What is the main purpose of the Chi-Square Test?

To test the independence or goodness of fit of categorical variables by comparing observed and expected frequencies.

What is the critical value in Chi-Square tests?

It depends on degrees of freedom and significance level (commonly 0.05). The chi-square distribution table provides critical values.

Can the Chi-Square Test be used for continuous data?

No, it is mainly applicable to categorical data.

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Chi-Square Test

The Chi-Square Test is a statistical technique used to assess whether there is a meaningful relationship between categorical variables. It compares observed data with expected data based on a hypothesis to check for independence or goodness of fit. Widely used in fields like business, science, and social research, the test helps analyze relationships in contingency tables. The calculated Chi-Square value is compared with a critical value to accept or reject the null hypothesis.

1.0What is the Chi-Square Test?

Chi-Square Test Definition:

The Chi-Square Test is a statistical method to determine whether there is a significant relationship between two categorical variables by comparing observed and expected frequencies.

$χ^{2} = \sum \frac{( O _{i} - E _{i} ) ^{2}}{E _{i}}$

Where:

$O_{i}$ = Observed frequency
$E_{i}$ = Expected frequency

2.0About the Chi-Square Test

The chi-square test helps answer questions such as:

Are two variables independent?
Is the observed distribution close to the expected one?

It is mainly of two types:

Chi-Square Test for Independence – Checks if two variables are related.
Chi-Square Goodness of Fit Test – Checks how well observed data fit a theoretical distribution.

3.0Calculation of the Chi-Square Test

Step-by-Step Process:

Step 1: Set up hypotheses.
- $H_{0}$ : Variables are independent.
- $H_{1}$ : Variables are dependent.
Step 2: Prepare the contingency table (Observed frequencies).
Step 3: Calculate Expected frequencies using:

$E_{ij} = \frac{( Row Total ) ( Column Total )}{Grand Total}$

Step 4: Compute $χ^{2}$ :

$χ^{2} = \sum \frac{( O - E ) ^{2}}{E}$

Step 5: Compare with critical chi-square value from the chi-square distribution table at a given significance level (e.g., 0.05).

4.0Example of Chi-Square Test Explained

Example: A survey records the choice of drink by 100 people:

Drink Choice	Male	Female	Total
Tea	30	20	50
Coffee	20	30	50
Total	50	50	100

Solution:

Step 1: Calculate Expected Frequencies:

$E_{(Male,Tea)} = \frac{50 \times 50}{100} = 25$

Similarly:

$E_{(F e ma l e, T e a)}$ = 25
$E_{(M a l e, C o ff ee)}$ = 25
$E_{(F e ma l e, C o ff ee)}$ = 25

Step 2: Compute $/ c h i^{2}$ :

$χ^{2} = \frac{( 30 - 25 ) ^{2}}{25} + \frac{( 20 - 25 ) ^{2}}{25} + \frac{( 20 - 25 ) ^{2}}{25} + \frac{( 30 - 25 ) ^{2}}{25}$ $χ^{2} = \frac{25 + 25 + 25 + 25}{25} = 4$

Step 3: Interpretation:

At 1 degree of freedom (df = (2-1)(2-1) =1) and significance level 0.05, the critical value is 3.841.
Since $χ^{2} = 4 > 3.841$ , we reject H_0.

Conclusion: There is a significant association between gender and drink choice.

5.0Bayesian Chi-Square Test

An advanced method that incorporates prior beliefs into the chi-square analysis. It is useful when sample sizes are small or data is uncertain. Bayesian methods provide a probabilistic interpretation of the relationship between variables.

6.0Chi-Square Test A/B Test in Business Example

In business, A/B testing uses chi-square to compare two marketing strategies. Example:

Outcome	Strategy A	Strategy B
Success	120	150
Failure	80	50

Using the chi-square test, businesses can analyze if the difference in performance between Strategy A and B is statistically significant.

