Question 1194885
**a) Critical Value of F at the 5% level of significance**

* **Degrees of Freedom:**
    * Between groups (df1): k - 1 = 2 groups - 1 = 1 
    * Within groups (df2): N - k = 20 - 2 = 18

* **F-distribution table:** 
    * Look up the F-critical value in an F-distribution table with df1 = 1 and df2 = 18, and α = 0.05. 
    * **F-critical ≈ 4.41**

**b) Null and Alternative Hypotheses**

* **Null Hypothesis (H0):** 
    * μ1 = μ2 
    * There is no significant difference in the mean production costs between the two processes.

* **Alternative Hypothesis (H1):** 
    * μ1 ≠ μ2 
    * There is a significant difference in the mean production costs between the two processes.

**c) Decision Rule**

* **Reject H0 if the calculated F-statistic is greater than the critical F-value (F > 4.41).**
* **Fail to reject H0 if the calculated F-statistic is less than or equal to the critical F-value (F ≤ 4.41).**

**d) Compute the Value of the Test Statistic**

1. **Calculate Mean Squares:**
    * **Between Groups Sum of Squares (SSB):** 
        * SSB = [Σ(n_i * (X̄_i - X̄)²)] / (k - 1) 
            * where:
                * n_i: sample size of group i
                * X̄_i: mean of group i
                * X̄: overall mean 
                * k: number of groups

        * X̄_1 (Process 1) = 137 / 10 = 13.7
        * X̄_2 (Process 2) = 108 / 10 = 10.8
        * X̄ (Overall) = 245 / 20 = 12.25

        * SSB = [(10 * (13.7 - 12.25)²) + (10 * (10.8 - 12.25)²)] / (2 - 1) 
        * SSB = 198.05

    * **Within Groups Sum of Squares (SSW):** 
        * SSW = Total Sum of Squares (SST) - Between Groups Sum of Squares (SSB) 
        * SSW = 3081 - 198.05 = 2882.95

    * **Mean Square Between Groups (MSB):** 
        * MSB = SSB / (k - 1) = 198.05 / 1 = 198.05

    * **Mean Square Within Groups (MSW):** 
        * MSW = SSW / (N - k) = 2882.95 / 18 = 160.16

2. **Calculate F-statistic:**
    * F = MSB / MSW = 198.05 / 160.16 = 1.236

**e) Decision**

* Since the calculated F-statistic (1.236) is less than the critical F-value (4.41), we **fail to reject the null hypothesis (H0)**.

**f) Interpretation**

* There is **not** enough evidence at the 5% level of significance to conclude that there is a significant difference in the mean production costs between the two processes.

**In summary:**

* The analysis suggests that there is no significant difference in the mean production costs between the two processes.