Testing for significant difference in mortality rate between multiple groups How do I test for a significant difference between the mortality rate of 4 groups (each group with a different n) 
group 1: mortality rate = 30.9%, n=55
group 2: mr = 0%, n = 4
group 3: mr = 23.3%, n = 30
group 4: mr = 24.6%, n = 69
is it sufficient to just do a t-test between each group? seems like that would be a convoluted way to present the results.
 A: Since you have binomial data, the modern way is to use a binomial generalized linear model. Then we can get an overall test of the null of no differences between groups. Your approach of t-tests runs into multiplicity problems. To do this we first need to recover the number of deaths from your percentage and n, in R:
perc <- c(30.9, 0.0, 23.3, 24.6)
n    <- c(55, 4, 30, 69)
x    <- round((perc/100)*n)
yourdata <- data.frame(x=x, n=n, perc=perc, Group=as.factor(1:4))  
yourdata
   x  n perc Group
1 17 55 30.9     1
2  0  4  0.0     2
3  7 30 23.3     3
4 17 69 24.6     4

Then we can set up a binomial regression:
mod0 <- glm( cbind(x, n-x)  ~ Group, data=yourdata, family=binomial)
mod0

Call:  glm(formula = cbind(x, n - x) ~ Group, family = binomial, data = yourdata)

Coefficients:
(Intercept)       Group2       Group3       Group4  
    -0.8044     -22.5610      -0.3852      -0.3137  

Degrees of Freedom: 3 Total (i.e. Null);  0 Residual
Null Deviance:      3.253 
Residual Deviance: 5.697e-10    AIC: 20.25

Then the test, which is analogous to anova:
anova(mod0, test="Chisq")
Analysis of Deviance Table

Model: binomial, link: logit

Response: cbind(x, n - x)

Terms added sequentially (first to last)


      Df Deviance Resid. Df Resid. Dev Pr(>Chi)
NULL                      3     3.2531         
Group  3   3.2531         0     0.0000   0.3542

But, our data have the form of a contingency table, so we can do a traditional chisquare test, in this case using simulation for the p-value:
with(yourdata, chisq.test(cbind(x, n-x), sim=TRUE, B=20000 ))

    Pearson's Chi-squared test with simulated p-value (based on 20000
    replicates)

data:  cbind(x, n - x)
X-squared = 2.2744, df = NA, p-value = 0.5335