7.0Importance of Chi-Square Test

Helps in decision making using categorical data
Useful in survey analysis and market research
Validates hypotheses in experimental research
Crucial in quality control and business strategies

8.0Solved Example on Chi-Square Test

Example 1: In a survey, 200 people were asked about their preference for three different products: A, B, and C. The results are shown below:

Product	Observed Frequency (O)
A	80
B	70
C	50

Assuming all products are equally popular, test the hypothesis at a 5% significance level.

Solution:

Step 1 – Null Hypothesis H_0:

All products are equally preferred. So, expected frequency E for each = _{\frac{200}{3} \approx 66.67.}

Product	O	E	(O−E)2/E(O - E)^2 / E
A	80	66.67	$\frac{( 80 - 66.67 ) ^{2}}{66.67} = \frac{177.78}{66.67} \approx 2.67$
B	70	66.67	$\frac{( 70 - 66.67 ) ^{2}}{66.67} = \frac{11.11}{66.67} \approx 0.17$
C	50	66.67	$\frac{( 50 - 66.67 ) ^{2}}{66.67} = \frac{277.78}{66.67} \approx 4.17$

Step 2 – Calculate $/ c h i^{2}$ :

$= 2.67 + 0.17 + 4.17 = 7.01 χ^{2} = 2.67 + 0.17 + 4.17 = 7.01$

Step 3 – Degrees of Freedom (df):

df = n - 1 = 3 - 1 = 2

From the chi-square table, the critical value at 5% significance level and df = 2 is 5.991.

Since 7.01 > 5.991, we reject H0H_0.

Conclusion: The product preferences are not equally distributed.

Example 2: A company studied the relationship between customer gender and their product choice. The data is:

	Product X	Product Y	Total
Male	40	60	100
Female	30	70	100
Total	70	130	200

Test whether product choice is independent of gender at 5% significance level.

Solution:

Step 1 – Compute Expected Frequencies:

	Product X	Product Y
Male	$E_{11} = \frac{100 \times 70}{200} = 35$	$E_{12} = \frac{100 \times 130}{200} = 65$
Female	$E_{21} = \frac{100 \times 70}{200} = 35$	$E_{22} = \frac{100 \times 130}{200} = 65$

Step 2 – Calculate $χ^{2}$ :

$χ^{2} = \frac{( 40 - 35 ) ^{2}}{35} + \frac{( 60 - 65 ) ^{2}}{65} + \frac{( 30 - 35 ) ^{2}}{35} + \frac{( 70 - 65 ) ^{2}}{65}$

$χ^{2} = \frac{25}{35} + \frac{25}{65} + \frac{25}{35} + \frac{25}{65} = \frac{50}{35} + \frac{50}{65}$

$χ^{2} = 1.428 + 0.769 = 2.197$

Step 3 – Degrees of Freedom (df):

df = (rows -1)(columns -1) = (2 -1)(2 -1) = 1

Critical chi-square value at 5% significance level and df =1 is 3.841.

Since 2.197 < 3.841, we accept H_0.

Conclusion: Product choice is independent of gender.

Example 3: A manufacturer claims that the proportion of defective items in a batch is 5%. From a sample of 200 items, 15 defective items were found. Test the manufacturer’s claim at a 5% significance level.

Solution:

Step 1 – Null Hypothesis $H_{0}$ :

The proportion of defective items is 5%.

Expected defective items:

$E = 0.05 \times 200 = 10$

Expected non-defective items:

E' = 200 - 10 = 190

Observed (O):

Defective: 15
Non-defective: 185

Step 2 – Calculate $χ^{2}$ :

$χ^{2} = \frac{( 15 - 10 ) ^{2}}{10} + \frac{( 185 - 190 ) ^{2}}{190} = \frac{25}{10} + \frac{25}{190}$

$χ^{2} = 2.5 + 0.1316 = 2.6316$

Step 3 – Degrees of Freedom (df):

df = number of categories – 1 = 2 – 1 = 1

Critical value at 5% significance level and df = 1 is 3.841.

