# Linear VS Quadratic Discriminant Analysis in R

In this post we will look at linear discriminant analysis (LDA) and quadratic discriminant analysis (QDA). Discriminant analysis is used when the dependent variable is categorical. Another commonly used option is logistic regression but there are differences between logistic regression and discriminant analysis. Both LDA and QDA are used in situations in which there is a clear separation between the classes you want to predict. If the categories are fuzzier logistic regression is often the better choice.

For our example, we will use the “Mathlevel” dataset found in the “Ecdat” package. Our goal will be to predict the sex of a respondent based on SAT math score, major, foreign language proficiency, as well as the number of math, physic, and chemistry classes a respondent took. Below is some initial code to start our analysis.

library(MASS);library(Ecdat)
data("Mathlevel")

The first thing we need to do is clean up the data set. We have to remove any missing data in order to run our model. We will create a dataset called “math” that has the “Mathlevel” dataset but with the “NA”s removed use the “na.omit” function. After this, we need to set our seed for the purpose of reproducibility using the “set.seed” function. Lastly, we will split the data using the “sample” function using a 70/30 split. The training dataset will be called “math.train” and the testing dataset will be called “math.test”. Below is the code

math<-na.omit(Mathlevel)
set.seed(123)
math.ind<-sample(2,nrow(math),replace=T,prob = c(0.7,0.3))
math.train<-math[math.ind==1,]
math.test<-math[math.ind==2,]

Now we will make our model and it is called “lda.math” and it will include all available variables in the “math.train” dataset. Next we will check the results by calling the modle. Finally, we will examine the plot to see how our model is doing. Below is the code.

lda.math<-lda(sex~.,math.train)
lda.math
## Call:
## lda(sex ~ ., data = math.train)
##
## Prior probabilities of groups:
##      male    female
## 0.5986079 0.4013921
##
## Group means:
##        mathlevel.L mathlevel.Q mathlevel.C mathlevel^4 mathlevel^5
## male   -0.10767593  0.01141838 -0.05854724   0.2070778  0.05032544
## female -0.05571153  0.05360844 -0.08967303   0.2030860 -0.01072169
##        mathlevel^6      sat languageyes  majoreco  majoross   majorns
## male    -0.2214849 632.9457  0.07751938 0.3914729 0.1472868 0.1782946
## female  -0.2226767 613.6416  0.19653179 0.2601156 0.1907514 0.2485549
##          majorhum mathcourse physiccourse chemistcourse
## male   0.05426357   1.441860    0.7441860      1.046512
## female 0.07514451   1.421965    0.6531792      1.040462
##
## Coefficients of linear discriminants:
##                       LD1
## mathlevel.L    1.38456344
## mathlevel.Q    0.24285832
## mathlevel.C   -0.53326543
## mathlevel^4    0.11292817
## mathlevel^5   -1.24162715
## mathlevel^6   -0.06374548
## sat           -0.01043648
## languageyes    1.50558721
## majoreco      -0.54528930
## majoross       0.61129797
## majorns        0.41574298
## majorhum       0.33469586
## mathcourse    -0.07973960
## physiccourse  -0.53174168
## chemistcourse  0.16124610
plot(lda.math,type='both')

Calling “lda.math” gives us the details of our model. It starts be indicating the prior probabilities of someone being male or female. Next is the means for each variable by sex. The last part is the coefficients of the linear discriminants. Each of these values is used to determine the probability that a particular example is male or female. This is similar to a regression equation.

The plot provides us with densities of the discriminant scores for males and then for females. The output indicates a problem. There is a great deal of overlap between male and females in the model. What this indicates is that there is a lot of misclassification going on as the two groups are not clearly separated. Furthermore, this means that logistic regression is probably a better choice for distinguishing between male and females. However, since this is for demonstrating purposes we will not worry about this.

We will now use the “predict” function on the training set data to see how well our model classifies the respondents by gender. We will then compare the prediction of the model with thee actual classification. Below is the code.

math.lda.predict<-predict(lda.math)
math.train$lda<-math.lda.predict$class
table(math.train$lda,math.train$sex)
##
##          male female
##   male    219    100
##   female   39     73
mean(math.train$lda==math.train$sex)
## [1] 0.6774942

As you can see, we have a lot of misclassification happening. A large amount of false negatives which is a lot of males being classified as female. The overall accuracy us only 59% which is not much better than chance.

We will now conduct the same analysis on the test data set. Below is the code.

lda.math.test<-predict(lda.math,math.test)
math.test$lda<-lda.math.test$class
table(math.test$lda,math.test$sex)
##
##          male female
##   male     92     43
##   female   23     20
mean(math.test$lda==math.test$sex)
## [1] 0.6292135

As you can see the results are similar. To put it simply, our model is terrible. The main reason is that there is little distinction between males and females as shown in the plot. However, we can see if perhaps a quadratic discriminant analysis will do better

QDA allows for each class in the dependent variable to have it’s own covariance rather than a shared covariance as in LDA. This allows for quadratic terms in the development of the model. To complete a QDA we need to use the “qda” function from the “MASS” package. Below is the code for the training data set.

math.qda.fit<-qda(sex~.,math.train)
math.qda.predict<-predict(math.qda.fit)
math.train$qda<-math.qda.predict$class
table(math.train$qda,math.train$sex)
##
##          male female
##   male    215     84
##   female   43     89
mean(math.train$qda==math.train$sex)
## [1] 0.7053364

You can see there is almost no difference. Below is the code for the test data.

math.qda.test<-predict(math.qda.fit,math.test)
math.test$qda<-math.qda.test$class
table(math.test$qda,math.test$sex)
##
##          male female
##   male     91     43
##   female   24     20
mean(math.test$qda==math.test$sex)
## [1] 0.6235955

Still disappointing. However, in this post we reviewed linear discriminant analysis as well as learned about the use of quadratic linear discriminant analysis. Both of these statistical tools are used for predicting categorical dependent variables. LDA assumes shared covariance in the dependent variable categories will QDA allows for each category in the dependent variable to have it’s own variance.

# Probability,Odds, and Odds Ratio

In logistic regression, there are three terms that are used frequently but can be confusing if they are not thoroughly explained. These three terms are probability, odds, and odds ratio. In this post, we will look at these three terms and provide an explanation of them.

Probability

Probability is probably (no pun intended) the easiest of these three terms to understand. Probability is simply the likelihood that a certain even will happen.  To calculate the probability in the traditional sense you need to know the number of events and outcomes to find the probability.

Bayesian probability uses prior probabilities to develop a posterior probability based on new evidence. For example, at one point during Super Bowl LI the Atlanta Falcons had a 99.7% chance of winning. This was base don such factors as the number points they were ahead and the time remaining.  As these changed, so did the probability of them winning. yet the Patriots still found a way to win with less then a 1% chance

Bayesian probability was also used for predicting who would win the 2016 US presidential race. It is important to remember that probability is an expression of confidence and not a guarantee as we saw in both examples.

Odds

Odds are the expression of relative probabilities. Odds are calculated using the following equation

probability of the event ⁄ 1 – probability of the event

For example, at one point during Super Bowl LI the odds of the Atlanta Falcons winning were as follows

0.997 ⁄ 1 – 0.997 = 332

This can be interpreted as the odds being 332 to 1! This means that Atlanta was 332 times more likely to win the Super Bowl then loss the Super Bowl.

Odds are commonly used in gambling and this is probably (again no pun intended) where most of us have heard the term before. The odds is just an extension of probabilities and the are most commonly expressed as a fraction such as one in four, etc.

Odds Ratio

A ratio is the comparison of of two numbers and indicates how many times one number is contained or contains another number. For example, a ration of boys to girls is 5 to 1 it means that there are five boys for every one girl.

By  extension odds ratio is the comparison of two different odds. For example, if the odds of Team A making the playoffs is 45% and the odds of Team B making the playoffs is 35% the odds ratio is calculated as follows.

0.45 ⁄ 0.35 = 1.28

Team A is 1.28 more likely to make the playoffs then Team B.

The value of the odds and the odds ratio can sometimes be the same.  Below is the odds ratio of the Atlanta Falcons winning and the New Patriots winning Super Bowl LI

0.997⁄ 0.003 = 332

As such there is little difference between odds and odds ratio except that odds ratio is the ratio of two odds ratio. As you can tell, there is a lot of confusion about this for the average person. However, understanding these terms is critical to the application of logistic regression.

# Best Subset Regression in R

In this post, we will take a look at best subset regression. Best subset regression fits a model for all possible feature or variable combinations and the decision for the most appropriate model is made by the analyst based on judgment or some statistical criteria.

Best subset regression is an alternative to both Forward and Backward stepwise regression. Forward stepwise selection adds one variable at a time based on the lowest residual sum of squares until no more variables continues to lower the residual sum of squares. Backward stepwise regression starts with all variables in the model and removes variables one at a time. The concern with stepwise methods is they can produce biased regression coefficients, conflicting models, and inaccurate confidence intervals.

Best subset regression bypasses these weaknesses of stepwise models by creating all models possible and then allowing you to assess which variables should be including in your final model. The one drawback to best subset is that a large number of variables means a large number of potential models, which can make it difficult to make a decision among several choices.

In this post, we will use the “Fair” dataset from the “Ecdat” package to predict marital satisfaction based on age, Sex, the presence of children, years married, religiosity, education, occupation, and number of affairs in the past year. Below is some initial code.

library(leaps);library(Ecdat);library(car);library(lmtest)
data(Fair)

We begin our analysis by building the initial model with all variables in it. Below is the code

fit<-lm(rate~.,Fair)
summary(fit)
##
## Call:
## lm(formula = rate ~ ., data = Fair)
##
## Residuals:
##     Min      1Q  Median      3Q     Max
## -3.2049 -0.6661  0.2298  0.7705  2.2292
##
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)
## (Intercept)  3.522875   0.358793   9.819  < 2e-16 ***
## sexmale     -0.062281   0.099952  -0.623  0.53346
## age         -0.009683   0.007548  -1.283  0.20005
## ym          -0.019978   0.013887  -1.439  0.15079
## childyes    -0.206976   0.116227  -1.781  0.07546 .
## religious    0.042142   0.037705   1.118  0.26416
## education    0.068874   0.021153   3.256  0.00119 **
## occupation  -0.015606   0.029602  -0.527  0.59825
## nbaffairs   -0.078812   0.013286  -5.932 5.09e-09 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.03 on 592 degrees of freedom
## Multiple R-squared:  0.1405, Adjusted R-squared:  0.1289
## F-statistic:  12.1 on 8 and 592 DF,  p-value: 4.487e-16

The initial results are already interesting even though the r-square is low. When couples have children the have less martial satisfaction than couples without children when controlling for the other factors and this is the strongest regression weight. In addition, the more education a person has there is an increase in marital satisfaction. Lastly, as the number of affairs increases there is also a decrease in martial satisfaction. Keep in mind that the “rate” variable goes from 1 to 5 with one meaning a terrible marriage to five being a great one. The mean marital satisfaction was 3.52 when controlling for the other variables.

We will now create our subset models. Below is the code.

sub.fit<-regsubsets(rate~.,Fair)
best.summary<-summary(sub.fit)

In the code above we create the sub models using the “regsubsets” function from the “leaps” package and saved it in the variable called “sub.fit”. We then saved the summary of “sub.fit” in the variable “best.summary”. We will use the “best.summary” “sub.fit variables several times to determine which model to use.

There are many different ways to assess the model. We will use the following statistical methods that come with the results from the “regsubset” function.

• Mallow’ Cp
• Bayesian Information Criteria

We will make two charts for each of the criteria above. The plot to the left will explain how many features to include in the model. The plot to the right will tell you which variables to include. It is important to note that for both of these methods, the lower the score the better the model. Below is the code for Mallow’s Cp.

par(mfrow=c(1,2))
plot(best.summary$cp) plot(sub.fit,scale = "Cp") The plot on the left suggest that a four feature model is the most appropriate. However, this chart does not tell me which four features. The chart on the right is read in reverse order. The high numbers are at the bottom and the low numbers are at the top when looking at the y-axis. Knowing this, we can conclude that the most appropriate variables to include in the model are age, children presence, education, and number of affairs. Below are the results using the Bayesian Information Criterion par(mfrow=c(1,2)) plot(best.summary$bic)
plot(sub.fit,scale = "bic")

These results indicate that a three feature model is appropriate. The variables or features are years married, education, and number of affairs. Presence of children was not considered beneficial. Since our original model and Mallow’s Cp indicated that presence of children was significant we will include it for now.

Below is the code for the model based on the subset regression.

fit2<-lm(rate~age+child+education+nbaffairs,Fair)
summary(fit2)
##
## Call:
## lm(formula = rate ~ age + child + education + nbaffairs, data = Fair)
##
## Residuals:
##     Min      1Q  Median      3Q     Max
## -3.2172 -0.7256  0.1675  0.7856  2.2713
##
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)
## (Intercept)  3.861154   0.307280  12.566  < 2e-16 ***
## age         -0.017440   0.005057  -3.449 0.000603 ***
## childyes    -0.261398   0.103155  -2.534 0.011531 *
## education    0.058637   0.017697   3.313 0.000978 ***
## nbaffairs   -0.084973   0.012830  -6.623 7.87e-11 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.029 on 596 degrees of freedom
## Multiple R-squared:  0.1352, Adjusted R-squared:  0.1294
## F-statistic: 23.29 on 4 and 596 DF,  p-value: < 2.2e-16

The results look ok. The older a person is the less satisfied they are with their marriage. If children are presence the marriage is less satisfying. The more educated the more satisfied they are. Lastly, the higher the number of affairs indicate less marital satisfaction. However, before we get excited we need to check for collinearity and homoscedasticity. Below is the code

vif(fit2)
##       age     child education nbaffairs
##  1.249430  1.228733  1.023722  1.014338

No issues with collinearity.For vif values above 5 or 10 indicate a problem. Let’s check for homoscedasticity

par(mfrow=c(2,2))
plot(fit2)

The normal qqplot and residuals vs leverage plot can be used for locating outliers. The residual vs fitted and the scale-location plot do not look good as there appears to be a pattern in the dispersion which indicates homoscedasticity. To confirm this we will use Breusch-Pagan test from the “lmtest” package. Below is the code

bptest(fit2)
##
##  studentized Breusch-Pagan test
##
## data:  fit2
## BP = 16.238, df = 4, p-value = 0.002716

There you have it. Our model violates the assumption of homoscedasticity. However, this model was developed for demonstration purpose to provide an example of subset regression.

# Principal Component Analysis in R

This post will demonstrate the use of principal component analysis (PCA). PCA is useful for several reasons. One it allows you place your examples into groups similar to linear discriminant analysis but you do not need to know beforehand what the groups are. Second, PCA is used for the purpose of dimension reduction. For example, if you have 50 variables PCA can allow you to reduce this while retaining a certain threshold of variance. If you are working with a large dataset this can greatly reduce the computational time and general complexity of your models.

Keep in mind that there really is not a dependent variable as this is unsupervised learning. What you are trying to see is how different examples can be mapped in space based on whatever independent variables are used. For our example, we will use the “Carseats” dataset form the “ISLR”. Our goal is to understanding the relationship among the variables when examining the shelve location of the car seat. Below is the initial code to begin the analysis

library(ggplot2)
library(ISLR)
data("Carseats")

We first need to rearrange the data and remove the variables we are not going to use in the analysis. Below is the code.

Carseats1<-Carseats
Carseats1<-Carseats1[,c(1,2,3,4,5,6,8,9,7,10,11)]
Carseats1$Urban<-NULL Carseats1$US<-NULL

Here is what we did 1. We made a copy of the “Carseats” data called “Careseats1” 2. We rearranged the order of the variables so that the factor variables are at the end. This will make sense later 3.We removed the “Urban” and “US” variables from the table as they will not be a part of our analysis

We will now do the PCA. We need to scale and center our data otherwise the larger numbers will have a much stronger influence on the results than smaller numbers. Fortunately, the “prcomp” function has a “scale” and a “center” argument. We will also use only the first 7 columns for the analysis  as “sheveLoc” is not useful for this analysis. If we hadn’t moved “shelveLoc” to the end of the dataframe it would cause some headache. Below is the code.

Carseats.pca<-prcomp(Carseats1[,1:7],scale. = T,center = T)
summary(Carseats.pca)
## Importance of components:
##                           PC1    PC2    PC3    PC4    PC5     PC6     PC7
## Standard deviation     1.3315 1.1907 1.0743 0.9893 0.9260 0.80506 0.41320
## Proportion of Variance 0.2533 0.2026 0.1649 0.1398 0.1225 0.09259 0.02439
## Cumulative Proportion  0.2533 0.4558 0.6207 0.7605 0.8830 0.97561 1.00000

The summary of “Carseats.pca” Tells us how much of the variance each component explains. Keep in mind that number of components is equal to the number of variables. The “proportion of variance” tells us the contribution each component makes and the “cumulative proportion”.

If your goal is dimension reduction than the number of components to keep depends on the threshold you set. For example, if you need around 90% of the variance you would keep the first 5 components. If you need 95% or more of the variance you would keep the first six. To actually use the components you would take the “Carseats.pca$x” data and move it to your data frame. Keep in mind that the actual components have no conceptual meaning but is a numerical representation of a combination of several variables that were reduce using PCA to fewer variables such as going form 7 variables to 5 variables. This means that PCA is great for reducing variables for prediction purpose but is much harder for explanatory studies unless you can explain what the new components represent. For our purposes, we will keep 5 components. This means that we have reduce our dimensions from 7 to 5 while still keeping almost 90% of the variance. Graphing our results is tricky because we have 5 dimensions but the human mind can only conceptualize 3 at the best and normally 2. As such we will plot the first two components and label them by shelf location using ggplot2. Below is the code scores<-as.data.frame(Carseats.pca$x)
pcaplot<-ggplot(scores,(aes(PC1,PC2,color=Carseats1$ShelveLoc)))+geom_point() pcaplot From the plot you can see there is little separation when using the first two components of the PCA analysis. This makes sense as we can only graph to components so we are missing a lot of the variance. However for demonstration purposes the analysis is complete. # Developing a Data Analysis Plan It is extremely common for beginners and perhaps even experience researchers to lose track of what they are trying to achieve or do when trying to complete a research project. The open nature of research allows for a multitude of equally acceptable ways to complete a project. This leads to an inability to make decision and or stay on course when doing research. One way to reduce and eliminate the roadblock to decision making and focus in research is to develop a plan. In this post we will look at one version of a data analysis plan. Data Analysis Plan A data analysis plan includes many features of a research project in it with a particular emphasis on mapping out how research questions will be answered and what is necessary to answer the question. Below is a sample template of the analysis plan. The majority of this diagram should be familiar to someone who has ever done research. At the top, you state the problem, this is the overall focus of the paper. Next comes the purpose, the purpose is the over-arching goal of a research project. After purpose comes the research questions. The research questions are questions about the problem that are answerable. People struggle with developing clear and answerable research questions. It is critical that research questions are written in a way that they can be answered and that the questions are clearly derived from the problem. Poor questions means poor or even no answers. After the research questions it is important to know what variables are available for the entire study and specifically what variables can be used to answer each research question. Lastly, you must indicate what analysis or visual you will develop in order to answer your research questions about your problem. This requires you to know how you will answer your research questions Example Below is an example of a completed analysis plan for simple undergraduate level research paper In the example above, the student want to understand the perceptions of university students about the cafeteria food quality and their satisfaction with the university. There were four research questions, a demographic descriptive question, a descriptive question about the two main variables, a comparison question, and lastly a relationship question. The variables available for answering the questions are listed of to the left side. Under that, the student indicates the variables needed to answer each question. For example, the demographic variables of sex, class level, and major are needed to answer the question about the demographic profile. The last section is the analysis. For the demographic profile the student found the percentage of the population in each sub group of the demographic variables. Conclusion A data analysis plan provides an excellent way to determine what needs to be done to complete a study. It also helps a researcher to clearly understand what they are trying to do and provides a visuals for those who the research wants to communicate with about the progress of a study. # Generalized Additive Models in R In this post, we will learn how to create a generalized additive model (GAM). GAMs are non-parametric generalized linear models. This means that linear predictor of the model uses smooth functions on the predictor variables. As such, you do not need to specific the functional relationship between the response and continuous variables. This allows you to explore the data for potential relationships that can be more rigorously tested with other statistical models In our example, we will use the “Auto” dataset from the “ISLR” package and use the variables “mpg”,“displacement”,“horsepower”,and “weight” to predict “acceleration”. We will also use the “mgcv” package. Below is some initial code to begin the analysis library(mgcv) library(ISLR) data(Auto) We will now make the model we want to understand the response of “accleration” to the explanatory variables of “mpg”,“displacement”,“horsepower”,and “weight”. After setting the model we will examine the summary. Below is the code model1<-gam(acceleration~s(mpg)+s(displacement)+s(horsepower)+s(weight),data=Auto) summary(model1) ## ## Family: gaussian ## Link function: identity ## ## Formula: ## acceleration ~ s(mpg) + s(displacement) + s(horsepower) + s(weight) ## ## Parametric coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 15.54133 0.07205 215.7 <2e-16 *** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Approximate significance of smooth terms: ## edf Ref.df F p-value ## s(mpg) 6.382 7.515 3.479 0.00101 ** ## s(displacement) 1.000 1.000 36.055 4.35e-09 *** ## s(horsepower) 4.883 6.006 70.187 < 2e-16 *** ## s(weight) 3.785 4.800 41.135 < 2e-16 *** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## R-sq.(adj) = 0.733 Deviance explained = 74.4% ## GCV = 2.1276 Scale est. = 2.0351 n = 392 All of the explanatory variables are significant and the adjust r-squared is .73 which is excellent. edf stands for “effective degrees of freedom”. This modified version of the degree of freedoms is due to the smoothing process in the model. GCV stands for generalized cross validation and this number is useful when comparing models. The model with the lowest number is the better model. We can also examine the model visually by using the “plot” function. This will allow us to examine if the curvature fitted by the smoothing process was useful or not for each variable. Below is the code. plot(model1) We can also look at a 3d graph that includes the linear predictor as well as the two strongest predictors. This is done with the “vis.gam” function. Below is the code vis.gam(model1) If multiple models are developed. You can compare the GCV values to determine which model is the best. In addition, another way to compare models is with the “AIC” function. In the code below, we will create an additional model that includes “year” compare the GCV scores and calculate the AIC. Below is the code. model2<-gam(acceleration~s(mpg)+s(displacement)+s(horsepower)+s(weight)+s(year),data=Auto) summary(model2) ## ## Family: gaussian ## Link function: identity ## ## Formula: ## acceleration ~ s(mpg) + s(displacement) + s(horsepower) + s(weight) + ## s(year) ## ## Parametric coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 15.54133 0.07203 215.8 <2e-16 *** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Approximate significance of smooth terms: ## edf Ref.df F p-value ## s(mpg) 5.578 6.726 2.749 0.0106 * ## s(displacement) 2.251 2.870 13.757 3.5e-08 *** ## s(horsepower) 4.936 6.054 66.476 < 2e-16 *** ## s(weight) 3.444 4.397 34.441 < 2e-16 *** ## s(year) 1.682 2.096 0.543 0.6064 ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## R-sq.(adj) = 0.733 Deviance explained = 74.5% ## GCV = 2.1368 Scale est. = 2.0338 n = 392 #model1 GCV model1$gcv.ubre
##   GCV.Cp
## 2.127589
#model2 GCV
model2$gcv.ubre ## GCV.Cp ## 2.136797 As you can see, the second model has a higher GCV score when compared to the first model. This indicates that the first model is a better choice. This makes sense because in the second model the variable “year” is not significant. To confirm this we will calculate the AIC scores using the AIC function. AIC(model1,model2) ## df AIC ## model1 18.04952 1409.640 ## model2 19.89068 1411.156 Again, you can see that model1 s better due to its fewer degrees of freedom and slightly lower AIC score. Conclusion Using GAMs is most common for exploring potential relationships in your data. This is stated because they are difficult to interpret and to try and summarize. Therefore, it is normally better to develop a generalized linear model over a GAM due to the difficulty in understanding what the data is trying to tell you when using GAMs. # Intro to Using Machine Learning Machine learning is about using data to take action. This post will explain common steps that are taking when using machine learning in the analysis of data. In general, there are five steps when applying machine learning. 1. Collecting data 2. Preparing data 3. Exploring data 4. Training a model on the data 5. Assessing the performance of the model 6. Improving the model We will go through each step briefly Step One, Collecting Data Data can come from almost anywhere. Data can come from a database in a structured format like mySQL. Data can also come unstructured such as tweets collected from twitter. However you get the data, you need to develop a way to clean and process it so that it is ready for analysis. There are some distinct terms used in machine learning that some coming from traditional research my not be familiar. • example-An example is one set of data. In an excel spreadsheet, an example would be one row. In empirical social science research we would call an example a respondent or participant. • Unit of observation-This is how an example is measured. The units can be time, money, height, weight, etc. • feature-A feature is a characteristic of an example. In other forms of research we normally call a feature a variable. For example, ‘salary’ would be a feature in machine learning but a variable in traditional research. Step Two, Preparing Data This is actually the most difficult step in machine learning analysis. It can take up to 80% of the time. With data coming from multiple sources and in multiple formats it is a real challenge to get everything where it needs to be for an analysis. Missing data needs to be addressed, duplicate records, and other issues are a part of this process. Once these challenges are dealt with it is time to explore the data. Step Three, Explore the Data Before analyzing the data, it is critical that the data is explored. This is often done visually in the form of plots and graphs but also with summary statistics. You are looking for some insights into the data and the characteristics of different features. You are also looking out for things that might be unusually such as outliers. There are also times when a variable needs to be transformed because there are issues with normality. Step Four, Training a Model After exploring the data, you should have an idea of what you want to know if you did not already know. Determining what you want to know helps you to decide which algorithm is most appropriate for developing a model. To develop a model, we normally split the data into a training and testing set. This allows us to assess the model for its strengths and weaknesses. Step Five, Assessing the Model The strength of the model is deteremined. Every model has certain biases that limits its usefulness. How to assess a model depends on what type of model is developed and the purpose of the analysis. Whenever suitable, we want to try and improve the model. Step Six, Improving the Model Improve can happen in many ways. You might decide to normalize the variables in a different way. Or you may choose to add or remove features from the model. Perhaps you may switch to a different model. Conclusion Success in data analysis involves have a clear path for achieving your goals. The steps presented here provide one way of tackling machine learning. # Wilcoxon Signed Rank Test in R The Wilcoxon Signed Rank Test is the non-parametric equivalent of the t-test. If you have questions whether or not your data is normally distributed the Wilcoxon Signed Rank Test can still indicate to you if there is a difference between the means of your sample. Th Wilcoxon Test compares the medians of two samples instead of their means. The differences between the median and each individual value for each sample is calculated. Values that come to zero are removed. Any remaining values are ranked from lowest to highest. Lastly, the ranks are summed. If the rank sum is different between the two samples it indicates statistical difference between samples. We will now do an example using r. We want to see if there is a difference in enrollment between private and public universities. Below is the code We will begin by loading the ISLR package. Then we will load the ‘College’ data and take a look at the variables in the “College” dataset by using the ‘str’ function. library(ISLR) data=College str(College) ## 'data.frame': 777 obs. of 18 variables: ##$ Private    : Factor w/ 2 levels "No","Yes": 2 2 2 2 2 2 2 2 2 2 ...
##  $Apps : num 1660 2186 1428 417 193 ... ##$ Accept     : num  1232 1924 1097 349 146 ...
##  $Enroll : num 721 512 336 137 55 158 103 489 227 172 ... ##$ Top10perc  : num  23 16 22 60 16 38 17 37 30 21 ...
##  $Top25perc : num 52 29 50 89 44 62 45 68 63 44 ... ##$ F.Undergrad: num  2885 2683 1036 510 249 ...
##  $P.Undergrad: num 537 1227 99 63 869 ... ##$ Outstate   : num  7440 12280 11250 12960 7560 ...
##  $Room.Board : num 3300 6450 3750 5450 4120 ... ##$ Books      : num  450 750 400 450 800 500 500 450 300 660 ...
##  $Personal : num 2200 1500 1165 875 1500 ... ##$ PhD        : num  70 29 53 92 76 67 90 89 79 40 ...
##  $Terminal : num 78 30 66 97 72 73 93 100 84 41 ... ##$ S.F.Ratio  : num  18.1 12.2 12.9 7.7 11.9 9.4 11.5 13.7 11.3 11.5 ...
##  $perc.alumni: num 12 16 30 37 2 11 26 37 23 15 ... ##$ Expend     : num  7041 10527 8735 19016 10922 ...
##  $Grad.Rate : num 60 56 54 59 15 55 63 73 80 52 ... We will now look at the Enroll variable and see if it is normally distributed hist(College$Enroll)

This variable is highly skewed to the right. This may mean that it is not normally distributed. Therefore, we may not be able to use a regular t-test to compare private and public universities and the Wilcoxon Test is more appropriate. We will now use the Wilcoxon Test. Below are the results

wilcox.test(College$Enroll~College$Private)
##
##  Wilcoxon rank sum test with continuity correction
##
## data:  College$Enroll by College$Private
## W = 104090, p-value < 2.2e-16
## alternative hypothesis: true location shift is not equal to 0

The results indicate a difference we will now calculate the medians of the two groups using the ‘aggregate’ function. This function allows us to compare our two groups based on the median. Below is the code with the results.

aggregate(College$Enroll~College$Private, FUN=median)
##   College$Private College$Enroll
## 1              No       1337.5
## 2             Yes        328.0


As you can see, there is a large difference in enrollment in private and public colleges. We can then make the conclusion that there is a difference in the medians of private and public colleges with public colleges have a much higher enrollment.

Conclusion

The Wilcoxon Test is used for a non-parametric analysis of data. This test is useful whenever there are concerns with the normality of the data.

# Kruskal-Willis Test in R

Sometimes when the data that needs to be analyzed is not normally distributed. This makes it difficult to make any inferences based on the results because one of the main assumptions of parametric statistical test such as ANOVA, t-test, etc is normal distribution of the data.

Fortunately, for every parametric test there is a non-parametric test. Non-parametric test are test that make no assumptions about the normality of the data. This means that the non-normal data can still be analyzed with a certain measure of confidence in terms of the results.

This post will look at non-parametric test that are used to test the difference in means. For three or more groups we used the Kruskal-Wallis Test. The Kruskal-Wallis Test is the non-parametric version of ANOVA.

Setup

We are going to use the “ISLR” package available on R to demonstrate the use of the Kruskal-Wallis test. After downloading this package you need to load the “Auto” data. Below is the code to do all of this.

install.packages('ISLR')
library(ISLR)
data=Auto

We now need to examine the structure of the data set. This is done with the “str” function below is code followed by the results

str(Auto)
'data.frame':	392 obs. of  9 variables:
$mpg : num 18 15 18 16 17 15 14 14 14 15 ...$ cylinders   : num  8 8 8 8 8 8 8 8 8 8 ...
$displacement: num 307 350 318 304 302 429 454 440 455 390 ...$ horsepower  : num  130 165 150 150 140 198 220 215 225 190 ...
$weight : num 3504 3693 3436 3433 3449 ...$ acceleration: num  12 11.5 11 12 10.5 10 9 8.5 10 8.5 ...
$year : num 70 70 70 70 70 70 70 70 70 70 ...$ origin      : num  1 1 1 1 1 1 1 1 1 1 ...
$name : Factor w/ 304 levels "amc ambassador brougham",..: 49 36 231 14 161 141 54 223 241 2 ... So we have 9 variables. We first need to find if any of the continuous variables are non-normal because this indicates that the Kruskal-Willis test is needed. We will look at the ‘displacement’ variable and look at the histogram to see if it is normally distributed. Below is the code followed by the histogram hist(Auto$displacement)


This does not look normally distributed. We now need a factor variable with 3 or more groups. We are going to use the ‘origin’ variable. This variable indicates were the care was made 1 = America, 2 = Europe, and 3 = Japan. However, this variable is currently a numeric variable. We need to change it into a factor variable. Below is the code for this

Auto$origin<-as.factor(Auto$origin)

The Test

We will now use the Kruskal-Wallis test. The question we have is “is there a difference in displacement based on the origin of the vehicle?” The code for the analysis is below followed by the results.

> kruskal.test(displacement ~ origin, data = Auto)

Kruskal-Wallis rank sum test

data:  displacement by origin
Kruskal-Wallis chi-squared = 201.63, df = 2, p-value < 2.2e-16

Based on the results, we know there is a difference among the groups. However, just like ANOVA, we do not know were. We have to do a post-hoc test in order to determine where the difference in means is among the three groups.

