Hw 4 Solutions

1. a) The GLM Procedure

Dependent Variable: y

                       Sum of
Source     DF         Squares     Mean Square    F Value    Pr > F

Model       4     48568.87638     12142.21909      78.08    <.0001

Error      20      3110.18912       155.50946

C. Total   24     51679.06550

F = 78.08 with 4 and 20 df.
p-value <0.0001

We reject the null hypothesis and state that there is significant evidence that 
the mean serum T3 concentration is not the same in the five stages considered in 
this experiment.

1b) The GLM Procedure
Tukey's Studentized Range (HSD) Test for y
NOTE: This test controls the Type I experimentwise error rate.

Alpha                                   0.05
Error Degrees of Freedom                  20
Error Mean Square                   155.5095
Critical Value of Studentized Range  4.23186
Minimum Significant Difference        23.601

Comparisons significant at the 0.05 level are indicated by ***.

             Difference
   stage        Between      Simultaneous 95%
Comparison        Means     Confidence Limits

3    - 2         93.304       69.703   116.905  ***
3    - 4         95.462       71.861   119.063  ***
3    - 5        118.184       94.583   141.785  ***
3    - 1        119.782       96.181   143.383  ***
2    - 4          2.158      -21.443    25.759
2    - 5         24.880        1.279    48.481  ***
2    - 1         26.478        2.877    50.079  ***
4    - 5         22.722       -0.879    46.323
4    - 1         24.320        0.719    47.921  ***
5    - 1          1.598      -22.003    25.199

Thus, at the 0.05 level, stage 3 is significantly different from all other 
stages, stage 2 is different from stages 5 and 1, and stage 4 is different from 
stage 1.  All other comparisons are not significantly different and may have the 
same mean serum T3 concentration.




1c) proc glm;
 	class stage;
	model y=stage;
      /*Conduct one-sided tests with alternative H_A:Treatment>Control*/ 
      means stage / alpha=0.05 dunnettu('1'); 
    run;

The GLM Procedure

Dunnett's One-tailed t Tests for y
NOTE: This test controls the Type I experimentwise error for comparisons of all 
treatments against a control.

Alpha                              0.05
Error Degrees of Freedom             20
Error Mean Square              155.5095
Critical Value of Dunnett's t   2.30445
Minimum Significant Difference   18.175
Comparisons significant at the 0.05 level are indicated by ***.

              Difference
   stage         Between     Simultaneous 95%
Comparison         Means     Confidence Limits

3    - 1         119.782      101.607 Infinity  ***
2    - 1          26.478        8.303 Infinity  ***
4    - 1          24.320        6.145 Infinity  ***
5    - 1           1.598      -16.577 Infinity

Stages 2, 3, and 4 have a significantly greater mean serum T3 concentration 
than the control, stage 1.

1d) We can be 95% confident that Stage 3 is the only stage in the set of stages
that could have the highest mean. All other stages are significantly smaller
than stage 3.

1e) Stages 1 and 5 are in the set of stages with the smallest mean.  Although 
stage 1 has the smallest observed mean, the mean for stage 5 is not 
significantly greater than the mean for stage 1.  Stages 2, 3, and 4 are 
significantly larger than the mean for stage 1.  Thus we can rule these stages
out with 95% confidence.

1f) No, there is no significant evidence (p-value = 0.7279) of a difference 
between the mean serum T3 concentration corresponding to the stages with the two 
smallest means.

2a) No, C1 and C3 do not sum to zero.

2b) No, the sum of (the product of the contrast coefficients divided by the 
number of  replications per treatment) does not sum to zero. That is 
sum[(ki*li)/ri] does not = 0.  Since the experiment is assumed to be balanced, 
the number of replications is the same for all treatments and we can just sum of 
the product of the contrast coefficients.

(1*1) + (3*-1) + (-1*0) + (-1*-1) + (-1*1) = -2 does not equal 0.

2c) There are infinitely many possibilities. A simple one is (1, -1, 0, 0, 0).
(0*1) + (0*-1) + (2*0) + (-1*0) + (-1*0) = 0.