Hello all,
How would one acquire an Odds ratio per standard deviation (OR per SD) increase/decrease of an explanatory variable?
The link from the IBM website below explains that one may use OR per SD if the value of exp(B) is too close to 1 to be meaningful, by "simply [standardizing] the variable, so that a unit change is one standard deviation."
I am confused how this might be implemented in SPSS. Let's say I have an explanatory variable (money) in a regression model with an outcome variable (happiness) for example. If I were to follow the definition of OR per SD above, that would mean a dollar change would be a change in SD.
But what if my money variable had a standard deviation value of, say, 200. How wold I use this value as the SD in a regression model?
http://www-01.ibm.com/support/docview.wss?uid=swg21479543
Here's one fairly quick & easy way. Use the following SPSS command to save a new variable Zmoney to your working dataset:
DESCRIPTIVES VARIABLES=money /SAVE.
Then estimate your logistic regression model with Zmoney in place of money. The Exp(B) column in the table of coefficients will show the odds ratio for a one-SD increase in money.
HTH.
Here's one fairly quick & easy way. Use the following SPSS command to save a new variable Zmoney to your working dataset:
DESCRIPTIVES VARIABLES=money /SAVE.
Then estimate your logistic regression model with Zmoney in place of money. The Exp(B) column in the table of coefficients will show the odds ratio for a one-SD increase in money.
HTH.
Bruce's solution is excellent and should solve your problem. I would clarify however that your new z-scores, or standard deviation units of money, are standardized to your particular dataset and not some external standard. You could alternatively standardize your z-scores to some external standard, for example the national income distribution. In that case, your z-scores would indicate how many national standard deviation units your subjects are away from the national mean income level, rather than from the mean of the incomes in your dataset in your dataset's SD units. This would make your z-scores comparable to other studies by other researchers, if you both used the national income as your reference standard. This is commonly done for some medical parameters such as weight and height, so study populations in different areas can be compared to each other in terms of SD units (or z-units) of the standard reference population. Otherwise, a standard deviation of income for people in Beijing will be different from a standard deviation of income for people in Shanghai.
Also, with the method that Bruce mentioned, it assumes that the change in Happiness units is the same from when you go from the mean income to 1 SD in Income as it is when you go from 1 to 2 SDs in income, or from 2 SDs to 3 SDs, which might not always be true.
Jerry, excellent point about the distinction between the SD computed from the sample data and some external (e.g., national) SD. I agree that it is often preferable to standardize to some such external standard if one is available. (I often make the same kind of argument about centering variables on some convenient value that can be used across studies rather than centering on the mean, or some other value, computed for each sample separately.)
Re the final point, using the Z-score variable has not changed anything in that regard. If you include a continuous variable in the model (with no polynomial terms etc.), the odds ratio for that variable is for a one-unit increase at any point along the scale. (By using a z-score variable, all you do is change one unit from the original units to SD units.) If this is not a reasonable assumption, then one has to do something else (e.g., introduce appropriate polynomial terms, splines, etc.)
HTH.
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