Python Proportion test similar to prop.test in R

Question

I am looking for a test in Python that does this:

> survivors <- matrix(c(1781,1443,135,47), ncol=2) > colnames(survivors) <- c('survived','died') > rownames(survivors) <- c('no seat belt','seat belt') > survivors              survived died no seat belt     1781  135 seat belt        1443   47 > prop.test(survivors)      2-sample test for equality of proportions with continuity correction  data:  survivors X-squared = 24.3328, df = 1, p-value = 8.105e-07 alternative hypothesis: two.sided 95 percent confidence interval:  -0.05400606 -0.02382527 sample estimates:    prop 1    prop 2  0.9295407 0.9684564  

I am mostly interested in p-value calculation.

The example is taken form here

Answer 1

I think I got it:

In [11]: from scipy import stats  In [12]: import numpy as np  In [13]: survivors = np.array([[1781,135], [1443, 47]])  In [14]: stats.chi2_contingency(survivors) Out[14]:  (24.332761232771361,       # x-squared  8.1048817984512269e-07,   # p-value  1,  array([[ 1813.61832061,   102.38167939],        [ 1410.38167939,    79.61832061]])) 

Answer 2

Adding to @Akavall's answer: If you don't explicitly have the "failure" counts (# of deaths in your example), R's prop.test lets you specify just the total number of trials e.g. prop.test(c(1781, 1443), c(1781+135, 1443+47)) would give you the same results as with the contingency table you built.

Scipy's chi2_contingency explicitly ask for the failure counts and the complete contingency tables. If you don't explicitly have the failure counts and just want to check whether the proportion of successes out of the total are equal for two samples, you can hack around scipy's function with

survivors = np.array([[1781, total1 - 1781], [1443, total2 - 47]]) chi2_contingency(survivors)  # Result: (24.332761232771361, 8.1048817984512269e-07, 1, array([[ 1813.61832061,   102.38167939],            [ 1410.38167939,    79.61832061]])) 

Took me some time to figure this one out. Hope it helps someone.