BIN510_Unit2_Exam
Gordon Ober
For this exam you will be using the file “tick.csv” found on canvas:
Ticks use large mammals as hosts and the presence of mammals represents an important vector for tick-borne diseases and humans. Tick presence may also be influenced by environmental factors. The tick.csv file has a number of different variables that can be analyzed, below are a list of the important variables for this study:
Treatment - what type of animals were allowed within the plot? CONT (control, no exclosure), LMH (total exclosure, no mammals), MESO (exclosion of large herbivores >15kg), and MEGA (exclusion of Elephants and Giraffes)
Rain - average rainfall (mm/year)
Total - total number of ticks found across 3 study species
PART 1a: How does treatment of mammals influence the total number of ticks found? Answer the questions below:
NOTE: for each prompt, I need to see code in order to give you credit! NOTE: you can perform all of these tasks in one code chunk or many code chunks, this is up to you
- Code: bring in the dataset and name it “tick”
tick <- read.csv("ticks.csv", header = TRUE)
str(tick)## 'data.frame': 180 obs. of 11 variables:
## $ Period : num 11 11 11 11 11 11 11 11 11 11 ...
## $ Date : chr "20/08/2014" "20/08/2014" "20/08/2014" "21/08/2014" ...
## $ Month : chr "Aug" "Aug" "Aug" "Aug" ...
## $ Level : int 2 2 2 3 3 3 1 1 1 2 ...
## $ Replicate: int 4 5 6 7 8 9 1 2 3 4 ...
## $ Treatment: chr "CONT" "CONT" "CONT" "CONT" ...
## $ Total : int 1 1 1 0 0 5 4 0 1 37 ...
## $ RHPU : int 0 0 0 0 0 4 3 0 0 0 ...
## $ RHPR : int 0 0 0 0 0 0 0 0 0 1 ...
## $ RHPV : int 1 1 1 0 0 1 1 0 1 36 ...
## $ Rain : int 580 580 580 440 440 440 640 640 640 580 ...table(tick$Treatment)##
## CONT LMH MEGA MESO
## 45 45 45 45- Question: what type of test would be appropriate to run in order to compare total ticks by treatment type? Why?
- Total ticks is a numerical value and treatment type is a categorical
variaible with four groups. The appropriate test to evaluate these two
columns is an ANOVA test because ANOVA allows us to test whether the
mean number of ticks differs between the four treatments.
- Question: what assumptions need to be tested before running the test?
- We need to test the data for normal distribution and for homogeneity
of variance.
- Code: test all assumptions from 3
#histogram
hist(tick$Total,
main = "Total Ticks Distribution",
xlab = "Total Number of Ticks",
col = "orchid")
#Shapiro-Wilk test
shapiro.test(tick$Total)##
## Shapiro-Wilk normality test
##
## data: tick$Total
## W = 0.64898, p-value < 2.2e-16#Levene's test
library(car)## Loading required package: carDataleveneTest(Total ~ Treatment, data = tick)## Warning in leveneTest.default(y = y, group = group, ...): group coerced to
## factor.## Levene's Test for Homogeneity of Variance (center = median)
## Df F value Pr(>F)
## group 3 7.506 9.346e-05 ***
## 176
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1- Question: interpret the results of 4. (tell me if assumptions are met or not)
- The histogram shows a heavy right-skew in the dataset. In addition to this, the Shapiro_Wilk test to evaluate normality had a p-value that was significantly less than 0.05, which indicates that we reject the null hypothesis which states that the data is normally distributed. As for Levene’s test of equal variance, the p-value was less than 0.05, so we reject the null hypothesis which states that the variance between the groups is equal. Therefore, neither of the assumptions have been met for this dataset.
- Code: Regardless of your answer to 5*, run the test you propose in 2
tickANOVA <- aov(Total ~ Treatment,
data = tick)
summary(tickANOVA)## Df Sum Sq Mean Sq F value Pr(>F)
## Treatment 3 3387 1128.9 8.67 2.13e-05 ***
## Residuals 176 22915 130.2
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1- Question: what does the output of 6 tell you about ticks and treatment?
- The null hypothesis states that the total number of ticks is the same across the four treatments, the alternate hypothesis states that at least one treatment differs. Because the p-value is less than 0.05, we reject the null hypothesis that all four groups have the same mean total tick population.
- Question: do you need to run post-hoc analysis? Why or why not?
- Yes, a post-hoc test is important because even though the ANOVA tells us that there is a significant difference in tick population between treatments, it does not specify which treatment differs.
- Code: if you answered YES to 8, run that analysis here
TukeyHSD(tickANOVA)## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = Total ~ Treatment, data = tick)
##
## $Treatment
## diff lwr upr p adj
## LMH-CONT 11.844444 5.6050266 18.083862 0.0000115
## MEGA-CONT 3.177778 -3.0616401 9.417196 0.5508095
## MESO-CONT 5.355556 -0.8838623 11.594973 0.1201486
## MEGA-LMH -8.666667 -14.9060845 -2.427249 0.0022968
## MESO-LMH -6.488889 -12.7283067 -0.249471 0.0380749
## MESO-MEGA 2.177778 -4.0616401 8.417196 0.8020299- Question: if you ran 9, tell me what the output tells us about the treatments
- The output indicates that LMH treatment differs most significantly in tick abundance between treatments.
