Solutions HW 3

1a. (1/2, 1/2, -1/2, -1/2) or (1, 1, -1, -1)

b. (1, -1, 0, 0)

c. (0, 0, 1, -1)

d. H0: (mu1 + mu2)/2  =  (mu3 + mu4)/2    or   mu1 + mu2 = mu3 + mu4    
or   mu1 + mu2 – mu3 – mu4 = 0

HA: (mu1 + mu2)/2  is not equal to  (mu3 + mu4)/2    or 
mu1 + mu2 not equal  mu3 + mu4    or   
mu1 + mu2 – mu3 – mu4 not equal 0

t = (mu1 + mu2 – mu3 – mu4)/sqrt(MSE*SUM(ki^2/ri))
t = (72 + 85 – 76 – 62)/sqrt(100.9*(1/6 + 1/6 + 1/6 + 1/6)) = 2.32

t = 2.32 with 20 df
two-sided p-value < 0.05

We reject the null hypothesis and say there is significant evidence that the 
mean of the animal fat means is greater than the mean of the vegetable fat means.

e. estimate: 85 – 72 = 13
t(0.025,20) = 2.086
SE: sqrt(MSE*(1/6 + 1/6)) = 5.8

95% CI: estimate +/- t*SE = 13 +/- 2.086*5.8 = 13 +/- 12.1

The 95% confidence interval for the difference between animal fats is 
0.9 to 25.1.

f. contrast a: [(72 + 85 + (-1)76 + (-1)62)^2]/[1/6 + 1/6 + (-1^2)/6 + (-1^2)/6] = 541.5
contrast b: [(72 – 85 + 0 + 0)^2]/[1/6 + (-1^2)/6 + 0 + 0] = 507
contrast c: [(0 + 0 + 76 – 62)^2]/[0 + 0 + 1/6 + (-1^2)/6] = 588

g. MS(a) = 541.5/1 = 541.5   F = 541.5/100.9 = 5.37     p-value = 0.0313
MS(b) = 507/1 = 507             F = 507/100.9 = 5.02        p-value = 0.0365
MS(c) = 588/1 = 588             F = 588/100.9 = 5.83        p-value = 0.0255
Note: The F df are 1 and 20.  Students are not expected to compute the exact p-
value but only to provide approximations based on the tables in the back of the 
text.  For example, it would be sufficient to say that all the p-values are 
between 0.01 and 0.05.

Thus, all the contrasts reject the null hypothesis at the 0.05 level.
(a) There is significant evidence that the mean of the animal fat means does not 
equal the mean of the vegetable fat means.
(b) There is significant evidence that the mean of animal fat 1 does not equal 
the mean of animal fat 2.
(c) There is significant evidence that the mean of vegetable fat 1 does not 
equal the mean of vegetable fat 2.

h. The test statistic t^2 = F.  (2.32^2 = 5.37)  These are equivalent tests.

i. (a) to (b): [(1*1)/6 + (1*-1)/6 + (-1*0)/6 + (-1*0)/6] = 0
(a) to (c): [(1*0)/6 + (1*0)/6 + (-1*1)/6 + (-1*-1)/6] = 0
(b) to (c); [(1*0)/6 + (-1*0)/6 + (0*1)/6 + (0*-1)/6] = 0

j. F = [(SS(a) + SS(b) + SS(c))/3]/MSE = [(541.5 + 507 + 588)/3]/100.9 = 5.41
F = 5.41 with 3 and 20df.  
p-value = 0.0069
Thus, reject the null hypothesis and say that at least one contrast does not 
equal zero.

k. These tests are equivalent, since the three orthogonal contrasts completely 
partitioned the treatment sum of squares.  Note that all three contrasts equal
0 if and only if all four means are equal to each other.

2a. The experimental units are ponds.

b. The observational units are fish or sets of 20 fish if fish are weighed in 
groups of 20 rather than individually.

c. proc glm;
	class density;
	model weight = density;
run;

The GLM Procedure

Dependent Variable: weight

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

Model                   4     1012.083000      253.020750      35.16    <.0001

Error                   15     107.935000        7.195667

Corrected Total         19    1120.018000

F = 35.16 with 4 and 15 df.
p-value < 0.0001
There is significant evidence that stocking density affects catfish growth.

d) data one;
	input density weight;
      cards;
	linear=density;
      quadratic=density**2;
      cubic=density**3;
      quartic=density**4;
      cards;
<DATA GOES HERE>
run;

proc glm;
   model weight=linear quadratic cubic quartic;
run;

 The GLM Procedure

Dependent Variable: weight



Source                 DF       Type I SS     Mean Square    F Value    Pr > F

linear                 1     912.9802500     912.9802500     126.88    <.0001
quadratic              1      93.8616071      93.8616071      13.04    0.0026
cubic                  1       2.4010000       2.4010000       0.33    0.5721
quartic                1       2.8401429       2.8401429       0.39    0.5393

The sum of squares, F-statistics and p-values are in the table above.

e) proc reg;
   model weight=linear quadratic;
run;

The REG Procedure
                                         Model: MODEL1
                                  Dependent Variable: weight
                                      Parameter Estimates

                       Parameter       Standard
  Variable     DF       Estimate          Error    t Value    Pr > |t|

  Intercept     1       36.66000        2.76695      13.25      <.0001
  linear        1        2.99036        2.10860       1.42      0.1742
  quadratic     1       -1.29464        0.34479      -3.75      0.0016

The regression function is Yhat = B0 + B1*X + B2*X^2.
The estimates of the partial regression coefficients are B0 = 36.7, B1 = 3, 
B2 = -1.3.

f) Intercept: t = 13.25, p-value < 0.0001, there is evidence that the intercept 
is significantly different from zero.
Linear: t = 1.42, p-value = 0.1742, there is no evidence that the linear 
coefficient is significantly different from zero.  B1 may equal zero.
Quadratic: t = 3.75, p-value = 0.0016, there is evidence that the quadratic 
coefficient is significantly different from zero.

g. SSE(reduced) = 113.18, df = 17
SSE(full) = 107.94, df = 15

F = [(113.18 – 107.94)/(17 – 15)]/(107.94/15) = 0.36 with 2 and 15 df.
p-value > 0.05
There is no evidence to reject the null that the quadratic regression function 
(reduced model) adequately fits the data.

This test could also be obtained automatically using the code

proc glm;
   class lack_of_fit;
   model weight=linear quadratic lack_of_fit;
run;

where lack_of_fit is defined to equal density in the data step.  See the 
"lack_of_fit" line in the output below.

Source            DF       Type I SS     Mean Square    F Value    Pr > F
linear             1     912.9802500     912.9802500     126.88    <.0001
quadratic          1      93.8616071      93.8616071      13.04    0.0026
lack_of_fit        2       5.2411429       2.6205714       0.36    0.7007