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This is more an extended comment than an answer, and anyway this is a stretch as there are certainly missing or confusing information in the instructions and answer key. However, one possibility is th …

It will affect the Chi-square test of independence, and in short, you should probably not do what you suggest. You can test it for yourself, by making the number of observations repescado similar to …

Since your question in 2014, the information has been removed from the Wikipedia article. If you read the discussion page, the person who originally added this information mentions that they didn't ha …

The sample size and the difference between each age category are both important enough that you get really tiny p-values. Under a certain point (2.22e-16), R assigns a value of 0 to p-values. For inst …

As Peter Flom points out in comments, a chi-square test requires to use counts, not proportions (that is, in your code, you should set the normalize parameter to False, not True). The fact that you ha …

Maybe if you're lucky, but the problem is that there is no guarantee that a unique table satisfies these constraints. I can't tell for analytical methods, but you could try numerical methods to find o …

Your multiple choice question isn't a single question, but multiple "Yes/No" questions disguised in a single question. You don't have a sample of 1400 observations, you have a sample of 1000 observati …

Your expectation is that no one in your population of interest would choose the last three options, and instead would choose equally between the two first options. Obviously, this is not the case, as …

In comments, you say you intend to compare the distributions of two groups, and you want to know if "a larger Cramer's V [...] implies a stronger degree of differences detected between the two populat …

I remember very clearly an academic paper that investigated thoroughly the "minimum of 5 expected counts per cell" rule of thumb when conducting a chi-square test on a contingency table. It explained …

I think your confusion comes from several things. In the first place, your calculations are theoretically correct, but apply to a two-way contingency table. The rub here seems to be that it looks like …

A common approach is to look at the standardized residuals (a.k.a. Pearson residuals), or to the adjusted standardized residuals. See Donald Sharpe's paper "Chi-Square Test is Statistically Significan …

To give some context, the table comes from p.381 of "Nonsymmetric Correspondence Analysis: A Tool for Analysing Contingency Tables With a Dependence Structure" by Kroonenberg & Lombardo (1999) https:/ …

This is not entirely a statistics question, and mainly a programming one. To answer the statistics aspect, this is very simple once you understand what sklearn is doing: the chi2 function performs a g …

I see at least two problems with your approach: the data, and your interpretation of the correspondence analysis plot. First, you shouldn't use a chi-square test like that with weighted data (NB: I as …