In statistics, there are two different types of Chi-Square tests:
1. The Chi-Square Goodness of Fit Test – Used to determine whether or not a categorical variable follows a hypothesized distribution.
2. The Chi-Square Test of Independence – Used to determine whether or not there is a significant association between two categorical variables.
In this article, we share several examples of how each of these types of Chi-Square tests are used in real-life situations.
Example 1: Chi-Square Goodness of Fit Test
Suppose a shop owner claims that an equal number of customers come into his shop each weekday.
To test this hypothesis, he records the number of customers that come into the shop on a given week and finds the following:
- Monday:Â 50 customers
- Tuesday:Â 60 customers
- Wednesday:Â 40 customers
- Thursday:Â 47Â customers
- Friday:Â 53 customers
He can use a Chi-Square Goodness of Fit Test to determine if the distribution of the customers that come in each day is consistent with his hypothesized distribution.
Using the Chi-Square Goodness of Fit Test Calculator, he can find that the p-value of the test is 0.359.
Since this p-value is not less than .05, there is not sufficient evidence to say that the true distribution of customers is different from the distribution that the shop owner claimed.
Example 2: Chi-Square Goodness of Fit Test
Suppose a biologist claims that an equal number of four different species of deer enter a certain wooded area in a forest each week.
To test this hypothesis, she records the number of each species of deer that enter the wooded area over the course of one week:
- Species #1: 22Â
- Species #2: 20
- Species #3: 23
- Species #4: 35
She can use a Chi-Square Goodness of Fit Test to determine if the distribution of the deer species that enter the wooded area in the forest each week is consistent with his hypothesized distribution.
Using the Chi-Square Goodness of Fit Test Calculator, she can find that the p-value of the test is 0.137.
Since this p-value is not less than .05, there is not sufficient evidence to say that the true distribution of deer is different from the distribution that the biologist claimed.
Example 3: Chi-Square Test of Independence
Suppose a policy maker in a certain town wants to know whether or not gender is associated with political party preference.
He decides to take a simple random sample of 500 voters and survey them on their political party preference. The following table shows the results of the survey:
| Â | Republican | Democrat | Independent | Total |
| Male | 120 | 90 | 40 | 250 |
| Female | 110 | 95 | 45 | 250 |
| Total | 230 | 185 | 85 | 500 |
He can use a Chi-Square Test of Independence to determine if there is a statistically significant association between the two variables.
Using the Chi-Square Test of Independence Calculator, he can find that the p-value of the test is 0.649.
Since the p-value is not less than .05, there is not sufficient evidence to say that there is an association between gender and political party preference.
Example 4: Chi-Square Test of Independence
Suppose a researcher wants to know whether or not marital status is associated with education level.
He decides to take a simple random sample of 300 individuals and obtains the following results:
|  | High School | Bachelor’s | Master’s or Higher | Total |
| Married | 20 | 100 | 35 | 155 |
| Single | 50 | 80 | 15 | 145 |
| Total | 70 | 180 | 50 | 300 |
He can use a Chi-Square Test of Independence to determine if there is a statistically significant association between the two variables.
Using the Chi-Square Test of Independence Calculator, he can find that the p-value of the test is 0.000011.
Since the p-value is less than .05, there is sufficient evidence to say that there is an association between marital status and education level.
Additional Resources
The following tutorials provide an introduction to the different types of Chi-Square Tests:
The following tutorials explain the difference between Chi-Square tests and other statistical tests: