What is stated by the null hypothesis for the chi-square test for independence?

what is stated by the null hypothesis for the chi square test for independen

This problem aims to familiarize us with the concept of the null hypothesis and the chi-square test for independence. This problem uses the basic concept of inferential statistics in which null-hypothesis helps us test different relationships among different phenomena whereas the chi-square test determines the relationship among the variables encountered in that phenomena.

In inferential statistics, the null hypothesis, referred to as $ H_o $, states that the two occurring possibilities are exact. The null hypothesis is that the experimental discrepancy is due to chance alone. Using statistical tests, it is possible to calculate the possibility that the null hypothesis is true. The term “null” in this context indicates that it’s a normally acknowledged reality that researchers work to nullify. It does not imply that the information is null itself.

Expert Answer

The Chi-square test of independence decides whether there is a statistically meaningful relationship among definite variables. This statistical hypothesis test answers the query—does the magnitude of one definite variable rely on the magnitude of other definite variables? This hypothetical test is also comprehended as the chi-square test of association.

The null hypothesis states there are no connections between the definite variables. If you know the magnitude of one variable, it does not enable you to forecast the magnitude of another variable, whereas the alternative hypothesis states that there are connections between the definite variables. Knowing the magnitude of one variable does enable you to forecast the magnitude of another variable.

Numerical Result

The null hypothesis for this chi-square test for independence states the interconnection/independence or the experimental frequencies between the two definite variables.


When should we use the chi-square test for independence?

The chi-square test can be used:

– To experiment with the goodness of fit of the variables when we are given their expected and experimental frequencies.

– To experiment with the independence of the definite variables.

– To experiment with the importance of the single variance with the assigned variance.

The goodness of fit test is used to review how nicely the obtained sample data serves the allocation of the selected population.
The chi-square statistic test can be computed using the formula:

\[ x^2 = \sum \dfrac{ \left( O_i – E_i \right)^ 2 }{E_i} \]


$O_i$ symbolizes the observed value,

$E_i$ illustrates the expected value.

In the test for independence, we experiment if there is a relationship between the definite variables using the same formula with some slight changes:

\[ x^2 = \sum \dfrac{ \left( O_{ij} – E_{ij} \right) ^2 }{E_{ij}} \]


$O_{ij}$ symbolizes the observed value in the $i^{th}$ column and $j^{th}$ row,

$E_{ij}$ illustrates the expected value in $i^{th}$ column and $j^{th}$ row.

The chi-square test can also be used to approximate the single sampling variance with the population variance using a slightly different formula from before:

\[ x^2 = \dfrac{ \left( n – 1 \right) \times s ^2 }{\sigma^2} \]

$n$ represents the sample size
$s ^2$ represents the sample variance
$\sigma ^2$ represents the population variance

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