To do this we need to install a new package and do a new analysis. We will download the “PCMR” package and run the code below

install.packages('PMCMR')
library(PMCMR)
data(Auto)
attach(Auto)
posthoc.kruskal.nemenyi.test(x=displacement, g=origin, dist='Tukey')

Here is what we did,

1. Installed the PMCMR package and loaded it
2. Loaded the “Auto” data and used the “attach” function to make it available
3. Ran the function “posthoc.kruskal.nemenyi.test” and place the appropriate variables in their place and then indicated the type of posthoc test ‘Tukey’

Below are the results

Pairwise comparisons using Tukey and Kramer (Nemenyi) test
with Tukey-Dist approximation for independent samples

data:  displacement and origin

1       2
2 3.4e-14 -
3 < 2e-16 0.51

P value adjustment method: none
Warning message:
In posthoc.kruskal.nemenyi.test.default(x = displacement, g = origin,  :
Ties are present, p-values are not corrected.

The results are listed in a table. When a comparison was made between group 1 and 2 the results were significant (p < 0.0001). The same when group 1 and 3 are compared (p < 0.0001).  However, there was no difference between group 2 and 3 (p = 0.51).

Do not worry about the warning message this can be corrected if necessary

Perhaps you are wondering what the actually means for each group is. Below is the code with the results

> aggregate(Auto[, 3], list(Auto$origin), mean) Group.1 x 1 1 247.5122 2 2 109.6324 3 3 102.7089 Cares made in America have an average displacement of 247.51 while cars from Europe and Japan have a displacement of 109.63 and 102.70. Below is the code for the boxplot followed by the graph boxplot(displacement~origin, data=Auto, ylab= 'Displacement', xlab='Origin') title('Car Displacement') Conclusion This post provided an example of the Kruskal-Willis test. This test is useful when the data is not normally distributed. The main problem with this test is that it is less powerful than an ANOVA test. However, this is a problem with most non-parametric test when compared to parametric test. # Data Mining Process Processes serve the purpose of providing people with a clear step-by-step procedures to accomplish a task. In many ways a process serves as a shortcut in solving a problem. As data mining is a complex situation with an endless number of problems there have been developed several processes for completing a data mining project. In this post we will look at the Cross-Industry Standard Process for Data Mining (CRISP-DM). CRISP-DM The CRISP-DM is an iterative process that has the following steps… 1. Organizational understanding 2. Data understanding 3. Data preparation 4. Modeling 5. Evaluation 6. Deployment We will look at each step briefly 1. Organizational Understanding Step 1 involves assessing the current goals of the organization and the current context. This information is then used to in deciding goals or research questions for data mining. Data mining needs to be done with a sense of purpose and not just to see what’s out there. Organizational understanding is similar to the introduction section of a research paper in which you often include the problem, questions, and even the intended audience of the research 2. Data Understanding Once a purpose and questions have been developed for data mining, it is necessary to determine what it will take to answer the questions. Specifically, the data scientist assess the data requirements, description, collection, and assesses data quality. In many ways, data understanding is similar to methodology of a standard research paper in which you assess how you will answer the research questions. It is particularly common to go back and forth between steps one and two. Organizational understanding influences data understanding which influences data under standing. 3. Data Preparation Data preparation involves cleaning the data. Another term for this is data mugging. This is the main part of an analysis in data mining. Often the data comes in a very messy way with information spread all over the place and incoherently. This requires the researcher to carefully deal with this problem. 4. Modeling A model provides a numerical explanation of something in the data. How this is done depends on the type of analysis that is used. As you develop various models you are arriving at various answers to your research questions. It is common to move back and forth between step 3 and 4 as the preparation affects the modeling and the type of modeling you may want to develop may influence data preparation. The results of this step can also be seen as being similar to the results section of an empirical paper. 5. Evaluation Evaluation is about comparing the results of the study with the original questions. In many ways it is about determining the appropriateness of the answers to the research questions. This experience leads to ideas for additional research. As such, this step is similar to the discussion section of a research paper. 6. Deployment The last step is when the results are actually used for decision-making or action. If the results indicate that a company should target people under 25 then this is what they do as an example. Conclusion The CRISP-DM process is a useful way to begin the data mining experience. The primary goal of data mining is providing evidence for making decisions and or taking action. This end goal has shaped the development of a clear process for taking action. # Big Data & Data Mining Dealing with large amounts of data has been a problem throughout most of human history. Ancient civilizations had to keep large amounts of clay tablets, papyrus, steles, parchments, scrolls etc. to keep track of all the details of an empire. However, whenever it seemed as though there would be no way to hold any more information a new technology would be developed to alleviate the problem. When people could not handle keeping records on stone paper scrolls were invented. When scrolls were no longer practical books were developed. When hand-copying books was too much the printing press came along. By the mid 20th century there were concerns that we would not be able to have libraries large enough to keep all of the books that were being developed. With this problem came the solution of the computer. One computer could hold the information of several dozen if not hundreds of libraries. Now even a single computer can no longer cope with all of the information that is constantly being developed for just a single subject. This has lead to computers working together in networks to share the task of storing information. With data spread across several computers it makes analyzing data much more challenging. It was now necessary to mine for useful information in a way that people used to mine for gold in the 19th century. Big data is data that is too large to fit within the memory of a single computer. Analyzing data that is spread across a network of databases takes skills different from traditional statistical analysis. This post will explain some of the characteristics of big data as well as data mining. Big Data Traits The three main traits of big data are volume, velocity, and variety. Volume describes the size of big data, which means data to big to be on only one computer. Velocity is about how fast the data can be processed. Lastly, variety different types of data. common sources of big data includes the following • Metadata from visual sources such as cameras • Data from sensors such as in medical equipment • Social media data such as information from google, youtube or facebook Data Mining Data mining is the process of discovering a model in a big dataset. Through the development of an algorithm, we can find specific information that helps us to answer our questions. Generally, there are two ways to mine data and these are extraction and summarization. Extraction is the process of pulling specific information from a big dataset. For example, if we want all the addresses of people who bought a specific book from Amazon the result would be an extraction from a big data set. Summarization is reducing a dataset to describe it. We might do a cluster analysis in which similar data is combine on a characteristic. For example, if we analyze all the books people ordered through Amazon last year we might notice that one cluster of groups buys mostly religious books while another buys investment books. Conclusion Big data will only continue to get bigger. Currently, the response has been to just use more computers and servers. As such, there is now a need for finding information across many computers and servers. This is the purpose of data mining, which is to find pertinent information that answers stakeholders questions. # Random Forest in R Random Forest is a similar machine learning approach to decision trees. The main difference is that with random forest. At each node in the tree, the variable is bootstrapped. In addition, several different trees are made and the average of the trees are presented as the results. This means that there is no individual tree to analyze but rather a ‘forest’ of trees The primary advantage of random forest is accuracy and prevent overfitting. In this post, we will look at an application of random forest in R. We will use the ‘College’ data from the ‘ISLR’ package to predict whether a college is public or private Preparing the Data First we need to split our data into a training and testing set as well as load the various packages that we need. We have run this code several times when using machine learning. Below is the code to complete this. library(ggplot2);library(ISLR) library(caret) data("College") forTrain<-createDataPartition(y=College$Private, p=0.7, list=FALSE)
trainingset<-College[forTrain, ]
testingset<-College[-forTrain, ]

Develop the Model

Next, we need to setup the model we want to run using Random Forest. The coding is similar to that which is used for regression. Below is the code

Model1<-train(Private~Grad.Rate+Outstate+Room.Board+Books+PhD+S.F.Ratio+Expend, data=trainingset, method='rf',prox=TRUE)

We are using 7 variables to predict whether a university is private or not. The method is ‘rf’ which stands for “Random Forest”. By now, I am assuming you can read code and understand what the model is trying to to. For a refresher on reading code for a model please click here.

If you type “Model1” followed by pressing enter, you will receive the output for the random forest

Random Forest

545 samples
17 predictors
2 classes: 'No', 'Yes'

No pre-processing
Resampling: Bootstrapped (25 reps)
Summary of sample sizes: 545, 545, 545, 545, 545, 545, ...
Resampling results across tuning parameters:

mtry  Accuracy   Kappa      Accuracy SD  Kappa SD
2     0.8957658  0.7272629  0.01458794   0.03874834
4     0.8969672  0.7320475  0.01394062   0.04050297
7     0.8937115  0.7248174  0.01536274   0.04135164

Accuracy was used to select the optimal model using
the largest value.
The final value used for the model was mtry = 4.

