Example Of Chi Square Test Of Independence

The Chi-Square Test of Independence is a powerful statistical tool used to determine if there is a significant association between two categorical variables. It's a cornerstone technique in fields like market research, sociology, and healthcare, allowing researchers to move beyond simple observation and infer relationships between different categories. This article delves into the Chi-Square Test of Independence, providing real-world examples to illustrate its application and interpretation.

Understanding the Chi-Square Test of Independence

Before diving into examples, it's crucial to grasp the fundamental concepts. The Chi-Square Test of Independence assesses whether the observed frequencies of two categorical variables differ significantly from the frequencies we would expect if there were no association between them.

• Null Hypothesis (H0): There is no association between the two categorical variables. They are independent.
• Alternative Hypothesis (H1): There is an association between the two categorical variables. They are dependent.

The test relies on calculating a chi-square statistic, which measures the discrepancy between the observed and expected frequencies. A large chi-square statistic suggests a strong association, leading to rejection of the null hypothesis. The p-value, derived from the chi-square statistic, indicates the probability of observing the data (or more extreme data) if the null hypothesis were true. A small p-value (typically less than 0.05) provides evidence against the null hypothesis, suggesting that the association is statistically significant.

Example 1: Smoking and Lung Cancer

Let's consider a classic example: investigating the relationship between smoking and lung cancer. Researchers want to determine if there's a statistically significant association between these two variables.

Data Collection:

A study is conducted on a sample of 500 individuals. Participants are classified based on their smoking status (Smoker or Non-Smoker) and whether they have been diagnosed with lung cancer (Yes or No). The observed frequencies are summarized in the following contingency table:

Calculating Expected Frequencies:

The expected frequencies represent the values we would expect in each cell if smoking and lung cancer were independent. To calculate the expected frequency for each cell, we use the following formula:

Expected Frequency = (Row Total * Column Total) / Grand Total

• Expected Frequency (Smoker, Lung Cancer Yes) = (200 * 80) / 500 = 32
• Expected Frequency (Smoker, Lung Cancer No) = (200 * 420) / 500 = 168
• Expected Frequency (Non-Smoker, Lung Cancer Yes) = (300 * 80) / 500 = 48
• Expected Frequency (Non-Smoker, Lung Cancer No) = (300 * 420) / 500 = 252

The table of expected frequencies is:

Calculating the Chi-Square Statistic:

The chi-square statistic is calculated using the following formula:

χ² = Σ [(Observed Frequency - Expected Frequency)² / Expected Frequency]

χ² = [(60 - 32)² / 32] + [(140 - 168)² / 168] + [(20 - 48)² / 48] + [(280 - 252)² / 252]

χ² = (28² / 32) + (-28² / 168) + (-28² / 48) + (28² / 252)

χ² = (784 / 32) + (784 / 168) + (784 / 48) + (784 / 252)

χ² = 24.5 + 4.67 + 16.33 + 3.11

χ² = 48.61

Determining Degrees of Freedom:

The degrees of freedom (df) are calculated as:

df = (Number of Rows - 1) * (Number of Columns - 1)

In this case, df = (2 - 1) * (2 - 1) = 1

Finding the P-Value:

Using a chi-square distribution table or statistical software with df = 1, we find that the p-value associated with a chi-square statistic of 48.61 is extremely small (p < 0.0001).

Interpreting the Results:

Since the p-value (p < 0.0001) is less than the significance level (typically 0.05), we reject the null hypothesis. This indicates that there is a statistically significant association between smoking and lung cancer. The data provides strong evidence that smoking is related to an increased risk of developing lung cancer.

Example 2: Gender and Political Affiliation

A political analyst wants to investigate whether there is a relationship between gender and political affiliation. They collect data from a random sample of 400 registered voters.

Data Collection:

The data is categorized by gender (Male or Female) and political affiliation (Democrat, Republican, or Independent). The observed frequencies are shown in the following contingency table:

Calculating Expected Frequencies:

Using the same formula as before:

• Expected Frequency (Male, Democrat) = (180 * 140) / 400 = 63
• Expected Frequency (Male, Republican) = (180 * 130) / 400 = 58.5
• Expected Frequency (Male, Independent) = (180 * 130) / 400 = 58.5
• Expected Frequency (Female, Democrat) = (220 * 140) / 400 = 77
• Expected Frequency (Female, Republican) = (220 * 130) / 400 = 71.5
• Expected Frequency (Female, Independent) = (220 * 130) / 400 = 71.5

The table of expected frequencies is:

Calculating the Chi-Square Statistic:

χ² = Σ [(Observed Frequency - Expected Frequency)² / Expected Frequency]

χ² = [(60 - 63)² / 63] + [(80 - 58.5)² / 58.5] + [(40 - 58.5)² / 58.5] + [(80 - 77)² / 77] + [(50 - 71.5)² / 71.5] + [(90 - 71.5)² / 71.5]

χ² = (9 / 63) + (462.25 / 58.5) + (342.25 / 58.5) + (9 / 77) + (462.25 / 71.5) + (342.25 / 71.5)

χ² = 0.14 + 7.90 + 5.85 + 0.12 + 6.46 + 4.79

χ² = 25.26

Determining Degrees of Freedom:

df = (Number of Rows - 1) * (Number of Columns - 1)

In this case, df = (2 - 1) * (3 - 1) = 2

Finding the P-Value:

Using a chi-square distribution table or statistical software with df = 2, we find that the p-value associated with a chi-square statistic of 25.26 is very small (p < 0.0001).

