Say you are interested in predicting whether somebody is a fan of Justin Bieber according to the amount of beer they have consumed as well as their gender. Bieber fever is coded 1 if the respondent is a fan and zero otherwise. Because the dependent variable is dichotomous, the appropriate method is logistic regression.

The logistic regression (or logit) model is linear in the log odds of the dependent variable.

Most people don’t think in terms of log odds, so it’s common to interpret the results either by exponentiating coefficients to yield odds ratios, or else by computing predicted probabilities. An odds ratio greater than one means that an increase in X leads to an increase in the odds that the dependent variable equals one; an odds ratio less than one means that the odds are decreasing. The predicted probabilities can be calculated using the formula for the cdf to the standard logistic distribution:

Dear Adriana Santos-Caballero, Thanks for your answer. I understand the concept and assumptions of logistic regression and how to run the test in SPSS. My question was how to deal with the results if one or more of the 2x2 cells have a zero value in SPSS??? The results (extremely high beta and odds ratio) seem inappropriate to use. I get the following error message under the "model summary" table: “Estimation terminated at iteration number 20 because maximum iterations has been reached. Final solution cannot be found.”

Dear Heba,

Have you considered using bootstrapping? It can give you a bias corrected estimate of your confidence intervals. It is recommended when your residuals are not normally distributed.

Hi please explain what one categorical predictor (2x2) means. Best, David

Dear Peter,

Thanks for your answer. Yes, I’ve tried bootstrap. Can I use the standard error (SE) and confidence interval limits from the bootstrap for variables in the equation? And what odd ratio value should I report?

Hi David, sorry for the confusion.

I mean an independent variable which is a dichotomous nominal variable.

It yields a table with one or two zero value cells which causes the error in the calculations. I've run the regression with this variable alone to show the zeros in the classification table (2x2 table).

I wonder if Firth Logistic Regression would work better? There is an extension command that can be used to perform it with SPSS. And you can read about it in some nice notes by Richard Williams. (He describes a Stata command for doing it, but his general discussion should still be helpful to you.)

• https://www.ibm.com/support/knowledgecenter/en/SSLVMB_24.0.0/spss/programmability_option/r_package_installed_extensions.html
• https://www3.nd.edu/~rwilliam/stats3/RareEvents.pdf

Dear Bruce, Thanks for your answer. I read the links you've kindly provided. Firth logistic regression is indeed a solution for the analysis of a 2x2 table with one zero cell count. However, I've been trying to install SPSS extensions for R but it seems so complicated. I'll try again though.

Hello Heba. For questions about how to do X in SPSS, you'll likely get more help if you post to the SPSS mailing list. One way to join is via the Nabble archive:

• http://spssx-discussion.1045642.n5.nabble.com/

Click on the more options link at the top of the page to join. HTH.

Dear Heba,

You need to take the exponent of the B coefficients to calculate your odds ratios. You need to do the same with the bootstrapped confidence intervals for these coefficients when reporting on your results. Bootstrapped confidence intervals are an alternative to standard errors.

I agree with Adriana

Dear Heba Yousry Raslan,

Could you please explain in more details or suggest me some references why you said that 'Firth logistic regression is indeed a solution for the analysis of a 2x2 table with one zero cell count'

I think zeros in 2x2 contingency table and data separation are two different issues and Firth method is only for data separation. How Firth method deals with zeros in 2x2 contingency table?

Dear Yên Trần,

You may read the links provided above by Bruce for clarification. Here's a link for a research paper "An alternative method for logistic regression on contingency tables with zero cell counts" that explains this situation in details:

https://www.google.com/url?sa=t&source=web&rct=j&url=http://rdo.psu.ac.th/sjstweb/journal/38-2/38-2-8.pdf&ved=2ahUKEwiK55Gf4fDjAhVN-aQKHUr3CasQFjAAegQIAxAB&usg=AOvVaw1aCEbqe5v8GGwsj2AsnIYT

Hope you find this useful.

Heba

Dear Heba

Thank you for your quick reply. I have read the paper you suggested and found the text 'The Firth procedure is the current method of choice for logistic regression in tables with zero cell counts (Heinze, 2009)'

As you mentioned 'one zero cell count' in your previous post, I would like to discuss this issue in a little more detail.

Given that we have a simple logistic regression model where the dependent variable y is predicted by the variable x and complete data separation occurs, such as in the contingency table below,

(X=0) (X=1)

(Y=0) 10 0

(Y=1) 0 5

It is intuitive that there will exist infinite sets of estimates as X perfectly predict Y (e.g., X always takes 1 when Y=1 and takes 0 when Y=0).

However, consider the following contingency table:

(X=0) (X=1)

(Y=0) 10 1

(Y=1) 0 5

The above table has only one cell with zero as you said.

When X=0, we can see that Y always takes 0 but when X=1, Y can takes both values 0 and 1. This case is a quasi-data separation, is it right? In this case, X nearly perfectly predict Y. So how can we know that there is the possibility of infinite sets of estimates in the case of quasi-data separation above?

Tran Viet Yen.