# 10 Statistical Summaries and Tests Figure 10.1: Contrasting distributions

## 10.1 Goals of statistical analysis

To frame how we might approach statistical analysis and modeling, there are various goals that are commonly involved:

• To understand our data

• nature of our data, through summary statistics and various graphics like histograms
• spatial statistical analysis
• time series analysis
• To group or classify things based on their properties

• using factors to define groups, and deriving grouped summaries
• comparing observed vs expected counts or probabilities
• To understand how variables relate to one another

• or maybe even explain variations in other variables, through correlation analysis
• To model behavior and maybe predict it

• various linear models
• To confirm our observations from exploration (field/lab/vis)

• inferential statistics e.g. difference of means tests, ANOVA, $$\chi^2$$
• To have the confidence to draw conclusions, make informed decisions

• To help communicate our work

These goals can be seen in the context of a typical research paper or thesis outline in environmental science:

• Introduction

• Literature Review

• Methodology

• Results

• field, lab, geospatial data
• Analysis

• statistical analysis
• qualitative analysis
• visualization
• Discussion

• making sense of analysis
• possibly recursive, with visualization
• Conclusion

• conclusion about what the above shows
• new questions for further research
• possible policy recommendation

The scope and theory of statistical analysis and models is extensive, and there are many good books on the subject that employ the R language. This chapter is a short review of some of these methods and how they apply to environmental data science.

## 10.2 Summary Statistics

Summary statistics such as mean, standard deviation, variance, minimum, maximum, and range are derived in quite a few R functions, commonly as a parameter or a sub-function (see mutate). An overall simple statistical summary is very easy to do in base R:

summary(tidy_eucoak)
##      site               site #          tree                Date
##  Length:180         Min.   :1.000   Length:180         Min.   :2006-11-08
##  Class :character   1st Qu.:2.000   Class :character   1st Qu.:2006-12-07
##  Mode  :character   Median :4.000   Mode  :character   Median :2007-01-30
##                     Mean   :4.422                      Mean   :2007-01-29
##                     3rd Qu.:6.000                      3rd Qu.:2007-03-22
##                     Max.   :8.000                      Max.   :2007-05-07
##
##     month              rain_mm      rain_subcanopy      slope
##  Length:180         Min.   : 1.00   Min.   : 1.00   Min.   : 9.00
##  Class :character   1st Qu.:16.00   1st Qu.:16.00   1st Qu.:12.00
##  Mode  :character   Median :28.50   Median :30.00   Median :24.00
##                     Mean   :37.99   Mean   :34.84   Mean   :20.48
##                     3rd Qu.:63.25   3rd Qu.:50.00   3rd Qu.:27.00
##                     Max.   :99.00   Max.   :98.00   Max.   :32.00
##                     NA's   :36      NA's   :4
##      aspect         runoff_L      surface_tension runoff_rainfall_ratio
##  Min.   :100.0   Min.   : 0.000   Min.   :28.51   Min.   :0.00000
##  1st Qu.:143.0   1st Qu.: 0.000   1st Qu.:37.40   1st Qu.:0.00000
##  Median :196.0   Median : 0.825   Median :62.60   Median :0.03347
##  Mean   :186.6   Mean   : 2.244   Mean   :55.73   Mean   :0.05981
##  3rd Qu.:221.8   3rd Qu.: 3.200   3rd Qu.:72.75   3rd Qu.:0.08474
##  Max.   :296.0   Max.   :16.000   Max.   :72.75   Max.   :0.42000
##                  NA's   :8        NA's   :44      NA's   :8