Interpreting the Results:

Since the p-value (p < 0.0001) is less than the significance level (0.05), we reject the null hypothesis. This indicates that there is a statistically significant association between gender and political affiliation. The data suggests that gender and political affiliation are not independent in this sample.

Example 3: Education Level and Income Bracket

A researcher wants to explore whether there is a relationship between education level and income bracket. They survey a sample of 1000 adults.

Data Collection:

The data is categorized by education level (High School, Bachelor's Degree, Graduate Degree) and income bracket (Low, Medium, High). The observed frequencies are shown in the following contingency table:

Calculating Expected Frequencies:

• Expected Frequency (High School, Low) = (400 * 350) / 1000 = 140
• Expected Frequency (High School, Medium) = (400 * 450) / 1000 = 180
• Expected Frequency (High School, High) = (400 * 250) / 1000 = 100
• Expected Frequency (Bachelor's Degree, Low) = (400 * 350) / 1000 = 140
• Expected Frequency (Bachelor's Degree, Medium) = (400 * 450) / 1000 = 180
• Expected Frequency (Bachelor's Degree, High) = (400 * 250) / 1000 = 100
• Expected Frequency (Graduate Degree, Low) = (200 * 350) / 1000 = 70
• Expected Frequency (Graduate Degree, Medium) = (200 * 450) / 1000 = 90
• Expected Frequency (Graduate Degree, High) = (200 * 250) / 1000 = 50

The table of expected frequencies is:

Calculating the Chi-Square Statistic:

χ² = Σ [(Observed Frequency - Expected Frequency)² / Expected Frequency]

χ² = [(200 - 140)² / 140] + [(150 - 180)² / 180] + [(50 - 100)² / 100] + [(100 - 140)² / 140] + [(200 - 180)² / 180] + [(100 - 100)² / 100] + [(50 - 70)² / 70] + [(100 - 90)² / 90] + [(50 - 50)² / 50]

χ² = (3600 / 140) + (900 / 180) + (2500 / 100) + (1600 / 140) + (400 / 180) + (0 / 100) + (400 / 70) + (100 / 90) + (0 / 50)

χ² = 25.71 + 5 + 25 + 11.43 + 2.22 + 0 + 5.71 + 1.11 + 0

χ² = 76.18

Determining Degrees of Freedom:

df = (Number of Rows - 1) * (Number of Columns - 1)

In this case, df = (3 - 1) * (3 - 1) = 4

Finding the P-Value:

Using a chi-square distribution table or statistical software with df = 4, we find that the p-value associated with a chi-square statistic of 76.18 is extremely small (p < 0.0001).

Interpreting the Results:

Since the p-value (p < 0.0001) is less than the significance level (0.05), we reject the null hypothesis. This indicates that there is a statistically significant association between education level and income bracket. The data suggests that higher levels of education are associated with different income brackets.

Important Considerations and Assumptions

While the Chi-Square Test of Independence is a valuable tool, it's essential to be aware of its limitations and assumptions:

• Categorical Data: The test is specifically designed for categorical variables. It should not be used with continuous data.
• Independence of Observations: Each observation should be independent of the others. This means that one participant's response should not influence another participant's response.
• Expected Frequencies: The test is most reliable when the expected frequencies are sufficiently large. A common rule of thumb is that all expected frequencies should be at least 5. If this assumption is violated, consider combining categories or using a different statistical test, such as Fisher's exact test.
• Sample Size: A sufficiently large sample size is necessary to ensure the validity of the test. Smaller sample sizes can lead to inaccurate results.
• Association, Not Causation: The Chi-Square Test of Independence can only demonstrate an association between two variables. It cannot prove causation. Even if a statistically significant association is found, it does not necessarily mean that one variable causes the other. There may be other confounding variables influencing the relationship.

Alternatives to the Chi-Square Test of Independence

When the assumptions of the Chi-Square Test of Independence are not met, or when dealing with different types of data, alternative statistical tests may be more appropriate:

• Fisher's Exact Test: This test is used when dealing with small sample sizes or when expected frequencies are less than 5. It provides an exact p-value, making it more accurate than the Chi-Square Test in these situations.
• Yates' Correction for Continuity: This correction is sometimes applied to the Chi-Square Test when dealing with 2x2 contingency tables (two rows and two columns) to improve the accuracy of the p-value, especially with smaller sample sizes. However, its use is debated, and some statisticians advise against it.
• McNemar's Test: This test is used when analyzing paired or matched data, where the same subjects are measured twice on a categorical variable (e.g., before and after an intervention).
• Correlation and Regression Analysis: For examining relationships between continuous variables, correlation and regression analysis are more suitable.

Conclusion

The Chi-Square Test of Independence is a fundamental statistical tool for analyzing the relationship between two categorical variables. By comparing observed and expected frequencies, researchers can determine whether an association is statistically significant. The examples provided illustrate the application and interpretation of the test in various contexts, from public health to political science. However, it's crucial to be aware of the assumptions and limitations of the test and to consider alternative methods when appropriate. Understanding these nuances ensures that the Chi-Square Test of Independence is used effectively and responsibly to draw meaningful conclusions from data. Remember that statistical significance does not necessarily imply practical significance or causation, and further investigation may be needed to fully understand the nature of the relationship between the variables.