ANOVA & Chi-Square Tests: Hypothesis, F-Test, Goodness-of-Fit

Key Concepts & Definitions

1. Analysis of Variance (ANOVA)

ANOVA is a statistical method used to compare the means of three or more groups to determine if at least one group mean is different from the others. It partitions total variability into:

• Between-group variability (due to differences between group means)
• Within-group variability (due to natural variation within each group)

Assumptions of ANOVA

mindmap
  root((ANOVA Assumptions))
    Normality["Data in each group is normally distributed"]
    Homogeneity["Variances (σ²) are equal across groups (homoscedasticity)"]
    Independence["Observations are independent"]
    Random["Samples are randomly selected"]

Steps in ANOVA

1. State Hypotheses:
   • $H_0$: All group means are equal (${\mu }_1={\mu }_2={\mu }_3$)
   • $H_1$: At least one group mean is different.
2. Calculate Sum of Squares (SS):
   • Total SS (SST): Total variation in data.
   • Between-group SS (SSB): Variation due to group differences.
   • Within-group SS (SSW): Variation within groups.
3. Compute Mean Squares (MS):
   • $M{S}_B=\frac{SSB}{d{f}_B}$ (Between-group mean square)
   • $M{S}_W=\frac{SSW}{d{f}_W}$ (Within-group mean square)
4. Calculate F-statistic: $$F=\frac{M{S}_B}{M{S}_W}$$
5. Compare with Critical F-value (from F-distribution table) or compute p-value.

2. Chi-Square Tests

Chi-Square tests analyze categorical data (frequencies) and include:

• Goodness-of-Fit Test: Checks if observed frequencies match expected frequencies.
• Test of Independence: Tests if two categorical variables are independent.

Assumptions of Chi-Square Tests

mindmap
  root((Chi-Square Assumptions))
    Independence["Observations are independent"]
    Expected["Expected frequency ≥5 in ≥80% of cells"]
    Categorical["Data is categorical (counts/frequencies)"]

Steps in Chi-Square Test

1. State Hypotheses:
   • $H_0$: Observed frequencies = Expected frequencies (or variables are independent).
   • $H_1$: They are not equal (or variables are dependent).
2. Compute Chi-Square Statistic: $${\chi }^2=\sum \frac{(O_i-E_i{)}^2}{E_i}$$ where $O_i$ = observed frequency, $E_i$ = expected frequency.
3. Determine Critical Value (from Chi-Square table) or compute p-value.

Worked Examples

Example 1: One-Way ANOVA (Comparing 3 Groups)

Scenario: A company tests the effectiveness of 3 different advertising methods (TV, Radio, Social Media) on sales. Sample data (in Rs. '000):

Step 1: Calculate SSB (Between-group SS) $$SSB=\sum n_i({\bar{X}}_i-\bar{X}{)}^2$$ where $\bar{X}=\frac{50+45+60}{3}=51.67$

$$SSB=10(50-51.67{)}^2+10(45-51.67{)}^2+10(60-51.67{)}^2=333.67$$

Step 2: Calculate SSW (Within-group SS) $$SSW=(n_1-1)s_1^2+(n_2-1)s_2^2+(n_3-1)s_3^2$$ $$SSW=9(100)+9(80)+9(120)=2430$$

Step 3: Degrees of Freedom $$d{f}_B=k-1=3-1=2(k=\text{number of groups})$$ $$d{f}_W=N-k=30-3=27(N=\text{total observations})$$

Step 4: Mean Squares $$M{S}_B=\frac{SSB}{d{f}_B}=\frac{333.67}{2}=166.835$$ $$M{S}_W=\frac{SSW}{d{f}_W}=\frac{2430}{27}=90$$

Step 5: F-statistic $$F=\frac{M{S}_B}{M{S}_W}=\frac{166.835}{90}=1.854$$

Step 6: Compare with Critical F-value From F-table ($d{f}_B=2,d{f}_W=27,\alpha =0.05$): Critical $F=3.35$.

Since $1.854<3.35$, fail to reject $H_0$. No significant difference in sales across advertising methods.

Example 2: Chi-Square Goodness-of-Fit Test

Scenario: A factory claims 20% of its products are defective. In a sample of 200, 50 were defective. Test the claim at $\alpha =0.05$.

Step 1: State Hypotheses

• $H_0$: $p=0.20$ (20% defective)
• $H_1$: $p\ne 0.20$

Step 2: Expected Frequencies $$E=np=200\times 0.20=40\text{ (defective)},160\text{ (non-defective)}$$

Step 3: Chi-Square Statistic $${\chi }^2=\frac{(50-40{)}^2}{40}+\frac{(150-160{)}^2}{160}=\frac{100}{40}+\frac{100}{160}=2.5+0.625=3.125$$

Step 4: Critical Value From Chi-Square table ($df=1,\alpha =0.05$): Critical ${\chi }^2=3.841$.

Since $3.125<3.841$, fail to reject $H_0$. The claim is not rejected.

In the Real World

1. ANOVA in E-Sewa & Khalti (Nepal)

• Scenario: E-Sewa tests 3 payment methods (Mobile Banking, Credit Card, UPI) to see which drives higher transaction success rates.
• How ANOVA Helps:
  • Compares mean success rates across methods.
  • If $F>\text{critical value}$, at least one method performs significantly better.
  • Example: If Mobile Banking has a higher mean success rate, E-Sewa can prioritize it.

2. Chi-Square in Daraz & Pathao (Market Segmentation)

• Scenario: Daraz wants to check if customer preferences for product categories (Electronics, Fashion, Grocery) are independent of age groups (18-25, 26-35, 36+).
• How Chi-Square Helps:
  • Tests if age group affects product choice.
  • If ${\chi }^2>\text{critical value}$, preferences are not independent (e.g., younger users prefer Electronics).
  • Example: Pathao uses this to tailor promotions (e.g., discounts on food for 26-35 age group).

3. ANOVA in NTC (Network Performance)

• Scenario: NTC tests 3 internet service providers (Ncell, NTC, Smart) for download speeds in Kathmandu.
• How ANOVA Helps:
  • Compares mean speeds across providers.
  • If $F$ is significant, NTC can identify the best-performing ISP for upgrades.

Comparison Table: ANOVA vs. Chi-Square

Exam Tip

1. ANOVA Questions:

   • Always state hypotheses clearly.
   • Show all calculations (SSB, SSW, MS, F).
   • Use F-table to compare critical values.
   • Interpret results: "Reject $H_0$" means at least one group is different.

2. Chi-Square Questions:

   • Expected frequencies must be calculated carefully ($E=\text{row total}\times \text{column total}/\text{grand total}$).
   • Check assumptions (e.g., no expected frequency <5).
   • Degrees of freedom for independence test: $(r-1)(c-1)$ (rows × columns).

3. Common Mistakes to Avoid:

   • Forgetting to subtract 1 for degrees of freedom in ANOVA.
   • Misapplying Chi-Square to continuous data (use ANOVA instead).
   • Ignoring assumptions (e.g., non-normal data in ANOVA).