### 10.2.1 Summarize by group: stratifying a summary

eucoakrainfallrunoffTDR %>%
group_by(site) %>%
summarize(
rain = mean(rain_mm, na.rm = TRUE),
rainSD = sd(rain_mm, na.rm = TRUE),
runoffL_oak = mean(runoffL_oak, na.rm = TRUE),
runoffL_euc = mean(runoffL_euc, na.rm = TRUE),
runoffL_oakMax = max(runoffL_oak, na.rm = TRUE),
runoffL_eucMax = max(runoffL_euc, na.rm = TRUE),
)
## # A tibble: 8 x 7
##   site   rain rainSD runoffL_oak runoffL_euc runoffL_oakMax runoffL_eucMax
##   <chr> <dbl>  <dbl>       <dbl>       <dbl>          <dbl>          <dbl>
## 1 AB1    48.4   28.2      6.80         6.03          6.80            6.03
## 2 AB2    34.1   27.9      4.91         3.65          4.91            3.65
## 3 KM1    48     32.0      1.94         0.592         1.94            0.592
## 4 PR1    56.5   19.1      0.459        2.31          0.459           2.31
## 5 TP1    38.4   29.5      0.877        1.66          0.877           1.66
## 6 TP2    34.3   29.2      0.0955       1.53          0.0955          1.53
## 7 TP3    32.1   28.4      0.381        0.815         0.381           0.815
## 8 TP4    32.5   28.2      0.231        2.83          0.231           2.83

### 10.2.2 Boxplot for visualizing distributions by group

A Tukey boxplot is a good way to visualize distributions by group. In this soil CO2 study of the Marble Mountains , some sites had much greater variance, and some sites tended to be low vs high:

soilCO2_97$SITE <- factor(soilCO2_97$SITE)
ggplot(data = soilCO2_97, mapping = aes(x = SITE, y = CO2pct)) +
geom_boxplot() Figure 10.2: Tukey boxplot by group Figure 10.3: Marble Mountains CO2 sampling sites Figure 10.4: Marble Mountains CO2 sites on orthophotography

### 10.2.3 Generating pseudorandom numbers

Functions commonly used in R books for quickly creating a lot of numbers to display are those that generate pseudorandom numbers. These are also useful in statistical methods that need a lot of these, such as in Monte Carlo simulation. The two most commonly used are:

• runif() generates a vector of n pseudorandom numbers ranging by default from min=0 to max=1.
• rnorm() generates a vector of n normally distributed pseudorandom numbers with a default mean=0 and sd=0.

To see both in action as x and y:

x <- as_tibble(runif(n=1000, min=10, max=20))
names(x) <- 'x'
ggplot(x, aes(x=x)) + geom_histogram() Figure 10.5: random uniform histogram

y <- as_tibble(rnorm(n=1000, mean=100, sd=10))
names(y) <- 'y'
ggplot(y, aes(x=y)) + geom_histogram() Figure 10.6: random normal histogram

ggplot(y, aes(x=y)) + geom_density() Figure 10.7: random normal density plot

xy <- bind_cols(x,y)
ggplot(xy, aes(x=x,y=y)) + geom_point() Figure 10.8: random normal plotted against random uniform

## 10.3 Correlation r and Coefficient of Determination r2

In the visualization chapter, we looked at creating scatter plots and also arrays of scatter plots where we could compare variables to visually see if they might be positively or negatively correlated. A statistic that is commonly used for this is the Pearson product-moment correlation coefficient or r statistic.

We’ll look at the formula for r below, but it’s easier to just use the cor function. You just need the two variables as vector inputs in R to return the r statistic. Squaring r to r2 is the coefficient of determination and can be interpreted as the amount of the variation in the dependent variable y that that is “explained” by variation in the independent variable x. This coefficient will always be positive, with a maximum of 1 or 100%.

If we create two random variables, uniform or normal, there shouldn’t be any correlation. We’ll do this 5 times each set:

for (i in 1:5) {print(paste(i, cor(rnorm(100), rnorm(100))))}
##  "1 -0.0483034983965455"
##  "2 0.0946839578200478"
##  "3 0.0240357737510824"
##  "4 0.024324951661481"
##  "5 -0.101852562111154"
for (i in 1:5) {print(paste(i, cor(runif(100), runif(100))))}
##  "1 0.0672906303525644"
##  "2 -0.144615305323489"
##  "3 -0.0231296740848959"
##  "4 0.0260569732832872"
##  "5 0.0280672096592171"
for (i in 1:5) {print(paste(i, cor(rnorm(100), runif(100))))}
##  "1 -0.0634233290468282"
##  "2 0.0521792859706672"
##  "3 0.161751299947425"
##  "4 0.0670344283905028"
##  "5 -0.0317114650439946"

