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

Question

4.) The following data was collected on nitrate concentrations in two stream:s Stream 1 15.4213.579.49 13.95 16.77 21.17 6.80 7.3415.93 Stream 2 13.6813.68 13.18 14.00 11.84 11.49 15.5113.2914.22 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. 5) The following data on nitrate concentrations was collected at specific points along the same stream before and after a potential pollution source was established on the banks of the stream Sample1 4 10 Point Time1 4.977.399.089.659.83 10.8410.86 10.93 11.78 12.01 Time 2 2.92 5.98 6.757.72 11.98 14.07 14.65 14.94 17.80 19.52 Use R to calculate: a) a paired sample t-test, b) a Wilcoxon Signed-Rank 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. Why couldn't you have used these tests in problem 4?

Explanation / Answer

4) Ans:

Stream1=c(15.42,13.57,9.49,13.95,16.77,21.17,6.8,7.35,15.93) ## Enter the data
Stream2=c(13.68,13.68,13.18,14.00,11.84,11.49,15.51,13.29,14.22)

Null hypothesis: Mean of Stream 1 and Stream 2 are equal.

Alternative Hypothesis: Mean of Stream 1 and Stream 2 are not equal.
t.test(Stream1, Stream2) # R command for test

Welch Two Sample t-test

data: Stream1 and Stream2
t = -0.030114, df = 9.0538, p-value = 0.9766
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-3.718086 3.620309
sample estimates:
mean of x mean of y
13.38333 13.43222

The estimated p-value is 0.9766. Hence, we can not reject the null hypothesis and conclude that there is no significant difference between means of Stream 1 and Stream 2 at 0.05 level of significance.

The assumption of these test statistic is that these two samples are coming from a normal distribution.

## wilcox.test
wilcox.test(Stream1, Stream2)

Wilcoxon rank sum test with continuity correction

data: Stream1 and Stream2
W = 45, p-value = 0.7238
alternative hypothesis: true location shift is not equal to 0

Warning message:
In wilcox.test.default(Stream1, Stream2) :
cannot compute exact p-value with ties

Conclusion:

The estimated p-value is 0.7238. Hence, we can not reject the null hypothesis and conclude that there is no significant difference between means of Stream 1 and Stream 2 at 0.05 level of significance.

There is no assumption of distribution for the test.

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