Statistical significance in categorical data I am trying to analyze a small data (n=30) which is the frequency of sites in a particular category.  Below is an example of the set up I have (sorry for the formatting, I am not sure how to add a table or image here). 
Bins:                       1, 2, 3, 4.
Frequency for group 1:      5, 5, 2, 1.
Frequency for group 2:      3, 2, 1, 6.
What statistic test would be good to comment on the statistical difference in these categories?
I hope someone can recommend something simple, I am also using excel for my data analysis.  I have been considering learning R so if there is a simple method using R. 
 A: What you have is two samples originating from the multinomial distribution with 4 classes. The question is if these are samples from the same distribution (i.e if the distributions have the same parameters). You can find more info on the wikipedia article for the Multinomial test
A normal approach to testing if two samples come from the same distribution is the $\chi^2$-test for homogeneity. 
$$
  Q_{obs} = \sum^s_{i=1}\sum^r_{j=1}\frac{(x_{ij} - n_ip_j^*)^2}{n_ip_j^*}
$$
where $r$ is the number of classes and $s$ the number of series (in your case $2$). $n_i$ is the number of trials in series $i$. $p^*_j$ is the estimated probability for class $j$ using all series. i.e
$$
   p^*_j = \sum^s_{i=i} x_{ij}/n
$$
Degrees of freedom is computed as 
$$
df = (r-1)(s-1)
$$
where $s$ is the number of series you want to compare (i.e 2 in your example but it could be more) and $r$ is the number of classes (i.e 4 in your example). 
So we have 
$$
df = (2-1)(4-1) = 3.
$$
This results in the following statistic in your case (I'll use R to compute it, see example below, but you should be able to do it in Excel as well)
$$
Q_{obs} = 5.659531
$$
which is not larger than the $95\%$ quantile $\chi^2_{0.05}(3) = 7.814728$ ergo you cannot say with certainty that they are not different. 
This test is normally done as an approximation of the multinomial test as it is less computationally taxing for larger samples. But, as in your case, if you have a small sample it might be better to compute the exact test. 
This is an example in R:
# Assigning the counts you specified (which do not add to 30 btw)
x1 <- c(5,5,2,1) # sum = 13
x2 <- c(3,2,1,6) # sum = 12

n <- c(sum(x1), sum(x2))
x <- matrix(c(x1,x2), nrow=2, byrow=T)
p <- (x1 + x2)/(n1 + n2)

Q <- sum((x - outer(n,p))^2 / outer(n,p)) # 5.659531
Q > qchisq(0.95, df=3)
# [1] FALSE 
1 - pchisq(Q, df=3) 
# [1] 0.1294022

# which is the same as you get using the built in function
chisq.test(x)
#
#   Pearson's Chi-squared test
# 
# data:  x
# X-squared = 5.6595, df = 3, p-value = 0.1294
# 
# Warning message:
# In chisq.test(x) : Chi-squared approximation may be incorrect

The warning is due to the low expected counts in the bins which is low because of the small amount of trails.
There is an Excel function CHITEST that should do this for you but I can't test it as I don't have Excel. 
