Statistical Analysis of the Number of Seed Capsules from the Portulaca
Measurements 1/30/2017
 |
| Seed capsules of Portulaca amilis. |
After determining the
method to evaluate and identify a difference in plant height between the P
(potting soil) and M (50/50 potting soil and sand), I will attempt to analyze
the seed pod measurements.
Here is the raw data
table:
Tag
|
Type
|
# of Plants
|
Tallest Plant
|
# of pods
|
Weight (nearest 5 g
|
TP2
|
P
|
3
|
22.5
|
27
|
335
|
AP1
|
P
|
5
|
21.5
|
27
|
335
|
LP1
|
P
|
10
|
22
|
33
|
335
|
EP2
|
P
|
1
|
24
|
22
|
330
|
RP2
|
P
|
2
|
22.5
|
17
|
330
|
CP1
|
P
|
1
|
22
|
19
|
335
|
LP2
|
P
|
6
|
22
|
37
|
335
|
FP2
|
P
|
8
|
23
|
56
|
330
|
AP2
|
P
|
6
|
21
|
35
|
335
|
YP1
|
P
|
5
|
22
|
30
|
335
|
CP2
|
P
|
3
|
24
|
30
|
330
|
UP2
|
P
|
2
|
22.5
|
29
|
330
|
RP1
|
P
|
1
|
19
|
16
|
330
|
YP2
|
P
|
8
|
20
|
24
|
330
|
TP1
|
P
|
1
|
14.5
|
3
|
330
|
FP1
|
P
|
4
|
20.5
|
34
|
330
|
UP1
|
P
|
1
|
21
|
14
|
330
|
BP2
|
P
|
4
|
21.5
|
25
|
330
|
BP1
|
P
|
9
|
23.5
|
40
|
330
|
BM2
|
M
|
5
|
19.5
|
20
|
360
|
TM2
|
M
|
3
|
17.5
|
8
|
360
|
AM2
|
M
|
4
|
18
|
18
|
360
|
UM2
|
M
|
4
|
18.5
|
27
|
360
|
YM2
|
M
|
5
|
18.5
|
25
|
360
|
FM1
|
M
|
6
|
17.5
|
25
|
360
|
FM2
|
M
|
4
|
18.5
|
15
|
360
|
WM1
|
M
|
1
|
19
|
6
|
360
|
LM1
|
M
|
4
|
18
|
22
|
360
|
RM1
|
M
|
1
|
18.5
|
8
|
360
|
AM1
|
M
|
6
|
18
|
18
|
360
|
LM2
|
M
|
7
|
19
|
20
|
360
|
RM2
|
M
|
2
|
19.5
|
9
|
360
|
TM1
|
M
|
2
|
17.5
|
12
|
360
|
UM1
|
M
|
1
|
19.5
|
9
|
360
|
YM1
|
M
|
7
|
16.5
|
27
|
360
|
BM1
|
M
|
4
|
16.5
|
17
|
360
|
The main columns of
interest are the Type and # of pods columns. I need to form the data into the
form to enter into the R program. Instead of typing the code out, I went to the
http://astatsa.com/WilcoxonTest/
website and copy and pasted my values. With the click of a button, the site
formatted my numbers, and I take the needed information.
 |
| Portulaca amilis plants of the experiment. Left: plants grown in the potting soil (P) portion of the experiment. Right: plants grown in the 50/50 potting soil and sand (M) portion of the experiment. |
P <- c(27, 27, 33,
22, 17, 19, 37, 56, 35, 30, 30, 29, 16, 24, 3, 34, 14, 25, 40)
M <- c(20, 8, 18,
27, 25, 25, 15, 6, 22, 8, 18, 20, 9, 12, 9, 27, 17)
Part 1. This is my work to figure out what to do. Part 1
is THE WRONG WAY.
I must check the assumptions
of a parametric test. This will test for equal variances (note the test code is
sensitive to capitalization):
var.test(P,M)
F test to compare
two variances
data: P and M
F = 2.5685, num df = 18, denom df = 16,
p-value = 0.06375
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.9453428 6.7817415
sample estimates:
ratio of variances
2.5685
From this result, my
data is just within equal variances. A thought came to me, the Levene’s test.
This was the name of the test to check for equal variances I was instructed to
use in my statistics class. The process could not be straight forward, rather I
must take another detour that may perhaps change my results of the tallest
plant. After investigation, I updated my tallest plant post to utilize this
form of the equal variance test. Now to continue. For additional reading: https://www.r-bloggers.com/example-9-36-levenes-test-for-equal-variances/
and http://stats.stackexchange.com/questions/15722/how-to-use-levene-test-function-in-r.
