blog/chi-square-test-of-independence-by-hand/

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Chi-square test of independence by hand - Stats and R

Test if two categorical variables are dependent via the Chi-square test of independence. See also how to compute it by hand and how to interpret the results

https://statsandr.com/blog/chi-square-test-of-independence-by-hand/

[AntoineSoetewey]

Comment written by Kuo Yao Hung on January 29, 2020 03:47:04:

A really nice article, thanks.

[AntoineSoetewey]

Comment written by Kuo Yao Hung on January 29, 2020 03:47:04:

A really nice article, thanks.

Comment written by Antoine Soetewey on January 29, 2020 08:58:47:

Glad you liked it Kuo!

[manoj1123]

Dear Antonie.
Thank you so much for the article.
I really added more understanding to the concept.
I had a small doubt, regarding the following statement. "If the test statistic is above the critical value, it means that the probability of observing such a difference between the observed and expected frequencies is unlikely"

My confusion is if the probability of such a difference is unlikely then there should be no relation between the two variable.
Instead can I read the statement as "If the test statistic is above the critical value, it means that the probability of observing such a difference between the observed and expected frequencies is unlikely BY RANDOM CHANCE".

Thanks again.
Thanks for your time.

[AntoineSoetewey]

Dear Antonie.
Thank you so much for the article.
I really added more understanding to the concept.
I had a small doubt, regarding the following statement. "If the test statistic is above the critical value, it means that the probability of observing such a difference between the observed and expected frequencies is unlikely"

My confusion is if the probability of such a difference is unlikely then there should be no relation between the two variable.
Instead can I read the statement as "If the test statistic is above the critical value, it means that the probability of observing such a difference between the observed and expected frequencies is unlikely BY RANDOM CHANCE".

Thanks again.
Thanks for your time.

Thanks for your question.

You're right; if the test statistic is above the critical value (determined by the Chi-square table), it means that the probability of observing such a difference between the observed and expected frequencies is unlikely BY RANDOM CHANCE.

However, remember that the null and alternative hypothesis of the Chi-square test of independence are:

• H0: the 2 variables are independent
• H1: the 2 variables are dependent

This means that: if the test statistic is above the critical value --> we reject the null hypothesis of independence because the probability of observing such a large difference between the expected and observed frequencies just by chance is small (i.e., the p-value is small) --> the 2 variables are dependent --> there is a significant relationship between the 2 variables.

Hope this helps. Let me know if it is still unclear.

Regards,
Antoine

[manoj1123]

I must say thak you so much for the clarity and response. Wishing a very happy new year. Best Regards, Manoj

[norberello]

Thanks a lot for the clear explanation of the mechanics and intuition of a chi square test of independence. I have repeated your exercise in R and post a related question about how to calculate the chi square distribution afterwards:

#preparing data matrix
m.smoke <- matrix(data = c(14,0,4,10),nrow = 2, ncol = 2)
rownames(m.smoke) = c("Athlete", "Non-Athlete")    
colnames(m.smoke) = c("Non-smoker", "Smoker") 
m.smoke#print contingency table

	Non-smoker	Smoker
Athlete	14	4
Non-Athlete	0	10

#mosaic plot
mosaicplot(m.smoke,color=c(2,1),main="mosaic plot of smoke x sport",cex=1.1)

#chi square test (is the Yates correction because of the 0 in one of cells?
(x.smoke<-chisq.test(m.smoke))
#Pearson's Chi-squared test with Yates' continuity
#	correction
#
#data:  m.smoke
#X-squared = 12.6, df = 1, p-value = 0.0003857
#https://www.statology.org/yates-continuity-correction/

Instead of the table of critical values, I will plot the actual chi square distribution

curve(dchisq(x, df = 1), from = 0, to = 15,
      main = 'Chi-Square Distribution (df = 1)',
      ylab = 'Density',
      lwd = 0.5,
      xlab='Chi-squared values',
      cex.lab=1.4)
x_vector <- seq(qchisq(0.95,1), 15)#create vector of x values starting from critical value (alpha = 0.05)
p_vector <- dchisq(x_vector, df = 1)#prob for each chi square value
#fill in portion of the density plot from the critical value to 15
polygon(c(x_vector, rev(x_vector)), c(p_vector, rep(0, length(p_vector))),
                           col = adjustcolor('red', alpha=0.8), border = NA)
arrows(x.smoke$statistic,0.85,x.smoke$statistic,0)#arrows indicating observed chi square value
legend("topright","observed chi square value",
       bty="n",cex=1.2)

My question is how to build such a chi square null distribution in which you constrast the observed value versus the critical one by hand (well, using R), it would be great exactly how this was made in the time when there were no computers. Is it that you calculate all chi square possible values (or do thousand of simulations) for one degree of freedom to find such an idealized distribution? This is, considering your sample size, n = 28, do we need to calculate the chi square for all possible combinations for each of the 4 possible cells?

