Chi Square Goodness Of Fit Test Example

The Chi-Square Goodness-of-Fit test is a powerful statistical tool used to determine if sample data aligns with a hypothesized distribution. It essentially compares observed frequencies with expected frequencies to assess whether any differences between them are due to chance or a real discrepancy. This test is particularly useful when dealing with categorical data and wanting to see if it follows a specific pattern. Let's explore this test with detailed examples.

Understanding the Chi-Square Goodness-of-Fit Test

At its core, the Chi-Square Goodness-of-Fit test assesses how well a set of observed data "fits" a theoretical distribution. It's a non-parametric test, meaning it doesn't rely on assumptions about the underlying distribution of the data being normally distributed. Instead, it compares observed frequencies (the actual counts of data in each category) with expected frequencies (the counts we'd expect if the data perfectly followed the hypothesized distribution).

• Observed Frequencies (O): The actual counts of observations in each category from your sample data.
• Expected Frequencies (E): The counts we'd anticipate in each category if the null hypothesis (the hypothesized distribution) is true. These are calculated based on the hypothesized distribution and the total sample size.

The test statistic, Chi-Square (χ²), is calculated using the following formula:

χ² = Σ [(O - E)² / E]

Where:

• Σ represents the summation across all categories.
• O is the observed frequency for a category.
• E is the expected frequency for a category.

A larger Chi-Square value indicates a greater discrepancy between observed and expected frequencies, suggesting that the data does not fit the hypothesized distribution well.

Key Assumptions

Before applying the Chi-Square Goodness-of-Fit test, it's crucial to ensure the following assumptions are met:

1. Random Sampling: The data must be obtained through random sampling to ensure that it is representative of the population.
2. Categorical Data: The data should consist of categorical variables, where observations are classified into distinct categories.
3. Expected Frequencies: Each category should have an expected frequency of at least 5. This is a common rule of thumb to ensure the Chi-Square approximation is valid. If expected frequencies are too low, consider combining categories.
4. Independence of Observations: Each observation should be independent of the others. One observation should not influence another.

Steps to Conduct a Chi-Square Goodness-of-Fit Test

1. State the Hypotheses:

   • Null Hypothesis (H₀): The observed frequencies follow the hypothesized distribution.
   • Alternative Hypothesis (H₁): The observed frequencies do not follow the hypothesized distribution.

2. Determine the Expected Frequencies: Calculate the expected frequency for each category based on the hypothesized distribution and the total sample size.

3. Calculate the Chi-Square Test Statistic: Use the formula χ² = Σ [(O - E)² / E] to calculate the Chi-Square value.

4. Determine the Degrees of Freedom: The degrees of freedom (df) are calculated as (number of categories - 1). This reflects the number of independent pieces of information used to calculate the Chi-Square statistic.

5. Find the P-value: Using the Chi-Square value and the degrees of freedom, find the p-value from a Chi-Square distribution table or statistical software. The p-value represents the probability of observing a Chi-Square value as extreme as, or more extreme than, the one calculated, assuming the null hypothesis is true.

6. Make a Decision: Compare the p-value to the significance level (alpha, typically 0.05).

   • If p-value ≤ alpha: Reject the null hypothesis. There is sufficient evidence to conclude that the observed frequencies do not follow the hypothesized distribution.
   • If p-value > alpha: Fail to reject the null hypothesis. There is not sufficient evidence to conclude that the observed frequencies do not follow the hypothesized distribution.

Example 1: Testing for Uniform Distribution (Fair Die)

Scenario: A researcher suspects that a six-sided die is not fair. They roll the die 60 times and observe the following frequencies:

Face	Observed Frequency (O)
1	7
2	11
3	8
4	13
5	12
6	9

Step 1: State the Hypotheses

• H₀: The die is fair (the observed frequencies follow a uniform distribution).
• H₁: The die is not fair (the observed frequencies do not follow a uniform distribution).

Step 2: Determine the Expected Frequencies

If the die is fair, we would expect each face to appear an equal number of times. With 60 rolls, the expected frequency for each face is 60 / 6 = 10.