But variables such as temperature and elevation in the Sierra will be strongly negatively correlated, as seen in the scatter plot and the r value close to -1, and have a high r2:

library(igisci); library(tidyverse)
sierra <- sierraFeb %>% filter(!is.na(TEMPERATURE))
elev <- sierra$ELEVATION; temp <- sierra$TEMPERATURE
plot(elev, temp) Figure 10.9: Scatter plot illustrating negative correlation

cor(elev, temp)
##  -0.9357801
cor(elev, temp)^2
##  0.8756845

While you don’t need to use this, since the cor function is easier to type, it’s interesting to know that the formula for Pearson’s correlation coefficient is something that you can actually code in R, taking advantage of its vectorization methods:

$r = \frac{\sum{(x_i-\overline{x})(y_i-\overline{y})}}{\sqrt{\sum{(x_i-\overline{x})^2\sum(y_i-\overline{y})^2}}}$

r <- sum((elev-mean(elev))*(temp-mean(temp)))/
sqrt(sum((elev-mean(elev))^2*sum((temp-mean(temp))^2)))
r
##  -0.9357801
r^2
##  0.8756845

Another version of the formula runs faster, so might be what R uses, but you’ll never notice the time difference:

$r = \frac{n(\sum{xy})-(\sum{x})(\sum{y})}{\sqrt{(n\sum{x^2}-(\sum{x})^2)(n\sum{y^2}-(\sum{y})^2)}}$

n <- length(elev)
r <- (n*sum(elev*temp)-sum(elev)*sum(temp))/
sqrt((n*sum(elev^2)-sum(elev)^2)*(n*sum(temp^2)-sum(temp)^2))
r
##  -0.9357801
r^2
##  0.8756845

… and as you can see, all three methods give the same results.

### 10.3.1 Displaying r in a pairs plot

We can use another type of pairs plot from the psych package to look at the correlation coefficient in the upper right part of the pairs plot, since correlation can be determined between x and y or y and x; the result is the same. In contrast to what we’ll see in regression models, there doesn’t have to be one explanatory (or independent) variable and one response (or dependent) variable; either one will do. The r value shows both the direction (positive or negative) and the magnitude of the correlation, with values closer to 1 or -1 being more correlated.

library(psych)
sierraFeb %>%
dplyr::select(ELEVATION:TEMPERATURE) %>%
pairs.panels(method = "pearson", # correlation method
hist.col = "#00AFBB",
density = TRUE, # show density plots
ellipses = F, smooth = F) # unneeded Figure 10.10: Pairs plot with r values

We can clearly see in the graphs the negative relationships between elevation and temperature and between latitude and longitude (what is that telling us?) and these correspond to strongly negative correlation coefficients.

One problem with Pearson: While there’s nothing wrong with the correlation coefficient, Pearson’s name, along with some other pioneers of statistics like Fisher, is tainted by an association with the racist field of eugenics. But the mathematics is not to blame, and the correlation coefficient is still as useful as ever. Maybe we can just say that Pearson discovered the correlation coefficient…

## 10.4 Statistical tests

Tests that compare our data to other data or look at relationships among variables are important statistical methods, and you should refer to statistical references to best understand how to apply the appropriate methods for your research.

### 10.4.1 Comparing samples and groupings with a t test and a non-parametric Kruskal-Wallis Rank Sum test

A common need in environmental research is to compare samples of a phenomenon or compare samples with an assumed standard population. The simplest application of this is the t-test, which can only involve comparing two samples or one sample with a population. After this, we’ll look at analysis of variance, extending this to allow for more than two groups.

XSptsPheno %>%
ggplot(aes(NDVI, fill=phenology)) +
geom_density(alpha=0.2) Figure 10.11: NDVI by phenology

t.test(NDVI~phenology, data=XSptsPheno) 
##
##  Welch Two Sample t-test
##
## data:  NDVI by phenology
## t = 5.4952, df = 52.03, p-value = 1.19e-06
## alternative hypothesis: true difference in means between group growing and group senescent is not equal to 0
## 95 percent confidence interval:
##  0.1421785 0.3057412
## sample estimates:
##   mean in group growing mean in group senescent
##               0.5901186               0.3661588

While these data sets appear reasonably normal, the Shapiro-Wilk test (which uses a null hypothesis of normal) has a p value < 0.05 for the senescent group, so the data can’t be assumed to be normal.

