Comparing proportions of 2 groups. Do I need to know the actual number affected? I have 2 groups of people.
Both groups have n=35
I go look in the hospital
I do not know how many people are in there, 
but I do know that 25% of the sick are from group 1 
and 75% of the sick are from group 2.
If things were random, you'd expect the same proportion from each group to exhibit some behavior that matches the population (getting sick, etc)
Clearly, there is some difference between the groups.
(Something in group 2 is making them get sick at a greater rate)
We use p-value to determine the odds that this variation is merely random chance.
But, do I need to know the total number of people who are sick in order to do this calculation?
B.c the 25/75 split can be several variations:
1 person from group 1, and 3 people from group 2
2 people from group 1, and 6 people from group 2
3 people from group 1, and 9 people from group 2
etc..
That will affect the actual p-value right?

Group 1 has n=35
20 of them are sick (.571)
Group 2 has n=39
33 of them are sick (.846)

I will now do the calculation for p-value to see if there is a difference between groups.
p-value = .0088 for 2-tailed.
So, there is a difference in the groups.
 A: *

*"Clearly, there is some difference between the groups." - Never ever do such statement before you calculate the actual probability of such event. Things can be extremely deceiving sometimes (and this is one of those cases). 

*Yes, you do need the actual number. Look at your first example - one out of 35 and 3 out of 35 is pretty much random (one sick person accidentally could get to group 1 and now you get 50/50 split - without even calculating anything you should know that such a result is not even close to proving anything, which adds to my first point above). It also could be that 33 out of 35 people in group 2 got sick and 11 in group 1 got sick. Now this certainly looks very suspicious and you can start getting the actual numbers and researching further. These are two border cases which show that your confidence could range wildly for different values, so you do need the actual numbers to draw any conclusion. The only thing you can really find is the range of p-values, but, in the described case it will be quite useless.
A: I have to assume that your 2 groups are mutually exclusive and together describe everyone in your population. Otherwise we have no hope to gain any insight from this situation. 
I also have to clarify that you have 70 people, split into 2 groups, and that some of these people are sick. You don't know how many; only that for every sick person from group 1 there are 3 sick people from group 2. Again, to make any sense of the question, I have to assume that your 70 people are representive of some population you want to make inferences about. 
Lets then assume that your 25/75 split is based on 1 sick person from group 1 and 3 from group 2. That means you have a total of 4 sick people, and 66 healthy people. Now you can make a nice 2X2 table. Amongst g1, the odds of sickness = 1/34; or the proportion is 1/35. Amongst g2, this is 3/32; or the proportion is 3/35. OR = .31; PR = .33.
Lets then assume that your 25/75 split is based on 10 sick people from group 1 and 30 from group 2. That means you have a total of 40 sick people, and 30 healthy people. Amongst g1, the odds of sickness = 10/25; or the proportion is 10/35. Amongst g2, this is 30/5; or the proportion is 30/35. OR = 0.067; PR = 0.33.
From this info, you can also calcuate the corresponding p-values for no effect (=1), or confidence intervals. Repeat for every possible scenario, in the manner that sashkello has suggested.  
I must add that if your interest really is in estimating the effect of group membership on diease risk, this scenario is not how you would go about it. Whether you should use the odds or the proportion is another topic. 
