options ls=78;
title "Split-Plot Analysis - Dog Data";

/* After reading in the dog1 data, the variables are stacked
  * into two columns, one named 'time' for the time points and
  * one named 'k' for the quantitative response values.
  * The original p1 through p4 responses are removed.
  */

data dogs;
  infile "D:\Statistics\STAT 505\data\dog1.csv" firstobs=2 delimiter=',';
  input treat dog p1 p2 p3 p4;
  time=1;  k=p1; output;
  time=5;  k=p2; output;
  time=9;  k=p3; output;
  time=13; k=p4; output;
  drop p1 p2 p3 p4;
  run;

 /* The class statement specifies treat, dog, and time 
  * as categorical variables.
  * The model statement specifies k as the response and
  * treat, time, and treat-by-time interaction as factors.
  * dog (nested within treat) is also specified as a factor.
  * The h= option in the test statement is used to specify over 
  * which groups the mean responses are to be compared.
  * The e= option specifies the error term for the test specified
  * in the test statement, which is treat here.
  */

proc glm data=dogs;
  class treat dog time;
  model k=treat dog(treat) time treat*time;
  test h=treat e=dog(treat);
  run;

Analysis

Run the SAS program inspecting how the program applies this procedure.

Hypotheses Tests

Now that we have the results from the analysis, the first thing that we want to look at is the interaction between treatment and time. We want to determine here if the effect of treatment depends on time. Therefore, we will start with:

1. The interaction between treatment and time, or:

   \(H_0\colon (\alpha\tau)_{ik} = 0 \) for all \( i = 1,2, \dots, a;\) \(k = 1,2, \dots, t\)

   Here we need to look at the treatment by interaction term whose F-value is reported at 2.06. We want to compare this to an F-distribution with  (a - 1)(t - 1) = 9 and (N - a)(t - 1) = 96 degrees of freedom. The numerator d.f. of 9 is tied to the source variation due to the interaction, while the denominator d.f. is tied to the source of variation due to error(b).

   We can reject \(H_0\) at level alpha; if

   \(F = \dfrac{MS_{\text{treat x time}}}{MS_{error(b)}} > F_{(a-1)(t-1), (N-a)(t-1), \alpha}\)

   Therefore, we want to compare this to an F with 9 and 96 degrees of freedom. Here we see that this is significant with a p-value of 0.0406.

   Result: We can conclude that the effect of treatment depends on time (F = 2.06; d. f. = 9, 96; p = 0.0406)

   Next Steps...

   • Because the interaction between treatment and time is significant, the next step in the analysis would be to further explore the nature of that interaction using something called profile plots, (we will look at this later...).
   • If the interaction between treatment and time was not significant, the next step in the analysis would be to test for the main effects of treatment and time.
2. Let's suppose that we had not found a significant interaction. Let's do this so that you can see what it would look like to consider the effects of treatment.

   Consider testing the null hypothesis that there are no treatment effects, or

   \(H_0\colon \alpha_1 = \alpha_2 = \dots = \alpha_a = 0\)

   To test this null hypothesis, we compute the F-ratio between the Mean Square for Treatment and Mean Square for Error (a). We then reject our H_o at level α if

   \(F = \dfrac{MS_{treat}}{MS_{error(a)}} > F_{a-1, N-a, a}\)

   Here, the numerator degrees of freedom is equal to the number of degrees of freedom a - 1 = 3 for treatment, while the denominator degrees of freedom are equal to the number of degrees of freedom N - a = 32 for Error(a).

   Result: We can conclude that the treatment significantly affects the mean coronary sinus potassium over the t = 4 sampling times (F = 6.00; d. f. = 3,32; p = 0.0023).
3. Consider testing the effects of time:

   \(H_0\colon \tau_1 = \tau_2 = \dots = \tau_t = 0\)

   To test this null hypothesis, we compute the F-ratio between Mean Square for Time and Mean Square for Error(b). We then reject H_o at level \(\alpha\); if

   \(F = \dfrac{MS_{time}}{MS_{error(b)}} > F_{t-1, (N-a)(t-1), \alpha}\)

   Here, the numerator degrees of freedom is equal to the number of degrees of freedom t - 1 = 3 for time, while the denominator degrees of freedom is equal to the number of degrees of freedom (N - a)(t - 1) = 96 for Error(b).

   Result: We can conclude that coronary sinus potassium varies significantly over time (F = 11.15; d. f. = 3, 96; p < 0.0001).