shapiro.test(XSptsPheno$NDVI[XSptsPheno$phenology=="growing"])
##
##  Shapiro-Wilk normality test
##
## data:  XSptsPheno$NDVI[XSptsPheno$phenology == "growing"]
## W = 0.93608, p-value = 0.07918
shapiro.test(XSptsPheno$NDVI[XSptsPheno$phenology=="senescent"])
##
##  Shapiro-Wilk normality test
##
## data:  XSptsPheno$NDVI[XSptsPheno$phenology == "senescent"]
## W = 0.88728, p-value = 0.004925

Therefore we should use a non-parametric alternative such as the Kruskal-Wallis Rank Sum test:

kruskal.test(NDVI~phenology, data=XSptsPheno)
##
##  Kruskal-Wallis rank sum test
##
## data:  NDVI by phenology
## Kruskal-Wallis chi-squared = 19.164, df = 1, p-value = 1.199e-05

A bit of a review on significance tests and p values

First, each type of test will be testing a particular summary statistic, like the t test is comparing means and the analysis of variance will be comparing variances. In confirmatory statistical tests, you’re always seeing if you can reject the null hypothesis that there’s no difference, so in the t test or the rank sum test above that compares two samples, it’s that there’s no difference between two samples.

There will of course nearly always be some difference, so you might think that you’d always reject the null hypothesis that there’s no difference. That’s where random error and probability comes in. You can accept a certain amount of error – say 5% – to be able to say that the difference in values could have occurred by chance. That 5% or 0.05 is often called the “significance level” and is the probability of the two values being the same that you’re willing to accept. (The remainder 0.95 is often called the confidence level, the extent to which you’re confident that rejecting the null hypothesis and accepting the working hypothesis is correct.)

When R reports the p (or Pr) probability value, you compare that to that significance level to see if it’s lower, and then use that to possibly reject the null hypothesis and accept the working hypothesis. R will simply report the p value and commonly show asterisks along with it to indicate if it’s lower than various common significance levels, like 0.1, 0.05, 0.01, and 0.001.

#### 10.4.1.1 Runoff under Eucalyptus vs Oaks – is there a difference?

For the question “Is the runoff under Eucalyptus canopy significantly different from that under oaks?” we’ll then start by test for normality of each of the two samples (euc and oak)

shapiro.test(tidy_eucoak$runoff_L[tidy_eucoak$tree == "euc"])
##
##  Shapiro-Wilk normality test
##
## data:  tidy_eucoak$runoff_L[tidy_eucoak$tree == "euc"]
## W = 0.74241, p-value = 4.724e-11
shapiro.test(tidy_eucoak$runoff_L[tidy_eucoak$tree == "oak"])
##
##  Shapiro-Wilk normality test
##
## data:  tidy_eucoak$runoff_L[tidy_eucoak$tree == "oak"]
## W = 0.71744, p-value = 1.698e-11

which shows clearly that both samples are non-normal. So we might apply the non-parametric Kruskal-Wallis test:

kruskal.test(runoff_L~tree, data=tidy_eucoak)
##
##  Kruskal-Wallis rank sum test
##
## data:  runoff_L by tree
## Kruskal-Wallis chi-squared = 2.2991, df = 1, p-value = 0.1294

and no significant difference can be seen. If we look at the data graphically, this makes sense:

tidy_eucoak %>%
ggplot(aes(log(runoff_L),fill=tree)) +
geom_density(alpha=0.2) Figure 10.12: Runoff under Eucalyptus and Oak in Bay Area sites

However, some of this may result from major variations among sites, which is apparent in this boxplot:

ggplot(data = tidy_eucoak) +
geom_boxplot(aes(x=site, y=runoff_L, color=tree)) Figure 10.13: Runoff at various sites contrasting euc and oak

We might restrict our analysis to Tilden Park sites in the East Bay.

