浙大生物统计样卷2010-2011 A答案
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Biostatistics
余思扬
3100100227
2012 年 6 月
浙江大学 2010–2011 学年 秋冬 学期 《生物统计(学)与实验设计》课程期末考试试卷 — A 卷
Problem 1 (25 points):
One experiment is conducted for studying the influence of plant condition on the content of Nicotine in leaf of tobacco. Two tobacco varieties and two different plant cultivations (fertilization, no fertilization) were arranged, 2 pieces of leaves were sampled for each variety. The obtained experiment data are shown in the following table. 叶片 1 施肥 品种 1 品种 2 9 4 15 12 叶片 2 施肥 5 7 17 11 叶片 1 不施肥 19 14 15 12 叶片 2 不施肥 20 16 17 11
DATA Rice; INPUT Variety $ Density $ Yield @@; DATALINES; A1 B1 8 A1 B2 8 A1 B3 8 A1 B1 7 A1 B2 7 A1 B3 6 A1 B1 6 A1 B2 5 A1 B3 6 A2 B1 9 A2 B2 9 A2 B3 8 A2 B1 7 A2 B2 9 A2 B3 6 A2 B1 8 A2 B2 7 A2 B3 6 ; PROC GLM; CLASS Variety Density; MODEL Yield = Variety | Density; MEANS Variety /TUKEY; MEANS Density /TUKEY; LSMEANS Variety*Density /STDERR PDIFF; RUN;
(1) What is this experiment design? This is a typical Crossed Nested Design. (2) Write out the ANOVA model of this experiment for analysis; define each factor in the model.
i 1,2
j 1,2,3 k 1,2,3
: Mean α : Variety : Density : Error
(3) Can the interaction of variety and planting density be analyzed？Why? Yes. Because in this experiment we had three replicates, so we can conduct the analysis of interaction. (4) Assume the variety is fixed, planting density is random; write out the SAS program to analysis this data.
CONTRAST 'Linear' Density 1 1 -2;
(6) For this set of experiment data, how to construct a regression model of rice yield on planting density, so that it can be used to predict rice yield for other planting density.
a b c i 1 j 1 k 1
2
Y
a b c r i 1 j 1 k 1 n1
ijkn
Yijk
2
Note: A, B, C and Error are referred to factor α, β, γ and Error ε. And a, b, c and r are the number tested in this design of factor α, β, γ and Error ε.
(1) Is it a factorial design or nested design? It is a typical Two-way Factorial Design. (2) Write out the ANOVA model for this experiment.
Yijk i j ij ijk
Hale Waihona Puke -2-浙江大学 2010–2011 学年秋冬学期《生物统计(学)与实验设计》课程期末考试试卷 A 卷
Problem 2 (30 points):
One experiment of rice variety is conducted for studying the relationship of yield and planting density. The experiment has 2 varieties (A1, A2) and 3 planting densities 10 (B1), 20 (B2), 30 (B3). The observation data are shown in the following table. B1 A1 A2 8 9 B2 8 9 B3 8 8 B1 7 7 B2 7 9 B3 6 6 B1 6 8 B2 5 7 B3 6 6
(5) According to the following output of SAS analysis, draw appropriate statistical conclusion.
From the SAS result, we could find that the Pr>F of model is 0.1861>0.05, the model is not reach significant level, need to be adjusted. And from Type III SS we can find that only the factor that significant is fertilization. So the conclusion is in the model, fertilization is significant but the model itself is not reach the significant level, we need to change model. And also we can find in this model, all the interaction factors are missing, so the change of model we can add all the interactions as I write in Question (4) above, and we may get a better result.
(5) How to use the SAS to test the difference between B3 and the average of B1 and B2? We should adapt Linear Contrast command in this test: Add this term into PROC:
bcr Yi Y
i 1 b
a
2
acr Y j Y
j 1 b c
2
ar Y jk Y
j 1 k 1
a b
2
cr Yij Yi Y j Y
i 1 j 1
2
r Yijk Yij Y jk Y
-3-
浙江大学 2010–2011 学年秋冬学期《生物统计(学)与实验设计》课程期末考试试卷 A 卷
Problem 3 (15 points):
The following data is about the height (X1), weight (X2) and body surface area (Y) of ten infants. Male X1 54 50.5 51 56.5 52 X2 3 2.25 2.5 3.5 3 Y 2446.2 1928.4 2094.5 2506.7 2121 X1 54 53 51.5 51 51 Female X2 3 2.25 2.5 3 3 Y 2117.3 2200.2 1906.2 1850.3 1632.5
-1-
浙江大学 2010–2011 学年秋冬学期《生物统计(学)与实验设计》课程期末考试试卷 A 卷
(4) Write out the SAS program for analysis of this data.
