### (a) Model for the Design
The model for the given design can be written as:
\[ Y_{ij} = \mu + \alpha_i + \beta_j + \epsilon_{ij} \]
where:
- \( Y_{ij} \) represents the sales (in K'000s) of the products in location \( j \) by salesman \( i \).
- \( \mu \) is the overall mean sales.
- \( \alpha_i \) is the effect of the \( i \)-th salesman (A, B, C, D) on the sales.
- \( \beta_j \) is the effect of the \( j \)-th location type (1, 2, 3) on the sales.
- \( \epsilon_{ij} \) is the random error term associated with the sales by salesman \( i \) in location \( j \).
### (b) Blocking Variable and Treatment
- **Blocking Variable**: The blocking variable is the 'Location' (1, 2, 3). It is used to reduce the variability in sales data by accounting for the differences due to location.
- **Treatment**: The treatment is the 'Salesman' (A, B, C, D), as this is the main factor being tested to determine if different sales approaches or skills among salesmen result in different sales outcomes.
### (c) Two-way Analysis of Variance (ANOVA) Table
Let's first calculate the sums and means required for the ANOVA table.
Assuming no interaction effect between salesmen and location.
#### Step 1: Calculate Grand Total (G) and Grand Mean (\(\bar{Y}\))
\[ G = 52.8 + 49.4 + 58.6 + 42.9 + 60.1 + 48.1 + 61.0 + 50.3 + 62.0 + 56.4 + 63.3 + 61.2 = 666.1 \]
\[ \bar{Y} = \frac{G}{n} = \frac{666.1}{12} = 55.5083 \]
#### Step 2: Calculate Sum of Squares
**Total Sum of Squares (SS_T)**:
\[ SS_T = \sum (Y_{ij} - \bar{Y})^2 \]
We'll compute this sum:
\[ SS_T = (52.8 - 55.5083)^2 + (49.4 - 55.5083)^2 + \ldots + (61.2 - 55.5083)^2\]
\[ SS_{Total} = 388.15 \]
**Sum of Squares for Salesmen (SS_A)**:
\[ SS_A = \frac{1}{3} \left( \sum_{i=1}^4 (\sum_{j=1}^3 Y_{ij})^2 \right) - 12\bar{Y}^2 \]
\[ SS_A = \frac{1}{3} ( (52.8 + 60.1 + 62.0)^2 + (49.4 + 48.1 + 56.4)^2 + (58.6 + 61.0 + 63.3)^2 + (42.9 + 50.3 + 61.2)^2) - 12\bar{Y}^2 \]
Using the respective sums:
\[ SS_A = \frac{1}{3} ( 208.9^2 + 153.9^2 + 182.9^2 + 154.4^2 ) - 12 (55.5083)^2 \]
\[ SS_A = \frac{1}{3} (43618.41 + 23683.21 + 33453.21 + 23840.16 ) - 3690.483 \]
\[ SS_A = 135595.0 - 3690.483 \]
\[ SS_A = 81.56 \]
**Sum of Squares for Locations (SS_B)**:
\[ SS_B = \frac{1}{4} \left( \sum_{j=1}^3 (\sum_{i=1}^4 Y_{ij})^2 \right) - 12 \bar{Y}^2 \]
\[ SS_B = \frac{1}{4} ( (52.8 + 49.4 + 58.6 + 42.9)^2 + (60.1 + 48.1 + 61.0 + 50.3)^2 + (62.0 + 56.4 + 63.3 + 61.2)^2) - 12 \bar{Y}^2 \]
Using the respective sums:
\[ SS_B = \frac{1}{4} ( 203.7^2 + 219.5^2 + 242.90^2) - 3690.483 \]
\[ SS_B = \frac{1}{4} (41483.69 + 48190.25 + 59005.61 ) - 3690.483 \]
\[ SS_B = 177679.55 - 3690.483 \]
\[ SS_B = 81.22 \]
**Sum of Squares for Error (SS_E)**:
\[ SS_E = SS_{Total} - SS_A - SS_B \]
\[ SS_E = 388.15 - 81.56 - 81.22 \]
\[ SS_E = 225.37 \]
### Two-way ANOVA Table:
| Source | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Square (MS) = SS/df | F-Value | P-Value |
|-------------|----------------------|-------------------------|--------------------------|---------------------|--------------------|
| Salesmen | 81.56 | 3 | 27.187 | \( \frac{27.187}{24.957} \approx 1.09 \) | Refer to F-Table |
| Location | 81.22 | 2 | 40.612 | \( \frac{40.612}{24.957} \approx 1.63 \) | Refer to F-Table |
| Error | 225.37 | 6 | 37.5624 | | |
| Total | 388.15 | 11 | | | |
### (d) Test Null Hypothesis for Salesmen
**Null Hypothesis \(H_{0A}\)**: The population mean sales are identical for all four salesmen.
**F-value for Salesman**:
\[ F = \frac{MS_A}{MS_E} = \frac{27.187}{37.5624} \approx 0.7241 \]
Using \(F_{critical}\) from F-table at \(\alpha = 0.05\), \(df_1 = 3\) and \(df_2 = 6\),
\(F_{critical, 0.05, 3, 6} \approx 4.76 \).
Since \( F = 0.7241 \) is significantly less than \( F_{critical} = 4.76 \), we fail to reject the null hypothesis \(H_{0A}\).
### (e) Test Null Hypothesis for Locations
**Null Hypothesis \(H_{0B}\)**: The population mean sales are the same for all three different types of locations.
**F-value for Location**:
\[ F = \frac{MS_B}{MS_E} = \frac{40.612}{37.5624} \approx 1.0813 \]
Using \(F_{critical}\) from F-table at \(\alpha = 0.05\), \(df_1 = 2\) and \(df_2 = 6\),
\(F_{critical, 0.05, 2, 6} \approx 5.14 \).
Since \( F = 1.0813 \) is significantly less than \( F_{critical} = 5.14 \), we fail to reject the null hypothesis \(H_{0B}\).
### Conclusion:
- For both the Salesmen and Locations, at the 5% significance level, we failed to reject the null hypothesis. This means that there is insufficient evidence to claim that the sales vary significantly among different salesmen or different locations.
Four salesmen in XYZ Company are competing for the little ‘Salesman of the year’. Each has a
task of selling milk in three different types of locations. Their resulting sales, in K’000, were as
follows:
Area Salesmen
A B C D
1 52.8 49.4 58.6 42.9
2 60.1 48.1 61.0 50.3
3 62.0 56.4 63.3 61.2
(a) Write down a model for the above design. Explain each term in the model in the context
of the given information
(b) What is the blocking variable and what is the treatment?
(c) Prepare a two –way analysis of variance table.
(d) Test at the 5% level of significance the null hypothesis that the population mean sales are
identical for all four salesmen.
(e) Test at the 5% level of significance the null hypothesis that the population mean yields
are the same for all three different types of location.
1 answer