Question 1167906
This scenario is perfectly suited for a **One-Way Analysis of Variance (ANOVA)**.

**Reasoning for choosing One-Way ANOVA:**

* **One Categorical Independent Variable:** The "teaching method" is a single independent variable with four distinct levels (Programmed Instruction, Lecture, TV Presentation, Group Discussion).
* **One Quantitative Dependent Variable:** The "test score" is a quantitative dependent variable.
* **Independent Samples:** Employees were *randomly assigned* to one of the four groups, meaning the observations in one group are independent of the observations in another group.
* **Comparing More Than Two Means:** We are comparing the means of four different groups. While multiple t-tests could compare pairs of groups, ANOVA is designed to compare three or more means simultaneously, controlling the Type I error rate (the chance of falsely rejecting a true null hypothesis) across all comparisons.

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## Sample Computation and Analysis: One-Way ANOVA

**Scenario:** A university is trying to determine the most effective way to teach safety principles. Four teaching methods are tested on separate groups of 20 employees each (though the provided data only shows 5 observations per group, which we will use for the sample calculation). A test out of 10 points is administered afterwards.

**Research Question:** Is there a significant difference in the mean test scores among the four different teaching methods?

**Hypotheses:**
* **Null Hypothesis ($H_0$):** There is no significant difference among the mean test scores of the four teaching methods. ($\mu_1 = \mu_2 = \mu_3 = \mu_4$)
* **Alternative Hypothesis ($H_1$):** At least one of the mean test scores is significantly different from the others. (Not all $\mu_i$ are equal)

**Significance Level ($\alpha$):** $\alpha = 0.05$

**Data (Test Grades of Employees):**

| Programmed Instruction (Group 1) | Lecture (Group 2) | TV Presentation (Group 3) | Group Discussion (Group 4) |
| :------------------------------- | :---------------- | :------------------------ | :------------------------- |
| 6                                | 8                 | 7                         | 8                          |
| 7                                | 5                 | 9                         | 5                          |
| 6                                | 8                 | 6                         | 6                          |
| 5                                | 6                 | 8                         | 6                          |
| 6                                | 8                 | 5                         | 5                          |

**Preliminary Calculations:**

| Group                | $n_i$ | Scores ($\boldsymbol{X_i}$) | $\sum X_i$ | $\bar{X_i}$ | $\sum X_i^2$ |
| :------------------- | :---- | :------------------------ | :--------- | :---------- | :----------- |
| Programmed Instr. (1)| 5     | 6, 7, 6, 5, 6             | 30         | 6.0         | 182          |
| Lecture (2)          | 5     | 8, 5, 8, 6, 8             | 35         | 7.0         | 253          |
| TV Presentation (3)  | 5     | 7, 9, 6, 8, 5             | 35         | 7.0         | 255          |
| Group Discussion (4) | 5     | 8, 5, 6, 6, 5             | 30         | 6.0         | 186          |
| **Total** | $N=20$|                           | $\sum X = 130$ | $\bar{X}_{\text{grand}} = 6.5$ | $\sum X^2 = 876$ |

**ANOVA Computation (Sum of Squares):**

1.  **Total Sum of Squares (SST):** Measures total variability in data.
    $SST = \sum X^2 - \frac{(\sum X)^2}{N}$
    $SST = 876 - \frac{(130)^2}{20} = 876 - \frac{16900}{20} = 876 - 845 = 31$

2.  **Sum of Squares Between Groups (SSB) / Sum of Squares Treatment (SSTreatment):** Measures variability between group means.
    $SSB = \sum \left( \frac{(\sum X_i)^2}{n_i} \right) - \frac{(\sum X)^2}{N}$
    $SSB = \left( \frac{30^2}{5} + \frac{35^2}{5} + \frac{35^2}{5} + \frac{30^2}{5} \right) - \frac{130^2}{20}$
    $SSB = \left( \frac{900}{5} + \frac{1225}{5} + \frac{1225}{5} + \frac{900}{5} \right) - 845$
    $SSB = (180 + 245 + 245 + 180) - 845$
    $SSB = 850 - 845 = 5$

3.  **Sum of Squares Within Groups (SSW) / Sum of Squares Error (SSError):** Measures variability within each group (error).
    $SSW = SST - SSB$
    $SSW = 31 - 5 = 26$

**Degrees of Freedom (df):**

1.  **Total df ($df_{Total}$):** $N - 1 = 20 - 1 = 19$
2.  **Between Groups df ($df_{Between}$):** $k - 1 = 4 - 1 = 3$ (where $k$ is the number of groups)
3.  **Within Groups df ($df_{Within}$):** $N - k = 20 - 4 = 16$
    (Check: $df_{Total} = df_{Between} + df_{Within} \Rightarrow 19 = 3 + 16$, which is correct)

**Mean Squares (MS):**

1.  **Mean Square Between Groups (MSB):**
    $MSB = \frac{SSB}{df_{Between}} = \frac{5}{3} \approx 1.667$

2.  **Mean Square Within Groups (MSW):**
    $MSW = \frac{SSW}{df_{Within}} = \frac{26}{16} = 1.625$

**F-statistic:**

$F = \frac{MSB}{MSW} = \frac{1.667}{1.625} \approx 1.026$

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**ANOVA Summary Table:**

| Source of Variation | Sum of Squares (SS) | df | Mean Square (MS) | F-statistic | p-value | Critical F (α=0.05) |
| :------------------ | :------------------ | :- | :--------------- | :---------- | :------ | :------------------ |
| Between Groups      | 5                   | 3  | 1.667            | 1.026       | 0.407   | 3.24                |
| Within Groups       | 26                  | 16 | 1.625            |             |         |                     |
| Total               | 31                  | 19 |                  |             |         |                     |

* **Critical F-value:** For $df_1 = 3$ and $df_2 = 16$ at $\alpha = 0.05$, the critical F-value from an F-distribution table is approximately $3.24$.
* **p-value:** Using statistical software or an F-distribution calculator for $F=1.026$ with $df_1=3, df_2=16$, the p-value is approximately $0.407$.

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**Analysis and Conclusion:**

* **Comparison of F-statistic to Critical F-value:** Our calculated F-statistic ($1.026$) is less than the critical F-value ($3.24$).
* **Comparison of p-value to Significance Level:** Our p-value ($0.407$) is greater than the significance level ($\alpha = 0.05$).

Since our calculated F-statistic does not fall into the rejection region (and our p-value is greater than $\alpha$), we **fail to reject the null hypothesis**.

**Conclusion:** At the 0.05 significance level, there is **no statistically significant difference** among the mean test scores of the four different safety teaching methods. This suggests that, based on this sample data, none of the teaching methods proved to be significantly more effective than the others in terms of test performance. It implies that any observed differences in average scores are likely due to random chance rather than the teaching method itself.