DATA Nicotine; INPUT Fertilization $ Variety $ Leaf $ Content @@; DATALINES; A1 B1 C1 9 A1 B1 C2 5 A2 B1 C1 19 A2 B1 C2 20 A1 B1 C1 4 A1 B1 C2 7 A2 B1 C1 14 A2 B1 C2 16 A1 B2 C1 15 A1 B2 C2 17 A2 B2 C1 15 A2 B2 C2 17 A1 B2 C1 12 A1 B2 C2 11 A2 B2 C1 12 A2 B2 C2 11 ; PROC GLM; CLASS Fertilization Variety Leaf; MODEL Content = Fertilization Variety Leaf(Variety) Fertilization*Variety Fertilization*Leaf(Variety); RANDOM Leaf(Variety) /TEST; CONTRAST 'Leaf in B1' Leaf(Variety) 1 -1 0 0; CONTRAST 'Leaf in B2' Leaf(Variety) 0 0 1 -1; MEANS Fertilization /TUKEY; MEANS Variety /TUKEY; LSMEANS Leaf(Variety) /STDERR PDIFF; LSMEANS Fertilization*Variety /STDERR PDIFF; LSMEANS Fertilization*Leaf(Variety) /STDERR PDIFF; RUN;
Yijkn i j k j ij ik j ijkn
i 1,2
j 1,2 k 1,2 n 1,2
: Mean α : Fertilization : Variety : Leaf
: Error
(3) For the above ANOVA model, write out the formula of degree freedom and corresponding sum of square for factors in the model. Source Degree of Freedom (df) Sum of Squares (SS) A B C(B) AB AC(B) Error a-1 = 1 b-1 = 1 b(c-1) = 2 (a-1)(b-1) = 1 (a-1)b(c-1) = 2 abc(r-1) = 8
余思扬
3100100227
2012 年 6 月
浙江大学 2010–2011 学年 秋冬 学期 《生物统计(学)与实验设计》课程期末考试试卷 — A 卷
Problem 1 (25 points):
One experiment is conducted for studying the influence of plant condition on the content of Nicotine in leaf of tobacco. Two tobacco varieties and two different plant cultivations (fertilization, no fertilization) were arranged, 2 pieces of leaves were sampled for each variety. The obtained experiment data are shown in the following table. 叶片 1 施肥 品种 1 品种 2 9 4 15 12 叶片 2 施肥 5 7 17 11 叶片 1 不施肥 19 14 15 12 叶片 2 不施肥 20 16 17 11
DATA Rice; INPUT Variety $ Density $ Yield @@; DATALINES; A1 B1 8 A1 B2 8 A1 B3 8 A1 B1 7 A1 B2 7 A1 B3 6 A1 B1 6 A1 B2 5 A1 B3 6 A2 B1 9 A2 B2 9 A2 B3 8 A2 B1 7 A2 B2 9 A2 B3 6 A2 B1 8 A2 B2 7 A2 B3 6 ; PROC GLM; CLASS Variety Density; MODEL Yield = Variety | Density; MEANS Variety /TUKEY; MEANS Density /TUKEY; LSMEANS Variety*Density /STDERR PDIFF; RUN;
(1) What is this experiment design? This is a typical Crossed Nested Design. (2) Write out the ANOVA model of this experiment for analysis; define each factor in the model.
i 1,2
j 1,2,3 k 1,2,3
: Mean α : Variety : Density : Error
(3) Can the interaction of variety and planting density be analyzed？Why? Yes. Because in this experiment we had three replicates, so we can conduct the analysis of interaction. (4) Assume the variety is fixed, planting density is random; write out the SAS program to analysis this data.
CONTRAST 'Linear' Density 1 1 -2;
(6) For this set of experiment data, how to construct a regression model of rice yield on planting density, so that it can be used to predict rice yield for other planting density.
a b c i 1 j 1 k 1
2
Y
a b c r i 1 j 1 k 1 n1
ijkn
Yijk
2
Note: A, B, C and Error are referred to factor α, β, γ and Error ε. And a, b, c and r are the number tested in this design of factor α, β, γ and Error ε.
