Exercises uses Deer dataset and random sets of data using r* functions

deer = read.csv("Deer.csv")
aragorn = rnorm(50, mean=180, sd=10)
gimli = rnorm(50, mean=132, sd = 15)
legolas = rnorm(50, mean=195, sd=15)

Exercises

Run a t-test to compare the Legolas actors to the set of Aragorns and then to the set of Gimlis. Do you find evidence for significant differences?

t.test(legolas, aragorn, alternative="two.sided")
## 
##  Welch Two Sample t-test
## 
## data:  legolas and aragorn
## t = 5.0087, df = 84.264, p-value = 2.974e-06
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##   8.954966 20.746866
## sample estimates:
## mean of x mean of y 
##  192.1601  177.3092

The low p value of the t-test means that there is evidence for significant differences in the heights of the sets of actors.

t.test(legolas, gimli, alternative="two.sided")
## 
##  Welch Two Sample t-test
## 
## data:  legolas and gimli
## t = 18.829, df = 89.492, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  51.72471 63.92864
## sample estimates:
## mean of x mean of y 
##  192.1601  134.3335

The low p value of the t-test means that there is evidence for significant differences in the heights of the sets of actors.

Rerun the variance test (F-test) to compare the group of Gimli and Legolas actors. Do these groups have different variance?

var.test(gimli, legolas)
## 
##  F test to compare two variances
## 
## data:  gimli and legolas
## F = 0.52866, num df = 49, denom df = 49, p-value = 0.02771
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
##  0.2999996 0.9315912
## sample estimates:
## ratio of variances 
##          0.5286558

As the p value is above 0.05, there is likely not a difference in variance.

Redo the correlation for the sepal length and sepal width for the Iris dataset, but for the three individual species. Are these correlated?

vers = subset(iris, Species == "versicolor")
cor(vers$Sepal.Length, vers$Sepal.Width)
## [1] 0.5259107

As the correlation value is 0.526, there is a positive correlation between the variables for the species versicolor

virg = subset(iris, Species == "virginica")
cor(virg$Sepal.Length, virg$Sepal.Width)
## [1] 0.4572278

As the correlation value is 0.457, there is a positive correlation between the variables for the species virginica

set = subset(iris, Species == "setosa")
cor(set$Sepal.Length, set$Sepal.Width)
## [1] 0.7425467

As the correlation value is 0.743, there is a positive correlation between the variables for the species setosa

Using the deer dataset and the chisq.test() function, test: (1) If there are significant differences in the number of deer caught per month

table(deer$Month)
## 
##   1   2   3   4   5   6   7   8   9  10  11  12 
## 256 165  27   3   2  35  11  19  58 168 189 188
chisq.test(table(deer$Month))
## 
##  Chi-squared test for given probabilities
## 
## data:  table(deer$Month)
## X-squared = 997.07, df = 11, p-value < 2.2e-16

As the p value is very low, there are significant differences in the number of deer caught per month. (2) If the cases of tuberculosis are uniformly distributed across all farms

table(deer$Farm, deer$Tb)
##       
##          0   1
##   AL    10   3
##   AU    23   0
##   BA    67   5
##   BE     7   0
##   CB    88   3
##   CRC    4   0
##   HB    22   1
##   LCV    0   1
##   LN    28   6
##   MAN   27  24
##   MB    16   5
##   MO   186  31
##   NC    24   4
##   NV    18   1
##   PA    11   0
##   PN    39   0
##   QM    67   7
##   RF    23   1
##   RN    21   0
##   RO    31   0
##   SAL    0   1
##   SAU    3   0
##   SE    16  10
##   TI     9   0
##   TN    16   2
##   VISO  13   1
##   VY    15   4
chisq.test(table(deer$Farm, deer$Tb))
## Warning in chisq.test(table(deer$Farm, deer$Tb)): Chi-squared approximation may
## be incorrect
## 
##  Pearson's Chi-squared test
## 
## data:  table(deer$Farm, deer$Tb)
## X-squared = 129.09, df = 26, p-value = 1.243e-15

As the p value is very low, the cases of tuberculosis are not uniformly distributed across all farms.