Since 2.6316<3.8412.6316 < 3.841, we accept $H_{0}$ .

Conclusion: The manufacturer’s claim holds.

Example 4: A study records the following data regarding study method and exam success:

Study Method	Passed	Failed	Total
Method A	80	20	100
Method B	50	50	100
Total	130	70	200

Is there an association between study method and exam success at a 1% significance level?

Solution:

Step 1 – Expected Frequencies:

	Passed	Failed
Method A	$\frac{100 \times 130}{200} = 65$	$\frac{100 \times 70}{200} = 35$
Method B	Same as above: 65 and 35

Step 2 – Compute $χ^{2}$ :

$χ^{2} = \frac{( 80 - 65 ) ^{2}}{65} + \frac{( 20 - 35 ) ^{2}}{35} + \frac{( 50 - 65 ) ^{2}}{65} + \frac{( 50 - 35 ) ^{2}}{35}$

$χ^{2} = \frac{225}{65} + \frac{225}{35} + \frac{225}{65} + \frac{225}{35}$

$χ^{2} = 3.46 + 6.43 + 3.46 + 6.43 = 19.78$

Step 3 – Degrees of Freedom (df):

df = (2 -1)(2 -1) =1

Critical chi-square value at 1% significance level and df =1 is 6.635.

Since 19.78 > 6.635, we reject $H_{0}$ .

Conclusion: Study method is significantly associated with exam success.

Example 5: A geneticist expects that a particular plant produces flowers in the ratio of Red:White:Pink as 9:3:4. In an experiment, out of 160 plants, the observed counts were Red – 90, White – 30, Pink – 40. Test the hypothesis at 5% significance level.

Solution:

Step 1 – Expected Frequencies:

Total parts = 9 + 3 + 4 = 16

Expected Red:

$E_{Red} = \frac{9}{16} \times 160 = 90$

Expected White:

$E_{White} = \frac{3}{16} \times 160 = 30$

Expected Pink:

$E_{Pink} = \frac{4}{16} \times 160 = 40$

Step 2 – Compute $/ c h i^{2}$ :

$χ^{2} = \frac{( 90 - 90 ) ^{2}}{90} + \frac{( 30 - 30 ) ^{2}}{30} + \frac{( 40 - 40 ) ^{2}}{40} = 0$

Step 3 – Degrees of Freedom (df):

df = 3 -1 = 2

Critical value at 5% significance level and df = 2 is 5.991.

Since $χ^{2} = 0 < 5.991$ we accept $H_{0}$ .

Conclusion: The observed data perfectly matches the expected ratio.

9.0Practice Questions on Chi-Square Test

In a dice-rolling experiment of 120 rolls, the observed frequencies of outcomes 1, 2, 3, 4, 5, and 6 are: 15, 20, 18, 22, 25, and 20 respectively. Test at 5% significance level whether the dice is fair.
A survey of 150 people records the choice of car brand (Brand A or Brand B) and gender (Male or Female):

	Brand A	Brand B	Total
Male	50	40	90
Female	30	30	60
Total	80	70	150

Test whether car brand preference is independent of gender at 5% significance level.

Also Read:

Bayes Theorem	Standard Deviation	Alternative Hypothesis
Kurtosis and Skewness	Coefficient of Correlation	Spearman’s Rank Correlation
Inferential Statistics	Statistics	Taylor Series

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Chi-Square Test

The Chi-Square Test is a statistical technique used to assess whether there is a meaningful relationship between categorical variables. It compares observed data with expected data based on a hypothesis to check for independence or goodness of fit. Widely used in fields like business, science, and social research, the test helps analyze relationships in contingency tables. The calculated Chi-Square value is compared with a critical value to accept or reject the null hypothesis.

1.0What is the Chi-Square Test?