tilden <- tidy_eucoak %>% filter(str_detect(tidy_eucoak$site,"TP")) tilden %>% ggplot(aes(log(runoff_L),fill=tree)) + geom_density(alpha=0.2) Figure 10.14: Easy Bay sites shapiro.test(tilden$runoff_L[tilden$tree == "euc"]) ## ## Shapiro-Wilk normality test ## ## data: tilden$runoff_L[tilden$tree == "euc"] ## W = 0.73933, p-value = 1.764e-07 shapiro.test(tilden$runoff_L[tilden$tree == "oak"]) ## ## Shapiro-Wilk normality test ## ## data: tilden$runoff_L[tilden$tree == "oak"] ## W = 0.59535, p-value = 8.529e-10 So once again, as is common with small sample sets, we need a non-parametric test. kruskal.test(runoff_L~tree, data=tilden) ## ## Kruskal-Wallis rank sum test ## ## data: runoff_L by tree ## Kruskal-Wallis chi-squared = 14.527, df = 1, p-value = 0.0001382 Analysis process from exploration to testing [eucoak] In the year runoff was studied, there were no runoff events sufficient to mobilize sediments. The next year, January had a big event, so we collected sediments and processed them in the lab. Questions: • Is there a difference between eucs and oaks in terms of fine sediment yield? • Is there a difference between eucs and oaks in terms of total sediment yield? (includes litter) eucoaksed <- read_csv(ex("eucoak/eucoaksediment.csv")) summary(eucoaksed) ## id site trtype slope ## Length:14 Length:14 Length:14 Min. : 9.00 ## Class :character Class :character Class :character 1st Qu.:12.00 ## Mode :character Mode :character Mode :character Median :21.00 ## Mean :20.04 ## 3rd Qu.:25.00 ## Max. :32.00 ## ## bulkDensity litter Jan08rain mean_runoff_ratio ## Min. :0.960 Min. : 25.00 Min. :228.1 Min. :0.00450 ## 1st Qu.:1.060 1st Qu.: 51.25 1st Qu.:290.8 1st Qu.:0.02285 ## Median :1.125 Median : 77.00 Median :301.1 Median :0.04835 ## Mean :1.156 Mean : 76.64 Mean :298.5 Mean :0.06679 ## 3rd Qu.:1.245 3rd Qu.: 95.75 3rd Qu.:317.0 3rd Qu.:0.12172 ## Max. :1.490 Max. :135.00 Max. :328.5 Max. :0.16480 ## ## med_runoff_ratio std_runoff_ratio fines_g litter_g ## Min. :0.00000 Min. :0.01070 Min. : 7.50 Min. :14.00 ## 1st Qu.:0.01105 1st Qu.:0.01642 1st Qu.:13.30 1st Qu.:18.80 ## Median :0.04950 Median :0.02740 Median :18.10 Median :40.40 ## Mean :0.06179 Mean :0.04355 Mean :27.77 Mean :41.32 ## 3rd Qu.:0.09492 3rd Qu.:0.05735 3rd Qu.:45.00 3rd Qu.:57.50 ## Max. :0.16430 Max. :0.11480 Max. :66.70 Max. :97.80 ## NA's :1 NA's :1 ## total_g fineTotalRatio fineRainRatio ## Min. : 23.50 Min. :0.1300 Min. :0.025 ## 1st Qu.: 35.00 1st Qu.:0.2700 1st Qu.:0.044 ## Median : 60.50 Median :0.3900 Median :0.064 ## Mean : 69.12 Mean :0.3715 Mean :0.097 ## 3rd Qu.: 95.50 3rd Qu.:0.4900 3rd Qu.:0.141 ## Max. :125.90 Max. :0.5500 Max. :0.293 ## NA's :1 NA's :1 NA's :1 eucoaksed %>% group_by(trtype) %>% summarize(meanfines = mean(fines_g, na.rm=T), sdfines = sd(fines_g, na.rm=T), meantotal = mean(total_g, na.rm=T), sdtotal = sd(total_g, na.rm=T)) ## # A tibble: 2 x 5 ## trtype meanfines sdfines meantotal sdtotal ## <chr> <dbl> <dbl> <dbl> <dbl> ## 1 euc 14.2 3.50 48.6 35.0 ## 2 oak 39.4 20.4 86.7 26.2 eucoakLong <- eucoaksed %>% pivot_longer(col=c(fines_g,litter_g), names_to = "sed_type", values_to = "sed_g") eucoakLong %>% ggplot(aes(trtype, sed_g, col=sed_type)) + geom_boxplot() Figure 10.15: Eucalyptus and Oak sediment runoff box plots eucoakLong %>% ggplot(aes(sed_g, col=sed_type)) + geom_density() + facet_grid(trtype ~ .) Figure 10.16: facet density plot of Eucalpytus and Oak sediment runoff shapiro.test(eucoaksed$fines_g[eucoaksed$trtype == "euc"]) ## ## Shapiro-Wilk