Question

This homework assignment should help you towards understanding one-way analysis of variance (ANOVA) statistical tests AND allow you to become familiar with using Rmarkdownand some basic LaTexprogramming. By using Rmarkdown, you can include your R code, the output and any written explanations within one output file.  

You should complete this entire assignment with Rmarkdown.  

1. Recently scientists have surveyed the local fish species in several nearby ponds. The scientists are trying to figure out if the fish are significantly different in length (cm) from one another.    Could you help them out?

species	length
red	53.2
red	44.6
red	60.2
red	50.7
red	44.8
red	54.0
red	51.4
red	51.5
red	53.3
red	47.5
red	44.5
red	49.0
red	50.2
blue	59.1
blue	60.3
blue	60.4
blue	60.1
blue	58.5
blue	59.4
blue	61.3
blue	60.4
blue	60.0
blue	61.6
blue	59.9
green	63.6
green	85.1
green	79.7
green	67.5
green	90.6
green	78.9
green	71.9
green	66.9
green	81.3
green	46.1
green	87.1
green	90.6

1. What are the null and alternative hypotheses?
2. How do the data look? (also please report basic summary statistics (a measure of central tendency and variance).  Are the data normally distributed (each group)?
3. Are the variances equal?
4. What kind of ANOVA? Run analysis and include a statement on the outcome of the ANOVA.
5. If the ANOVA indicates that you may reject the null hypothesis, what fish are different from one another and how do you know?

Answer 1

##First make a ".txt" file of data named "data.txt"
d=read.table("data.txt",header=TRUE)

##Question of interest: The scientists are trying to figure out if the fish are significantly different in length (cm) from one another.
##Null Hypothesis-H0: the species of fish are not significantly different in length (cm) from one another.
##Alternative Hypothesis-H1: the species of fish are significantly different in length (cm) from one another.

summary(d)
species length   
blue :11 Min. :44.50
green:12 1st Qu.:51.48
red :13 Median :60.05
Mean :61.81
3rd Qu.:67.05
Max. :90.60

var(d[,2])
[1] 177.697

y=d$length
d1=subset(y,species=="red")
d2=subset(y,species=="blue")
d3=subset(y,species=="green")
> summary(d1) ##red
Min. 1st Qu. Median Mean 3rd Qu. Max.
44.50 47.50 50.70 50.38 53.20 60.20
> summary(d2) ##blue
Min. 1st Qu. Median Mean 3rd Qu. Max.
58.50 59.65 60.10 60.09 60.40 61.60
> summary(d3) ##green
Min. 1st Qu. Median Mean 3rd Qu. Max.
46.10 67.35 79.30 75.78 85.60 90.60

> var(d1)
[1] 19.81359
> var(d2)
[1] 0.8009091
> var(d3)
[1] 172.8693
##The variances of different species are not equal.

##Shapiro-Wilks test for Normality
##H00: Data are normally distributed.
shapiro.test(d1)
Shapiro-Wilk normality test
data: d1
W = 0.9338, p-value = 0.3818

shapiro.test(d2)
Shapiro-Wilk normality test
data: d2
W = 0.96771, p-value = 0.8625

shapiro.test(d3)
Shapiro-Wilk normality test
data: d3
W = 0.9174, p-value = 0.2651

##p-value for all the three species is greater than alpha=0.05. Thus, we fail to Reject H00.
##Therefore, Data (each group) is normally distributed.

species=factor(d$species)
species
[1] red red red red red red red red red red red red red blue blue blue blue blue blue blue blue blue blue blue green green green green green green green green green green green green
Levels: blue green red

ANOVA=aov(y~species)
ANOVA
Call:
aov(formula = y ~ species)
Terms:
species Residuals
Sum of Squares 4072.061 2147.335
Deg. of Freedom 2 33
Residual standard error: 8.066644
Estimated effects may be unbalanced

summary(ANOVA)
Df Sum Sq Mean Sq F value Pr(>F)
species 2 4072 2036.0 31.29 2.4e-08 ***
Residuals 33 2147 65.1
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

##From the output of summary, p-value=2.4e-08=0 (approx.) which is less than alpha=0.05. We Reject H0.
##Thus, the species of fish are significantly different in length (cm) from one another.

###Finding which species is different
TukeyHSD(ANOVA,'species',conf.level=0.95)
Tukey multiple comparisons of means
95% family-wise confidence level
Fit: aov(formula = y ~ species)
$species
diff lwr upr p adj
green-blue 15.684091 7.421656 23.94653 0.0001459
red-blue -9.713986 -17.823012 -1.60496 0.0160087
red-green -25.398077 -33.321973 -17.47418 0.0000000

##From output of TukeyHSD test,p-value for all 3 pairs is less than alpha=0.05.
##Therefore, it is clear that at 5% l.o.s, all pairs of species are significant.