Most of this is self-explanatory. The main focus is on the mtry, accuracy, and Kappa.

The shows several different models that the computer generated. Each model reports the accuracy of the model as well as the Kappa. The accuracy states how well the model predicted accurately whether a university was public or private. The kappa shares the same information but it calculates how well a model predicted while taking into account chance or luck. As such, the Kappa should be lower than the accuracy.

At the bottom of the output, the computer tells whech mtry was the best. For our example, the best mtry was number 4. If you look closely, you will see that mtry 4 had the highest accuracy and Kappa as well.

Confusion Matrix for the Training Data

Below is the confusion matrix for the training data using the model developed by the random forest. As you can see, the results are different from the random forest output. This is because this model is predicting without bootstrapping

> predNew<-predict(Model1, trainingset)

> trainingset$predRight<-predNew==trainingset$Private

> table(predNew, trainingset$Private) predNew No Yes No 149 0 Yes 0 396 Results of the Testing Data We will now use the testing data to check the accuracy of the model we developed on the training data. Below is the code followed by the output pred <- predict(Model1, testingset) testingset$predRight<-pred==testingset$Private table(pred, testingset$Private)
pred   No Yes
No   48  11
Yes  15 158



For the most part, the model we developed to predict if a university is private or not is accurate.

How Important is a Variable

You can calculate how important an individual variable is in the model by using the following code

Model1RF<-randomForest(Private~Grad.Rate+Outstate+Room.Board+Books+PhD+S.F.Ratio+Expend, data=trainingset, importance=TRUE)
importance(Model1RF)

The output tells you how much the accuracy of the model is reduce if you remove the variable. As such, the higher the number the more valuable the variable is in improving accuracy.

Conclusion

This post exposed you to the basics of random forest. Random forest is a technique that develops a forest of decisions trees through resampling. The results of all these trees are then averaged to give you an idea of which variables are most useful in prediction.

# Decision Trees in R

Decisions tress are useful for splitting data based into smaller distinct groups based on a a criteria you establish. This post will attempt to explain how to develop decision trees in R.

We are going to use the ‘College’ dataset found in the “ISLR” package. Once you load the package you need to split the data into a training and testing set as shown in the code below. We want to divide the data based on education level, age, and income

library(ISLR); library(ggplot2); library(caret)
data("College")
inTrain<-createDataPartition(y=College$education, p=0.7, list=FALSE) trainingset <- College[inTrain, ] testingset <- College[-inTrain, ] Visualize the Data We will now make a plot of the data based on education as the groups and age and wage as the x and y variable. Below is the code followed by the plot. Please note that education is divided into 5 groups as indicated in the chart. qplot(age, wage, colour=education, data=trainingset) Create the Model We are now going to develop the model for the decision tree. We will use age and wage to predict education as shown in the code below. TreeModel<-train(education~age+income, method='rpart', data=trainingset) Create Visual of the Model We now need to create a visual of the model. This involves installing the package called ‘rattle’. You can install ‘rattle’ separately yourself. After doing this, below is the code for the tree model followed by the diagram. fancyRpartPlot(TreeModel$finalModel)

Here is what the chart means

1. At the top is node 1 which is called ‘HS Grad” the decimals underneath is the percentage of the data that falls within the “HS Grad” category. As the highest node, everything is classified as “HS grad” until we begin to apply our criteria.
2. Underneath nod 1 is a decision about wage. If a person makes less than 112 you go to the left if the make more you go to the right.
3. Nod 2 indicates the percentage of the sample that was classified as HS grade regardless of education. 14% of those with less than a HS diploma were classified as a HS Grade based on wage. 43% of those with a HS diploma were classified as a HS grade based on income. The percentage underneath the decimals indicates the total amount of the sample placed in the HS grad category. Which was 57%.
4. This process is repeated for each node until the data is divided as much as possible.

Predict

You can predict individual values in the data set by using the ‘predict’ function with the test data as shown in the code below.

predict(TreeModel, newdata = testingset)

Conclusion

Prediction Trees are a unique feature in data analysis for determining how well data can be divided into subsets. It also provides a visual of how to move through data sequentially based on characteristics in the data.

# Z-Scores

A z-score indicates how closely related one given score is to mean of the sample. Extremely high or low z-scores indicates that the given data point is unusually above or below the mean of the sample.

In order to understand z-scores you need to be familiar with distribution. In general, data is distributed in a bell shape curve. With the mean being the exact middle of the graph as shown in the picture below.

The Greek letter μ is the mean. In this post, we will go through an example that will try to demonstrate how to use and interpret the z-score. Notice that a z-score + 1 takes of 68% of the potential values a z-score + 2 takes of 95%, a z-score + 3 takes of 99%.

Imagine you know the average test score of students on a quiz. The average is 75%. with a standard deviation of 6.4%. Below is the equation for calculating the z-score.

Let’s say that one student scored 52% on the quiz. We can calculate the likelihood for this data point by using the formula above.

(52 – 75) / 6.4 = -3.59

Our value is negative which indicates that the score is below the mean of the sample. Are score is very exceptionally low from the mean. This makes sense given that the mean is 75% and the standard deviation is 6.4%. To get a 52% on the quiz was really bad performance.

We can convert the z-score to a percentage to indicate the probability of get such a value. To do this you would need to find a z-score conversion table on the internet. A quick glance at the table will show you that the probability of getting a score of 52 on the quiz is less than 1%.

Off course, this is based on the average score of 75% with a standard deviation of 6.4%. A different average and standard deviation would change the probability of getting a 52%.

Standardization

Z-scores are also used to standardize a variable. If you look at our example, the original values were in percentages. By using the z-score formula we converted these numbers into a different value. Specifically, the values of a z-score represent standard deviations from the mean.

In our example, we calculated a z-score of -3.59. In other words, the person who scored 52% on the quiz had  a score 3.59 standard deviations below the mean. When attempting to interpret data the z-score is a foundational piece of information that is used extensively in statistics.

# Using Regression for Prediction in R

In the last post about R, we looked at plotting information to make predictions. We will now look at an example of making predictions using regression.

We will used the same data as last time with the help of the ‘caret’ package as well. The code below sets up the seed and the training and testing sets we need.

> library(caret); library(ISLR); library(ggplot2)
> data("College");set.seed(1)
> PracticeSet<-createDataPartition(y=College$Grad.Rate, + p=0.5, list=FALSE) > TrainingSet<-College[PracticeSet, ]; TestingSet<- + College[-PracticeSet, ] > head(TrainingSet) The code above should look familiar from previous post. Make the Scatterplot We will now create the scatterplot showing the relationship between “S.F. Ratio” and “Grad.Rate” with the code below and the scatterplot. > plot(TrainingSet$S.F.Ratio, TrainingSet$Grad.Rate, pch=5, col="green", xlab="Student Faculty Ratio", ylab="Graduation Rate") Here is what we did 1. We used the ‘plot’ function to make this scatterplot. The x variable was ‘S.F.Ratio’ of the ‘TrainingSet’ the y variable was ‘Grad.Rate’. 2. We picked the type of dot to use using the ‘pch’ argument and choosing ’19’ 3. Next we chose a color and labeled each axis Fitting the Model We will now develop the linear model. This model will help us to predict future models. Furthermore, we will compare the model of the Training Set with the Test Set. Below is the code for developing the model. > TrainingModel<-lm(Grad.Rate~S.F.Ratio, data=TrainingSet) > summary(TrainingModel)  How to interpret this information was presented in a previous post. However, to summarize, we can say that when the student to faculty ratio increases one the graduation rate decreases 1.29. In other words, an increase in the student to faculty ratio leads to decrease in the graduation rate. Adding the Regression Line to the Plot Below is the code for adding the regression line followed by the scatterplot > plot(TrainingSet$S.F.Ratio, TrainingSet$Grad.Rate, pch=19, col="green", xlab="Student Faculty Ratio", ylab="Graduation Rate") > lines(TrainingSet$S.F.Ratio, TrainingModel$fitted, lwd=3) Predicting New Values With are model complete we can now predict values. For our example, we will only predict one value. We want to know what the graduation rate would be if we have a student to faculty ratio of 33. Below is the code for this with the answer > newdata<-data.frame(S.F.Ratio=33) > predict(TrainingModel, newdata) 1 40.6811 Here is what we did 1. We made a variable called ‘newdata’ and stored a data frame in it with a variable called ‘S.F.Ratio’ with a value of 33. This is x value 2. Next, we used the ‘predict’ function from the ‘caret’ package to determine what the graduation rate would be if the student to faculty ratio is 33. To do this we told caret to use the ‘TrainingModel’ we developed using regression and to run this model with the information in the ‘newdata’ dataframe 3. The answer was 40.68. This means that if the student to faculty ratio is 33 at a university then the graduation rate would be about 41%. Testing the Model We will now test the model we made with the training set with the testing set. First, we will make a visual of both models by using the “plot” function. Below is the code follow by the plots. par(mfrow=c(1,2)) plot(TrainingSet$S.F.Ratio,
TrainingSet$Grad.Rate, pch=19, col=’green’, xlab=”Student Faculty Ratio”, ylab=’Graduation Rate’) lines(TrainingSet$S.F.Ratio,  predict(TrainingModel), lwd=3)
plot(TestingSet$S.F.Ratio, TestingSet$Grad.Rate, pch=19, col=’purple’,
xlab=”Student Faculty Ratio”, ylab=’Graduation Rate’)
lines(TestingSet$S.F.Ratio, predict(TrainingModel, newdata = TestingSet),lwd=3) In the code, all that is new is the “par” function which allows us to see to plots at the same time. We also used the ‘predict’ function to set the plots. As you can see, the two plots are somewhat differ based on a visual inspection. To determine how much so, we need to calculate the error. This is done through computing the root mean square error as shown below. > sqrt(sum((TrainingModel$fitted-TrainingSet$Grad.Rate)^2)) [1] 328.9992 > sqrt(sum((predict(TrainingModel, newdata=TestingSet)-TestingSet$Grad.Rate)^2))
[1] 315.0409

