1. ANOVA test

2. ANOVA
A specific statement or hypothesis is generated about a population parameter, and sample statistics are used to assess the likehood that the hypothesis is true.
ANOVA is a test of hypothesis that is appropriate to compare means of a continuous variable in two or more independent comparison groups.
The test statistic must take into account the sample sizes, sample means or sample standard deviations in each of the comparison groups.

3. ANOVA The ANOVA approach:
Consider an example with 3 independent groups and continuous outcome measure.
The sample data are organized as follows:
Group 1
Group 2
Group 3
Group4
Sample size n1 n2 n3 n4
Sample Mean
𝑥 1
𝑥 2
𝑥 3
𝑥 4
Sample SD s1 s2 s3 s4

4. ANOVA Hypothesis building: Ho: μ1 = μ2 = μ3 = μ4
Ha: The means are not equal.
for ANOVA, the alternative hypothesis captures any differences in means and include, for example the situation where all the means are unique, when one is different from the other three, where two are different…and so on.

5. ANOVA
2. The test statistic for testing Ho: (μ1 = μ2 = μ3 = μ4 )

6. ANOVA
3. The critical value is found in a table of probability values for the F distribution with (degrees of freedom) df1 = k-1, df2=N-k.

7. ANOVA
4. Decision We reject H0 if Ft < Fc We accept H0 if Ft < Fc

8. Example
A study is designed to test whether there is a difference in the mean for the content of N in the soil for three different samples (of 6 observations each) taken in three different positions.
The data are shown below:
Point 1
Point 2
Point 3
1200
1000
890
1100
650
980
700
900
800
750
500
400
350

9. ANOVA
Is there a statistically difference in mean content of N when comparing three samples for three different positions?
We will run the ANOVA using the 4-steps approach.

10. library(MASS) # load the MASS package data<-c(1200,1000,980,900,750,800,1000,1100,700,800,500,700,890,650,1100,900,400,350) f = c("Point1","Point2","Point3") k = 3 # number of treatment levels n = 6 # observations per treatment tm = gl(k, 1, n*k, factor(f)) # matching treatments av = aov(data ~ tm) summary(av)

11. ANOVA Analysis of Variance
Comparison of means/variance between more than two samples
Can also compare within a multi-factorial design
Assumptions of linear model apply
Can be followed up with a Post-Hoc test to determine which groups are significantly different (e.g. Tukey HSD)

12. ANOVA

13. Exercise 5
Does the quantity of N fertilizer applied influence crop growth?
1 kg.ha
10 kg.ha
100 kg.ha
1.00
1.10
0.87
0.95
1.15
0.88
0.98
1.13
0.92
0.94
1.18
0.97
1.16
0.90

14. Implementation in R A two step process: Use the function:
Run a linear model
Apply the outcome of the linear model to ANOVA
Use the function:
aov(response~terms)
This runs the linear model followed by ANOVA
lm(response~terms)
anova(lm)
BALANCED DESIGN

15. Exercise 6
In this exercise we will use the internal dataset “ChickWeight”
Ho: Diet does not effect 20 day old chicken weight
Create a boxplot of the data
Consider we are looking at differences in the mean, can you see if there are any means that look different?
Conduct an ANOVA analysis
Conduct a post-hoc test (TukeyHSD)

16. Exercise 6 – R code Data(ChickWeight) Attach(ChickWeight)
boxplot(weight[Time==20]~Diet[Time==20])
summary(lm(weight[Time==20]~Diet[Time==20]))
summary(aov(weight[Time==20]~Diet[Time==20]))
TukeyHSD(aov(weight[Time==20]~Diet[Time==20]))

17. Exercise 7 Confirm Exercise 2 excel test within R
The file can be imported into R (“crop.RData”)

18. References References: http://www.r-tutor.com
Chun and Griffith (2013). Spatial Statistics and Geostatistics (Book)
(Dr. Sullivan Notes)
Master of Photogrammetry and Geoinformatics (Dr. Rawirl Notes)