(精品)浙大生物统计样卷2010-2011 A答案

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 叶片2 叶片1 叶片2

施肥 施肥 不施肥 不施肥 品种1 9 5 19 20

4 7 14 16 品种2

15 17 15 17

12

11

12

11

(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.

()()()

Error

: Leaf : Variety :ion Fertilizat :Mean :2,1 2,1 2,1 2,1 εγβμεαγαβγβαμαn k j i Y ijkn j ik ij j k j i ijkn ====++++++=

(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 a-1 = 1 ()∑=∙∙∙∙∙∙∙-a

i i Y Y bcr 12

B b-1 = 1 ()∑=∙∙∙∙∙∙∙-b

j j Y Y acr 1

2

C(B) b(c-1) = 2 ()∑∑==∙∙∙∙∙∙-b j c

k jk Y Y ar 112

AB (a-1)(b-1) = 1 ()∑∑==∙∙∙∙∙∙∙∙∙∙∙∙+--a

i b

j j i ij Y Y Y Y cr 11

2

AC(B) (a-1)b(c-1) = 2 ()∑∑∑===∙∙∙∙∙∙∙∙∙+--a i b j c

k jk ij ijk Y Y Y Y r 111

2

Error

abc(r-1) = 8

()∑∑∑∑====∙-a i b j c k r

n ijk ijkn Y Y

1111

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.

(4) Write out the SAS program for analysis of this data.

(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.

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

It is a typical Two-way Factorial Design.

(2) Write out the ANOVA model for this experiment.

()

Error

: Density : Variety :Mean :3,2,1 3,2,1 2,1 εβμεαββαμαk j i Y ijk ij j i ijk ===+++++=

(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.

(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:

(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.