Midterm 2
Biometry 333
Spring 2005
Name:____________
Show all work. Ask the instructor if a question is not clear. All problems are
worth 5 points unless otherwise stated.
The “Alfalfa” output is for the analysis of alfalfa yield produced by 6 different varieties of alfalfa.
A randomized block design was used where each variety was measured once in each of four
fields (blocks). The word formula for the model is Yield = Variety + Field.
(1a) What is the coefficient value for variety 6? Show your work.
(1b. 6pts) Calculate the expected yield for variety 2 in field 3. Show your work.
(1c) Did the different fields have statistically different yields? Explain how you reached your
conclusion.
(1d) Which alfalfa variety had the largest yield?
(1e) The fitted model was Yield = Variety + Field + N(0,  2 ), where N(0,  2 ) is the normally
distributed error term. Estimate the standard deviation  and explain how you derived your
answer.
(1f) Is the analysis orthogonal? Explain how you reached your answer.
1
The “Xs and Y” analysis is performed using 3 continuous explanatory X variables to model the Y
variable. The word formula for the model is y = x1 + x2 + x3. Or more specifically,
y   0  x1 1  x2  2  x3  3 . Furthermore, in general, recall how residual sum of squares (SS)
n
is calculated via
 y
i 1
 yˆ i  .
2
i
(2a. 6pts) What is the expected value of y when x1=7, x2=8, and x3=10.
(2b. 7pts) Calculate a 95% confidence interval for the x2 coefficient  2 .
(2 practice) Using model equations for this analysis, describe how the 643.02 for the x1
sequential SS was calculated. Answer: It is the reduction in the residual sum of squares going
from the model y  ̂ 0 to the model y  ˆ 0  x1 ˆ1 .
(2c. 6pts) Using model equations for this analysis, describe how the 3890.16 for Total SS was
calculated. (note: this is not a sequential SS)
(2d. 6pts) Using model equations for this analysis, describe how the 1143.71 for Error SS was
calculated. (note: this is not a sequential SS)
(2e) Using model equations for this analysis, describe how the 742.28 for the x2 Sequential SS
was calculated.
(2f) Using model equations for this analysis, describe how the 425.34 for the x2 Adjusted SS
was calculated.
(2g) Explain what the R-square describes. Be specific.
(2bonus, 3 pts) Calculate the R-square.
2
The “Sparrow, no interaction” analysis is trying to predict sparrow weight (wt) from the
birds’ wing spans (continuous, “wing”) and age-class (2-level factor, “age”, coded: adult,
juvenile). The word model is wt = wing + age.
(3a. 3pts) What are the estimated coefficients for the effect of age? That is, give the
coefficient for the effect of being an adult and the coefficient for the effect of being a
juvenile.
(3b. 4pts) What is the estimated difference in weight between adult and juveniles which
have the same wing span? Explain if the difference is statistically significant.
(3c. 7pts) Write the two equations for the lines that predict adult and juvenile weights.
Another model was fit allowing an interaction between age class and wing span. The
word model for this equation is wt = age + wing + age*wing. The output is found in
“Sparrow with interaction”.
(4a. 7pts) Calculate the two equations for the lines that predict adult and juvenile weights.
(4b. 4pts) Explain what is meant by an interaction in terms of sparrow weights, age class,
and wing span. (No numbers are required to answer this question.)
(4c. 4pts) Fundamentally (with regards to lines, so numbers are not needed) how does this
model differ from the model in problem 3?
3
Problem 1: ALFALFA
General Linear Model: Yield versus Variety, Field
Factor
Variety
Field
Type
fixed
fixed
Levels
6
4
Values
1, 2, 3, 4, 5, 6
1, 2, 3, 4
Analysis of Variance for Yield, using Adjusted SS for Tests
Source
Variety
Field
Error
Total
Term
Constant
Variety
1
2
3
4
5
Field
1
2
3
DF
5
3
15
23
Seq SS
2.43507
0.01878
0.13715
2.59100
Adj SS
2.43507
0.01878
0.13715
Adj MS
0.48701
0.00626
0.00914
F
53.27
0.68
Coef
2.95542
SE Coef
0.01952
T
151.42
P
0.000
0.30458
0.16958
0.10708
-0.30542
0.27708
0.04364
0.04364
0.04364
0.04364
0.04364
6.98
3.89
2.45
-7.00
6.35
0.000
0.001
0.027
0.000
0.000
-0.01542
0.04292
-0.03208
0.03381
0.03381
0.03381
-0.46
1.27
-0.95
0.655
0.224
0.358
P
0.000
0.575
Problem 2: Xs and Y
General Linear Model: y versus
Factor
Type
Analysis of
Source DF
x1
1
x2
1
x3
1
Error
36
Total
39
S = 5.63646
Term
Constant
x1
x2
x3
Levels
Values
Variance for y, using Adjusted SS
Seq SS
Adj SS
Adj MS
F
643.02
3.75
3.75
0.12
742.28
425.34
425.34 13.39
1361.16 1361.16 1361.16 42.84
1143.71 1143.71
31.77
3890.16
R-Sq = ZZ%
Coef
40.632
-0.461
3.734
-4.1162
SE Coef
7.969
1.341
1.021
0.6289
for Tests
P
0.733
0.001
0.000
R-Sq(adj) = ZAZA%
T
5.10
-0.34
3.66
-6.55
P
0.000
0.733
0.001
0.000
4
Problem 3: Sparrows, no interaction
General Linear Model: wt versus age
Factor
age
Type
fixed
Levels
2
Values
adult, juvenile
Analysis of Variance for wt, using Adjusted SS for Tests
Source
age
wing
Error
Total
DF
1
1
84
86
Term
Constant
age
adult
wing
Seq SS
0.010
43.420
130.610
174.040
Adj SS
0.085
43.420
130.610
Adj MS
0.085
43.420
1.555
F
0.05
27.92
Coef
-13.920
SE Coef
7.519
T
-1.85
P
0.068
-0.0334
0.16056
0.1431
0.03038
-0.23
5.28
0.816
0.000
P
0.816
0.000
Problem 4: Sparrow with interaction
General Linear Model: wt versus age
Factor
age
Type
fixed
Levels
2
Values
adult, juvenile
Analysis of Variance for wt, using Adjusted SS for Tests
Source
age
wing
age*wing
Error
Total
Term
Constant
age
adult
wing
wing*age
adult
DF
1
1
1
83
86
Seq SS
0.010
43.420
0.032
130.578
174.040
Adj SS
0.034
43.393
0.032
130.578
Adj MS
0.034
43.393
0.032
1.573
F
0.02
27.58
0.02
Coef
-13.989
SE Coef
7.579
T
-1.85
P
0.068
-1.113
0.16084
7.579
0.03062
-0.15
5.25
0.884
0.000
0.00436
0.03062
0.14
0.887
P
0.884
0.000
0.887
5