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.

## Example

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:

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

Where:

$O_i$ 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:

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

Where:

$O_{ij}$ 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:

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

Where:

$n$ represents the **sample size**

$s ^2$ represents the **sample variance**

$\sigma ^2$ represents the **population variance**