Question 1170624
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.**