Step 1: Design the Experiment

The epxperiment was performed to measure the biodiversity of a river. The biodiversity was measured by seeing if there was an equal number of fish in the pools as in the riffles. There were 15 different pools and riffles that were measured.

Step 2: Data

fish<-read.csv("fish.csv")
fish
##    ï..location pool riffle
## 1            1    6      3
## 2            2    6      3
## 3            3    3      3
## 4            4    8      4
## 5            5    5      2
## 6            6    2      2
## 7            7    6      2
## 8            8    7      2
## 9            9    1      2
## 10          10    3      2
## 11          11    4      3
## 12          12    5      1
## 13          13    4      3
## 14          14    6      2
## 15          15    4      3
colnames(fish)<-c("location", "pool", "riffle")

Step 3: Describe the Data

The data has 15 rows and 2 columns. The rows represent each of the fifteen locations, while the columns represent which type of location (pool vs. riffle)

Step 4: Identify the Purpose of the Study

The purpose of the study is to measure if there are different amounts of fish in pools compared to riffles.

Step 5: Visualize Data

library(ggplot2) 
ggplot(data=fish, mapping=aes(x=riffle, y=pool)) + geom_point() + geom_abline(slope=1, intercept=0) + annotate("text", x=1.25, y=4, label="More in Pool") + annotate("text", x=3.5, y=2, label="More in Riffle")

Step 6: Interpret the Data

In the graph shown, each point represents one of the fifteen locations where a pool and riffle where measured. The straight line going through the graph is where points would be plotted if the amount of fish in pools and riffles was equal. The plot suggests that more fish live in pools than in riffles because more points are above the line than below the line. To verify these results a t-test was performed.

Step 7: Null Hypotheses

The null hypothesis for this study would be that there are not equal amounts of fish in the two pools.

Step 8: Alternative Hypothesis

The alternative hypothesis would be that the mean number of fish in the pools is different to the mean number of fish in the riffles.

Step 9: Type of Test

To analyze the data a t-test was used to compare the means of the two groups.

Step 10: Number of samples

To compare the two groups of fish a two-sample t-procedure should be used because there are two distinctive groups, but individuals within the groups are not matched in any way.

Step 11: Check Assumptions

The provided qq-plot will help judge the assumptions of the test.

ggplot(data=fish) + geom_qq(mapping=aes(sample=pool))

If the data is normal, it will follow this trend.

Step 12: Level of Significance

The level of significance is 0.05 becasue it is a 95% confidence interval.

Step 13: Perform test

t.test(fish$pool,fish$riffle, paired=TRUE)
## 
##  Paired t-test
## 
## data:  fish$pool and fish$riffle
## t = 4.5826, df = 14, p-value = 0.0004264
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  1.170332 3.229668
## sample estimates:
## mean of the differences 
##                     2.2

Step 14: Interpret the p-value

The p-value is 0.0004264. For a 95 percent confidence interval the alpha value is 0.05, so the null hypothesis can be rejected because it is smaller than the alpha value.This means that there are different amounts of fish in the pools compared to the riffles.

Step 15: Interpret Confidence Intervals

It is not possible for the difference in means to be zero because as stated earlier, the mean number of fish in the pools and in the riffles is not the same.

Step 16: Interpret the Sample estimates

The mean of differences is 2.2, which suggests that there are 2.2 more fish in pools than in riffles.

Step 17: Conclusion

As specified earlier, the null hypothesis can be rejected, which suggests that there is a different amount of fish in the pools compared to the riffles. On average, there are 2.2 more fish in the pools than in the rffles. Since there are different amounts of fish in the pools and riffles, there is a lower biodiversity in the river.