Question

1. A study to compare the rates of growth of vegetation at four swampy undeveloped sites involved measuring the leaf lengths of a particular plant species at a particular date. Six plants were randomly selected at each of the four sites and the mean leaf length per plant (in centimeters) for a random sample of 10 leaves per plant were recorded. The following are the data:

    Location                                      Mean leaf length

1	5.7	6.3	6.1	6	5.8	6.2
2	6.2	5.3	5.7	6	5.2	5.5
3	5.4	5	6	5.6	4.9	5.2
4	3.7	3.2	3.9	4	3.5	3.6

   

1. What is the factor of interest in this experiment? _________________________

2. What is an experimental unit ______________ and the response variable? __________________

3. [R] Construct the ANOVA table for this experiment. Copy/paste it here.

4. Do we have enough evidence to conclude that the mean growth rates at the four sites are not the same? ______ Use α=0.05.

5. [R] Carry out the Shapiro test of normality for the residuals. Can we conclude that the normality assumption is not satisfied? _____ Give the p-value: ________

6. [R] Carry out Bartlett’s test for variances. Can we conclude that the population variances are different? _____ Give the p-value: ________

7. [R] Use Tukey’s method to perform pairwise comparisons among treatment means. Which sites have equal growth rates? ___________________ Copy/paste the R output that supports your answer.

Answer 1

i) the factor of interest in this experiment: location

ii)an experimental unit: plant of each of four swampy undeveloped sites

the response variable: leaf length

iii)ANOVA table:

	Df	Sum Sq	Mean Sq	F value	Pr(>F)
group	3	19.74	6.58	57.38	5.25E-10
Residuals	20	2.293	0.115
Total	23	22.033

Since p-value is almost 0, we can claim that we have enough evidence to conclude that the mean growth rates at the four sites are not the same.

iv)p-value = 0.4236>0.05 hence we can claim that the normality assumption for the residuals is satisfied.

v)p-value = 0.5999>0.05 hence we can say that there is insufficient evidence to support the claim that the population variances are not same. Hence assumption of homogeneity is satisfied.

vi)Pair of locations (1,2) and (1,3) have same growth rates since p-value for each pair is quite high.

R CODE and OUTPUT:

> data=read.csv("C:/Users/cssc/Desktop/b1.csv",header=TRUE)
> # the analysis of variance
> a <- aov(length ~ group, data = data)
> # Summary of analysis
> summary(a)

> # Extracting residuals
> res <- residuals(object = a )
> # Shapiro test
> shapiro.test(x = res)

Shapiro-Wilk normality test

data: res
W = 0.95926, p-value = 0.4236

> # Bartlett test
> bartlett.test(length ~ group, data = data)

Bartlett test of homogeneity of variances

data: length by group
Bartlett's K-squared = 1.8695, df = 3, p-value = 0.5999

> #Tukey's comparison
> TukeyHSD(a)
Tukey multiple comparisons of means
95% family-wise confidence level

Fit: aov(formula = length ~ group, data = data)

$group
diff lwr upr p adj
l2-l1 -0.3666667 -0.9138728 0.1805395 0.2696656
l3-l1 -0.6666667 -1.2138728 -0.1194605 0.0135850
l4-l1 -2.3666667 -2.9138728 -1.8194605 0.0000000
l3-l2 -0.3000000 -0.8472061 0.2472061 0.4366710
l4-l2 -2.0000000 -2.5472061 -1.4527939 0.0000000
l4-l3 -1.7000000 -2.2472061 -1.1527939 0.0000002