Chi-Square Test Definition:

The Chi-Square Test is a statistical method to determine whether there is a significant relationship between two categorical variables by comparing observed and expected frequencies.

$χ^{2} = \sum \frac{( O _{i} - E _{i} ) ^{2}}{E _{i}}$

Where:

$O_{i}$ = Observed frequency
$E_{i}$ = Expected frequency

2.0About the Chi-Square Test

The chi-square test helps answer questions such as:

Are two variables independent?
Is the observed distribution close to the expected one?

It is mainly of two types:

Chi-Square Test for Independence – Checks if two variables are related.
Chi-Square Goodness of Fit Test – Checks how well observed data fit a theoretical distribution.

3.0Calculation of the Chi-Square Test

Step-by-Step Process:

Step 1: Set up hypotheses.
- $H_{0}$ : Variables are independent.
- $H_{1}$ : Variables are dependent.
Step 2: Prepare the contingency table (Observed frequencies).
Step 3: Calculate Expected frequencies using:

$E_{ij} = \frac{( Row Total ) ( Column Total )}{Grand Total}$

Step 4: Compute $χ^{2}$ :

$χ^{2} = \sum \frac{( O - E ) ^{2}}{E}$

Step 5: Compare with critical chi-square value from the chi-square distribution table at a given significance level (e.g., 0.05).

4.0Example of Chi-Square Test Explained

Example: A survey records the choice of drink by 100 people:

Drink Choice	Male	Female	Total
Tea	30	20	50
Coffee	20	30	50
Total	50	50	100

Solution:

Step 1: Calculate Expected Frequencies:

$E_{(Male,Tea)} = \frac{50 \times 50}{100} = 25$

Similarly:

$E_{(F e ma l e, T e a)}$ = 25
$E_{(M a l e, C o ff ee)}$ = 25
$E_{(F e ma l e, C o ff ee)}$ = 25

Step 2: Compute $/ c h i^{2}$ :

$χ^{2} = \frac{( 30 - 25 ) ^{2}}{25} + \frac{( 20 - 25 ) ^{2}}{25} + \frac{( 20 - 25 ) ^{2}}{25} + \frac{( 30 - 25 ) ^{2}}{25}$ $χ^{2} = \frac{25 + 25 + 25 + 25}{25} = 4$

Step 3: Interpretation:

At 1 degree of freedom (df = (2-1)(2-1) =1) and significance level 0.05, the critical value is 3.841.
Since $χ^{2} = 4 > 3.841$ , we reject H_0.

Conclusion: There is a significant association between gender and drink choice.

5.0Bayesian Chi-Square Test

An advanced method that incorporates prior beliefs into the chi-square analysis. It is useful when sample sizes are small or data is uncertain. Bayesian methods provide a probabilistic interpretation of the relationship between variables.

6.0Chi-Square Test A/B Test in Business Example

In business, A/B testing uses chi-square to compare two marketing strategies. Example:

Outcome	Strategy A	Strategy B
Success	120	150
Failure	80	50

Using the chi-square test, businesses can analyze if the difference in performance between Strategy A and B is statistically significant.

7.0Importance of Chi-Square Test

Helps in decision making using categorical data
Useful in survey analysis and market research
Validates hypotheses in experimental research
Crucial in quality control and business strategies

8.0Solved Example on Chi-Square Test

Example 1: In a survey, 200 people were asked about their preference for three different products: A, B, and C. The results are shown below:

Product	Observed Frequency (O)
A	80
B	70
C	50

Assuming all products are equally popular, test the hypothesis at a 5% significance level.

