SOLUTION: A manufacturer of automobile transmissions uses two different processes. Management ordered a study of the production costs to see if there is a difference between the two processe
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Question 1194885: A manufacturer of automobile transmissions uses two different processes. Management ordered a study of the production costs to see if there is a difference between the two processes. A summary of the findings is shown next.
Process 1 Process 2 Total Process totals ($100s) 137 108 245
Sample size 10 10 20
Sum of squares 1,893 1,188 3,081
a. What is the critical value of F at the 5% level of significance?
b. State the null hypothesis and the alternate hypothesis
c. State the decision rule
d. Compute the value of the test statistic
e. What is your decision regarding H0
f. Interpret the result
Answer by ElectricPavlov(122) (Show Source): You can put this solution on YOUR website!
**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.
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