Face	Observed Frequency (O)	Expected Frequency (E)
1	7	10
2	11	10
3	8	10
4	13	10
5	12	10
6	9	10

Step 3: Calculate the Chi-Square Test Statistic

χ² = Σ [(O - E)² / E]

χ² = [(7-10)² / 10] + [(11-10)² / 10] + [(8-10)² / 10] + [(13-10)² / 10] + [(12-10)² / 10] + [(9-10)² / 10]

χ² = [9/10] + [1/10] + [4/10] + [9/10] + [4/10] + [1/10]

χ² = 0.9 + 0.1 + 0.4 + 0.9 + 0.4 + 0.1

χ² = 2.8

Step 4: Determine the Degrees of Freedom

df = (number of categories - 1) = (6 - 1) = 5

Step 5: Find the P-value

Using a Chi-Square distribution table or statistical software, with χ² = 2.8 and df = 5, the p-value is approximately 0.729.

Step 6: Make a Decision

Since the p-value (0.729) is greater than the significance level (alpha = 0.05), we fail to reject the null hypothesis.

Conclusion: There is not sufficient evidence to conclude that the die is not fair. The observed frequencies are not significantly different from what we would expect if the die were fair.

Example 2: Testing Genetic Ratios (Mendelian Genetics)

Scenario: In a genetics experiment, a researcher crosses two heterozygous pea plants for a single trait (e.g., flower color) where purple (P) is dominant over white (p). According to Mendelian genetics, the expected phenotypic ratio in the offspring is 3:1 (3 purple, 1 white). The researcher observes the following results in 200 offspring:

Phenotype	Observed Frequency (O)
Purple	160
White	40

Step 1: State the Hypotheses

• H₀: The observed phenotypic ratio follows the Mendelian ratio of 3:1.
• H₁: The observed phenotypic ratio does not follow the Mendelian ratio of 3:1.

Step 2: Determine the Expected Frequencies

Based on the 3:1 ratio, we expect 3/4 of the offspring to be purple and 1/4 to be white. With 200 offspring:

• Expected Purple: (3/4) * 200 = 150
• Expected White: (1/4) * 200 = 50

Phenotype	Observed Frequency (O)	Expected Frequency (E)
Purple	160	150
White	40	50

Step 3: Calculate the Chi-Square Test Statistic

χ² = Σ [(O - E)² / E]

χ² = [(160-150)² / 150] + [(40-50)² / 50]

χ² = [100/150] + [100/50]

χ² = 0.667 + 2

χ² = 2.667

Step 4: Determine the Degrees of Freedom

df = (number of categories - 1) = (2 - 1) = 1

Step 5: Find the P-value

Using a Chi-Square distribution table or statistical software, with χ² = 2.667 and df = 1, the p-value is approximately 0.102.

Step 6: Make a Decision

Since the p-value (0.102) is greater than the significance level (alpha = 0.05), we fail to reject the null hypothesis.

Conclusion: There is not sufficient evidence to conclude that the observed phenotypic ratio does not follow the Mendelian ratio of 3:1. The observed results are consistent with Mendelian genetics.

Example 3: Testing Consumer Preferences (Color Preference)

Scenario: A marketing company wants to determine if there is a preference for different colors of a new product. They survey 300 consumers and ask them to choose their favorite color from four options: Red, Blue, Green, and Yellow. The observed frequencies are:

Color	Observed Frequency (O)
Red	85
Blue	70
Green	65
Yellow	80

Step 1: State the Hypotheses

• H₀: There is no preference for any color (the observed frequencies follow a uniform distribution).
• H₁: There is a preference for at least one color (the observed frequencies do not follow a uniform distribution).

Step 2: Determine the Expected Frequencies

If there is no preference, we would expect each color to be chosen equally. With 300 consumers and 4 colors, the expected frequency for each color is 300 / 4 = 75.

Color	Observed Frequency (O)	Expected Frequency (E)
Red	85	75
Blue	70	75
Green	65	75
Yellow	80	75

Step 3: Calculate the Chi-Square Test Statistic

χ² = Σ [(O - E)² / E]

χ² = [(85-75)² / 75] + [(70-75)² / 75] + [(65-75)² / 75] + [(80-75)² / 75]

χ² = [100/75] + [25/75] + [100/75] + [25/75]

χ² = 1.333 + 0.333 + 1.333 + 0.333

χ² = 3.332

Step 4: Determine the Degrees of Freedom

df = (number of categories - 1) = (4 - 1) = 3

Step 5: Find the P-value

Using a Chi-Square distribution table or statistical software, with χ² = 3.332 and df = 3, the p-value is approximately 0.343.

