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ID: 3361129 • Letter: I
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iPad 10:14 PM 93% Document HomeI Insert Draw Layout Review View Table Times New Roma 10 A. #3 As a measure of impact of pollution along a river, the diversity of species along two (2) river floodplains (Vermillion River and the Black River) were studied. Randomly placed 50 x 20 meter plots were sampled along each river and the number of species of trees in each plot was recorded. The data are presented in the following table. Vermillion Black River River Identify the type of analysis to be carried out on these data. Explain. Formulate the appropriate hypotheses for these data (and explain your rational) List & discuss the appropriate evaluation and/or assumptions you will use for these data. Evaluate these assumptions using appropriate methods; present your results. Carry out the appropriate statistical procedures (alpha .05) and present your conclusions No IfExplanation / Answer
a)We need to test if there is any significant difference in the number of tree species between the two rivers. To do this, we need to carry out Two – sample t – test.
b) Here, our null hypothesis is -> H0 : mean difference = 0.
And our alternative hypothesis is -> H1 : mean difference 0.
c) Assumption is that the data from both the rivers follows normal distribution and the variance of the data of both the rivers are assumed to be equal.
d) For the second assumption, we use pooled variance.
For the first one, we plot the histogram of the data of both the rivers.
e)
t-Test: Two-Sample Assuming Equal Variances
Vermillion river
Black river
Mean
11.15384615
9.444444444
Variance
5.641025641
16.02777778
Observations
13
9
Pooled Variance
9.795726496
Hypothesized Mean Difference
0
df
20
t Stat
1.25952676
P(T<=t) one-tail
0.111170578
t Critical one-tail
1.724718243
P(T<=t) two-tail
0.222341156
t Critical two-tail
2.085963447
Since p-value > 0.05, we accept the null hypothesis and conclude that there is no significant difference in the number of tree species near both the rivers.
t-Test: Two-Sample Assuming Equal Variances
Vermillion river
Black river
Mean
11.15384615
9.444444444
Variance
5.641025641
16.02777778
Observations
13
9
Pooled Variance
9.795726496
Hypothesized Mean Difference
0
df
20
t Stat
1.25952676
P(T<=t) one-tail
0.111170578
t Critical one-tail
1.724718243
P(T<=t) two-tail
0.222341156
t Critical two-tail
2.085963447
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