Solution:

Step 1 – Null Hypothesis H_0:

All products are equally preferred. So, expected frequency E for each = _{\frac{200}{3} \approx 66.67.}

Product	O	E	(O−E)2/E(O - E)^2 / E
A	80	66.67	$\frac{( 80 - 66.67 ) ^{2}}{66.67} = \frac{177.78}{66.67} \approx 2.67$
B	70	66.67	$\frac{( 70 - 66.67 ) ^{2}}{66.67} = \frac{11.11}{66.67} \approx 0.17$
C	50	66.67	$\frac{( 50 - 66.67 ) ^{2}}{66.67} = \frac{277.78}{66.67} \approx 4.17$

Step 2 – Calculate $/ c h i^{2}$ :

$= 2.67 + 0.17 + 4.17 = 7.01 χ^{2} = 2.67 + 0.17 + 4.17 = 7.01$

Step 3 – Degrees of Freedom (df):

df = n - 1 = 3 - 1 = 2

From the chi-square table, the critical value at 5% significance level and df = 2 is 5.991.

Since 7.01 > 5.991, we reject H0H_0.

Conclusion: The product preferences are not equally distributed.

Example 2: A company studied the relationship between customer gender and their product choice. The data is:

	Product X	Product Y	Total
Male	40	60	100
Female	30	70	100
Total	70	130	200

Test whether product choice is independent of gender at 5% significance level.

Solution:

Step 1 – Compute Expected Frequencies:

	Product X	Product Y
Male	$E_{11} = \frac{100 \times 70}{200} = 35$	$E_{12} = \frac{100 \times 130}{200} = 65$
Female	$E_{21} = \frac{100 \times 70}{200} = 35$	$E_{22} = \frac{100 \times 130}{200} = 65$

Step 2 – Calculate $χ^{2}$ :

$χ^{2} = \frac{( 40 - 35 ) ^{2}}{35} + \frac{( 60 - 65 ) ^{2}}{65} + \frac{( 30 - 35 ) ^{2}}{35} + \frac{( 70 - 65 ) ^{2}}{65}$

$χ^{2} = \frac{25}{35} + \frac{25}{65} + \frac{25}{35} + \frac{25}{65} = \frac{50}{35} + \frac{50}{65}$

$χ^{2} = 1.428 + 0.769 = 2.197$

Step 3 – Degrees of Freedom (df):

df = (rows -1)(columns -1) = (2 -1)(2 -1) = 1

Critical chi-square value at 5% significance level and df =1 is 3.841.

Since 2.197 < 3.841, we accept H_0.

Conclusion: Product choice is independent of gender.

Example 3: A manufacturer claims that the proportion of defective items in a batch is 5%. From a sample of 200 items, 15 defective items were found. Test the manufacturer’s claim at a 5% significance level.

Solution:

Step 1 – Null Hypothesis $H_{0}$ :

The proportion of defective items is 5%.

Expected defective items:

$E = 0.05 \times 200 = 10$

Expected non-defective items:

E' = 200 - 10 = 190

Observed (O):

Defective: 15
Non-defective: 185

Step 2 – Calculate $χ^{2}$ :

$χ^{2} = \frac{( 15 - 10 ) ^{2}}{10} + \frac{( 185 - 190 ) ^{2}}{190} = \frac{25}{10} + \frac{25}{190}$

$χ^{2} = 2.5 + 0.1316 = 2.6316$

Step 3 – Degrees of Freedom (df):

df = number of categories – 1 = 2 – 1 = 1

Critical value at 5% significance level and df = 1 is 3.841.

Since 2.6316<3.8412.6316 < 3.841, we accept $H_{0}$ .

Conclusion: The manufacturer’s claim holds.

Example 4: A study records the following data regarding study method and exam success:

Study Method	Passed	Failed	Total
Method A	80	20	100
Method B	50	50	100
Total	130	70	200

Is there an association between study method and exam success at a 1% significance level?

