Chi-Square Test Examples | Statistics Examples

Explore practical examples of Chi-Square tests to understand their applications in statistics.
By Jamie

Understanding the Chi-Square Test

The Chi-Square Test is a statistical method used to determine whether there is a significant association between categorical variables. It helps researchers understand if the observed frequencies in different categories differ from expected frequencies. Here, we present three diverse examples to illustrate how the Chi-Square Test can be applied in real-world scenarios.

Example 1: Assessing Gender Preferences in Product Choice

In a market research study, a company wants to evaluate whether there is a gender preference for a new beverage product. They survey 200 customers, recording their gender and their preferred beverage (Product A or Product B).

Gender Product A Product B Total
Male 70 30 100
Female 40 60 100
Total 110 90 200

To determine if gender influences product choice, we conduct a Chi-Square Test:

  1. Null Hypothesis (H0): There is no significant association between gender and product preference.
  2. Expected Frequencies: Calculate expected counts based on the total proportions:
  • Expected for Male - Product A: (100 * 110) / 200 = 55
  • Expected for Male - Product B: (100 * 90) / 200 = 45
  • Expected for Female - Product A: (100 * 110) / 200 = 55

  • Expected for Female - Product B: (100 * 90) / 200 = 45

    1. Chi-Square Calculation:

  • χ² = Σ((Observed - Expected)² / Expected)

  • χ² = ((70-55)² / 55) + ((30-45)² / 45) + ((40-55)² / 55) + ((60-45)² / 45)
  • χ² ≈ 11.11
    1. Degrees of Freedom (df): (rows - 1) * (columns - 1) = (2-1)*(2-1) = 1
    2. Significance Level: Compare χ² with the critical value from the Chi-Square distribution table for df = 1 at α = 0.05.

If χ² exceeds the critical value, we reject H0, indicating a significant association between gender and product preference.

Notes

This example helps marketers understand customer preferences, guiding them in product development and advertising strategies.

Example 2: Evaluating Treatment Effectiveness in Medicine

A clinical trial is conducted to compare the effectiveness of two treatments for a health condition. Researchers categorize patients into two groups based on their treatment (Treatment 1 or Treatment 2) and their recovery status (Recovered or Not Recovered).

Treatment Recovered Not Recovered Total
Treatment 1 30 10 40
Treatment 2 20 20 40
Total 50 30 80

The researchers want to determine if the type of treatment affects recovery rates using a Chi-Square Test:

  1. Null Hypothesis (H0): There is no significant association between treatment type and recovery status.
  2. Expected Frequencies: Calculate expected counts:
  • Expected for Treatment 1 - Recovered: (40 * 50) / 80 = 25
  • Expected for Treatment 1 - Not Recovered: (40 * 30) / 80 = 15
  • Expected for Treatment 2 - Recovered: (40 * 50) / 80 = 25

  • Expected for Treatment 2 - Not Recovered: (40 * 30) / 80 = 15

    1. Chi-Square Calculation:

  • χ² = ((30-25)² / 25) + ((10-15)² / 15) + ((20-25)² / 25) + ((20-15)² / 15)

  • χ² ≈ 3.33
    1. Degrees of Freedom (df): (2-1)*(2-1) = 1
    2. Significance Level: Compare χ² with the critical value at df = 1 and α = 0.05.

If the calculated χ² is greater than the critical value, it suggests a significant effect of treatment type on recovery.

Notes

This example illustrates how Chi-Square tests can help assess the effectiveness of medical treatments, informing healthcare decisions and practices.

Example 3: Analyzing Voting Behavior by Age Group

A political analyst wants to determine if voting behavior is influenced by age. They survey a sample of 300 voters, categorizing them into three age groups (18-30, 31-50, 51+) and recording their voting choices (Candidate X or Candidate Y).

Age Group Candidate X Candidate Y Total
18-30 60 40 100
31-50 70 50 120
51+ 30 50 80
Total 160 140 300

To investigate the relationship between age group and voting choice, a Chi-Square Test is performed:

  1. Null Hypothesis (H0): There is no significant association between age group and voting choice.
  2. Expected Frequencies: Calculate expected counts based on total proportions:
  • Expected for 18-30 - Candidate X: (100 * 160) / 300 = 53.33
  • Expected for 18-30 - Candidate Y: (100 * 140) / 300 = 46.67
  • Expected for 31-50 - Candidate X: (120 * 160) / 300 = 64
  • Expected for 31-50 - Candidate Y: (120 * 140) / 300 = 56
  • Expected for 51+ - Candidate X: (80 * 160) / 300 = 42.67

  • Expected for 51+ - Candidate Y: (80 * 140) / 300 = 37.33

    1. Chi-Square Calculation:

  • χ² = Σ((Observed - Expected)² / Expected)

  • This involves calculating for each cell and summing the values.
    1. Degrees of Freedom (df): (3-1)*(2-1) = 2
    2. Significance Level: Compare χ² with the critical value at df = 2 and α = 0.05.

If the calculated χ² exceeds the critical value, it suggests a significant association between age group and voting choice.

Notes

This analysis can provide valuable insights for political campaigns, helping them tailor strategies to different demographic segments.