Step 6: Make a Decision

Since the p-value (0.343) is greater than the significance level (alpha = 0.05), we fail to reject the null hypothesis.

Conclusion: There is not sufficient evidence to conclude that there is a preference for any particular color. The observed frequencies are not significantly different from what we would expect if consumers had no color preference.

Example 4: Testing a Claim About Population Proportions

Scenario: A political analyst claims that in a certain city, 40% of voters are registered as Democrats, 35% as Republicans, and 25% as Independents. A survey of 500 registered voters reveals the following:

Affiliation	Observed Frequency (O)
Democrat	180
Republican	160
Independent	160

Step 1: State the Hypotheses

• H₀: The distribution of voter affiliations in the city is 40% Democrat, 35% Republican, and 25% Independent.
• H₁: The distribution of voter affiliations in the city is different from the analyst's claim.

Step 2: Determine the Expected Frequencies

Based on the analyst's claim and the sample size of 500:

• Expected Democrats: 0.40 * 500 = 200
• Expected Republicans: 0.35 * 500 = 175
• Expected Independents: 0.25 * 500 = 125

Affiliation	Observed Frequency (O)	Expected Frequency (E)
Democrat	180	200
Republican	160	175
Independent	160	125

Step 3: Calculate the Chi-Square Test Statistic

χ² = Σ [(O - E)² / E]

χ² = [(180-200)² / 200] + [(160-175)² / 175] + [(160-125)² / 125]

χ² = [400/200] + [225/175] + [1225/125]

χ² = 2 + 1.286 + 9.8

χ² = 13.086

Step 4: Determine the Degrees of Freedom

df = (number of categories - 1) = (3 - 1) = 2

Step 5: Find the P-value

Using a Chi-Square distribution table or statistical software, with χ² = 13.086 and df = 2, the p-value is approximately 0.001.

Step 6: Make a Decision

Since the p-value (0.001) is less than the significance level (alpha = 0.05), we reject the null hypothesis.

Conclusion: There is sufficient evidence to conclude that the distribution of voter affiliations in the city is different from the analyst's claim. The observed frequencies deviate significantly from the expected frequencies based on the analyst's percentages.

Considerations and Limitations

While the Chi-Square Goodness-of-Fit test is a valuable tool, it's important to be aware of its limitations:

• Sensitivity to Sample Size: With large sample sizes, even small deviations from the hypothesized distribution can lead to statistically significant results. Conversely, with small sample sizes, the test may lack the power to detect real differences.
• Expected Frequency Rule: The rule of thumb of having expected frequencies of at least 5 in each category is crucial for the validity of the test. If this assumption is violated, consider combining categories or using an alternative test.
• Doesn't Indicate the Nature of the Difference: The test only indicates whether there is a difference between the observed and expected frequencies; it doesn't tell you how they differ. Further analysis is needed to understand the specific patterns of deviation.

Alternatives to the Chi-Square Goodness-of-Fit Test

If the assumptions of the Chi-Square Goodness-of-Fit test are not met, or if you are interested in different types of analyses, alternative tests may be more appropriate:

• Kolmogorov-Smirnov Test: This test can be used for continuous data to compare the empirical distribution function of the sample data to a hypothesized cumulative distribution function.
• Anderson-Darling Test: Another test for continuous data that is more sensitive to differences in the tails of the distribution than the Kolmogorov-Smirnov test.
• Fisher's Exact Test: This test is suitable for small sample sizes and 2x2 contingency tables when the expected frequencies are low.

Conclusion

The Chi-Square Goodness-of-Fit test is a versatile statistical tool for evaluating whether observed categorical data aligns with a hypothesized distribution. By understanding the underlying principles, assumptions, and steps involved in conducting the test, researchers and analysts can effectively use it to draw meaningful conclusions from their data. Remember to carefully consider the limitations of the test and explore alternative methods when necessary to ensure the validity and robustness of your findings. Through careful application and interpretation, the Chi-Square Goodness-of-Fit test can provide valuable insights into the nature of your data and its adherence to expected patterns.