To conform to the code
I changed P and M to sample1 and sample2 as exampled:
sample1 <- c(27, 27, 33, 22, 17, 19, 37, 56, 35, 30, 30, 29, 16, 24,
3, 34, 14, 25, 40)
sample2 <- c(20, 8, 18, 27, 25, 25, 15, 6, 22, 8, 18, 20, 9, 12, 9,
27, 17)
y <- c(sample1, sample2)
group <- as.factor(c(rep(1, length(sample1)), rep(2,
length(sample2))))
levene.test(y, group)
It returned:
modified robust
Brown-Forsythe Levene-type
test based on the
absolute deviations from
the median
data: y
Test Statistic = 1.3307, p-value = 0.2567
The results from the
Levene’s test would indicate the samples are of equal variances. This differs
from the var.test. I will use and accept the Levene’s test as the better
method.
Next I check the data
for normal distributions using shapiro.test(). Note sample1 is P, the data for
potting soil samples; sample2 is M, the data for the 50/50 potting soil and
sand samples.
shapiro.test(sample1)
Shapiro-Wilk
normality test
data: sample1
W = 0.97128, p-value = 0.8017
shapiro.test(sample2)
Shapiro-Wilk
normality test
data: sample2
W = 0.92503, p-value = 0.1796
Both sets of data meet
the assumption of normal distribution.
The data meets both
assumptions of equal variances and normal distribution so that a parametric
test can be performed. I am testing two group of data that can use a parametric
test, I use the two sample t test. I will mention for tallest plants, I had two
groups of data that required the use of a non-parametric test, that is why I
used the Mann-Whitney U test. The t test is as follows:
t.test(sample1,sample2, var.equal = TRUE)
Two Sample t-test
data: sample1 and sample2
t = 3.2435, df = 34, p-value = 0.002648
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
3.898681 16.980576
sample estimates:
mean of x mean of y
27.26316 16.82353
 |
| Portulaca amilis Plants from the potting soil (P) portion of the experiment. |
The t.test results have
a p-value of 0.002648 indicating the groups have a statistically significant
difference. Another way to determine the difference, looking at a t Table,
going to the .05 column for a two-tailed test, then going down to the row of 34
for DF is a critical value around 2.02 to 2.04; my number is 3.2435 which is
larger than the critical value and larger means statistically significant
difference.
Some notes on the t
test: https://www.cliffsnotes.com/study-guides/statistics/univariate-inferential-tests/two-sample-t-test-for-comparing-two-means.
This provided me the explanation for partial degrees of freedom with the fix
for equal variances “var.equal=TRUE” http://www.talkstats.com/showthread.php/49431-Fractional-degrees-of-freedom-when-running-t.test().
The above is incorrect
because I have unequal sample sizes (see https://en.wikipedia.org/wiki/Student's_t-test
for independent two sample t test, unequal sample sizes, equal variance for
equation).
The post is gold: http://daniellakens.blogspot.com/2015/01/always-use-welchs-t-test-instead-of.html
again from the master I referenced in my permits work. He explains always using
the Welch’s test. He also provides the citation that should be used when such
use of the test is put into action: http://www.jstor.org/stable/2684403?seq=1#page_scan_tab_contents.
This is basically what
I would input before. I change the P and M to sample1 and sample2 so that I
could perform the Levene’s test.
sample1 <- c(27, 27, 33, 22, 17, 19, 37, 56, 35, 30, 30, 29, 16, 24,
3, 34, 14, 25, 40)
sample2 <- c(20, 8, 18, 27, 25, 25, 15, 6, 22, 8, 18, 20, 9, 12, 9,
27, 17)
y <- c(sample1, sample2)
group <- as.factor(c(rep(1, length(sample1)), rep(2,
length(sample2))))
levene.test(y, group)
shapiro.test(sample1)
shapiro.test(sample2)
t.test(sample1,sample2)
After the literature
review, the Levene’s test is not necessary.
PART 2. THE RIGHT WAY
P <- c(27, 27, 33, 22, 17, 19, 37, 56, 35, 30, 30, 29, 16, 24, 3,
34, 14, 25, 40)
M <- c(20, 8, 18, 27, 25, 25, 15, 6, 22, 8, 18, 20, 9, 12, 9, 27,
17)
shapiro.test(P)
shapiro.test(M)
t.test(P,M)
 |
| Portulaca amilis plants from the 50/50 potting soil and sand soil (M) portion of the experiment. |
Performing the
statistics by hand.
I followed the equation
listed in https://en.wikipedia.org/wiki/Welch's_t-test,
but it is incorrect. It states use the sample variances. Reading http://www.sthda.com/english/wiki/welch-t-test,
I found I should use standard deviation in the equation rather than variances,
then my results started to match with the R program, yet it too had a flaw to
calculate the degrees of freedom that had be go back to the other website citation.
For each group data I
calculated the average, standard deviation, and variances. I also counted the
sample size.
