[Meng Yu]

So if the coefficient is 0.46 now, after the unit of the predictor changes from 0.001 to 0.01, the coefficient will become 4.6? And the odds ratio will be 99.48 (e to the power of 4.6).

[Clyde Schechter]

Yes.

[Meng Yu]

Thank you.

[Meng Yu]

According to one creator of this variable, the score should be considered as a continuous variable and should not be viewed as a discrete variable with a given unit. I wonder in interpreting the regression results, whether I can say something like "The increase in the score is associated with a xx% higher/lower odds of (whatever 1 stands for in the outcome)." Thank you.

[Clyde Schechter]

the score should be considered as a continuous variable and should not be viewed as a discrete variable with a given unit

I don't understand this. Discrete variables are simply numbers being used as designators of levels of the variable. When working with these variables we usually create dichotomous variables for each level (except one reference level) and in that case the meaning of a unit change in one of those dichotomous variables is the difference between the level that variable represents and the reference levels.

Continuous variables usually have units unless they are specifically constructed to be dimensionless. So for example, a variable denoting a velocity would have units like km/sec or mi/hr or something like that. But you can make a dimensionless velocity variable by using the ratio (in whatever units) of the velocity to the speed of light. Continuous variables that denote the probability of an event are also dimensionless: they are a pure number between 0 and 1.

The increase in the score is associated with a xx% higher/lower odds of (whatever 1 stands for in the outcome).

The odds of an outcome, like a probability, is a dimensionless variable, a pure number, this time with a range from 0 to infinity. The odds ratio for a logistic regression coefficient of a continuous variable is the rate of change of the odds of a positive outcome per unit of the continuous variable.

[Meng Yu]

esttab model* using "P:\....rtf", se(2) b(2) nomtitles eform stats (ll N aic)

I used the above syntax to produce logit regression results tables. I believe it gives me the odds ratio of the coefficients.

If the predictor is a categorical variable, say the odd ratio is 0.72 for one category, can I interpret the result as "Being in this category of the predictor is associated with a 28% decrease in the odds of (what 1 stands for in the outcome)?"

If the predictor is a continuous variable, say the odd ratio is 2.61, can I interpret the result as "With each unit of increase in the predictor, their odds of (what 1 stands for in the outcome) is 2.61 times higher?" Or "A one unit increase in the predictor is associated with a 161% increase in the odds of (what 1 stands for in the outcome).

This language score variable is derived from several ordinal and binary variables. They were rescaled to fall between 0 and 1. I believe this continuous variable has a unit although I don't know what the unit is.

[Clyde Schechter]

If the predictor is a categorical variable, say the odd ratio is 0.72 for one category, can I interpret the result as "Being in this category of the predictor is associated with a 28% decrease in the odds of (what 1 stands for in the outcome)?"

Being in this category of the predictor is associated with a 28% decrease in the odds of (what 1 stands for in the outcome) compared to being in (whatever the reference category for that predictor is).

If the predictor is a continuous variable, say the odd ratio is 2.61, can I interpret the result as "With each unit of increase in the predictor, their odds of (what 1 stands for in the outcome) is 2.61 times higher?" Or "A one unit increase in the predictor is associated with a 161% increase in the odds of (what 1 stands for in the outcome).

This would be correct given that you are referring to a logistic model. But it may not be the best way to say it. You indicate in your final paragraph that the language score variable is rescaled to fall between 0 and 1. So a 1 unit change in that variable corresponds to going from the lowest possible score to the highest possible score. My guess is that there are few, possibly no, observations with scores at these extremes of the range. (If there are, the measure is probably deeply flawed by ceiling or floor effects.) So differences of a unit magnitude will be very unusual events. It would probably be better to present the change in odds associated with a smaller difference. If, for example, a difference of 0.5 units in the score is fairly common in the data, then you could say that a difference of 0.5 units in the language score variable is associated with a 62% increase in the odds of (whatever 1 stands for in the outcome). The 62% figure is arrived at as follows. Given that a 1.0 difference in score is associated with an OR of 2.61, a 0.5 difference will be associated with an OR of 2.61^0.5, which is 1.62. An odds ratio of 1.62 means a 62% increase in the odds.

