STEM Problem Detail

Problem Statement

Prompt

Provide the solution of the Example 5.5(a), (b), (c), and (d)

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## Solution to Example 5.5

**Null Hypothesis ($$H_0$$):** $$\mu_1 = \mu_2 = \mu_3$$, i.e., there is no significant difference between the average scores of the three groups.
**Alternative Hypothesis ($$H_1$$):** The average scores of the three groups are not all equal.

To simplify the calculations, we subtract a constant $$A = 10$$ from each observation. The coded data and preliminary calculations are as follows:

| **S. No.** | **Group 1** ($x_1$) | **Group 2** ($x_2$) | **Group 3** ($x_3$) | **Total** |
|---|---|---|---|---|
| 1 | -3 | -2 | -2 | |
| 2 | -4 | -3 | -3 | |
| 3 | -1 | -4 | -1 | |
| 4 | -5 | -1 | -4 | |
| 5 | -2 | -5 | -2 | |
| Total ($T_i$) | $T_1 = -15$ | $T_2 = -15$ | $T_3 = -12$ | $G = -42$ |
| Mean ($\bar{x}_i$) | -3 | -3 | -2.4 | |
| $\sum x_i^2$ | 55 | 55 | 34 | $\sum \sum x_{ij}^2 = 144$ |

**Step 1: Correction Factor (C.F.)**
\[ \text{C.F.} = \frac{G^2}{N} = \frac{(-42)^2}{15} = \frac{1764}{15} = 117.6 \]

**Step 2: Total Sum of Squares (T.S.S.)**
\[ \text{T.S.S.} = \sum \sum x_{ij}^2 - \text{C.F.} = 144 - 117.6 = 26.4 \]

**Step 3: Sum of Squares Between Groups (S.S.G.)**
\[ \text{S.S.G.} = \left( \frac{T_1^2}{n_1} + \frac{T_2^2}{n_2} + \frac{T_3^2}{n_3} \right) - \text{C.F.} \]
\[ \text{S.S.G.} = \left( \frac{(-15)^2}{5} + \frac{(-15)^2}{5} + \frac{(-12)^2}{5} \right) - 117.6 \]
\[ \text{S.S.G.} = (45 + 45 + 28.8) - 117.6 = 118.8 - 117.6 = 1.2 \]

**Step 4: Sum of Squares Within Groups (S.S.E.)**
\[ \text{S.S.E.} = \text{T.S.S.} - \text{S.S.G.} = 26.4 - 1.2 = 25.2 \]

**Step 5: ANOVA Table**

| Source of Variation | d.f. | S.S. | M.S.S. | $F$-ratio |
|---|---|---|---|---|
| Between Groups | $3-1=2$ | 1.2 | $s_G^2 = \frac{1.2}{2} = 0.6$ | $F = \frac{2.1}{0.6} = 3.5$ |
| Within Groups (Error) | $15-3=12$ | 25.2 | $s_E^2 = \frac{25.2}{12} = 2.1$ | (Note: $F = s_E^2 / s_G^2$) |
| Total | 14 | 26.4 | | |

**Conclusion:**
The tabulated value of $$F$$ for $$(12, 2)$$ degrees of freedom at a 5

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Problem #82