(1) Is it a factorial design or nested design? It is a typical Two-way Factorial Design. (2) Write out the ANOVA model for this experiment.
Yijk i j ij ijk
Hale Waihona Puke -2-浙江大学 2010–2011 学年秋冬学期《生物统计(学)与实验设计》课程期末考试试卷 A 卷
Problem 2 (30 points):
One experiment of rice variety is conducted for studying the relationship of yield and planting density. The experiment has 2 varieties (A1, A2) and 3 planting densities 10 (B1), 20 (B2), 30 (B3). The observation data are shown in the following table. B1 A1 A2 8 9 B2 8 9 B3 8 8 B1 7 7 B2 7 9 B3 6 6 B1 6 8 B2 5 7 B3 6 6
(5) According to the following output of SAS analysis, draw appropriate statistical conclusion.
From the SAS result, we could find that the Pr>F of model is 0.1861>0.05, the model is not reach significant level, need to be adjusted. And from Type III SS we can find that only the factor that significant is fertilization. So the conclusion is in the model, fertilization is significant but the model itself is not reach the significant level, we need to change model. And also we can find in this model, all the interaction factors are missing, so the change of model we can add all the interactions as I write in Question (4) above, and we may get a better result.
(5) How to use the SAS to test the difference between B3 and the average of B1 and B2? We should adapt Linear Contrast command in this test: Add this term into PROC:
bcr Yi Y
i 1 b
a
2
acr Y j Y
j 1 b c
2
ar Y jk Y
j 1 k 1
a b
2
cr Yij Yi Y j Y
i 1 j 1
2
r Yijk Yij Y jk Y
-3-
浙江大学 2010–2011 学年秋冬学期《生物统计(学)与实验设计》课程期末考试试卷 A 卷
Problem 3 (15 points):
The following data is about the height (X1), weight (X2) and body surface area (Y) of ten infants. Male X1 54 50.5 51 56.5 52 X2 3 2.25 2.5 3.5 3 Y 2446.2 1928.4 2094.5 2506.7 2121 X1 54 53 51.5 51 51 Female X2 3 2.25 2.5 3 3 Y 2117.3 2200.2 1906.2 1850.3 1632.5
-1-
浙江大学 2010–2011 学年秋冬学期《生物统计(学)与实验设计》课程期末考试试卷 A 卷
(4) Write out the SAS program for analysis of this data.
DATA Nicotine; INPUT Fertilization $ Variety $ Leaf $ Content @@; DATALINES; A1 B1 C1 9 A1 B1 C2 5 A2 B1 C1 19 A2 B1 C2 20 A1 B1 C1 4 A1 B1 C2 7 A2 B1 C1 14 A2 B1 C2 16 A1 B2 C1 15 A1 B2 C2 17 A2 B2 C1 15 A2 B2 C2 17 A1 B2 C1 12 A1 B2 C2 11 A2 B2 C1 12 A2 B2 C2 11 ; PROC GLM; CLASS Fertilization Variety Leaf; MODEL Content = Fertilization Variety Leaf(Variety) Fertilization*Variety Fertilization*Leaf(Variety); RANDOM Leaf(Variety) /TEST; CONTRAST 'Leaf in B1' Leaf(Variety) 1 -1 0 0; CONTRAST 'Leaf in B2' Leaf(Variety) 0 0 1 -1; MEANS Fertilization /TUKEY; MEANS Variety /TUKEY; LSMEANS Leaf(Variety) /STDERR PDIFF; LSMEANS Fertilization*Variety /STDERR PDIFF; LSMEANS Fertilization*Leaf(Variety) /STDERR PDIFF; RUN;
Yijkn i j k j ij ik j ijkn
i 1,2
j 1,2 k 1,2 n 1,2
: Mean α : Fertilization : Variety : Leaf
: Error
(3) For the above ANOVA model, write out the formula of degree freedom and corresponding sum of square for factors in the model. Source Degree of Freedom (df) Sum of Squares (SS) A B C(B) AB AC(B) Error a-1 = 1 b-1 = 1 b(c-1) = 2 (a-1)(b-1) = 1 (a-1)b(c-1) = 2 abc(r-1) = 8