Potting
|
50/50
|
|
27
|
20
|
|
27
|
8
|
|
33
|
18
|
|
22
|
27
|
|
17
|
25
|
|
19
|
25
|
|
37
|
15
|
|
56
|
6
|
|
35
|
22
|
|
30
|
8
|
|
30
|
18
|
|
29
|
20
|
|
16
|
9
|
|
24
|
12
|
|
3
|
9
|
|
34
|
27
|
|
14
|
17
|
|
25
|
||
40
|
||
27.26316
|
16.82353
|
Average
|
19
|
17
|
Count
|
11.42046
|
7.125967
|
Stdev
|
130.4269
|
50.77941
|
Variance
|
Note: Excel does
calculate the standard deviation “=stdev()” and variance “=var” correctly. I
performed these by hand to double check. Count and averages are also calculated
correctly.
This is the equation
for the Welch’s Test I obtained from http://www.sthda.com/english/wiki/welch-t-test:
Here is the Excel
machine:
Mean1
|
X1
|
27.26316
|
Mean2
|
X2
|
16.82353
|
Variance1
|
s1
|
11.42046
|
Variance2
|
s2
|
7.125967
|
samples size1
|
n1
|
19
|
samples size2
|
n2
|
17
|
X1-X2
|
10.43963
|
|
s21/n1
|
6.864574
|
|
s22/n2
|
2.987024
|
|
Added
|
9.851598
|
|
Sqrt
|
3.138726
|
|
Test statistic =
|
top/bottom
|
3.326072
|
This test statistic
matches that of the R Program.
I calculated the
degrees of freedom from the equation found at https://en.wikipedia.org/wiki/Welch's_t-test
with help from http://www.sthda.com/english/wiki/welch-t-test.
These equations are not easy and both websites had flaws in equations. Between
the two of them, I was able to piece together the correct equations. I hope I
write it correctly (this used the same symbols for values listed in the
previous equation I hand wrote):
Again I hope I wrote it
correctly. I added my calculations to the machine:
Mean1
|
X1
|
27.26316
|
Mean2
|
X2
|
16.82353
|
Variance1
|
s1
|
11.42046
|
Variance2
|
s2
|
7.125967
|
samples size1
|
n1
|
19
|
samples size2
|
n2
|
17
|
X1-X2
|
10.43963
|
|
s21/n1
|
6.864574
|
|
s22/n2
|
2.987024
|
|
Added
|
9.851598
|
|
Sqrt
|
3.138726
|
|
Test statistic =
|
top/bottom
|
3.326072
|
Degrees of Freedom
|
Top added
|
9.851598
|
top squared
|
97.05398
|
|
Bottom Left
|
2.61791
|
|
Bottom Right
|
0.557645
|
|
Added
|
3.175554
|
|
top/bottom
|
30.56285
|
To obtain a p-value, I
tried Excel using “=T.DIST.2T()” that returned 0.0023349788208801. I went to http://www.socscistatistics.com/pvalues/tdistribution.aspx
and entered the t statistic and df I calculated above. I then selected at the
0.05 significance level as well as two-tailed with the result of 0.002301. I
trust the website version more than excel. My results: t = 3.326072; df =
30.56285; p = 0.002301.
After completing the
work by hand, I will use the R Program:
P <- c(27, 27, 33, 22, 17, 19, 37, 56, 35, 30, 30, 29, 16, 24, 3,
34, 14, 25, 40)
M <- c(20, 8, 18, 27, 25, 25, 15, 6, 22, 8, 18, 20, 9, 12, 9, 27,
17)
shapiro.test(P)
shapiro.test(M)
t.test(P,M)
The results are:
> shapiro.test(P)
Shapiro-Wilk
normality test
data: P
W = 0.97128, p-value = 0.8017
> shapiro.test(M)
Shapiro-Wilk
normality test
data: M
W = 0.92503, p-value = 0.1796
> t.test(P,M)
Welch Two Sample
t-test
data: P and M
t = 3.3261, df = 30.563, p-value =
0.002301
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
4.034442 16.844815
sample estimates:
mean of x mean of y
27.26316 16.82353
Looks good to me.
 |
| Portulaca amilis plants of the experiment. Left: plants grown in the potting soil (P) portion of the experiment. Right: plants grown in the 50/50 potting soil and sand (M) portion of the experiment. |
Conclusion:
Plants grown in potting
soil had more seed capsules than those grown in a 50/50 mixture of potting soil
and sand (potting soil average capsules per jar 27.3 cm (n=19) and 50/50
mixture potting soil and sand mean average at 16.8 cm (n=17) Welch’s two sample
test: t=3.326072, df=30.56285, p = 0.002301).
Figure. Seed capsules
per jar by soil type (values are significantly different: Welch’s unpaired t
test: t=3.326072, df=30.56285, p = 0.002301; error bars are ± one standard error).
How to read the results: Looking at the bars, potting soil had more seed capsules. The standard error-error bars do not touch, so there could be a significant different, which is confirmed in the text and caption with a p-value below 0.05.
Thank you for viewing. Comment if you want. Have a nice day!