[Meng Yu]

Thank you very much for your reply. According to the accompanying document of the score variable, the creators assigned equal weight (a weight of 1) to all components of the score and they used Cronbach’s alpha coefficient to test the internal consistency of the components. The average of one score at one wave is 0.62 and there are some people with 0 or 1 point. Since this is a host country language score for immigrants, it is possible for some people to get 0 or 1. But you are definitely right, the majority fall between 0 and 1.

Using 0.5 as a unit is a good idea. But I don't understand why 2.61^0.5? Why not 2.61/2?

[Clyde Schechter]

But I don't understand why 2.61^0.5? Why not 2.61/2?

OK, this is a logistic model. So log odds outcome | x1 = b0 + b1*x1 + ..., right? Here x1 is the score variable, OR for x1 is 2.61, and so b1 = log(2.61). Suppose we have two observations whose values of x1 differ by 0.5. Say x1' = x1 + 0.5. Then:

log odds outcome | x1' = b0 +log(2.61)*(x1+0.5) + ...

    = b0 +log( 2.61)*x1 + log(2.61)*0.5

    = (log odds outcome | x1) + log(2.61)*0.5

Now we exponentiate both sides applying the rules for log of a product and log of a power:

odds outcome | x1' = (odds outcome | x1) * exp(log(2.61)*0.5)

    = (odds outcome | x1) * exp(log(2.610.5))

    = (odds outcome | x1) * 2.610.5

[Meng Yu]

(log odds outcome | x1) + log(2.61)*0.5 = log (odds outcome | x1 * (2.61)*0.5) But in your second code, log appeared before (2.61)*0.5. I don't get that part.

Also in logistic regression, the dependent variable is the natural log of the odds. Then should it be ln odds outcome | x1'? I am confused by the difference between log and natural log in this situation.

[Clyde Schechter]

(log odds outcome | x1) + log(2.61)*0.5 = log (odds outcome | x1 * (2.61)*0.5)

That does not appear in my post. I do recall that after hitting "Post Reply" I noticed a typographical error in one of the equations, which I immediately changed in an edit. Perhaps that was the one, and perhaps you read the post before the edit appeared? I don't know. Anyway, there is no such text in the corrected post.

Also in logistic regression, the dependent variable is the natural log of the odds. Then should it be ln odds outcome | x1'? I am confused by the difference between log and natural log in this situation.

Sorry for the confusion. I do not use the ln notation. When I write log, I always mean the natural logarithm. In the event I needed to refer to logarithm base 10, I would write it as log₁₀. I realize that this is contrary to the conventions that many others follow, and I regret the confusion it sometimes causes.

[Meng Yu]

In your post #69, exponentiate your equation in code 1, we get

 odds outcome | x' = exp (log odds outcome | x1 +(log 2.61)*0.5)

Then you are right in your equations in code 2.

But why does (odds outcome | x1) equal 1?

[Meng Yu]

Is it because the outcome is a binary variable and equals 1 in this situation? The odds of 100% is 1.

[Clyde Schechter]

But why does (odds outcome | x1) equal 1?

It doesn't. I left it to you to finish the derivation of the odds ratio. But let me write it out:

The last equation in #69 says:
odds outcome | x1'= (odds outcome | x1) * 2.610.5

Now divide both sides by (odds outcome | x1):
Odds Ratio = (odds outcome  | x1')/(odds outcome | x1) = 2.610.5

[Meng Yu]

My understanding is if 2.61 is the odds ratio of the outcome when there is a one unit change in the predictor, then for a half a unit change, the log odds outcome should be

 log odds outcome | x1' = b0 +log(2.61)*(x1-0.5)