Lecture 18: Chi Square Goodness of Fit Test Notes

Overview of Chi-Square Distribution

• Definition: The Chi-Square distribution is a probability distribution characterized by being positively skewed.
• Parameter: It features an associated degrees of freedom ($df$) parameter which determines the specific shape of the distribution.
• Primary Applications in Statistical Inference:     * Test of Independence: Used to determine if there is a significant relationship between two categorical variables.     * Goodness of Fit Test: Used to determine how well an observed set of data fits a theoretical distribution.
• Source Reference: Weiss, p. 599.

Chi-Square Goodness of Fit Test Fundamentals

• Purpose: This test is utilized to assess whether the observed sample distribution of a variable (which can be either quantitative or qualitative) agrees with a pre-specified theoretical population pattern or distribution.
• Comparison Basis: The test relies on comparing the observed frequency ($O$) of a specific value or category to the expected frequency ($E$).
• Expected Frequency Calculation: The expected frequency is calculated under the assumption that the null hypothesis ($H_0$) is true, using the formula:     * $E = n imes p$     * Where $n$ is the total sample size and $p$ is the probability or proportion of the category under the assumed distribution ($0 \le p \le 1$).
• Logic of the Test: If the observed frequencies and expected frequencies differ significantly, the test leads to the rejection of the assumed distribution as a valid model for the observed pattern.
• Null and Alternative Hypotheses:     * $H_0$: The observed distribution matches the theoretical/past distribution.     * $H_1$: The observed distribution differs from the theoretical/past distribution.
• The Test Statistic: The Chi-Square test statistic is calculated as:     * $\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$     * This statistic follows a Chi-Square distribution with degrees of freedom $df = k - 1$, where $k$ represents the number of different groups or categories in the distribution.

Requirements and Assumptions for the Goodness of Fit Test

• The Rule of Five: Generally, the test is considered valid if each of the expected frequencies is at least $5$.
• Statistical Caveat: Dr. Javed Iqbal notes that on p. 603 of Weiss, it is mentioned that the statistician Cochran suggests the "rule of 5" is often too restrictive for practical application.

Example 1: Market Share for Fabric Softener (Company A)

• Context: Company A launched an aggressive advertising campaign to maintain or increase its market share.
• Prior Market Distribution:     * Company A: $45\,\%$     * Company B (Main Competitor): $40\,\%$     * Other Competitors: $15\,\%$
• Sample Data: A random sample of $n = 200$ customers was surveyed after the campaign.     * Observed Preference for Company A: $102$     * Observed Preference for Company B: $82$     * Observed Preference for Others: $16$
• Hypotheses:     * $H_0$: Market share distribution after campaign is same as the past market share distribution.     * $H_1$: Market share distribution after campaign is different from the past market share distribution.
• Calculation Table ($n = 200$):     * Category A: $O = 102$; $E = 200 \times 0.45 = 90$; $\frac{(O - E)^2}{E} = \frac{(102 - 90)^2}{90} = 1.60$     * Category B: $O = 82$; $E = 200 \times 0.40 = 80$; $\frac{(O - E)^2}{E} = \frac{(82 - 80)^2}{80} = 0.05$     * Others: $O = 16$; $E = 30$ (derived as $200 - 90 - 80$ to minimize rounding error); $\frac{(O - E)^2}{E} = \frac{(16 - 30)^2}{30} = 6.53$
• Summation of Test Statistic: $\chi^2 = 1.60 + 0.05 + 6.53 = 8.183$
• Decision Criteria:     * Significance Level ($\alpha$): $5\,\%$     * Degrees of Freedom ($df$): $3 - 1 = 2$     * Critical Value: $\chi^2(0.05, 2) = 5.991$
• Conclusion: Since the calculated statistic ($8.183$) is greater than the critical value ($5.991$), the null hypothesis is rejected. The market share distribution has changed.
• Observation: Company A successfully increased its market share, seemingly at the expense of "Other" competitors rather than its main competitor, Company B.

Weiss Example 13.2: Violent Crime Patterns

• Objective: To determine if the observed pattern of violent crimes in a recent year (Weiss Table 13.2) is the same as the pattern from the year 2010 (Weiss Table 13.1).
• Hypotheses:     * $H_0$: Crime distribution of last year is same as the 2010 crime distribution.     * $H_1$: Crime distribution of last year has changed from the 2010 crime distribution.
• Statistical Data:     * Calculated Test Statistic: $6.529$     * Significance Level: $5\,\%$     * Degrees of Freedom ($df$): $3$     * Critical Value (from Anderson Table 3): $\chi^2(0.05, 3) = 7.815$
• Conclusion: There is insufficient evidence in the data to conclude that the violent crime distribution in the last year differs from the 2010 distribution (calculated value 6.529 < 7.815).

Example 2: ABO Blood Type Distribution

• Prior Population Data (Believed Distribution):     * Type A: $34\,\%$     * Type B: $15\,\%$     * Type AB: $23\,\%$     * Type O: $28\,\%$
• Sample Data: $n = 100$ students from a campus.     * Observed Frequencies: A = $29$, B = $17$, AB = $20$, O = $34$.
• Hypotheses:     * $H_0$: Observed blood group distribution of campus students agrees with population distribution.     * $H_1$: Observed blood group distribution of campus students does not agree with population distribution.
• Calculation Table ($n = 100$):     * Type A: $O = 29$; $E = 34$; $\frac{(O - E)^2}{E} = 0.7352$     * Type B: $O = 17$; $E = 15$; $\frac{(O - E)^2}{E} = 0.2666$     * Type AB: $O = 20$; $E = 23$; $\frac{(O - E)^2}{E} = 0.3913$     * Type O: $O = 34$; $E = 28$; $\frac{(O - E)^2}{E} = 1.2857$
• Summation of Test Statistic: $\chi^2 = 2.6789$
• Decision Criteria:     * Degrees of Freedom ($df$): $4 - 1 = 3$     * Significance Level: $5\,\%$     * Critical Value: $\chi^2(0.05, 3) = 7.815$
• Conclusion: The calculated test statistic ($2.6789$) is less than the critical value ($7.815$). Therefore, the null hypothesis is not rejected. The campus blood group distribution matches the broad population.

Weiss Example 13.27: Color Distribution

• Overview: This data set analyzes color percentages and observed frequencies for a sample where the sum of observed frequencies is $509$.
• Calculation Table:     * Brown: Percentage = $30\,\%$, $E = 152.7$, $O = 152$, $\frac{(O - E)^2}{E} = 0.0032$     * Yellow: Percentage = $20\,\%$, $E = 101.8$, $O = 114$, $\frac{(O - E)^2}{E} = 1.4620$     * Red: Percentage = $20\,\%$, $E = 101.8$, $O = 106$, $\frac{(O - E)^2}{E} = 0.1732$     * Orange: Percentage = $10\,\%$, $E = 50.9$, $O = 51$, $\frac{(O - E)^2}{E} = 0.0002$     * Green: Percentage = $10\,\%$, $E = 50.9$, $O = 43$, $\frac{(O - E)^2}{E} = 1.226$     * Blue: Percentage = $10\,\%$, $E = 50.9$, $O = 43$, $\frac{(O - E)^2}{E} = 1.226$
• Sums: Total Expected = $509$, Total Observed = $509$, Calculated $\chi^2$ statistic = $4.091$.

Reference Exercises

• Anderson (pdf p. 611): Exercises 22, 23, and 24 are listed for further study of these concepts.