This question involves hypothesis testing to determine if the three training methods lead to significantly different productivity levels. Since the data involves three groups (Methods 1, 2, and 3), a one-way Analysis of Variance (ANOVA) test is appropriate.

Step-by-step solution:

1. State the hypotheses:

• Null hypothesis ($H_0$): The means of the three methods are equal ($\mu_1 = \mu_2 = \mu_3$).
• Alternative hypothesis ($H_a$): At least one mean is different.

2. Significance level:

• The level of significance is $\alpha = 0.05$.

3. Calculate ANOVA components:

a. Grand mean ($\bar{X}$): The problem states that the grand mean is 44.94.

b. Total Sum of Squares (SST): $SST = \sum_{i=1}^k \sum_{j=1}^{n_i} (X_{ij} - \bar{X})^2$ where $k = 3$ (number of methods) and $n_i$ is the sample size for each method.

c. Between-group Sum of Squares (SSB): $SSB = \sum_{i=1}^k n_i (\bar{X}_i - \bar{X})^2$ Using the means and sample sizes for each method:

• Method 1: $n_1 = 6, \bar{X}_1 = 45.17$
• Method 2: $n_2 = 6, \bar{X}_2 = 48.00$
• Method 3: $n_3 = 6, \bar{X}_3 = 41.67$

d. Within-group Sum of Squares (SSW): $SSW = \sum_{i=1}^k \sum_{j=1}^{n_i} (X_{ij} - \bar{X}_i)^2$ This is also equivalent to summing the variances within each group multiplied by their respective degrees of freedom.

4. Calculate the F-statistic:

$F = \frac{\text{MSB}}{\text{MSW}}$ where:

• $\text{MSB} = \frac{SSB}{k-1}$ (Mean Square Between)
• $\text{MSW} = \frac{SSW}{N-k}$ (Mean Square Within)

$N = 18$ is the total number of observations ($6 + 6 + 6$).

5. Compare the F-statistic to the critical value:

Use the F-distribution table with $df_1 = k-1 = 2$ and $df_2 = N-k = 15$ at $\alpha = 0.05$.

6. Conclusion:

• If $F > F_{\text{critical}}$, reject $H_0$: There is evidence that the training methods lead to different productivity levels.
• Otherwise, fail to reject $H_0$: No evidence of significant differences among the methods.

Would you like me to calculate the exact ANOVA values step by step or assist with a specific part of this?