normality test ## ## data: eucoaksed$fines_g[eucoaksed$trtype == "euc"] ## W = 0.9374, p-value = 0.6383 shapiro.test(eucoaksed$fines_g[eucoaksed$trtype == "oak"]) ## ## Shapiro-Wilk normality test ## ## data: eucoaksed$fines_g[eucoaksed$trtype == "oak"] ## W = 0.96659, p-value = 0.8729 t.test(fines_g~trtype, data=eucoaksed)  ## ## Welch Two Sample t-test ## ## data: fines_g by trtype ## t = -3.2102, df = 6.4104, p-value = 0.01675 ## alternative hypothesis: true difference in means between group euc and group oak is not equal to 0 ## 95 percent confidence interval: ## -44.059797 -6.278299 ## sample estimates: ## mean in group euc mean in group oak ## 14.21667 39.38571 shapiro.test(eucoaksed$total_g[eucoaksed$trtype == "euc"]) ## ## Shapiro-Wilk normality test ## ## data: eucoaksed$total_g[eucoaksed$trtype == "euc"] ## W = 0.76405, p-value = 0.02725 shapiro.test(eucoaksed$total_g[eucoaksed$trtype == "oak"]) ## ## Shapiro-Wilk normality test ## ## data: eucoaksed$total_g[eucoaksed$trtype == "oak"] ## W = 0.94988, p-value = 0.7286 kruskal.test(total_g~trtype, data=eucoaksed)  ## ## Kruskal-Wallis rank sum test ## ## data: total_g by trtype ## Kruskal-Wallis chi-squared = 3.449, df = 1, p-value = 0.06329 So we used a t test for the fines_g, and the test suggests that there’s a significant difference in sediment yield for fines, but the Kruskal-Wallis test on total sediment (including litter) did not show a significant difference. Both results support the conclusion that oaks in this study produced more soil erosion, largely because the Eucalyptus stands generate so much litter cover, and that litter also made the total sediment yield not significantly different. ### 10.4.2 Analysis of Variance The purpose of Analysis of Variance (ANOVA) is to compare groups based upon continuous variables. Can be thought of as an extension of a t test where you have more than two groups, or as a linear model where one variable is a factor. In a confirmatory statistical test, you’ll want to see if you can reject the null hypothesis that there’s no difference between the within-sample variances and the between-sample variances. • Response variable is a continuous variable • Explanatory variable is the grouping – categorical (a factor in R) From a study of a karst system in Tennessee : “Are water samples from streams draining sandstone, limestone, and shale different based on solutes measured as total hardness?” Figure 10.17: Water sampling Figure 10.18: Cave Cove Sink on limestone Figure 10.19: Helen Highwater Spring on shale wChemData <- read_excel(ex("SinkingCove/SinkingCoveWaterChem.xlsx")) %>% mutate(siteLoc = str_sub(Site,start=1L, end=1L)) wChemTrunk <- wChemData %>% filter(siteLoc == "T") %>% mutate(siteType = "trunk") wChemDrip <- wChemData %>% filter(siteLoc %in% c("D","S")) %>% mutate(siteType = "dripwater") wChemTrib <- wChemData %>% filter(siteLoc %in% c("B", "F", "K", "W", "P")) %>% mutate(siteType = "tributary") wChemData <- bind_rows(wChemTrunk, wChemDrip, wChemTrib) sites <- read_csv(ex("SinkingCove/SinkingCoveSites.csv")) wChem <- wChemData %>% left_join(sites, by = c("Site" = "site")) %>% st_as_sf(coords = c("longitude", "latitude"), crs = 4326) library(terra) tmap_mode("plot") DEM <- rast(ex("SinkingCove/DEM_SinkingCoveUTM.tif")) slope <- terrain(DEM, v='slope') aspect <- terrain(DEM, v='aspect') hillsh <- shade(slope/180*pi, aspect/180*pi, angle=40, direction=330) bounds <- st_bbox(wChem) xrange <- bounds$xmax - bounds$xmin yrange <- bounds$ymax - bounds$ymin xMIN <- as.numeric(bounds$xmin - xrange/10)
