Binomial Logistic Regression.


I’m officially a Kaggler.

Cut to the good ol’ Titanic challenge. Ol’ is right – It’s been running since 2012 and ends in 3 months! I showed up late to the party. Oh well, I guess it’s full steam ahead from now on.

The competition  ‘Machine Learning from Disaster’ asks you to apply machine learning to analyse and predict  which passengers survived the Titanic tragedy. It is placed as knowledge competition.

Since I am still inching my way up the learning curve, I tried to see what I could achieve with my current tool set. For my very quick first attempt, it seemed like a no-brainer to apply out of the box logistic regression. For those in the competition, this approach, got me at around 75.something% and  placed me at 4,372 of 5172 entries. I have 3 months to better this score. And better it, I shall!

So essentially how this works is that you download the data from Kaggle. 90% of it (889 rows)  is flagged as training data and the rest is test data(418 rows). You need to build your model, predict survival on the test set and pass the data back to Kaggle which computes a score for you and places you accordingly on the ‘Leaderboard’.

The data:

Since we’re working with real world data, we’ll need to take into account the NAs, improper formatting, missings values et al.

After reading in the data, I ran a check to see how many entries had missing values. The simple sapply() sufficed in this case.

dat<-read.csv("C:/Users/Amita/Downloads/train (1).csv",header=T,sep=","
      ,na.strings = c(""))

The column ‘Cabin’ seems to have the most missing values – like a LOT, so I ditched it. ‘Age’ had quite a few missing values as well , but it seems like a relevant column.

I went ahead and dealt with the missing values by replacing them with the mean of the present values in that column.Easy peasy.

dat[$Age),][6]<- mean(dat$Age,na.rm=T)
dat <- dat[,-11]


Next, I divided the training dataset into two – ‘traindat’ and ‘testdat’. The idea was to train the prospective model on the traindat dataset and the predict using the rows in testdat. Computing the RMSE would then give us an idea about the performance of the model.

indices<- rnorm(nrow(dat))>0
traindat<- dat[indices,]


Structure wise, except for a couple of columns that had to be converted into factors,  the datatypes were on point.

testdat$Pclass<- factor(testdat$Pclass)
testdat$SibSp <- factor(testdat$SibSp)
testdat$Parch <- factor(testdat$Parch)
traindat$Pclass<- factor(traindat$Pclass)
traindat$SibSp <- factor(traindat$SibSp)
traindat$Parch <- factor(traindat$Parch)


The model:

Since, the response variable is a categorical variable with only two outcomes, and the predictors are both continuous and categorical, it makes it a candidate for conducting binomial logistic regression.

mod <- glm(Survived ~ Pclass + Sex + Age + SibSp+ Parch + Embarked + Fare ,



The result shows significance values for only ‘Pclass’, ‘Sex’, ‘Age’, ‘SibSp’ , so we’ll build a second model with just these variables and use that for further analysis.

mod2 <- glm(Survived ~ Pclass + Sex + Age + SibSp ,

Let’s visualize the relationships between the response and predictor variables.


  • The log odds (since we used link= ‘logit’) of survival seems to decline as the passenger’s class decreases.
  • Women have a higher log odds of survival than men.
  • Higher the age gets, lower the log odds of survival get.
  • The number of siblings/spouses aboard* also affects the log odds of survival. The log odds for numbers 5 and 8 can go either way indicated by the wide CIs. (*needs to be explored more).


Model Performance:

We will test the peformance of mod2 in predicting ‘Survival’ on a new set of data.

pred <- predict(mod2,newdata=testdat[,-2],type="response")
pred <- ifelse(pred > 0.5,1,0)
testdat$Survived <- as.numeric(levels(testdat$Survived))[testdat$Survived]
rmse <- sqrt((1/nrow(testdat)) * sum( (testdat$Survived - pred) ^ 2))
rmse  #0.4527195

error <- mean(pred != testdat$Survived)
print(paste('Accuracy',1-error)) #81% accuracy

81.171% passenger has been correctly classified.

But when I used the same model to run predictions on Kaggle’s test dataset, the uploaded results fetched me a 75.86 %. I’m guessing the reason could be the arbit split ratio between the ‘traindat’ and ‘testdat’. Maybe next time I’ll employ some sort of bootstrapping.

Well, this is pretty much it for now. I will attempt to better my score in the upcoming weeks (time permitting) and in the event I am successful, I shall add subsequent blog posts and share my learnings (Is that even a word?!).

One thing’s for sure though –  This road is loooooong, is long , is long . . . . .









Linear mixed-effects model with bootstrapping.


