Question

3. An ecologist studied the spatial distribution of tree species in a wooded area. From a total area of 21 acres, he randomly selected 144 quadrats (plots), each 38 feet square, and noted the presence or absence of maples and hickories in each quadrat. The data are as follows Maples Absent 63 26 Hickories Present Absent Present 26 29 Let pı - Pr{hickories present|maples present} and p2 - Prfhickories present|maples absent in the entire 21 acres (a) Based on the data, compute pi and p2, the sample proportions. Find a point estimate for the difference in proportion pı - p2 (b) Find a point estimate for the relative risk pı/p2 then interpret (c) Find a point estimate for the odds ratio P1/(1 -p) p2/(1 p2) then interpret. (d) Use the following R code to obtain a 95% confidence interval for the odds ratio θ then interpret. fisher.test(tree,conf.int-TRUE)

Answer 1

mop Pruned. 29 26 63 29 29 A 5514A imila 144 0.3699

The odds ratio is less than 1. Hence, the present of Hickories is better than absent.

d)

## R command

Input =("
Injection.area Present Absent   
Present 26 63
Absent 29 26
")

Matrix = as.matrix(read.table(textConnection(Input),
header=TRUE,
row.names=1))
fisher.test(Matrix,conf.int=TRUE)

########

## Output

> Input =("
+ Injection.area Present Absent   
+ Present 26 63
+ Absent 29 26
+ ")
>
> Matrix = as.matrix(read.table(textConnection(Input),
+ header=TRUE,
+ row.names=1))
> fisher.test(Matrix,conf.int=TRUE)

Fisher's Exact Test for Count Data

data: Matrix
p-value = 0.007783
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
0.1733713 0.7893954
sample estimates:
odds ratio
0.3727507
###

Interpretation: The p-value is 0.007783 and less than 0.05. Hence, we can conclude that there is a significant association between Maples and Hickories at the 0.05 significance level.