Lab 6 More Linear Regression

Transcription

1 Lab 6 More Linear Regression Corrections from last lab 5: Last week we produced the following plot, using the code shown below. plot(sat$verbal, sat$math,, col=c(1,2)) legend("bottomright", legend=c("male", "Female"), pch=c(1,2), col=c(1,2)) This plot is incorrect for a few reasons. First, I recommend, when you have the need to distinguish your dots by categories (like Male, Female), pick one way, either changing symbols, or changing colors. We tried to do both, and ended up with 4 categories unknowingly (black/red circles, black/red triangles)!!! To choose different dot symbols, use argument pch=as.integer(sat$sex) To choose different dot colors, use col=as.integer(sat$sex) Next, R goes alphabetical in assigning symbols/colors to categories. My legend should have had the argument legend=c("female", "Male") instead of legend=c("male", "Female") This resulted in getting the symbols backwards!! Circles/black indicated Females, not Males, but I forced the legend to name the symbols backwards. R Tips: [1] Last week a student asked me how to tell R what to do when a data set variable (like V1) contains blank or 9999 symbols instead of NA or legal values for the variable that R sometimes gives error messages when trying to obtain plots or statistics when it encounters these cell anomalies in the variable. My general answer, given to me by a student of STAT 452, is to do the following, Say you have 3 variables, V1, V2, V3, which probably have bad/blank data values in them, and you want to correct that. Use the command na.omit(data.frame)v1, V2, V3)) or dataname <- na.omit(data.frame)v1, V2, V3)) if you want to name the new data frame constructed, and this will take care of these data values which mess up the R function you are trying to use. Be careful in your assignment symbol <-, Leave a space on either side of it, and put the 2 symbols together, in your code to avoid the following confusions. -1-

2 x < -5 (Is x less than -5?) x <-5 (is x assigned 5 or is x less than -5?) x < - 5 (is x less than or minus of 5?) etc... Transforming Variables: Last week we had the following scatter plot in our pairs() display, resulting from the STAT 216 data. # transforming variables stat216 <- read.csv("c:/users/michael/desktop/stat216.csv") plot(stat216$text, stat216$facebook, main="facebook friends vs TEXT per day") We will do a transformation of x, then y, then both, to see what results. # transforming variables logface <- log(stat216$facebook) ; logtext <- log(stat216$text) # note that log() finds the natural log, base e # log10() would find thr common log, base 10 plot(logtext, stat216$facebook, main="logging x variable") plot(stat216$text, logface, main="logging y variable") plot(logtext, logface, main="logging both variables") next -2-

3 As demonstrated, if you have many data points clustered in the bottom left (low values of x and y), and a few high values (of x, y, or both), then possibly a log transform is worth inspecting, in order to get the low points spread out along the x and y axes without moving the high end values very much from their relative positions. We see that logging both x and y gives us our best chance to model a linear fit to the data. Let's inspect our data. logtext ;logface We have to clean up the NA, NAN, and Inf cells in our results before we can properly model and graph. See below. So, forming data1 got rid of the NA and NAN. -3-

4 remove1 <- apply(data1,1, function(x) all(is.finite(x))) data2 <- data1[remove1,] data2 Doing this little routine above got rid of all Inf cells, resulting in data2, shown below. I found this little R routing online and posted the information in hint1.pdf in Moodle. See results below. We now have clean data to finish our regression analysis. See below. # make lm model now that NA, NAN, and Inf removed from data set model1 <- lm(data2$logface ~ data2$logtext) model1 summary(model1) scatter.smooth(data2$logtext,data2$logface, lwd=2, main="face on TEXT") abline(model1, lty=2, col="red", lwd=3) # make residual plot plot(predict(model1), resid(model1), main="residual plot of FACEBOOK=0.26TEXT ") abline(h=0) Results are shown below. -4-

5 Next We see that our line of best fit matches the smooth line well, and our residual plot seems to confirm that the linear model seems appropriate. So, our line of fit (model) is loge(face) = loge(text) So, if we have 100 text messages per day, we predict ln(face) = ln(100) = (4.6052) = so, FACE = e = Facebook friends. How accurate is this model? The multiple R-squared = r2 = from the summary() statement indicates an r = , a rather weak correlation. So, we do not have much confidence in our prediction, since our relationship between facebook friends and text messages/day is so weak. -5-

6 Another useful package: I have found the package UsingR to be very useful when doing linear regression problems because of the commands simple.lm() and plot(simple.lm()). See below for our use of this package in our problem. # install UsingR library(usingr) model2 <- simple.lm(data2$logtext, data2$logface) model2 Results are shown below. You get model coefficients and scatter plot with fit line superimposed. Next code. summary(model2) results in next code. plot(model2) results in 4 plots (residuals plot, normal Q-Q plot, standardized residuals root plot, and residuals vs leverage plot). Each of these has to be entered one at a time on the Console (see output below). -6-

7 Next Console queries (having you hit return key for each graph) is shown below. The Q-Q plot, standardized residuals plot, and leverage plot are tools used in more indepth regression analysis, which Dave will talk about sometime in his STAT sequence. We are mostly interested in the original scatter plot and residuals plot now. If you want 4 main plots to appear at one time without hitting ENTER type simple.lm(data2$logtext, data2$logface, show.residuals=true) to get -7-

