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1 Tutorial 7: Correlation and Regression Correlation Used to test whether two variables are linearly associated. A correlation coefficient (r) indicates the strength and direction of the association. A correlation coefficient ranges from 1 (perfect negative linear association) to 1 (perfect positive linear association). An r-value of 0 indicates no association between the two variables. Correlation does not indicate causation. To obtain a correlation coefficient (r) for two variables, use the cor() command. Pearson s correlation is the default method designated with the method= argument. Kendall ( kendall ) and Spearman ( spearman ) are also available. Pearson method is intended for use on normally distributed data (parametric). Spearman and Kendall are non-parametric alternatives. cor(datavector1, datavector2, method= pearson ) cor(dataset$measurement1, dataset$measurement2, method= pearson ) In order to obtain significance values for correlation coefficients, use the cor.test() command. Pearson s correlation is the default method; Kendall ( kendall ) and Spearman ( spearman ) are also available. cor.test(datavector1, datavector2, method= pearson ) cor.test(dataset$measurement1, dataset$measurement2, method= pearson ) Linear Regression Used to find a best fit line through the data that can be used to predict one variable based on another. Data inputs can be either vectors or columns from a data frame. The response variable (dependent variable; Y-axis) is listed first, then the explanatory variable (independent variable; X-axis). Linear regression between two variables requires the creation of a linear model: lm(responsevector~explanatoryvector) lm(dataset$response~dataset$explanatory) 1

2 Use the summary command to get more information about the linear model: summary(lm(responsevector~explanatoryvector)) summary(lm(dataset$response~dataset$explanatory)) You can also assign the linear model to a name and run the summary command on that. model1=lm(responsevector~explanatoryvector)) summary(model1) model2=lm(dataset$response~dataset$explanatory)) summary(model2) Look for the p-value and adjusted R-squared value in the output (highlighted red in the example below): Call: lm(formula = example$richness ~ example$diversity) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) * example$diversity e-07 *** --- Signif. codes: 0 *** ** 0.01 * Residual standard error: on 30 degrees of freedom Multiple R-squared: , Adjusted R-squared: F-statistic: on 1 and 30 DF, p-value: 1.278e-07 The strength of the relationship between the two variables is expressed as an R 2 value. This value tells us what percentage of the variation in the response variable (dependent variable; Y-axis) is explained by the explanatory variable (independent variable; X-axis). R 2 values range from 0 (no variation explained) to 1 (variation explained perfectly, 100%). The P-value tells whether there is a significant relationship between these variables. If less than 0.05, you can be fairly sure the explanatory variable is influencing the response variable in a meaningful way. 2

3 Adding a Trendline to a Scatterplot Regression and correlation data are usually presented as a scatterplot with a trendline (best fit line). A trendline can be added to a scatterplot (as created with plot() command) by using the abline() command. The response variable is the Y-axis (dependent) variable, and the explanatory variable is the X-axis (independent) variable. Pay close attention to the order of the variables. The abline() command has arguments for line type (lty=), color (col=), and line width (lwd=). plot(explanatory, response) abline(lm(response~explanatory), lty=1, col= red, lwd=1) plot(dataset$explanatory, dataset$response) abline(lm(dataset$response~dataset$explanatory), lty=1, col= red, lwd=1) Testing for Significant Difference between Slopes The simba package contains a command that will calculate the difference in slope between two lines and test for a significant difference it compares linear regression models. The variables for the first regression model are specified in the x1 (explanatory) and y1 (response) arguments. These two vectors or data frame columns must be the same length (contain same number of values). The variables for the second regression model are specified in the x2 (explanatory) and y2 (response) arguments. These two vectors or data frame columns must be the same length (contain same number of values). Adding the argument ic = TRUE will also calculate the difference in intercept and test for significance. library(simba) diffslope(x1, y1, x2, y2, ic=true) diffslope(dataset$explanatory1, dataset$response1, dataset$explanatory2, dataset$response2, ic=true) 3

4 Quadratic (Polynomial) Regression To create a quadratic regression model, the poly() command is used to indicate that a secondorder polynomial should be used to fit the regression. The number 2 (second argument) designates this as a quadratic function, but you can also use higher order functions. lm(dataset$response~poly(dataset$explanatory, 2, raw=true)) More information about model can be obtained with the summary() command. summary(lm(dataset$response~poly(dataset$explanatory, 2, raw=true))) Again, the model can be assigned to a name for easier use with other commands. fit=lm(dataset$response~poly(dataset$explanatory, 2, raw=true)) summary(fit) To generate a plot showing a polynomial regression line, first generate the scatter plot. plot(dataset$explanatory, dataset$response) Then generate a function for drawing the regression line. Assign the function to a name. pol2<-function(x) fit$coefficient[3]*x^2 + fit$coefficient[2]*x + fit$coefficient[1] Then add the regression line to the scatter plot. The add=true argument adds the result of the curve() command to the previously created plot. curve(pol2, add=true) 4

