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.

${\chi }^2=\sum \frac{(O_i-E_i{)}^2}{E_i}$

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:

1. Chi-Square Test for Independence – Checks if two variables are related.
2. 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:

1. Step 1: Set up hypotheses.
2. • $H_0$: Variables are independent.
   • $H_1$: Variables are dependent.
2. Step 2: Prepare the contingency table (Observed frequencies).
3. Step 3: Calculate Expected frequencies using:

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

4. Step 4: Compute ${\chi }^2$ :

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

5. 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:

Solution: 

Step 1: Calculate Expected Frequencies:

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

Similarly:

• $E_{(Female,Tea)}$ = 25
• $E_{(Male,Coffee)}$ = 25
• $E_{(Female,Coffee)}$ = 25

Step 2: Compute $/ch{i}^2$:

${\chi }^2=\frac{(30-25{)}^2}{25}+\frac{(20-25{)}^2}{25}+\frac{(20-25{)}^2}{25}+\frac{(30-25{)}^2}{25}$${\chi }^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 ${\chi }^2=4>3.841$, 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:

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:

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.}

Step 2 – Calculate $/ch{i}^2$:

$=2.67+0.17+4.17=7.01{\chi }^2=2.67+0.17+4.17=7.01$

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:

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

Solution:

Step 1 – Compute Expected Frequencies:

Step 2 – Calculate ${\chi }^2$ :

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

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

${\chi }^2=1.428+0.769=2.197$

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 ${\chi }^2$ : 

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

${\chi }^2=2.5+0.1316=2.6316$

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:

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

Solution:

Step 1 – Expected Frequencies:

Step 2 – Compute ${\chi }^2$ : 

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

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

${\chi }^2=3.46+6.43+3.46+6.43=19.78$

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:

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

Expected White:

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

Expected Pink:

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

Step 2 – Compute $/ch{i}^2$ :

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

Step 3 – Degrees of Freedom (df):

df = 3 -1 = 2 

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

Since ${\chi }^2=0<5.991$ we accept $H_0$.

Conclusion: The observed data perfectly matches the expected ratio.

9.0Practice Questions on Chi-Square Test

1. 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.
2. A survey of 150 people records the choice of car brand (Brand A or Brand B) and gender (Male or Female):

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

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