4.) The following data was collected on nitrate concentrations in two streams: S

Question

4.) The following data was collected on nitrate concentrations in two streams: Stream 1 15.42 13.57 9.49 13.95 16.77 21.17 6.80 7.34 15.93 Stream 2 13.68 13.68 13.18 14.0011.84 11.49 15.51 13.29 14.22 Use R to calculate: a) a t-test and b) Wilcoxon Rank Sum test. Show the R syntax and results for each procedure in your write-up. Additionally, interpret the results of each test. In your write-up describing and interpreting the results of each test, specify the value of each test statistic, the null-hypothesis tested by each statistic, whether or not you rejected the null-hypothesis, and what assumptions of the tests might not have been met

Explanation / Answer

a.) t-test:

syntax:

y1 <- c(15.42, 13.57, 9.49, 13.95, 16.77, 21.17, 6.8, 7.34, 15.93)
y2 <- c(13.68, 13.68, 13.18, 14, 11.84, 11.49, 15.51, 13.29, 14.22)

t.test(y1,y2)

O/p:

data: y1 and y2
t = -0.030789,

df = 9.0531,

p-value = 0.9761

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:
-3.720408 3.620408
sample estimates:
mean of x mean of y
13.38222 13.43222

The above test checks for below Hypotheses:

Null: true difference in means of y1 and y2 is equal to 0

Alternate: true difference in means is not equal to 0

t-statistic: -0.03

Resulting in p-value: 0.97

Since p > 0.05 (at alpha = 0.05), We will reject the alternate Hypotheses

b.) Wilcoxon rank sum test:

syntax

y1 <- c(15.42, 13.57, 9.49, 13.95, 16.77, 21.17, 6.8, 7.34, 15.93)
y2 <- c(13.68, 13.68, 13.18, 14, 11.84, 11.49, 15.51, 13.29, 14.22)

wilcox.test(y1, y2, paired = FALSE, conf.level = 0.95, conf.int = TRUE, alternative = "two.sided", mu = 0)

output:

data: y1 and y2
W = 45,

p-value = 0.7238
alternative hypothesis: true location shift is not equal to 0
95 percent confidence interval:
-4.689963 3.090021
sample estimates:
difference in location
0.4199932

Warning messages:
1: In wilcox.test.default(y1, y2, paired = FALSE, conf.level = 0.95, :
cannot compute exact p-value with ties

The above test checks for below Hypotheses:

Null: true location shift is equal to 0

Alternate: true location shift is equal to 0

W = 45

Resulting in p-value: 0.72

Since p > 0.05 (at alpha = 0.05), We will reject the alternate Hypotheses

note that the "Warning" from wilcox.test was just that: a warning rather than an indication that your results are incorrect. The test was OK, reporting a p-value based on a normal approximation rather than an exact p-value based on the data values. If you had 50 or more cases, your call to wilcox.test would not even have tried to calculate exact p-values. The coin package in R has a wilcox_test function that can calculate exact p-values in the presence of ties, but I see no need here for an exact p-value.

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