5 Logistic Regression Used when the response (dependent; Y-axis) variable is binary possible outcomes are 0 or 1 (categorical; like yes/no or dead/alive). The explanatory (independent; X-axis) variable is numerical. Predicts the probability of having a 0 or 1 response based on a given explanatory value. For example, what is the probability of a tree surviving to the next year based on height? Commonly used for survival data obtained from population monitoring. Note the use of the glm() command instead of lm(). You must specify a binomial (logistic) model using the family= argument. The summary() command provides more information. glm(formula=dataset$response~dataset$explanatory, family=binomial) summary(glm(formula=dataset$response~dataset$explanatory, family=binomial)) There is an alternate way to code this, where the dataset itself is specified using the data= argument, which means the response and explanatory variables can be named without the preceeding dataset$ summary(glm(formula=response~explanatory, data=dataset, family=binomial)) In the output, look for the P value, Pr(> z ), in the Coefficients section that corresponds to the response variable ( Richness in this example output). Call: glm(formula = Age ~ Richness, family = binomial, data = example) Deviance Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error z value Pr(> z ) (Intercept) Richness (Dispersion parameter for binomial family taken to be 1) Null deviance: on 31 degrees of freedom Residual deviance: on 30 degrees of freedom AIC:

6 Odds Ratio For a logistic regression model, the odds ratio is a commonly calculated statistic. It indicates that for every unit increase in the explanatory variable, there is an n-fold increase in the probability of getting a particular response. exp(coeff(glm(formula=response~explanatory, data=dataset, family=binomial))) The output value under the response variable ( Richness in this example) is the odds ratio. (Intercept) Richness You can also assign the regression model to a name ( model.name in this example). model.name=glm(formula=response~explanatory, data=dataset, family=binomial) exp(coeff(model.name)) 6

7 Tutorial Code setwd("/users/johndoe/desktop/") example=read.csv("r_example_dataframe.csv") #Correlation cor(example$richness, example$diversity, method="pearson") cor.test(example$richness, example$diversity, method="pearson") cor.test(example$richness, example$diversity, method="spearman") #Linear Regression lm(example$richness~example$diversity) summary(lm(example$richness~example$diversity)) #Regression models can be assigned a name; for example, model1 model1=lm(example$richness~example$diversity) summary(model1) #Make a scatterplot of the data plot(example$diversity, example$richness, ylab="richness", xlab="diversity", ylim=c(0,14), xlim=c(0,10), pch=16, col="blue", cex=1.5, las=1) #Add the trendline to the scatterplot; Note two ways to do this, depending on whether regression model was named or not abline(lm(example$richness~example$diversity)) abline(model1, col="red", lwd=3, lty=2) #Add text to scatterplot with R 2 text(1.5, 13, "R^2 = ") text(1.5, 12, "P = 1.278e-07") and P values #Test for difference in slope between young and old plots #Create old and young subsets of the data young=example[grep("young", example$age),] old=example[grep("old", example$age),] 7

8 #Load the simba package library(simba) #Test whether the linear regressions of Richness vs. Diversity differ between young and old plots; Look for the Significance value in the output diffslope(young$richness, young$diversity, old$richness, old$diversity) #Quadratic Regression quadfit=lm(example$richness~poly(example$diversity, 2, raw=true)) summary(quadfit) #Create a scatterplot of the data plot(example$diversity, example$richness, ylab="richness", xlab="diversity", ylim=c(0,14), xlim=c(0,10), pch=16, col="blue", cex=1.5, las=1) #Specify the quadratic function for the polynomial fit pol2=function(x) quadfit$coefficient[3]*x^2 + quadfit$coefficient[2]*x + quadfit$coefficient[1] #Add the quadratic trendline to the scatterplot curve(pol2, add=true) #Logistic Regression #Specify the binomial regression model glm(formula=age~richness, data=example, family=binomial) summary(glm(formula=age~richness, data=example, family=binomial)) g=glm(formula=age~richness, data=example, family=binomial) summary(g) #Odds Ratio exp(coef(glm(formula=age~richness, data=example, family=binomial))) exp(coef(g)) 8

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