Solution:

Step 1 – Expected Frequencies:

	Passed	Failed
Method A	$\frac{100 \times 130}{200} = 65$	$\frac{100 \times 70}{200} = 35$
Method B	Same as above: 65 and 35

Step 2 – Compute $χ^{2}$ :

$χ^{2} = \frac{( 80 - 65 ) ^{2}}{65} + \frac{( 20 - 35 ) ^{2}}{35} + \frac{( 50 - 65 ) ^{2}}{65} + \frac{( 50 - 35 ) ^{2}}{35}$

$χ^{2} = \frac{225}{65} + \frac{225}{35} + \frac{225}{65} + \frac{225}{35}$

$χ^{2} = 3.46 + 6.43 + 3.46 + 6.43 = 19.78$

Step 3 – Degrees of Freedom (df):

df = (2 -1)(2 -1) =1

Critical chi-square value at 1% significance level and df =1 is 6.635.

Since 19.78 > 6.635, we reject $H_{0}$ .

Conclusion: Study method is significantly associated with exam success.

Example 5: A geneticist expects that a particular plant produces flowers in the ratio of Red:White:Pink as 9:3:4. In an experiment, out of 160 plants, the observed counts were Red – 90, White – 30, Pink – 40. Test the hypothesis at 5% significance level.

Solution:

Step 1 – Expected Frequencies:

Total parts = 9 + 3 + 4 = 16

Expected Red:

$E_{Red} = \frac{9}{16} \times 160 = 90$

Expected White:

$E_{White} = \frac{3}{16} \times 160 = 30$

Expected Pink:

$E_{Pink} = \frac{4}{16} \times 160 = 40$

Step 2 – Compute $/ c h i^{2}$ :

$χ^{2} = \frac{( 90 - 90 ) ^{2}}{90} + \frac{( 30 - 30 ) ^{2}}{30} + \frac{( 40 - 40 ) ^{2}}{40} = 0$

Step 3 – Degrees of Freedom (df):

df = 3 -1 = 2

Critical value at 5% significance level and df = 2 is 5.991.

Since $χ^{2} = 0 < 5.991$ we accept $H_{0}$ .

Conclusion: The observed data perfectly matches the expected ratio.

9.0Practice Questions on Chi-Square Test

In a dice-rolling experiment of 120 rolls, the observed frequencies of outcomes 1, 2, 3, 4, 5, and 6 are: 15, 20, 18, 22, 25, and 20 respectively. Test at 5% significance level whether the dice is fair.
A survey of 150 people records the choice of car brand (Brand A or Brand B) and gender (Male or Female):

	Brand A	Brand B	Total
Male	50	40	90
Female	30	30	60
Total	80	70	150

Test whether car brand preference is independent of gender at 5% significance level.

Also Read:

Bayes Theorem	Standard Deviation	Alternative Hypothesis
Kurtosis and Skewness	Coefficient of Correlation	Spearman’s Rank Correlation
Inferential Statistics	Statistics	Taylor Series

Frequently Asked Questions

What is the main purpose of the Chi-Square Test?

What is the critical value in Chi-Square tests?

Can the Chi-Square Test be used for continuous data?

Frequently Asked Questions

What is the main purpose of the Chi-Square Test?

What is the critical value in Chi-Square tests?

Can the Chi-Square Test be used for continuous data?

Join ALLEN!

Chi-Square Test

1.0What is the Chi-Square Test?

2.0About the Chi-Square Test

3.0Calculation of the Chi-Square Test

4.0Example of Chi-Square Test Explained

5.0Bayesian Chi-Square Test

6.0Chi-Square Test A/B Test in Business Example

7.0Importance of Chi-Square Test

8.0Solved Example on Chi-Square Test

9.0Practice Questions on Chi-Square Test

Table of Contents

Join ALLEN!

Chi-Square Test

1.0What is the Chi-Square Test?

2.0About the Chi-Square Test

3.0Calculation of the Chi-Square Test

4.0Example of Chi-Square Test Explained

5.0Bayesian Chi-Square Test

6.0Chi-Square Test A/B Test in Business Example

7.0Importance of Chi-Square Test

8.0Solved Example on Chi-Square Test

9.0Practice Questions on Chi-Square Test

Table of Contents