probably resolved by now.

Thanks for comments!

For confusion matrix you may refer to: ru-vid.com/group/PL34t5iLfZddvv-L5iFFpd_P1jy_7ElWMG Also note that when response is a factor variable, we do not use R-square.

You can use ordered() for independent ordinal variable. Some researchers also recommend changing then to numeric variable as it leads to much simpler model.

Yes, that would work fine.

The method depends on the dependent variable and not much on the independent variable.

Note that warning messages in R are ok. It's not an error.

Yes it is high. In the same example when we made a model with three variables, it was over 1700. By adding more variables it came down to about 1038, which is a significant improvement.

When you add more variables the AIC goes down, but then you select variables which have a significant level >0.1 and the AIC goes back up, isn’t it? Wouldn’t you use the model with the lowest AIC, and if not why use the AIC at all? Can I compare models with the AIC as well when in some models variables are log transformed as in others they are not log transformed?

1 | 2 means level-1 given level-2, and 2 | 3 means level-2 given level-3.

Dr. Bharatendra Could you please explain how to interpret the outcome of the dependent variable combined with | For example here is the summary and p-value of my model, I am struggling to interpreter the dependent variable outcome, TIA. Coefficients: Value Std. Error t value H 0.10955 0.06687 1.6381 AGR 0.05929 0.06825 0.8687 NP2 -1.00909 0.30407 -3.3186 NP3 -1.69956 0.40289 -4.2184 NP4 -0.28106 0.44589 -0.6303 Intercepts: Value Std. Error t value 1|2 -1.1571 0.6301 -1.8363 2|3 -0.0505 0.6090 -0.0829 3|4 0.9036 0.6022 1.5005 4|5 2.2627 0.7164 3.1584 5|6 5.1148 1.5859 3.2253 6|7 16.5213 9.1049 1.8145 Residual Deviance: 631.3888 AIC: 653.3888 Value Std. Error t value p-value H 0.10954539 0.06687426 1.6380799 0.1014 AGR 0.05928751 0.06825109 0.8686676 0.3850 NP2 -1.00909459 0.30407139 -3.3186107 0.0009 NP3 -1.69956102 0.40288860 -4.2184390 0.0000 NP4 -0.28105858 0.44589078 -0.6303306 0.5285 1|2 -1.15712803 0.63014735 -1.8362817 0.0663 2|3 -0.05048673 0.60902379 -0.0828978 0.9339 3|4 0.90356996 0.60219631 1.5004575 0.1335 4|5 2.26273192 0.71641548 3.1584073 0.0016 5|6 5.11484231 1.58585762 3.2252847 0.0013 6|7 16.52126027 9.10488998 1.8145480 0.0696

is goodness for fit test is ap plied in stata

It already includes test of significance.

Here is the link: goo.gl/Xc4G7J

It is just a warning message, not an error.

You are welcome!

I've not checked it in SPSS. But I guess results should be same.

I saw this today, hope it's taken care of.

Thanks!

@@bkrai Hello again Dr Rai, I tried to perform an OLR but the brant test assumption did not hold. Omnibus plus other variable were less than 0.05. What else should I do? is there any alternative test for ordinal dependent variables? Your kind advice will be greatly appreciated.

Thanks!

When a independent variable is categorical and takes three values, the correct way to represent it in a regression based model is with the help of 3-1=2 dummy variables. That's what you see here. When Tendency0 & Tendency1 are both zero, then Tendency = -1. When Tendency0 =1 & Tendency1 = 0, then Tendency = 0. When Tendency0 = 0 & Tendency1 = 1, then Tendency = 1. Note that in the equation Tendency0 & Tendency1 can only 0 or 1.

Bharatendra Rai in my case I used dummy variables of 1,2,3 for the three levels my independent categorical data. (Probably I should start with zero?) I converted them to factors. With some independent variables which were continous or categorical and the dependent variable, I ran the model using polr. The output gave me always a coffeficient value for the continous independent variables whereas the categorical ones had different coffeficient for each level. Like with yours Tendency 0 had different coffeficient and p values from Tendency 1 and both were significant. However, when I found the significancy of my data from their p values. I observed that the p value of the various levels differ in some variable (say e.g. edu with levels 1,2,3. R choose level 1 as reference level and so level 3 had value greater than 0.05 while level 2 had p value less than 0.05). I should remove the level 3 too as I remove the non significant variables from the equation I suppose. How can I do so and what may be the following interpretation. Thanks for your kind offer to help.

For categorical variables, even if one level is significant, do not drop the variable from the model.

Bharatendra Rai I sincerly appreciate your explanation. It is noted.

+Nasamu Bawa great 👍

+lauualb it was based on trial and error.

It is similar to steps shown in the link below at 4:13, ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-fDjKa7yWk1U.html

I have the same question. Only z-statistics' p-value can be calculated by pnorm() while hereby it is t-statistic.

Thanks for the suggestion, I'm adding it to my list for future.

See if this research paper helps: www.researchgate.net/publication/260273192_Does_Consumer_Behaviour_on_Meat_Consumption_Increase_Obesity_-_Empirical_Evidence_from_European_Countries

You can try this: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-dJclNIN-TPo.html

Send the codes that you used to look at.

mod

@Rahul Kadge could you find a solution for this error, I got the same and would like to know how you solved it. Thanks,

It estimates model related error. It is lower the better type of metric and helps to assess model quality. It is used for model selection or comparison.

Thanks for your comment!

Ordinal logistic regression is one type of ordinal regression.

ok.. what kind of ordinal regression you would suggest to a situation where, i have 15 features with 3 features integer, 3, numeric and 8 categorical (binary) and 1 count variable (dependent).. i followed logistic ordinal but not a better result.. i have zero inflated count and tried ZIP model too.. not that great.. ..and cumulative link model(clm) is not fitting as well..kindly suggest

what is your response variable?

@@bkrai it's count and also I tried with ranking it .. I have more zeros

Which part of the video are you referring to?

It must be some other issue. In this example I've used 21 variables without any problem.

@@bkrai Thanks! But the error I get is: attempt to find suitable starting values failed In addition: Warning messages: 1: glm.fit: algorithm did not converge 2: glm.fit: fitted probabilities numerically 0 or 1 occurred

Try Random Forest: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-dJclNIN-TPo.html

@@bkrai thanks again sir, grateful forever

You are welcome!

@@bkrai sir I am getting much error, can I have your mail id, please.

seemabharat@gmail.com

That's ok, we should still keep it.