# What is the difference between z-value and the Wald statistic in the summary function of the Cox Proportional Hazards model of the “survival” package?

When I run the code posted at the bottom (using the summary() function of the R survival package, I get the output shown immediately below:

Some sources (http://www.sthda.com/english/wiki/cox-proportional-hazards-model) state that the z-value is the “Wald statistic value” and continues “It corresponds to the ratio of each regression coefficient to its standard error (z = coef/se(coef)). The Wald statistic evaluates whether the beta (β) coefficient of a given variable is statistically significantly different from 0. From the output above, we can conclude that the variable sex have highly statistically significant coefficients.

On the other hand, other sources state “The z-value is a standardized score that measures the number of standard deviations a parameter estimate is from its null hypothesis value. It is calculated by dividing the estimated coefficient by its standard error. The z-value is used to calculate p-values and to assess the statistical significance of the coefficient. The Wald statistic, on the other hand, is a measure of the overall significance of a variable in the Cox proportional hazards model. It is calculated by dividing the squared coefficient estimate by its estimated variance. The Wald statistic is used to test the null hypothesis that the coefficient of a variable is equal to zero, which indicates that the variable is not a significant predictor of the outcome. In summary, the z-value is used to assess the statistical significance of individual coefficients, while the Wald statistic is used to test the overall significance of a variable in the Cox proportional hazards model. Both are important measures in assessing the validity and usefulness of a Cox proportional hazards model, but they serve different purposes.”

Which, if either, description of the z-value and Wald statistic is correct?

Code:

library(survival)
library(survminer)

res.cox <- coxph(Surv(time, status) ~ sex, data = lung)
summary(res.cox)

• Wald = Z^2 and is compared to a chisq(1) DF instead of a standard normal. Commented Mar 15, 2023 at 21:38

"[D]ividing the squared coefficient estimate by its estimated variance" gives a statistic evaluated against a chi-square distribution with 1 degree of freedom. That's just the square of the z-statistic in your display, which is evaluated against a standard normal distribution. $$(-3.176)^2=10.09$$