Question

Complete the following problems using R.  Please turn in your R code and answers to all questions all in one document, with each problem clearly labeled

1.Now consider the Height data.  

a.Make a box plot by speciesto show the variability of the height data  

b.Make a histogram for each species’ heights to show the distribution of the data.  Is the distribution roughly bell-shaped, or skewed?

c.Compute the mean value of height for each species.

d.Compute the standard deviation of height for each species.

e.What is the 90% confidence interval for height for each species?  

f.What do these intervals mean, and how do they compare to each other?

treecode,location,species,dbh,height
1-W-7,eastern,MR,46.8,39.2
1-N-1,eastern,MR,47.1,37.4
1-E-4,eastern,MR,60.6,41.1
2-M-2,eastern,MR,50.1,26.9
3-E-4,eastern,MR,16.4,25.2
3-M-6,eastern,MR,29.4,27.3
3-S-4,eastern,MR,33.3,38
3-N-5,eastern,MR,37.4,22.5
3-N-6,eastern,MR,36.4,25
4-M-3,eastern,MR,7.2,3.9
4-M-2,eastern,MR,8,6.1
5-E-3,eastern,MR,19.6,14.5
7-N-5,eastern,MR,41.6,34.1
8-W-5,eastern,MR,31.7,31
8-S-7,eastern,MR,39.4,31.2
11-M-3,eastern,MR,16.3,24.3
11-M-5,eastern,MR,22.8,26.2
11-M-8,eastern,MR,23.4,30.8
11-M-4,eastern,MR,29.5,25.6
11-N-7,eastern,MR,37.5,27.6
11-S-6,eastern,MR,40.4,34.8
13-M-3,eastern,MR,31.8,27.7
13-M-2,eastern,MR,67.4,37.8
14-M-5,eastern,MR,31.1,28.7
14-M-4,eastern,MR,35.4,30.6
14-W-2,eastern,MR,34.3,33.3
14-E-3,eastern,MR,41.5,32.4
14-N-3,eastern,MR,39.1,36.8
14-M-3,eastern,MR,51,43.4
42-N-1,western,MR,57.6,35.9
43-M-4,western,MR,2.5,4.8
43-M-5,western,MR,3.4,4.8
43-M-7,western,MR,7,7
43-M-9,western,MR,10.5,9.7
44-M-3,western,MR,18.2,22.3
44-E-3,western,MR,25.6,25.3
45-M-1,western,MR,62.7,40
46-M-1,western,MR,25.6,28.3
46-M-2,western,MR,28.7,26.6
46-M-5,western,MR,43.9,26.8
46-N-9,western,MR,50.9,36.1
46-M-7,western,MR,56.3,35.2
47-N-1,western,MR,8.6,8.7
47-E-4,western,MR,22.6,19.3
48-M-2,western,MR,26,21.6
48-S-5,western,MR,30.7,23.5
49-M-7,western,MR,38.3,26.8
49-W-1,western,MR,45.4,29.6
50-M-5,western,MR,27,19
3-W-5,eastern,OW,6,6.5
3-W-3,eastern,OW,12.5,7.8
3-M-4,eastern,OW,19.1,25.2
3-M-3,eastern,OW,34.4,25.2
7-M-3,eastern,OW,31.7,24.2
7-S-1,eastern,OW,40.2,23
7-N-6,eastern,OW,45.5,25.2
7-N-7,eastern,OW,52.4,31.3
7-W-5,eastern,OW,54.3,28.3
8-M-2,eastern,OW,37.8,30.9
8-E-1,eastern,OW,65,36.4
8-M-4,eastern,OW,68.1,32.2
9-S-6,eastern,OW,59.8,46.9
11-M-2,eastern,OW,12.1,18.6
11-E-2,eastern,OW,17.7,12.4
11-M-6,eastern,OW,26.4,22.6
13-M-5,eastern,OW,98.9,36.4
14-N-5,eastern,OW,62.2,29.6
15-M-1,eastern,OW,26.4,26.8
15-M-6,eastern,OW,43.6,35.7
15-S-3,eastern,OW,48.2,28.9
15-M-3,eastern,OW,61,33.1
42-M-2,western,OW,41.3,21.1
45-W-5,western,OW,87.2,41.2
46-M-6,western,OW,33.1,21.9
47-M-9,western,OW,4.8,3.2
47-M-3,western,OW,2.5,2.9
48-W-5,western,OW,61.1,29.7
48-M-7,western,OW,62.5,27.6
49-M-8,western,OW,3.7,4.4
49-N-4,western,OW,47.8,19.5
49-M-6,western,OW,53.4,29.6
49-M-4,western,OW,56,25.3
49-M-3,western,OW,67.4,27.4
50-N-2,western,OW,45.4,20.8
50-M-7,western,OW,48.1,23.9
50-M-4,western,OW,45.4,23.4
50-M-2,western,OW,48.7,22.7
52-N-6,western,OW,5.2,7
53-E-2,western,OW,44.3,30.6
53-N-2,western,OW,45.3,23.7
53-M-5,western,OW,76.9,28.6
54-M-6,western,OW,32.7,23.3
54-M-2,western,OW,40.2,24.1
54-S-6,western,OW,47.8,26.1
55-M-1,western,OW,49,27
55-M-2,western,OW,66.3,26.7
56-M-3,western,OW,29.3,18.8
57-M-3,western,OW,37.9,37.4
58-M-6,western,OW,41.9,22.4
59-M-2,western,OW,24.7,15.1
61-M-6,western,OW,57.3,30.6
62-M-10,western,OW,4.1,8.3
62-M-2,western,OW,9.1,2.7
62-S-1,western,OW,16.2,8.9
63-M-6,western,OW,16.2,13.8
63-M-2,western,OW,49.6,24.8
64-W-3,western,OW,30.2,15.1
64-M-6,western,OW,54.5,22.7
64-M-2,western,OW,60.6,24.9
64-M-5,western,OW,63,27.9
64-S-2,western,OW,72.9,28.8

Answer 1

First the data is stored in excel and saved as a csv file.

Trees = read.csv("Trees.scv")
splitTrees = split(Trees, Trees$species)
> TreesOW= splitTrees$OW
> TreesMR= splitTrees$MR

#The data is split according to the species

boxplot(height~species, data = Trees)

Box plot of heights with respect to species:

Histogram of height of OW species: (This is approximately normally distributed)

histogram of height of the other MR species:

This histogram is slightly skewed

For finding the mean the code used is:

> mean(TreesMR$height)
[1] 26.42245
> mean(TreesOW$height)
[1] 23.40484

For calculating the standard deviations this code is used:

> sd(TreesOW$height)
[1] 9.557667
> sd(TreesMR$height)
[1] 10.275