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I am running a binomial logistic regression in R. My colleague is running the same regression in SPSS. Our DV is dichotomous and data is non-normally distributed. The IVs are a mix of categorical and continuous.

When I compare our outputs, I notice that our estimate coefficients, standard error, and p values are the same. However, I get a z value in R, where she gets a Wald value. I have been reading and think that these are similar/the same with the exception that the Wald does not assume normality where the z does.

First question: Do I understand this relationship correctly?

Second question depending on answer to first question: If so, is there a way to compute the Wald statistic in R to assume non-normality when running the model?

If not, why would our outputs differ in this manner?

Example of what differs in our outputs:

My R:

Coefficients:
            Estimate Std. Error z value Pr(>|z|)  
(Intercept)  2.48462    1.17860   2.108   0.0350 *
age         -0.02332    0.01087  -2.145   0.0319 *
edu1         0.76923    1.16412   0.661   0.5087  
edu2         0.60235    1.12535   0.535   0.5925  
edu3         1.23647    1.13438   1.090   0.2757 

Colleagues SPSS:

        B     S.E.  Wald    df  Sig.    Exp(B)
age   -.022   .011  4.270   1   .039    .978
edu    .296   .201  2.173   1   .140    1.344
Constant2.661 .513  26.905  1   .000    14.314

Note: I understand the "scripting" nature of this question (R vs SPSS). However, I feel that the question more concerns the analytic output and the Wald/z statistic rather than the specifics of the programs.

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    $\begingroup$ See stats.stackexchange.com/questions/60074/… . It might be helpful. In this situation z^2 should equal Wald. Also in this case I believe that they are making the same assumptions. I think that the slight difference here is resulting from slight differences in B and SE if you care these out more digits. $\endgroup$
    – dbwilson
    Dec 16 '20 at 13:41

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