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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?

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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 :-)

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    $\begingroup$ (+1) The problem with using a t-test here is not that it assumes a continuous outcome - you're assuming the same when comparing the value of your test statistic against the chi-squared distribution function - but that it doesn't take into account the fixed relation of variance to mean for a binomial proportion: see stats.stackexchange.com/q/81975/17230. Note that the asymptotic generalized likelihood ratio test & Fisher's Exact Test are often used too. $\endgroup$ – Scortchi Mar 26 at 14:46
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    $\begingroup$ I think my subconscious main motivation to answer questions here is to fact-check my current understanding of the topic and to receive comments to deepen it :-) $\endgroup$ – LuckyPal Mar 26 at 14:50
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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

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  • $\begingroup$ HI! SO do you feel its 2 sample T Test? ( I am not worried of the result: ie if Sydney is better or vice versa). I want to know which is the right test to be applied here $\endgroup$ – user242468 Mar 26 at 13:28
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    $\begingroup$ Z-test for two proportions and Chi-squared-test for 2x2 contingency tables are the same. In R, prop.test(x = c(200, 210), n = c(300, 300)) actually gives you X-squared instead of Z. See also stats.stackexchange.com/questions/173415/… $\endgroup$ – LuckyPal Mar 26 at 13:30
  • $\begingroup$ @LuckyPal OK, now I understand what he meant with $\chi^2$ It is probably the right approach, since it is easier to understand what you are doing in my opinion. Anyway, both methods are fine $\endgroup$ – David Mar 26 at 14:27

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