Expected Values for Male/Female Populations I've got a population of individuals who went through a weight loss program. I'm trying to figure out if gender had an effect on whether they lost a certain threshold of weight by the end of the program or not. I think I should be using a Chi Square test, but how do I come up with the expected values?
Something like 74% of participants were female, and the rest were male, and overall across the entire population, 40.57% hit the weight loss threshold, and 52.11% did not.
Would I do something like the US ratio is 0.9787 M : 1.00 F, calculate how many participants would be male or female using that ratio, then apply the 40.57% and 52.11%, and use those as my expected values? I'm a little lost here, any help would be much appreciated.
Or is there another statistical test I should be using in this situation? Would it be better to set my two groups as male and female and compare total weight lost, rather than if they made it above a threshold or not? 
I think maybe I'm confused because it's two categorical groups - Male/Female, and Crossed Weight Threshold / Did Not Cross Threshold
 A: Comment continued: Perhaps the 2-by-2 table of my comment
is as follows
           Met Goal
Gender     Y      N    Total
----------------------------
  M       10     15      25
  W       40     35      75
----------------------------
Total     50     50     100

From the table it looks as if Women meet goals slightly more
often, but Fisher's Exact test as implemented in R does not find this difference to be statistically significant at the 5% level.
TABL = rbind(c(10,15),c(40,35))
fisher.test(TABL)

    Fisher's Exact Test for Count Data

data:  TABL
p-value = 0.3558
alternative hypothesis: 
 true odds ratio is not equal to 1
95 percent confidence interval:
 0.206663 1.603530
sample estimates:
odds ratio 
 0.5865061 

Perhaps with 10 times as much data the pattern will be
sufficiently strongly supported to give a significant result.
TABL2 = rbind(c(100,150),c(400,350))
fisher.test(TABL2)$p.value
[1] 0.0003347878

Perhaps with actual data the proportions of successes would differ more greatly
between the genders, so that 100 subjects would provide
a significant result.
TABL3 = rbind(c(10,15),c(60,15))
fisher.test(TABL3)$p.value
[1] 0.0003244721