xMAX <- as.numeric(bounds$xmax + xrange/10) yMIN <- as.numeric(bounds$ymin - yrange/10)
yMAX <- as.numeric(bounds$ymax + yrange/10) #st_bbox(c(xmin = 16.1, xmax = 16.6, ymax = 48.6, ymin = 47.9), crs = st_crs(4326)) newbounds <- st_bbox(c(xmin=xMIN, xmax=xMAX, ymin=yMIN, ymax=yMAX), crs= st_crs(4326)) tm_shape(hillsh,bbox=newbounds) + tm_raster(palette="-Greys",legend.show=F,n=20, alpha=0.5) + tm_shape(wChem) + tm_symbols(size="TH", col="Lithology", scale=2, shape="siteType") + tm_legend() + tm_layout(legend.position = c("left", "bottom")) + tm_graticules(lines=F) Figure 10.20: Upper Sinking Cove summary(aov(TH~siteType, data = wChemData)) ## Df Sum Sq Mean Sq F value Pr(>F) ## siteType 2 79172 39586 20.59 1.07e-07 *** ## Residuals 67 128815 1923 ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 summary(aov(TH~Lithology, data = wChemData)) ## Df Sum Sq Mean Sq F value Pr(>F) ## Lithology 3 98107 32702 19.64 3.28e-09 *** ## Residuals 66 109881 1665 ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 wChemData %>% ggplot(aes(x=TH, fill=siteType)) + geom_histogram() + facet_grid(Lithology ~ .) Figure 10.21: Sinking Cove dissolved carbonates as total hardness by lithology Some observations and caveats from the above: • There’s pretty clearly a difference between surface waters (trunk and tributary) and cave dripwaters (from stalactites) in terms of solutes. Analysis of variance simply confirms the obvious. • There’s also pretty clearly a difference among lithologies on the basis of solutes, not surprising since limestone is much more soluble than sandstones. Similarly, analysis of variance confirms the obvious. • The data may not be sufficiently normally distributed, and limestone hardness values are bimodal (largely due to the inaccessibility of waters in the trunk cave passages travelling 2 km through the Bangor limestone),3 though analysis of variance is less sensitive to this than a t test. • While shale creates springs, shale strata are very thin, with most of them in the “mix” category, or form the boundary between the two major limestone formations. Tributary streams appear to cross the shale in caves that were inaccessible for sampling. We visually confirmed this in one cave, but this exploration required some challenging rappel work to access, so we were not able to sample. • The geologic structure here is essentially flat, with sandstones on the plateau surface and the most massive limestones – the Bangor and Monteagle limestones – all below 400 m elevation. Figure 10.22: Upper Sinking Cove stratigraphy • While the rapid increase in solutes happens when Cave Cove Creek starts draining through the much more soluble limestone until reaches saturation, the distance traveled by the water (reflected by a drop in elevation) can be seen: wChemData %>% ggplot(aes(x=Elevation, y=TH, col=Lithology)) + geom_point() + geom_smooth(method= "lm") Figure 10.23: Sinking Cove dissolved carbonates as TH and elevation by lithology ### 10.4.3 Testing a correlation We earlier looked at the correlation coefficient r. One test we can do is whether that correlation is significant: cor.test(sierraFeb$TEMPERATURE, sierraFeb$ELEVATION) ## ## Pearson's product-moment correlation ## ## data: sierraFeb$TEMPERATURE and sierraFeb\$ELEVATION
## t = -20.558, df = 60, p-value < 2.2e-16
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  -0.9609478 -0.8952592
## sample estimates:
##        cor
## -0.9357801

