Using SAS. A horticulturist was investigating the phosphorus content of tree lea
ID: 3062269 • Letter: U
Question
Using SAS. A horticulturist was investigating the phosphorus content of tree leaves from 4 varieties of redbud.Random samples of 5 leaves from each of the different varieties were analyzed for phosphorus content.(I made up this data, but thought that something like this might be interesting because the leaves on these varieties are different colors, ranging from purple to golden to variegated to regular old green!).
Phosphorus content of leaves from each redbud variety
Forest Pansy
Silver Cloud
Hearts of Gold
Rubye Atkinson
0.35
0.32
0.65
0.60
0.40
0.51
0.70
0.80
0.58
0.45
0.90
0.75
0.50
0.49
0.84
0.76
0.47
0.39
0.79
0.73
Create a SAS program to analyze the data.
What are the null and alternative hypotheses?
Summarize the results in an ANOVA table
Are there significant differences among varieties i.e. do you reject the null hypothesis? Choose a level of significance that you feel is appropriate and be sure to include what level you are using. Support your answer with information from the SAS output.
Use the Ryan-Einot-Gabriel-Welsch multiple range test to perform comparisons among the treatment means (this can be done at the same time as the above ANOVA).
Are these comparisons appropriate for this data? Why or why not?
Summarize the results from the multiple range test in a table in your Word file.
Based on the results from the analysis, which variety(ies) has/have the highest leaf phosphorus content?Justify your answer using the results of the analysis.
Phosphorus content of leaves from each redbud variety
Forest Pansy
Silver Cloud
Hearts of Gold
Rubye Atkinson
0.35
0.32
0.65
0.60
0.40
0.51
0.70
0.80
0.58
0.45
0.90
0.75
0.50
0.49
0.84
0.76
0.47
0.39
0.79
0.73
Explanation / Answer
Let us code the variety as:
1: Forest Pansi, 2: Silver cloud, 3: Hearts of Gold, 4: Rubye Atkinson
SAS program:
Title1 ‘phosphorus content of tree leaves from 4 varieties of redbud’;
data Clover;
input Phosphorous_content factor;
datalines;
0.35 1
0.40 1
0.58 1
0.50 1
0.47 1
0.32 2
0.51 2
0.45 2
0.49 2
0.39 2
0.65 3
0.70 3
0.90 3
0.84 3
0.79 3
0.60 4
0.80 4
0.75 4
0.76 4
0.73 4
;
proc anova data = DATA;
class factor;
model Phosphorous_content=factor;
means factor / REGWQ;
run;
The hypothesis of testing is H0: Means of phosphorous contents in all varieties are same Vs. H1: Phosphorous contents in atleast two varites are different.
For testing above hypothesis, we use one way ANOVA, the output of SAS code is as follows
The ANOVA Procedure
Dependent Variable: Phosphorous_content
Sum of
Source DF Squares Mean Square F Value Pr > F
Model 3 0.47590000 0.15863333 21.10 <.0001
Error 16 0.12028000 0.00751750
Corrected Total 19 0.59618000
R-Square Coeff Var Root MSE Phosphorous_content Mean
0.798249 14.47471 0.086704 0.599000
From above ANOVA table it is clear that, we reject null hypothesis at 5% level of significance i.e. The phosphorus contents of tree leaves in the 4 varieties is different
The ANOVA Procedure
Ryan-Einot-Gabriel-Welsch Multiple Range Test for Phosphorous_content
NOTE: This test controls the Type I experimentwise error rate.
Alpha 0.05
Error Degrees of Freedom 16
Error Mean Square 0.007517
Number of Means 2 3 4
Critical Range 0.1352547 0.1414953 0.1568872
Means with the same letter are not significantly different.
REGWQ Grouping Mean N factor
A 0.77600 5 3
A
A 0.72800 5 4
B 0.46000 5 1
B
B 0.43200 5 2
From the above multiple comparison test we say that:
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