Comparing the means of 6 groups, binary dependent variable I accomplished a field experiment which results in six differently treated groups (including one control group, each n = 300). Due to randomization all groups are expected to be equal in terms of their characteristics (average price, average customer age, etc.). The only difference is supposed to be the treatment each group received. The resulting dependent variable (outcome) is binary (1/0). 
How can I now compare the six groups in terms of their binary outcome? I want to know the respective probability of a treatment leading to outcome 1.
Is a Kruskal-Wallis test adequate or should I rather use logistic regression? Using logistic regression also involves control variables although these should be distributed equally among all groups?
Thank you very much!
 A: The way to do this is with logistic regression with the group being the independent variable. You probably will want to control for other variables too.
A: You can organize the data as a 2 by 6 contingency table. Descriptively you can present this table (absolute counts) and the proportion of "1" (relative counts) per group. Different proportions highlight treatment differences. If it makes sense to test the null hypothesis of equal group proportions, you can e.g. use a classic chi-squared test of association (which is similar to the likelihood ratio test of the logistic regression). Maybe you rather want to compare each of the five test groups against the control. Then you can go for five tests. Another strategy is to test each treatment against each other (again e.g. with chi-squared test). Anyway you will need a prespecified strategy to correct for multiple testing (or not), like Bonferroni-Holm. It really depends on the research question.
Some example R code
# input: prepare data and check contingency table
set.seed(39)
n <- 300*6
group <- rep(LETTERS[1:6], n/6)
outcome <- sample(0:1, n, T)

(tab <- table(outcome, group))

# output
       group
outcome   A   B   C   D   E   F
      0 158 156 144 146 160 151
      1 142 144 156 154 140 149

# input, cont.: Proportions of "1" across groups
round(prop.table(tab, 2), 3)

# output
       group
outcome     A     B     C     D     E     F
      0 0.527 0.520 0.480 0.487 0.533 0.503
      1 0.473 0.480 0.520 0.513 0.467 0.497

# input: test for association
chisq.test(tab)

# output
X-squared = 2.8741, df = 5, p-value = 0.7194

In this data set, the proportions of "1" do not differ much across groups (always between 47% and 52%). No true group differences are detected at the 5% level.