The main take away from this complicated calculation is the number 328.9992 and 315.0409. These numbers tell you the amount of error in the training model and testing model. The lower the number the better the model. Since the error number in the testing set is lower than the training set we know that our model actually improves when using the testing set. This means that our model is beneficial in assessing graduation rates. If there were problems we may consider using other variables in the model.

Conclusion

This post shared ways to develop a regression model for the purpose of prediction and for model testing.

# Simple Prediction

Prediction is one of the key concepts of machine learning. Machine learning is a field of study that is focused on the development of algorithms that can be used to make predictions.

Anyone who has shopped online at has experienced machine learning. When you make a purchase at an online store, the website will recommend additional purchases for you to make. Often these recommendations are based on whatever you have purchased or whatever you click on while at the site.

There are two common forms of machine learning, unsupervised and supervised learning. Unsupervised learning involves using data that is not cleaned and labeled and attempts are made to find patterns within the data. Since the data is not labeled, there is no indication into what is right or wrong

Supervised machine learning is using cleaned and properly labeled data. Since the data is labeled there is some form of indication whether the model that is developed is accurate or not. If the is incorrect then you need to make adjustments to it. In other words, the model learns based on its ability to accurately predict results. However, it is up to the human to make adjustments to the model in order to improve the accuracy of it.

In this post we will look at using R for supervised machine learning. The definition presented so far will make more sense with an example.

The Example

We are going to make a simple prediction about whether emails are spam or not using data from kern lab.

The first thing that you need to do is to install and load the “kernlab” package using the following code

install.packages("kernlab")
library(kernlab)

If you use the “View” function to examine the data you will see that there are several columns. Each column tells you the frequency of a word that kernlab found in a collection of emails. We are going to use the word/variable “money” to predict whether an email is spam or not. First we need to plot the density of the use of the word “money” when the email was not coded as spam. Below is the code for this.

plot(density(spam$money[spam$type=="nonspam"]),
col='blue',main="", xlab="Frequency of 'money'")



This is an advance R post so I am assuming you can read the code. The plot should look like the following.

As you can see, money is not used to frequently in emails  that are not spam in this dataset. However, you really cannot say this unless you compare the times ‘money’ is labeled nonspam to the times that it is labeled spam. To learn this we need to add a second line that explains to us when the word ‘money’ is used and classified as spam. The code for this is below with the prior code included.

plot(density(spam$money[spam$type=="nonspam"]),
col='blue',main="", xlab="Frequency of 'money'")
lines(density(spam$money[spam$type=="spam"]), col="red")

Your new plot should look like the following

If you look closely at the plot doing a visual inspection, where there is a separation between the blue line for non spam and the red line for spam is the cutoff point for whether an email is spam or not. In other words, everything inside the arc is labeled correctly while the information outside the arc is not.

The next code and graph show that this cutoff point is around 0.1. This means that any email that has on average more than 0.1 frequency of the word ‘money’ is spam. Below is the code and the graph with the cutoff point indicated with a black line.

plot(density(spam$money[spam$type=="nonspam"]),
col='blue',main="", xlab="Frequency of 'money'")
lines(density(spam$money[spam$type=="spam"]), col="red")
abline(v=0.1, col="black", lw= 3)

Now we need to calculate the accuracy of the use of the word ‘money’ to predict spam. For our current example we will simply use in “ifelse” function. If the frequency is greater than 0.1.

We then need to make a table to see the results. The code for the “ifelse” function and the table are below followed by the table.