28 0 0 0 calculate Chi Square 1
27 1 0 0 calculate Chi Square 2
26 1 1 0 calculate Chi Square 3
and so on
(...) calculate Chi square n

Then, we plot all these chi square values in a histogram (or in a density density plot) and then “invent” an idealized distribution of chi squares resulting. My problem is with that "invent" as well, how is this done exactly? I would appreciate very much a step-by-step explanation in R, but I cannot find it anywhere, and it seems to me that this would complete our understanding of the pilars of inferential testing. Thanks a lot in advance!

[norberello]

Will take the momentum for an associated question. Is the chi square test a parametric test or a non-parametric test? Since the test assumes a distribution of how frequencies are distributed by chance for k degrees of freedom. I would say that it should be a parametric test, but I have read some places where it is categorized as non-parametric, which is confusing. Thanks!

[AntoineSoetewey]

Hello @norberello,

To be honest I have never tried to draw a chi-square distribution by hand, nor "by hand in R". All I can say is that the chi-square distribution is linked to the normal distribution, since a chi-square distribution with $k$ degrees of freedom is the distribution of a sum of the squares of $k$ independent standard normal random variables.

For instance, let $X \sim N(\mu, \sigma^2)$, then

$$Z^2 = \frac{(X - \mu)^2}{\sigma^2} \sim \chi^2_1.$$

And a chi-square distribution with, say 10 degrees of freedom, indicates that this particular chi-square distribution is the sum of 10 independent, squared standard normal variables:

Let $X_1, X_2, \ldots, X_{10}$ be $n$ independent random variables $N(\mu, \sigma^2)$, then

$$\sum_{i = 1}^{10} Z^2_i = \sum_{i = 1}^{10} \frac{(X_i - \mu)^2}{\sigma^2} \sim \chi^2_{10} .$$

Perhaps this will help you to build the chi-square distribution from scratch.

Regarding your second question, I confirm that the chi-square test of independence is a non-parametric test. It is robust with respect to the distribution of the data, that is, no assumptions are made about the underlying distribution (this is the reason it's also referred as a distribution free test, see the abstract and introduction of this article). It is true that some assumptions must still be met (e.g., independent observations and expected frequencies above 5), but this has nothing to do with the underlying distribution.

[norberello]

Thanks a lot for the reply!! Perhaps I am not understanding it correctly, but isn't the realization that the chi-square distribution fits the sum of k squared normal distributions came a posteriori? I mean, after looking at the distribution patterns of thousands of chi-square simulated calculations for diferent degrees of freedom, mathemathicians idealized the chi square distribution and, by doing so (or afterwards), they realized that the underlying distribution was the normal distribution squared. Is it possible to know how was it done the first time? I personally feel that without knowing the steps taken from the chi square simulations to the idealized distribution my understanding of how the statistical test works is not complete.

[AntoineSoetewey]

isn't the realization that the chi-square distribution fits the sum of k squared normal distributions came a posteriori?

In my opinion (as a practician of statistics rather than a pure mathematician, so don’t take my words for granted), I don’t think the realization that the chi-square distribution fits the sum of k squared normal distributions came a posteriori. I believe mathematicians and statisticians formalized the relationship between the normal and chi-squared distribution before it was verified in practice with simulations. This belief (which must be verified because I’m not an expert on this matter) is based on my personal supposition that many mathematical concepts are usually discovered and formalized from a theoretical point of view before it is possible to implement it in code and/or verify it through simulations (maths existed before computers, and distributions were formalized before we could run thousands of simulated calculations).

Is it possible to know how was it done the first time?

I will refrain from saying anything further than that because, at the risk of disappointing you, I must say that I honestly don’t know. I hope you’ll find someone who is more knowledgeable than me on this matter (and feel free to let me know here so I can learn).

[norberello]

Thanks for the reply, Antoine! It is a very good point that distributions were mathematically demonstrated or built before doing any simulation, I was completely unware of that possibility, cause it is somewhat couninterintuitive to me. It seems that mathematicians have a particular mindset and my doubt is only touching the surface of the story.

[mos790]

Hi, can I ask you what techno is used for your blog ?

[AntoineSoetewey]

Hello @mostafa790,

Here are all informations: https://statsandr.com/faq/#3-what-technology-and-theme-do-you-use-to-write-this-blog-and-the-articles

Feel free if you have any other questions!

Regards,
Antoine