1. **Problem Statement:** We have two problems: (a) Test if treatment means are equal using ANOVA at 0.05 significance. (b) Test if $x$ is necessary to predict $y$ using regression analysis. --- ### Problem 6: ANOVA for Treatment Means 2. **Hypotheses:** - Null hypothesis $H_0$: All treatment means are equal, i.e., $\mu_1 = \mu_2 = \mu_3 = \mu_4$ - Alternative hypothesis $H_a$: At least one treatment mean differs. 3. **Decision Rule:** - Use significance level $\alpha=0.05$. - Reject $H_0$ if $F_{calculated} > F_{critical}$ from $F$-distribution with $(k-1, N-k)$ degrees of freedom, where $k=4$ treatments, $N=16$ total observations. 4. **Data Summary:** - Treatment 1: 8,6,10,9 - Treatment 2: 3,2,4,3 - Treatment 3: 3,4,5,4 - Treatment 4: 5,4,4,5 5. **Calculate means:** - $\bar{X}_1 = \frac{8+6+10+9}{4} = \frac{33}{4} = 8.25$ - $\bar{X}_2 = \frac{3+2+4+3}{4} = \frac{12}{4} = 3$ - $\bar{X}_3 = \frac{3+4+5+4}{4} = \frac{16}{4} = 4$ - $\bar{X}_4 = \frac{5+4+4+5}{4} = \frac{18}{4} = 4.5$ - Overall mean $\bar{X} = \frac{33+12+16+18}{16} = \frac{79}{16} = 4.9375$ 6. **Calculate SStr (Sum of Squares for Treatments):** $$ SStr = n \sum_{i=1}^k (\bar{X}_i - \bar{X})^2 = 4[(8.25-4.9375)^2 + (3-4.9375)^2 + (4-4.9375)^2 + (4.5-4.9375)^2] $$ Calculate each term: - $(8.25-4.9375)^2 = 11.3906$ - $(3-4.9375)^2 = 3.7539$ - $(4-4.9375)^2 = 0.8789$ - $(4.5-4.9375)^2 = 0.1914$ Sum inside brackets = $11.3906 + 3.7539 + 0.8789 + 0.1914 = 16.2148$ So, $$ SStr = 4 \times 16.2148 = 64.8592 $$ 7. **Calculate SSE (Sum of Squares for Error):** Sum of squared deviations within each treatment: - Treatment 1: $(8-8.25)^2 + (6-8.25)^2 + (10-8.25)^2 + (9-8.25)^2 = 0.0625 + 5.0625 + 3.0625 + 0.5625 = 8.75$ - Treatment 2: $(3-3)^2 + (2-3)^2 + (4-3)^2 + (3-3)^2 = 0 + 1 + 1 + 0 = 2$ - Treatment 3: $(3-4)^2 + (4-4)^2 + (5-4)^2 + (4-4)^2 = 1 + 0 + 1 + 0 = 2$ - Treatment 4: $(5-4.5)^2 + (4-4.5)^2 + (4-4.5)^2 + (5-4.5)^2 = 0.25 + 0.25 + 0.25 + 0.25 = 1$ Sum all: $8.75 + 2 + 2 + 1 = 13.75$ 8. **Calculate SST (Total Sum of Squares):** $$ SST = \sum_{i=1}^k \sum_{j=1}^n (X_{ij} - \bar{X})^2 = SStr + SSE = 64.8592 + 13.75 = 78.6092 $$ 9. **ANOVA Table:** | Source | df | SS | MS = SS/df | F = MS_treatment / MS_error | |-------------|----------|----------|------------------|-----------------------------| | Treatment | $k-1=3$ | 64.8592 | $\frac{64.8592}{3} = 21.6197$ | $\frac{21.6197}{\frac{13.75}{12}} = \frac{21.6197}{1.1458} = 18.87$ | | Error | $N-k=12$ | 13.75 | $\frac{13.75}{12} = 1.1458$ | | | Total | 15 | 78.6092 | | | 10. **Decision:** - Critical $F$ at $(3,12)$ df and $\alpha=0.05$ is approximately 3.49. - Since $F_{calculated} = 18.87 > 3.49$, reject $H_0$. - Conclusion: There is significant evidence that not all treatment means are equal. --- ### Problem 7: Regression Analysis 11. **Data:** $x_i = [5.0, 2.0, 7.4, 10.2, 14.8, 7.5, 20.1, 4.4, 5.2, 24.9, 30.2, 28.0]$ $y_i = [12, 11, 12, 14, 18, 13, 22, 13, 12, 28, 30, 36]$ 12. **Calculate means:** $$\bar{x} = \frac{5.0 + 2.0 + \cdots + 28.0}{12} = \frac{159.7}{12} = 13.3083$$ $$\bar{y} = \frac{12 + 11 + \cdots + 36}{12} = \frac{201}{12} = 16.75$$ 13. **Calculate regression coefficients:** $$S_{xy} = \sum (x_i - \bar{x})(y_i - \bar{y}) = 682.68$$ $$S_{xx} = \sum (x_i - \bar{x})^2 = 2041.44$$ Slope: $$b = \frac{S_{xy}}{S_{xx}} = \frac{682.68}{2041.44} = 0.3345$$ Intercept: $$a = \bar{y} - b \bar{x} = 16.75 - 0.3345 \times 13.3083 = 16.75 - 4.45 = 12.30$$ 14. **Regression equation:** $$\hat{y} = 12.30 + 0.3345 x$$ 15. **Calculate SSR, SSE, SST:** - Total sum of squares: $$SST = \sum (y_i - \bar{y})^2 = 530.25$$ - Regression sum of squares: $$SSR = b \times S_{xy} = 0.3345 \times 682.68 = 228.3$$ - Error sum of squares: $$SSE = SST - SSR = 530.25 - 228.3 = 301.95$$ 16. **Calculate $R^2$:** $$R^2 = \frac{SSR}{SST} = \frac{228.3}{530.25} = 0.43$$ 17. **Calculate F-ratio:** Degrees of freedom: - Regression df = 1 - Error df = $n-2 = 10$ Mean squares: $$MSR = SSR/1 = 228.3$$ $$MSE = SSE/10 = 30.195$$ F-ratio: $$F = \frac{MSR}{MSE} = \frac{228.3}{30.195} = 7.56$$ 18. **ANOVA Table:** | Source | df | SS | MS | F | |-------------|-----|--------|---------|------| | Regression | 1 | 228.3 | 228.3 | 7.56 | | Error | 10 | 301.95 | 30.195 | | | Total | 11 | 530.25 | | | 19. **Decision:** - Critical $F$ at $(1,10)$ df and $\alpha=0.05$ is about 4.96. - Since $7.56 > 4.96$, reject $H_0$. - Conclusion: $x$ is significant in predicting $y$.