8 along with display For now, these 4 graphs may be the most useful collection for your regression studies. Finally, we might be interested in producing prediction bands and 90% confidence interval of parameters bands. See below. par(fig=c(0,1,0,1)) simple.lm(data2$logtext, data2$logfac,show.ci=true,conf.level=0.90) with output and -8-

9 Note that the par() command used before plotting reset the plotting regions from 4 plots per window back to the default 1 plot per window, from the 4 plots per window which was automatically set up when we evoked simple.lm() we want to avoid using only ¼ of the window for future plotting, so reset the par() partitioning of graphics view. Back to the lm() command results: When you set up a model with lm(), you generate a lot of variables in the R background which you should be made aware of and possibly utilize for various purposes. See below for a list of many of these automatically generated variables summary() returns summary information about the regression plot() makes diagnostic plots coef() returns the coefficients residuals() returns the residuals (can be abbreviated resid()) fitted() returns fitted values, yi -predicted deviance() returns RSS predict() performs model predictions anova() finds various sums of squares AIC() is used for model selection Some of these topics will be discussed later in Dave's class. Let us use some of these variables to produce by hand a confidence interval/prediction interval band plot, which simple.lm() produced rather automatically above. See code below, using data set patients.txt (which contains various physical characteristics of patients). It is a nice linearly modeled data set with very strong linear relationship between height and weight. -9-

10 # prediction and CI bands # ======================= patients <- read.delim("c:/users/michael/desktop/patients.txt") patients height <- patients[,1] ; weight <- patients[,2] catheter <- patients[,3] model3 <- lm(weight ~ height) plot(height, weight, main="weight on height") abline(model3) xval <- data.frame(height=seq(20, 65,1)) conf <- predict(model3, xval, interval="confidence") pred <- predict(model3, xval, interval="prediction") matlines(xval, conf[,c("lwr", "upr")], col="red", lty=1, lwd=3) matlines(xval, pred[,c("lwr", "upr")], col="blue", lty=2, lwd=3) text(35, 80, labels="red solid is CI, blue dashed is prediction") summary(model3) conf pred Results are and

11 and Homework [1]: Perform a linear regression analysis on AIRFARE vs DISTANCE from the dataset Airfares.csv (no transformation needed). Be sure to include the scatter plot with line of fit, smooth line superimposed, model summary(), and statement of final linear model appropriate for the result, along with correlation values and other elements you deem necessary to this study. Homework [2]: Remove any point or two which seems not to fit in the model of HW[1], but rather appears as outlier(s), and rerun your linear study. Why might the point or two be legitimately removed from this listing of airline pricing per mile, since this pricing assumes equal pricing from city to city? Homework [3]: Look through the matrix of scatter plots from stat216.csv and pick 2 other variables to develop a linear regression model from transformed variables (like logging x or y or both), as I did above with FACE on TEXT. Be sure to produce a model summary(), scatter plot with line of best fit superimposed on it, and residual plot. Homework[4]: The data set cloudseed.csv contains rain generating data done in Texas to overcome a summer drought. Perform a transformation (hint: try log()) on variable(s) and produce an acceptable linear model of SEEDED vs UNSEEDED (SEEDED is response) cloud variable results (units unknown). Produce whatever statistics and graphics you deem acceptable for a report on the modeling for this problem. Extra Information not required for this lab: Transforming Variables

12 When we look at scatter plots to find regression model fits for y on x, we would like to model a linear fit, if appropriate. We can determine linearity as appropriate by looking at the scatter plot, or more appropriately, by looking at the residual plot. We could also introduce a curve of best fit (like some sort of a lowess line, etc.), to help us believe that a linear fit is at least not inappropriate. When a linear fit is not appropriate, we can transform the x, the y or both variables, in order to get some form of a line to fit the resulting transformed (x, y) points. We are not changing the data by transforming, if we use transformation functions which are called mathematically one-to-one and onto. It is like transforming a distance from feet to meters in units--we have not changed the data, just changed the description numbers describing the data. Below is a nice chart of ideas how to transform variables, depending upon what you originally have. To do transform #1, use the following commands and formula: y-hat = βo +b1 x + ε lm(y ~ sqrt(x)) plot(sqrt(x), y) y-hat = βo + β1ln(x) + ε lm(y ~ log(x)) plot(log(x), y) -1 y-hat = βo +β1 x + ε lm(y ~ 1/x) plot(1/x,y) -12-

13 To do transform #2, use the following commands and formula: y-hat = βo + β1x + β2x2 + ε lm(y ~ x+i(x^2)) plot(x, y) To do transform #4, use the following commands and formula: (y-hat)2 = βo + β1x + ε lm(i(y^2) ~ x) plot(x, y^2) To do transform #5, use the following commands and formula: ln(y-hat) = βo + β1x +ε lm(log(y) ~ x) ln(y-hat) = βo + β1ln(x) + ε lm(log(y) ~ log(x)) plot (x, log(y)) plot(log(x), log(y)) there are other ways to do these formulas, for example, by making up a variable consisting of log(x), sqrt(x), etc. and call it var1, using var 1 in the formula of the lm() and plot () commands, as needed. -13-