- Create a visual that helps tell the story of how treatment influences ticks
boxplot(Total ~ Treatment,
data = tick,
main = "Tick Population based on Treatment",
xlab = "Treatment",
ylab = "Total Number of Ticks",
col = c("lightsalmon", "orangered2", "red3", "red4"))
PART 1b: violations of assumptions. Answer the questions below:
- Code: choose a transformation method of your choice and apply it to the “total” column
tick$logTotal <- log(tick$Total + 1)- Code: re-test the assumptions from Part 1a Q4
#histogram
hist(tick$logTotal,
main = "Total Ticks Distribution",
xlab = "Total Number of Ticks",
col = "darkorchid")
#Shapiro-Wilk test
shapiro.test(tick$logTotal)##
## Shapiro-Wilk normality test
##
## data: tick$logTotal
## W = 0.98087, p-value = 0.01424#Levene's test
library(car)
leveneTest(logTotal ~ Treatment, data = tick)## Warning in leveneTest.default(y = y, group = group, ...): group coerced to
## factor.## Levene's Test for Homogeneity of Variance (center = median)
## Df F value Pr(>F)
## group 3 2.3584 0.0733 .
## 176
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1- Question: are the assumptions met?
- After log transformation, the Shapiro-Wilk test produced a p-value =
0.0142. While this p-value is closer to 0.05, we still reject the null
hypothesis that the data is normally distributed. The Levene’s test
produced a p-value = 0.0733. This is greater than 0.05 so we fail to
reject the null hypothesis which states that the variance is equal among
treatments. Therefore, the assumption of equal variance has been met
post-log transformation.
- Code: Run the alternative non-parametric test appropriate for this data
kruskal.test(Total ~ Treatment,
data = tick)##
## Kruskal-Wallis rank sum test
##
## data: Total by Treatment
## Kruskal-Wallis chi-squared = 23.869, df = 3, p-value = 2.66e-05- Question: what does the output of 4 tell us about ticks and treatment?
- The Kruskal-Wallis test produced a p-value < 0.05. Based on this, I reject the null hypothesis that states there is no difference in tick population among treatment groups. In other words, there is evidence to support that the population of ticks across treatments differs among treatment groups.
PART 2: does environment influence ticks?
- Question: what type of test would be appropriate to run in order to compare total ticks by rain? Why?
- A linear regression would be appropriate because the variables in question are both quantitative and the question distinguishes a predictor variable, the rain, and a response variable, tick abundance. This test allows us to evaluate the direction and strength of the relationship between rain and tick abundance.
- Code: run the test
tickReg <- lm(Total ~ Rain, data = tick)
summary(tickReg)##
## Call:
## lm(formula = Total ~ Rain, data = tick)
##
## Residuals:
## Min 1Q Median 3Q Max
## -12.471 -6.593 -2.959 2.163 76.041
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 23.50781 5.95874 3.945 0.000115 ***
## Rain -0.02508 0.01065 -2.356 0.019565 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 11.97 on 178 degrees of freedom
## Multiple R-squared: 0.03024, Adjusted R-squared: 0.02479
## F-statistic: 5.55 on 1 and 178 DF, p-value: 0.01957- Question: using the output from 2, tell me what the model says about ticks and rainfall. Include in your answer an interpretation of the p-value and an interpretation of the R^2 value
- The Multiple R-Squared value of 0.03024 implies that approximately 3% of the observed variation in total tick abundance is accounted for by its linear relationship with rainfall. The p-value < 0.05, therefore I reject the null hypothesis which states there is no relationship between rainfall and tick abundance.
- Code: create a visual that shows the relationship between rainfall and ticks
plot(x = tick$Rain,
y = tick$Total,
pch = 16,
col = "blue",
xlab = "Average Rainfall (mm/year)",
ylab = "Tick Abundance",
main = "Relationship Between Rainfall and Tick Abundance")
5. Code: add a best fit line to help visualize the pattern
plot(x = tick$Rain,
y = tick$Total,
pch = 16,
col = "blue",
xlab = "Average Rainfall (mm/year)",
ylab = "Tick Abundance",
main = "Relationship Between Rainfall and Tick Abundance")
abline(tickReg,
col = "red",
lwd = 2)
6. Question: Do ticks prefer wet or dry climates? - According to the
linear regression line of best fit, ticks prefer drier climates due to
the indirect relationship between tick abundance and rainfall. 7. Code:
predict the number of ticks observed in a region that experiences 500
mm/year mean rainfall
newtick <- data.frame(Rain = 500)
predict(tickReg,
newdata = newtick)## 1
## 10.96561