chi-Squared and its types - Probability and Statistic

What is Chi-Square Test?

A chi-square test is a statistical test used to determine whether there is a significant association between categorical variables. It is widely used in hypothesis testing to examine the distribution of categorical data and to test the independence of two variables or the goodness of fit of an observed distribution to an expected distribution.

Need for Chi-Square Test

The chi-square test is essential in various research scenarios:

Testing Independence: To determine if there is a significant relationship between two categorical variables.

Goodness of Fit: To see if an observed frequency distribution matches an expected distribution.

Analyzing Categorical Data: To analyze data that are categorical in nature (e.g., yes/no, red/blue/green).

Types of Chi-Square Tests

1. Chi-Square Test for Independence: Tests whether two categorical variables are independent of each other.

2. Chi-Square Goodness of Fit Test: Tests whether an observed frequency distribution fits an expected distribution.

1. Chi-Square Test for Independence: Tests whether two categorical variables are independent of each other.
$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$
Where:
$O_i$ is the observed frequency,
$E_i$ is the expected frequency.

For Example: A researcher wants to test if there is an association between gender (male/female) and preference for a new product (like/dislike). The observed frequencies are recorded in a contingency table.

Data:

Solution:

1. Calculate the expected frequencies:
$E_{ij} = \frac{\text{row total} \times \text{column total}}{\text{grand total}}$

For males who like the product:
$E_{11} = \frac{30 \times 50}{100} = 15$

For males who dislike the product:
$E_{12} = \frac{30 \times 50}{100} = 15$

For females who like the product:
$E_{21} = \frac{70 \times 50}{100} = 35$

For females who dislike the product:
$E_{22} = \frac{70 \times 50}{100} = 35$

2. Calculate the chi-square statistic:
$\chi^2 = \frac{(20-15)^2}{15} + \frac{(10-15)^2}{15} + \frac{(30-35)^2}{35} + \frac{(40-35)^2}{35}$
$\chi^2 = \frac{25}{15} + \frac{25}{15} + \frac{25}{35} + \frac{25}{35}$
$\chi^2 = 1.67 + 1.67 + 0.71 + 0.71$
$\chi^2 = 4.76$

Therefore, the chi-square statistic is approximately 4.76. This value would be compared to the critical chi-square value from the chi-square distribution table at the desired significance level (e.g., 0.05) and degrees of freedom (df = (rows-1) * (columns-1)) to determine if there is a significant association between gender and product preference.

2. Chi-Square Goodness of Fit Test: Tests whether an observed frequency distribution fits an expected distribution.
$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$
Where:
$O_i$ is the observed frequency,
$E_i$ is the expected frequency.

For Example: A dice manufacturer claims that their dice are fair, meaning each number (1 through 6) should appear with equal probability. A dice is rolled 60 times, and the observed frequencies are recorded as follows:

Data:
1: 8
2: 10
3: 9
4: 12
5: 11
6: 10

Solution:

1. Calculate the expected frequencies:
If the dice are fair, each number should appear approximately $60 / 6 = 10$ times.

2. Calculate the chi-square statistic:
$\chi^2 = \frac{(8-10)^2}{10} + \frac{(10-10)^2}{10} + \frac{(9-10)^2}{10} + \frac{(12-10)^2}{10} + \frac{(11-10)^2}{10} + \frac{(10-10)^2}{10}$
$\chi^2 = \frac{4}{10} + \frac{0}{10} + \frac{1}{10} + \frac{4}{10} + \frac{1}{10} + \frac{0}{10}$
$\chi^2 = 0.4 + 0 + 0.1 + 0.4 + 0.1 + 0$
$\chi^2 = 1.0$

Therefore, the chi-square statistic is approximately 1.0. This value would be compared to the critical chi-square value from the chi-square distribution table at the desired significance level (e.g., 0.05) and degrees of freedom (df = number of categories - 1) to determine if the dice are fair.