Question 1170624: A manager at the head office of a company is considering 3 branch office managers for promotion. Branch reports include records of sales volume per agent for each branch. A random sample of records was selected for agents at each branch. All branches are located in cities with similar demographics (per capita income, population etc.) Using the samples, the manager wants to see if there is a significant difference in performance of agents at the three branches. If there is a difference, the information will be used to help determine which branch manager to promote; otherwise it will not be included in the decision. (All values are in hundreds of thousands of dollars.)
branch managed by Harrison: 7.2 ,6.4, 10.1, 11, 9.9, 10.6
branch managed by Dale: 8.8, 10.7, 11.1, 9.8
branch managed by Stevenson: 6.9, 8.7, 10.5, 11.4
For the single measurement problem, use an a = 0.01 level of significance. Conduct an appropriate hypothesis test and conclude whether to reject or not reject the claim that there is no difference among the agents at the different branches.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's conduct an ANOVA (Analysis of Variance) test to determine if there's a significant difference in sales performance among the three branches.
**1. Define Hypotheses**
* **Null Hypothesis (H0):** There is no significant difference in the mean sales performance among the three branches. (μ1 = μ2 = μ3)
* **Alternative Hypothesis (H1):** There is a significant difference in the mean sales performance among the three branches. (At least one mean is different)
**2. Set Significance Level (α)**
* α = 0.01
**3. Data**
* Harrison (H): 7.2, 6.4, 10.1, 11, 9.9, 10.6 (n1 = 6)
* Dale (D): 8.8, 10.7, 11.1, 9.8 (n2 = 4)
* Stevenson (S): 6.9, 8.7, 10.5, 11.4 (n3 = 4)
**4. Calculate Sample Statistics**
* **Harrison:**
* Sum (Σx1) = 7.2 + 6.4 + 10.1 + 11 + 9.9 + 10.6 = 55.2
* Mean (x̄1) = 55.2 / 6 = 9.2
* Sum of squares (Σx1^2) = 7.2^2 + 6.4^2 + 10.1^2 + 11^2 + 9.9^2 + 10.6^2 = 517.98
* **Dale:**
* Sum (Σx2) = 8.8 + 10.7 + 11.1 + 9.8 = 40.4
* Mean (x̄2) = 40.4 / 4 = 10.1
* Sum of squares (Σx2^2) = 8.8^2 + 10.7^2 + 11.1^2 + 9.8^2 = 410.58
* **Stevenson:**
* Sum (Σx3) = 6.9 + 8.7 + 10.5 + 11.4 = 37.5
* Mean (x̄3) = 37.5 / 4 = 9.375
* Sum of squares (Σx3^2) = 6.9^2 + 8.7^2 + 10.5^2 + 11.4^2 = 361.71
* **Total:**
* N = n1 + n2 + n3 = 6 + 4 + 4 = 14
* Grand Sum (ΣX) = 55.2 + 40.4 + 37.5 = 133.1
* Grand Mean (x̄) = 133.1 / 14 = 9.507
**5. Calculate Sum of Squares**
* **SST (Total Sum of Squares):**
* SST = Σ(X^2) - (ΣX)^2 / N
* Σ(X^2) = 517.98 + 410.58 + 361.71 = 1290.27
* SST = 1290.27 - (133.1)^2 / 14 = 1290.27 - 1264.38 = 25.89
* **SSB (Sum of Squares Between Groups):**
* SSB = Σ[n_i * (x̄_i - x̄)^2]
* SSB = 6(9.2 - 9.507)^2 + 4(10.1 - 9.507)^2 + 4(9.375 - 9.507)^2
* SSB = 6(-0.307)^2 + 4(0.593)^2 + 4(-0.132)^2
* SSB = 6(0.0942) + 4(0.3516) + 4(0.0174) = 0.5652 + 1.4064 + 0.0696 = 2.0412
* **SSW (Sum of Squares Within Groups):**
* SSW = SST - SSB = 25.89 - 2.0412 = 23.8488
**6. Calculate Mean Squares**
* **MSB (Mean Square Between Groups):**
* MSB = SSB / (k - 1) = 2.0412 / (3 - 1) = 1.0206
* **MSW (Mean Square Within Groups):**
* MSW = SSW / (N - k) = 23.8488 / (14 - 3) = 2.1681
**7. Calculate F-statistic**
* F = MSB / MSW = 1.0206 / 2.1681 = 0.4707
**8. Determine Degrees of Freedom**
* df1 (between groups) = k - 1 = 3 - 1 = 2
* df2 (within groups) = N - k = 14 - 3 = 11
**9. Find Critical F-value**
* Using an F-distribution table with α = 0.01, df1 = 2, and df2 = 11, the critical F-value is approximately 7.21.
**10. Make a Decision**
* Since the calculated F-statistic (0.4707) is less than the critical F-value (7.21), we fail to reject the null hypothesis.
**11. Conclusion**
* There is not enough evidence to reject the claim that there is no significant difference in the mean sales performance among the three branches at the α = 0.01 significance level.
**Therefore, the manager should not use the sales performance data as a factor in the promotion decision.**
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