predict<-ifelse(spam$money > 0.1, "spam","nonspam") table(predict, spam$type)/length(spam$type) predict nonspam spam nonspam 0.596392089 0.266898500 spam 0.009563138 0.127146273 Based on the table that I am assuming you can read, our model accurately calculates that an email is spam about 71% (0.59 + 0.12) of the time based on the frequency of the word ‘money’ being greater than 0.1. Off course, for this to be true machine learning we would repeat this process by trying to improve the accuracy of the prediction. However, this is an adequate introduction to this topic. # Survey Design Survey design is used to describe the opinions, beliefs, behaviors, and or characteristics of a population based on the results of a sample. This design involves the use of surveys that include questions, statements, and or other ways of soliciting information from the sample. This design is used for descriptive purpose primarily but can be combined with other designs (correlational, experimental) at times as well. In this post we will look at the following. • Types of Survey Design • Characteristics of Survey Design Types of Survey Design There are two common forms of survey design which are cross-sectional and longitudinal. A cross-sectional survey design is the collection of data at one specific point in time. Data is only collected once in a cross-sectional design. A cross-sectional design can be used to measure opinions/beliefs, compare two or more groups, evaluate a program, and or measure the needs of a specific group. The main goal is to analyze the data from a sample at a given moment in time. A longitudinal design is similar to a cross-sectional design with the difference being that longitudinal designs require collection over time.Longitudinal studies involve cohorts and panels in which data is collected over days, months, years, and even decades. Through doing this, a longitudinal study is able to expose trends over time in a sample. Characteristics of Survey Design There are certain traits that are associated with survey design. Questionnaires and interviews are a common component of survey design. The questionnaires can happen by mail, phone, internet, and in person. Interviews can happen by phone, in focus groups, or one-on-one. The design of a survey instrument often includes personal, behavioral and attitudinal questions, and open/closed questions. Another important characteristic of survey design is monitoring the response rate. The response rate is the percentage of participants in the study compared to the number of surveys that were distributed. The response rate varies depending on how the data was collected. Normally, personal interviews have the highest rate while email request have the lowest. It is sometimes necessary to report the response rate when trying to publish. As such, you should at the very least be aware of what the rate is for a study you are conducting. Conclusion Surveys are used to collected data at one point in time or over time. The purpose of this approach is develop insights into the population in order to describe what is happening or to be used to make decisions and inform practice. # Simple Regression in R In this post, we will look at how to perform a simple regression using R. We will use a built-in dataset in R called ‘mtcars.’ There are several variables in this dataset but we will only use ‘wt’ which stands for the weight of the cars and ‘mpg’ which stands for miles per gallon. Developing the Model We want to know the association or relationship between weight of a car and miles per gallon. We want to see how much of the variance of ‘mpg’ is explained by ‘wt’. Below is the code for this. > Carmodel <- lm(mpg ~ wt, data = mtcars) Here is what we did 1. We created the variable ‘Carmodel’ to store are information 2. Inside this variable, we used the ‘lm’ function to tell R to make a linear model. 3. Inside the function we put ‘mpg ~ wt’ the ‘~’ sign means ’tilda’ in English and is used to indicate that ‘mpg’ is a function of ‘wt’. This section is the actually notation for the regression model. 4. After the comma, we see ‘data = mtcars’ this is telling R to use the ‘mtcar’ dataset. Seeing the Results Once you pressed enter you probably noticed nothing happens. The model ‘Carmodel’ was created but the results have not been displayed. Below is various information you can extract from your model. The ‘summary’ function is useful for pulling most of the critical information. Below is the code with the output. > summary(Carmodel) Call: lm(formula = mpg ~ wt, data = mtcars) Residuals: Min 1Q Median 3Q Max -4.5432 -2.3647 -0.1252 1.4096 6.8727 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 37.2851 1.8776 19.858 < 2e-16 *** wt -5.3445 0.5591 -9.559 1.29e-10 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 3.046 on 30 degrees of freedom Multiple R-squared: 0.7528, Adjusted R-squared: 0.7446 F-statistic: 91.38 on 1 and 30 DF, p-value: 1.294e-10 We are not going to explain the details of the output (please see simple regression). The results indicate a model that explains 75% of the variance or has an r-square of 0.75. The model is statistically significant and the equation of the model is -5.45x + 37.28 = y Plotting the Model We can also plot the model using the code below > coef_Carmodel<-coef(Carmodel) > plot(mpg ~ wt, data = mtcars) > abline(a = coef_Carmodel[1], b = coef_Carmodel[2]) Here is what we did 1. We made the variable ‘coef_Carmodel’ and stored the coefficients (intercept and slope) of the ‘Carmodel’ using the ‘coef’ function. We will need this information soon. 2. Next, we plot the ‘mtcars’ dataset using ‘mpg’ and ‘wt’. 3. To add the regression line we use the ‘abline’ function. To add the intercept and slope we use a = the intercept from our ‘coef_Carmodel’ variable which is subset [1] from this variable. For slope we following the same process but use a [2]. This will add the line and your graph should look like the following. From the visual, you can see that as weight increases there is a decrease in miles per gallon. Conclusion R is capable of much more complex models than the simple regression used here. However, understanding the coding for simple modeling can help in preparing you for much more complex forms of statistical modeling. # Experimental Design: Within Groups Within groups experimental designs is the use of only one group in an experiment. This is in contrast to a between group design which involves two or more groups. Within group design is useful when the number of is two low in order to split them into different groups. There are two common forms of within group experimental design, time series and repeated measures. Under time series there are interrupted times series and equivalent time series. Under repeated measure there is only repeated measure design. In this post we will look at the following forms of within group experimental design. • interrupted time series • equivalent time series • repeated measures design Interrupted Time Series Design Interrupted time series design involves several pre-test followed by an intervention and then several post-test of one group. By measuring the several times, many threats to internal validity are reduced, such as regression, maturation, and selection. The pre-test results are also used as covariates when analyzing the post-tests. Equivalent Time Series Design Equivalent time series design involves the use of a measurement followed by intervention followed by measurement etc. In many ways, this design is a repeated post-test only design. The primary goal is to plot the results of the post-test and determine if there is a pattern that develops over time. For example, if you are tracking the influence of blog writing on vocabulary acquisition, the intervention is blog writing and the dependent variable is vocabulary acquisition. As the students write a blog, you measure them several times over a certain period. If a plot indicates an upward trend you could infer that blog writing made a difference in vocabulary acquisition. Repeated Measures Repeated measures is the use of several different treatments over time. Before each treatment, the group is measured. Each post-test is compared to other post-test to determine which treatment was the best. For example, lets say that you still want to assess vocabulary acquisition but want to see how blog writing and public speaking affect it. First, you measure vocabulary acquisition. Next, you employ the first intervention followed by a second assessment of vocabulary acquisition. Third, you use the public speaking intervention followed by a third assessment of vocabulary acquisition. You now have three parts of data to compare 1. The first assessment of vocabulary acquisition (a pre-test) 2. The second assessment of vocabulary acquisition (post-test 1 after the blog writing) 3. The third assessment of vocabulary acquisition (post-test 2 after the public speaking) Conclusion Within group experimental designs are used when it is not possible to have several groups in an experiment. The benefits include needing fewer participants. However, one problem with this approach is the need to measure several times which can be labor intensive. # ANOVA with R Analysis of variance (ANOVA) is used when you want to see if there is a difference in the means of three or more groups due to some form of treatment(s). In this post, we will look at conducting an ANOVA calculation using R. We are going to use a dataset that is already built in to R called ‘InsectSprays.’ This dataset contains information on different insecticides and their ability to kill insects. What we want to know is which insecticide was the best at killing insects. In the dataset ‘InsectSprays’, there are two variables, ‘count’, which is the number of dead bugs, and ‘spray’ which is the spray that was used to kill the bug. For the ‘spray’ variable there are six types label A-F. There are 72 total observations for the six types of spray which comes to about 12 observations per spray. Building the Model The code for calculating the ANOVA is below > BugModel<- aov(count ~ spray, data=InsectSprays) > BugModel Call: aov(formula = count ~ spray, data = InsectSprays) Terms: spray Residuals Sum of Squares 2668.833 1015.167 Deg. of Freedom 5 66 Residual standard error: 3.921902 Estimated effects may be unbalanced Here is what we did 1. We created the variable ‘BugModel’ 2. In this variable, we used the function ‘aov’ which is the ANOVA function. 3. Within the ‘aov’ function we told are to determine the count by the difference sprays that is what the ‘~’ tilda operator does. 4. After the comma, we told R what data set to use which was “InsectSprays.” 5. Next we pressed ‘enter’ and nothing happens. This is because we have to make R print the results by typing the name of the variable “BugModel” and pressing ‘enter’. 6. The results do not tell us anything to useful yet. However, now that we have the ‘BugModel’ saved we can use this information to find the what we want. Now we need to see if there is any significant results. To do this we will use the ‘summary’ function as shown in the script below > summary(BugModel) Df Sum Sq Mean Sq F value Pr(>F) spray 5 2669 533.8 34.7 <2e-16 *** Residuals 66 1015 15.4 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 These results indicate that there are significant results in the model as shown by the p-value being essentially zero (Pr(>F)). In other words, there is at least one mean that is different from the other means statistically. We need to see what the means are overall for all sprays and for each spray individually. This is done with the following script > model.tables(BugModel, type = 'means') Tables of means Grand mean 9.5 spray spray A B C D E F 14.500 15.333 2.083 4.917 3.500 16.667 The ‘model.tables’ function tells us the means overall and for each spray. As you can see, it appears spray F is the most efficient at killing bugs with a mean of 16.667. However, this table does not indicate the statistically significance. For this we need to conduct a post-hoc Tukey test. This test will determine which mean is significantly different from the others. Below is the script > BugSpraydiff<- TukeyHSD(BugModel) > BugSpraydiff Tukey multiple comparisons of means 95% family-wise confidence level Fit: aov(formula = count ~ spray, data = InsectSprays)$spray
diff        lwr       upr     p adj
B-A   0.8333333  -3.866075  5.532742 0.9951810
C-A -12.4166667 -17.116075 -7.717258 0.0000000
D-A  -9.5833333 -14.282742 -4.883925 0.0000014
E-A -11.0000000 -15.699409 -6.300591 0.0000000
F-A   2.1666667  -2.532742  6.866075 0.7542147
C-B -13.2500000 -17.949409 -8.550591 0.0000000
D-B -10.4166667 -15.116075 -5.717258 0.0000002
E-B -11.8333333 -16.532742 -7.133925 0.0000000
F-B   1.3333333  -3.366075  6.032742 0.9603075
D-C   2.8333333  -1.866075  7.532742 0.4920707
E-C   1.4166667  -3.282742  6.116075 0.9488669
F-C  14.5833333   9.883925 19.282742 0.0000000
E-D  -1.4166667  -6.116075  3.282742 0.9488669
F-D  11.7500000   7.050591 16.449409 0.0000000
F-E  13.1666667   8.467258 17.866075 0.0000000

There is a lot of information here. To make things easy, wherever there is a p adj of less than 0.05 that means that there is a difference between those two means. For example, bug spray F and E have a difference of 13.16 that has a p adj of zero. So these two means are really different statistically.This chart also includes the lower and upper bounds of the confidence interval.

The results can also be plotted with the script below

> plot(BugSpraydiff, las=1)

Below is the plot

Conclusion

ANOVA is used to calculate if there is a difference of means among three or more groups. This analysis can be conducted in R using various scripts and codes.