Solved – Test of independence on contingency table

categorical datachi-squared-testcontingency tableshypothesis testingstatistical significance

I have a simple contingency table with two nominal variables. Let's say Age and Gender. The software that I'm using reports the relationship between the two variables using Pearson's Chi-Square Test of Independence, to which I understand.

But when I select only two cells in the table and try to do a test for that, the software reports it still uses Pearson's Chi-Square Test of Independence, but use it to test the difference in the population (see below).

Q: Why does the software use the same Chi-Square test to test the difference in the population for the two cells I select. Shouldn't that be a t-test? Shouldn't that the chi-square be used to test the relationship for the two variables,

Q: Statistically, what's the interpretation if I try to do a hypothesis testing on only two cells in a table?

Total sample
Unweighted
base n = 327

Pearson's Chi-Square Test of Independence
Chi-Square = 0.023
Degrees of Freedom = 1
n = 327
Effective Sample Size = 327
Agresti, Alan (1990): Categorical Data Analysis, John Wiley & Sons: New York, pp 24-25, 47-48.
p = 0.88

Not significant

Null hypothesis: There is no difference in the population between the proportion of '18 to 24' of people in 'Male' than 'Female'.

At the 0.05 level of significance, the null hypothesis is not rejected.

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In the case of 2x2 contingency tables, the $z$-test of whether two proportions are equal is equivalent to the Pearson $\chi^2$ test of independence: the quantity $z^2$ is the Pearson $\chi^2$ statistic (Agresti, Introduction to Categorical Data Analysis, p. 26).

Keep in mind that the $t$-test is, essentially, a heavier-tailed $z$-test, and is used to counterbalance the (inappropriately narrow) confidence bands in the case of smaller sample sizes; asymptotically, as one increases the $t$ distribution's degrees of freedom to infinity (which happens as sample size increases), the tails of the $t$ distribution diminish and it approaches the normal.

Your results are interpreted in the ordinary way as for a Pearson $\chi^2$ test.

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Solved – Is the Chi-Square Test of Independence the best option for 3×2 contingency table

As a previous comment pointed out, you need count, instead of the percentages, for these sort of analyses. What I answered below is based on count. It seems that you have 20 per row. So I assume that you indeed know the counts for each cell in the table.

No, they should be equivalent when testing associations between the rows and columns. There shouldn't be any difference.
You can certainly use Chi-square test for association. Another option is to use the G test. G test appears to perform better than Chi-square test when sample size is small (see here). More generally, you can fit a log-linear model to the data and test for independence. In fact, the G-test is the likelihood ratio test for independence in the log-linear model. In R, the model can be fit as glm(cell count~category+example+category*example, family=poisson). Testing if the interaction is significant is the test for independence. If you use likelihood ratio test, it will be the same as the G test. I think the log-linear model would be the best tool for this as it allows you to test other things in additional to independence.
No, it can be shown that Poisson sampling (i.e. each cell count is a Poisson random variable), multinomial sampling (i.e. total count of the table is fixed), or product multinomial sampling (row total or column total is fixed) are equivalent when testing the independence between row and column variables. So the log-linear model should apply in your experimental design. Chapter 3 of the categorical data analysis textbook by Agresti has a more detailed discussion on this if you wish to read in more depth about it.
I think the log-linear model would be suitable for this. After fitting the model, you can test differences between any two cell counts or any marginal means if you wish.

Edit based on the new information in the question:

The null deviance is the deviance that compares the model with only an intercept and the full model. In your case, the null deviance is huge, meaning that the data cannot be explained by just an intercept. The null deviance is testing that category and example and their interaction do not influence the count. In other words, the probability of all categories are the same, the probability in all examples are the same and consequently the probability in one category does not depend on example. This is not the hypothesis you want to test.
The model you fit is the full model, i.e. you fit a mean for each cell. Thus, you would expect deviance residual to be 0.
If you have any predictor in the model, the null deviance will always be greater or equal to the residual deviance. The null deviance tells you whether having predictors in the model gives you significantly better fit than just an intercept. The residual deviance give you ideas of whether the model has significant lack of fit (For grouped data, you can test lack of fit but for data that cannot be grouped, you cannot test lack of fit, but some suggests that you can still look at residual deviance as a hint of lack of fit).
You do not use null deviance to test for independence. Instead, you want to test if the interaction term, i.e. category*example, is significant. This can be done by anova(fit, test="Chisq") and looking at the p-value for the interaction term. Alternatively, you can fit a second model without interaction and compare it to the first model. This will achieve the same thing.

   fit2 = gum(count~example+category, family=poisson)
   anova(fit, fit2)

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Solved – Is the Chi-Square Test of Independence the best option for 3×2 contingency table

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