ISSUE: Should normality be tested for probit model?

PROBIT: Probit (probability + unit = probit) is used when the Y (dependent variable) is nominal. The relationship is proved through the following:

(1)   Y = XB

… where Y = categorical answer (Yes | no), X = vector regressor; and B = parameter of X estimated by maximum likelihood.

TEST FOR RELATIONSHIP:  The decision rule is: H0: B = 0 and Ha: B > 0 or B < 0.

Assume that B > 0 or B < 0, then reject the null hypothesis. This is not the end of the analysis. Proceed to testing for the intensity of B by calculating the correlation coefficient.

TEST FOR INTENSITY OF RELATIONSHIP: The testing for correlation coefficient depends on the classification of the data on dependent (Y) and dependent (X) variables. There are three possible combination pairs: quantitative (Q), ordinal (O), and nominal (N). Therefore, the possible pairs are and their corresponding correlation coefficient types are given as:

(Qy,Qx) = Pearson r

(Oy,Qx) = Biserial r

(Ny,Qx) = Point biserial r

(Qy,Ox) = Biserial r

(Oy,Ox) = Spearman rho

(Ny,Ox) = Rank biserial

(Qy,Nx) = Point biserial

(Oy,Nx) = Rank biserial

(Ny,Nx) = Phi, lambda, or Pearson contingency C

Of all 9 possible pairs, only QQ scenario requires normality verification. The rest mostly accomplished through frequency probability do not require normality test.

   The test for the intensity of the relationship is accomplished through correlation coefficient. The type of correlation coefficient to be used depends on the type of data classification for Y and X. In Probit model, Y is binary data. Binary data may be either nominal or ordinal. If Y is nominal, the data sought is the type that identifies, i.e. sex, age, education, etc. If Y is ordinal, the data sought is a specific category, i.e. first place winner. In Probit, the Y could be only one of these two types: ordinal or nominal. Therefore, the test of relationship depends on the X-combination: quantitative (Q), ordinal (O) or nominal (N).

   In a case of QQ crossing where both X and Y are quantitative, the Pearson r is used and the t-test for the Pearson r is based on t-test---which requires a verification of normal distribution. However, Probit is a non-QQ scenario.

TEST FOR STATISTICAL SIGNIFICANCE OF THE INTENSITY LEVEL: In general, if the Y (dependent variable) is quantitative and the X (independent) is also quantitative, the test for significance in the correlation is given by a t-test, thus:

(2)   t = r (sqrt(n – 2)) / (Sqrt(1 – r2)

… where r = Pearson correlation coefficient; and n = sample size. In this case, in order to use the t-test, a normal distribution verification is required.

However, in probit modeling, the Y (dependent variable) is not quantitative; it is either “nominal” or “ordinal” in a form of binary data: (Yes | No) where the score is assigned as (Yes = 1; No = 0). Therefore, the type of correlation used to verified the intensity level of relationship in the model is one of the followings depending on the classification of X (independent variable): (i) biserial if Y ordinal and X quantitative; (ii) point biserial if Y nominal and X quantitative; (iii) spearman rho if both Y and X are ordinal; (iv) rank biserial if Y nominal and X ordinal; (v) rank iserial if Y ordinal and X nominal; (vi) phi, lambda or Pearson contingency C if Y nominal and X nominal. These six cases do not require a test for normality.

TEST COMMONLY NEGLECTED: Many researchers do not confirm the test for normality after finding there is a relationship and there is a moderate or high level of correlation coefficient. This is an incomplete analysis. The correlation must be statistically significant. Reporting R-squared is not a substitute either. However, this confirmation is required only in QQ scenario. In Probit where Y = nominal, it is not required.

Thanks Paul. I realized that probably my question was not so clear. My question was on the assumptions of normality of the link function in the Probit context. Some authors sustain that it must be verified, other as Wooldrige say (semplifying a lot the argument) look if we run a probit, but instead the true model is a logit, we have for sure an inconsistent B, but if we are interested in the partial effect of a predictor on the probability that our observed dependent is equal to one, this inconsistency is in practice irrelevant. Therefore, I was wondering if is it fair to simply offer different model specification (logit, probit, lpm) to show that our results do not depend on link function choice or we have to offer a more formal test as for instance Jarque-Bera-Lee LM test?

You are very unlikely to be able to distinguish between a logit and probit link as the functional form of both is so similar. Indeed a common rule of thumb is to take the logit estimated coefficients and divide by 1.6 to get probit coefficients see Amemiya (1981) Qualitataive response variables: a summary; download from

http://webkuliah.unimedia.ac.id/ebook/files/amemiya%281981%29%20logit%20tobit.pdf

Birnbaum show that the logistic function differs by less than 0.01 from the normal cdf with mean zero and standard deviation 1.7.Consequently in item response models the two are used interchangeably - the probit being called the normal ogive model

Birnbaum, A. (1968). Some latent trait models and their use in inferring an examinee's ability. In F.M. Lord and M.R. Novick (Eds.), Statistical theries of mental test scores. Addison-Wesly. 395-479.

The logistic has the advantage of being easily tuned into odds ratio while the probit (based as it is on an underlying normal distribution) makes for easier simulation in Bayesian MCMC estimation of complex discrete outcome models - the so called underlying normal trick

The original paper on this is  James H. Albert and Siddhartha Chib 1993, Bayesian Analysis of Binary and Polychotomous Response Data, Journal of the American Statistical Association Vol. 88, No. 422 (Jun., 1993), pp. 669-679

Thank you very much Kelvin and Rami. 

...although, it is imperative to understand a Probit is based upon a normal distribution, hence, if your data set is not normal, then you should think about abandoning such models. It is all-too-often that statistical methods are applied to non-normal sets, hence, all inferences derived therefrom are invalid.

In other words, you ask, "Should I test for normality...," and the answer is "Yes." The reason is to ensure the inferences drawn from the model are valid. If normality or some other essential statistical condition should fail, then it is an indication that all statistical inferences drawn from that set are not valid and should be ignored. 

Let me add that I am very familiar with Probit models, for I applied them to skin and ocular damage caused by laser energy. It is quite often that the Probit is misinterpreted, where most interpret such metrics to mean, "The probability for x to occur;" but, the correct interpretation is, "The probability to observe or detect x." These two interpretations may seem the same at first, but careful consideration will prove them dramatically different. The implication for the latter is that damage may have occurred, but I was unable to detect it. The former implies you know the entire probability space in its entirety.  

Now, it is true that the distinction between the two interpretations collapse onto each other in the case for lethal doses, for death is not uncertain; but, for estimated dose, it is critical to make this distinction. Your study involves nominal, qualitative differences and is not so cut and dry as life or death. 

You might find this useful: http://davegiles.blogspot.com/2015/06/specification-testing-in-ordered-probit.html

Thank you Paul. The blog of David Giles is very useful, he considers a lot of topic and all are explained in a very clear and concise way.