In Graph 2, the lines have been forced parallel. It is likely that in the population that, once corrected for initial score, the treatment mean increase in score would not all be the same.
FORMAL ANALYSIS: To formally test whether the population mean changes in the score under the three conditions, once adjustment has been made for the initial score, the following hypotheses are assumed:
Null hypothesis: After correction for initial score, the population mean changes in score are the same for all three conditions.
Alternative hypothesis: After correction for initial score, the population mean changes are not all the same.
The standard approach to Analysis of Covariance is based on the parallel lines assumption. That is, we assume that the relationship between the response and initial score in the populations may be represented by straight lines; furthermore we assume that these straight lines have all the same slope. The first thing to be checked is whether it is OK to assume that the population slopes are equal. This is done by testing whether or not the interaction between the initial and type is statistically significant. A general linear model is used, this shown below:
General Linear Model: Response versus Type
Factor Type Levels Values
Type fixed 3 Control Exercise Reading
Analysis of Variance for Response, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F P
Initial 1 754.09 641.10 641.10 31.01 0.000
Type 2 695.03 20.02 10.01 0.48 0.618
Type*Initial 2 65.97 65.97 32.98 1.60 0.210
Error 69 1426.43 1426.43 20.67
Total 74 2941.52
We see that the p-value for the interaction term is 0.210, this being much larger than the significant level of 0.05. Therefore the hypothesis that the population slopes are equal is not rejected.
Because the interaction is not statistically significant we may then make the assumption of equal slopes in the populations and proceed to perform a one-way analysis of covariance.
Firstly the following model assumptions are made:
- The random errors have mean zero. This means that the linear model which allows for a linear relationship between the response and initial shows no sign of bias; the assumption of linearity is OK.
- The random errors all have the same variance; in particular this is means that the variability of the data about the fitted lines is the same for all of the populations.
- The random errors follow a Normal probability model.
- The slopes of the population regression lines are all equal.
The residuals are plotted against the fitted values in the graph below:
It appears that the points are evenly scattered around zero for each condition and so we may assume that the random errors have a mean zero and therefore that linearity holds.
To assess normality, we can produce a normal probability plot, shown below:
The plot is approximately linear, and the points lie within the bands. Therefore it is reasonable to assume that the random errors follow a Normal probability model. Since the assumptions are OK, we can now make conclusions from the test.
General Linear Model: Response versus Type
Factor Type Levels Values
Type fixed 3 Control Exercise Reading
Analysis of Variance for Response, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F P
Initial 1 754.09 655.60 655.60 31.19 0.000
Type 2 695.03 695.03 347.52 16.53 0.000
Error 71 1492.40 1492.40 21.02
Total 74 2941.52
Term Coef SE Coef T P
Constant -7.566 3.884 -1.95 0.055
Initial 0.27523 0.04928 5.58 0.000
The p-value associated with the factor type (of condition) is 0.000. This is mush smaller than the significance level of 0.05 and we conclude that the condition has a significant effect on the average change in memory recall. The p-value for the initial score is less than 0.0005, which shows that there is a significant linear relationship between the change in score and the initial score. Therefore it is worthwhile to correct for the initial score by using the one-way ancova model.
The significant differences are investigated by using a multiple comparison procedure. This is shown below:
Bonferroni 95.0% Simultaneous Confidence Intervals
Response Variable Response
All Pairwise Comparisons among Levels of Type
Type = Control subtracted from:
Type Lower Center Upper ---------+---------+---------+-------
Exercise -2.839 0.3453 3.529 (--------*--------)
Reading 3.456 6.6378 9.819 (--------*--------)
---------+---------+---------+-------
0.0 3.5 7.0
Type = Exercise subtracted from:
Type Lower Center Upper ---------+---------+---------+-------
Reading 3.101 6.292 9.484 (--------*--------)
---------+---------+---------+-------
0.0 3.5 7.0
CONCLUSION: One of the intervals contains zero, so there is no evidence of a difference in the population adjusted mean memory recall between the control and exercise groups. All other intervals do not include zero. Reading is likely to have a higher mean recall than exercise by about 9 and 3 units, and a higher mean recall than the control group by about 10 and 4 units.
