Chi-Squared Independence Test is used when there are two categorical variables for a single population; it tests if the two variables are independent.

Requirements for using Chi-Squared Independence Test:

Here, the goal is to test if two attributes within a population are independent of one another.

The data provided are formatted as a Matrix with at least two rows and two columns. The rows represent the levels of one attribute and the columns represent the levels of the other attribute; the entries in the Matrix are counts of observations with the given combination of levels.

This test is performed within a single population.

The formula is: $\underset{j=1..K}{\sum _{i=1..N}}\frac{{\left(M_{i,j}-E_{i,j}\right)}^2}{E_{i,j}}$, where $M$ is the matrix of observations, and $E$ is the matrix of expected data, which is computed as: $E_{i,j}=\frac{{\mathrm{rowsum}}_i{\mathrm{columnsum}}_j}{\mathrm{matrixsum}}$.

In turn, ${\mathrm{rowsum}}_i$ is computed as $\sum _{j=1}^K{M}_{i,j}$; ${\mathrm{columnsum}}_j$ is computed as $\sum _{i=1}^N{M}_{i,j}$; and $\mathrm{matrixsum}$ is computed as $\underset{j=1..K}{\sum _{i=1..N}}{M}_{i,j}$.

where $N$ is the sample size of the observed and the expected samples, and $X^2$ follows a Chi-Squared distribution with $N-1$ degrees of freedom.

$\left[\begin{array}{rrr}250& 120& 180\\ 150& 300& 150\end{array}\right]$

Determine the null hypothesis:

Null Hypothesis: Gender and preferences to these three faculties are independent.

Compare the expected data and the observed data:

Substitute the information into the formula:

x = $\frac{{\left(250 - 191.30435\right)}^2}{191.30435} \+ \frac{{\left(120 - 200.86957\right)}^2}{200.86957 } \+ \frac{{\left(180 - 157.82609\right)}^2}{157.82609 }\+\frac{{\left(150 - 208.69565\right)}^2}{208.69565} \+\frac{{\left(300 - 219.13043\right)}^2 }{219.13043}\+ \frac{{\left(150 - 172.17391\right)}^2}{172.17391}$ = 102.891

Compute the p-value:

p-value = $\mathrm{Probability}\left(X^2>102.891\right)$ = 0   (a small value very close to 0)

$X^2˜\mathrm{ChiSquare}\left(3\right)$.

Draw the conclusion:

This statistical test provides evidence that the null hypothesis is false, so we reject the null hypothesis.