T Test or Chi Square? There are 2 variables, City and No Of Matches won.  
Variables:
1City


*

*Sydney

*Canberra


2No Of Matches Won?


*

*Sydney: 200 out of 300

*Canberra: 210 out of 300     


Question: Is team Canberra significantly better than team Sydney?   
Should we perform Chi Square test or 2 sample t test?
My wife says it is 2 sample T Test. According to her, we can check if Mean matches won by Canberra is greater than mean matches won by Sydney?
According to her
Null Hypothesis is: Mean(Canberra)- Mean(Sydney) = 0
Alternate: Mean(Canberra) > Mean(Sydney)      
According to me its Chi Square Test.. Can you please suggest which is right?
 A: You have a categorical outcome: either win or lose (1 vs. 0). You have two groups (Canberra and Sidney) and you have 300 samples from each group. The usual appropriate tests here are either Z-test for two proportions or Chi-squared-test for 2x2 contingency tables (both are equivalent).
With increasing N, the t-test will approximate the results of the above mentioned tests but I disagree with asdf that it is already equivalent for N=300 per group. Since the t-test does assume a continuous outcome I would not consider it to analyze the data here.
If we play around with following R code, we see that the p-values are quite different even for N=1000 per group.
prob1 <- 210/300
prob2 <- 200/300

set.seed(16)
sample1 <- rbinom(1e3, 1, prob = prob1)
sample2 <- rbinom(1e3, 1, prob = prob2)

t.test(sample1, sample2)$p.value # p = 0.567
chisq.test(cbind(table(sample1), table(sample2)))$p.value # p = 0.599

There is also an exact test available: Barnard's test. This is the most powerful test for equality of proportions and may be considered as the best test for this situation.
To answer your question: you are right and your wife is wrong. However, even from a statistical perspective I would not recommend to insist on it too stubbornly :-)
A: From a statistically rigurous perspective, her test is not the thing to do, but the result will be so close that if will work just fine (Z-test  and T-test with such amount of degrees of freedom will be almost the same) You can estimate the standard deviation as $\sqrt{np(1-p)}$ where n is the number of matches played and p is the proportion of won matches.
I have no idea, however, about how to proceed with a $\chi^2$ here. Please explain what's in your mind
Anyway, just from looking at the data and without doing the math, the 10-wins difference is pretty much nothing, statistical significance has not been achieved here
