- Example 20.7 Consider again the code size comparison study of Example 20.1. Suppose after the measurements have been completed, it is discovered that three of the observations had not been done correctly and their observations should not be used in the analysis. Of the three incorrect observations, suppose one is for system V and two are for system Z. This is now a case of single-factor design with unequal sample sizes. The resulting data and its analysis is shown in Table 20.8 Column sums, obtained by adding all available observations in each column, are divided by the number of observations in that column to produce the column mean. The grand mean is 172.25 and is obtained by dividing the grand sum (2067, the total of column sums) by 12, which is the total number of observations. The column effects are 2.15, 13.75, and -21.92. A breakdown of the observations into components corresponding to the grand mean, column effects, and residual terms of the model is as follows:
TABLE 20.8 Code Size Comparison Data with Unequal sample Sizes
|
|
| R
| V
| Z
|
|
| 144
| 101
| 130
|
| 120
| 144
| 180
|
| 176
| 211
| 141
|
| 288
| 288
|
|
| 144
|
|
|
Column sum
| 872
| 744
| 451
|
Column mean
| 174.40
| 186.00
| 150.33
|
Column effect
| 2.15
| 13.75
| -21.92
|
|

The sums of squares, obtained by squaring and adding the available terms in each array, are
- SSY =
= 397,375
- SS0 = Nµ2 = 356,040.75
- SSA = 5α21 + 4α22 + 3α23 = 2220.38
- SSE = (-30.40)2 + (-54.40)2 + ... + (-9.33)2 = 39,113.87
- SST = SSY - SS0 = 41,334.25
By counting the number of independent terms in each array, the degrees of freedom corresponding to various sums of square are
TABLE 20.9 ANOVA Table for the Code Size Comparison Data
|
|
Component
| Sum of Squares
| Percentage of Variation
| Degrees of Freedom
| Mean Square
| F- Computed
| F- Table
|
|
y
| 397,375.00
|
|
|
|
|
|
| 356,040.75
|
|
|
|
|
|
| 41,334.25
| 100.00
| 11
|
|
|
|
A
| 2,220.38
| 5.37
| 2
| 1110.19
| 0.26
| 3.01
|
Errors
| 39,113.87
| 94.63
| 9
| 4345.99
|
|
|
|
|