So we can reject the null hypothesis that the correlation is not equal to zero: the probability of the getting a correlation of -0.936 is less than $$2.2\times 10^{-16}$$ and thus the “true correlation is not equal to 0.” So we can accept an alternative working hypothesis that they’re negatively correlated, and that – no surprise – it gets colder as we go to higher elevations, at least in February in the Sierra, where our data come from.

In the next chapter, we’ll use these data to develop a linear model and get a similar result comparing the slope of the model predicting temerature from elevation…

## 10.5 Exercises

1. Build a soilvegJuly data frame.
• Create a new RStudio project named ‘Meadows’
• Create a soilveg tibble from “meadows/SoilVegSamples.csv” in the extdata.
• Have a look at this data frame and note that there is NA for SoilMoisture and NDVI for 8 of the records. These represent observations made in August, while the rest are all in July. While we have drone imagery and thus NDVI for these sites, but as they are during the senescent period we don’t want to compare these with the July samples.
• Filter the data frame to only include records that are not NA (!is.na) for SoilMoisture, and assign that to a new data frame ‘soilvegJuly.’
## # A tibble: 21 x 7
##    Meadow   id    DominantVegetation veg   BulkDensity SoilMoisture  NDVI
##    <chr>    <chr> <chr>              <chr>       <dbl>        <dbl> <dbl>
##  1 Loney    L1    JUNNEV             JUN         1.84         0.505 0.804
##  2 Loney    L2    CARNEB             CAR         0.916        0.711 0.849
##  3 Loney    L3    upland             UPL         3.08         0.296 0.838
##  4 Loney    L4    JUNBAL             JUN         3.90         0.239 0.756
##  5 Loney    L5    JUNBAL             JUN         1.63         0.508 0.865
##  6 Loney    L6    grass              GRA         2.59         0.366 0.84
##  7 Knuthson Q1    CARPEL             CAR         0.31         0.433 0.833
##  8 Knuthson Q2    grass              GRA         0.867        0.183 0.692
##  9 Knuthson Q3    JUNNEV             JUN         0.942        0.114 0.665
## 10 Knuthson Q4    JUNNEV             JUN         0.805        0.327 0.729
## # ... with 11 more rows
1. Visualizations:
• Create a scatter plot of Soil Moisture vs NDVI, colored by veg (the 3-character abbreviation of major vegetation types – CAR for Carex/sedge, JUN for Juncus/rush, GRA for mesic grasses and forbs, and UPL for more elevated (maybe by a meter) areas of sagebrush), for soilvegJuly. What we can see is that this is a small sample. This study involved a lot of drone imagery, probably 100s of GB of imagery, which was the main focus of the study – to detect channels – but a low density of soil and vegetation ground samples.
• Create a histogram of soil moisture colored by veg, also for the July data. We see the same story, with not very many samples, though suggestive of a multimodal distribution overall.
• Create a density plot of the same, using alpha=0.5 to see everything with transparency.
1. Tests, July data

Using either aov() or anova(lm()), run an analysis of variance test of soil moisture ~ veg. Remember to specify the data, which should be just the July data., then do the same test for NDVI ~ veg.

• Now compare the meadows based on soil moisture in July, using a boxplot

• Then run an ANOVA test on this meadow grouping

1. For the meadow data, create a pairs plot to find which variables are correlated along with their r values, and test the significance of that correlation and provide percentage of variation of one variable is explained by that of the other.

2. Bulk density test, all data

We’ll now look at bulk density for all samples (soilveg), including both July and August. Soil moisture and NDVI won’t be a part of this analysis, only bulk density and vegetation.
Look at the distribution of all bulk density values, using both a histogram with 20 bins and a density plot, then run an ANOVA test of bulk density predicted by (~) veg. What does the Pr(>F) value indicate?

1. Carex or not?

Derive a bulkDensityCAR data frame by mutating a new variable CAR derived as a boolean result of veg == “CAR.” This will group the vegetation points as either Carex or not, then use that in another ANOVA test to predict bulk density. What does the Pr(>F) value indicate?

### References

Davis, J. D., P. Amato, and R. Kiefer. 2001. “Soil Carbon Dioxide in a Summer-Dry Subalpine Karst, Marble Mountains, California, USA.” Zeitschrift Für Geomorphologie N.F. 45. https://www.researchgate.net/publication/258333952_Soil_carbon_dioxide_in_a_summer-dry_subalpine_karst_Marble_Mountains_California_USA.
Davis, Jerry D, and George A Brook. 1993. “Geomorphology and Hydrology of Upper Sinking Cove, Cumberland Plateau, Tennessee.” Earth Surface Processes and Landforms 18 (4): 339–62.
Thompson, A., J. D. Davis, and A. J. Oliphant. 2016. “Surface Runoff and Soil Erosion Under Eucalyptus and Oak Canopy.” Earth Surface Processes and Landforms. https://doi.org/10.1002/esp.3881.

1. We tried very hard to get into that cave that must extend from upper Cave Cove then under Farmer Cove to a spring in Wolf Cove – we have die traces to prove it↩︎