Dataset here.

We are going to perform a linear mixed effects analysis of the relationship between height and treatment of trees, as studied over a period of time.

Begin by reading in the data and making sure that the variables have the appropriate datatype.

tree<- read.csv(path, header=T,sep=",")
tree$ID<- factor(tree$ID)
tree$plotnr <- factor(tree$plotnr)

Plot the initial graph to get a sense of the relationship between the variables.

bwplot(height ~ Date|treat,data=tree)


There seems to be a linear-ish relationship between height and time . The height increases over time.The effect of the treatments(I,IL,C,F) is not clear at this point.

Based on the linear nature of relationship, we can model time to be numeric, i.e. , number of days.

tree$Date<- as.Date(tree$Date)
tree$time<- as.numeric(tree$Date - min(tree$Date))

The data shows us that the data was collected from the same plot on multiple dates, making it  a candidate for repeated measures.

Let’s visually check for the by-plot variability.

boxplot(height ~ plotnr,data=tree)


The variation between plots isn’t big – but there are still noticeable differences which will need to be accounted for in our model.




Let’s fit two models : one with interaction between the predictors and one without.


#fit a model

mod <- lmer(height ~ time + treat + (1|plotnr),data=tree) #without interaction <- lmer(height ~ time * treat + (1|plotnr),data=tree) #with interaction

Upon comparing the two,  it’s evident that the model with the interaction would perform better at predicting responses.

> anova(mod,
refitting model(s) with ML (instead of REML)
Data: tree
object: height ~ time + treat + (1 | plotnr)
..1: height ~ time * treat + (1 | plotnr)
       Df AIC BIC logLik deviance Chisq Chi Df Pr(>Chisq) 
object 7 22654 22702 -11320 22640 
..1   10 20054 20122 -10017 20034 2605.9 3 < 2.2e-16 ***
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

So far, we have a random intercept model, we should also check whether adding a random slope can help make the model more efficient. We do this by adding a random slope to the model and comparing that with the  random intercept model from before. <- lmer(height ~ time * treat + (treat|plotnr),data=tree)

> anova(, # we don't need the random slope
refitting model(s) with ML (instead of REML)
Data: tree
object: height ~ time * treat + (1 | plotnr)
..1: height ~ time * treat + (treat | plotnr)
 Df AIC BIC logLik deviance Chisq Chi Df Pr(>Chisq)
object 10 20054 20122 -10017 20034 
..1    19 20068 20197 -10015 20030 3.6708 9 0.9317 #AIC goes up

Nope. Don’t need the slope. Dropping it.

Continuing with Let’s determine the p-values using Anova() from the car package.


We can see that the main effects of time and treat are highly significant, as is the interaction between the two.

Running a summary() spews out the following log.


As seen,  time:treatF is not  significantly different from the first level , time:treatC , that is, there is no difference between F treatment and C treatment in terms of the interaction with time.Treatments I and IL are significantly different from treatment C in terms of interaction with time.Although there is a significant main effect of treatment, none of the levels are actually different from the first level.

Graphically represented, the affect of time on height, by treatment would like this.



Let’s check the data for adherence to model assumptions.

The plot of residuals versus fitted values doesn’t reflect any obvious pattern in the residuals.So, the assumption of linearity has not been violated and looking at the ‘blob’-like nature of the plot suggests the preservation of Homoskedasticity (One day, I shall pronounce this right).



We’ll check for the assumption of normality of residuals by plotting a histogram. Looks okay to me.


There are several methods for dealing with multicollinearity. In this case, we’ll follow the simplest one, which is to consult the correlation matrix from the output of summary( from before. Since none of the correlations between predictors approach 1, we can say that the assumption of  multicollinearity has not been violated.


Moving on.

Bootstrapping is a great resampling technique to arrive at CIs when dealing with mixed effect models where the degree of  various output from them is not clearly known.While one would want to aim  for as high a sampling number as possible to get tighter CIs , we also need to make allowance for longer associated processing times.

So here, we want predictions for all values of height. We will create a test dataset, newdat, which will go as input to predict().

newdat <- subset(tree,treat=="C")
bb <- bootMer(, FUN=function(x)predict(x, newdat, re.form=NA),

#extract the quantiles and the fitted values.
lci <- apply(bb$t, 2, quantile, 0.025)   
uci <- apply(bb$t, 2, quantile, 0.975)   
pred <- predict(,newdat,re.form=NA)


Plotting the confidence intervals.







Using R and lme4 (Bates, Maechler & Bolker, 2012) We performed a linear mixed effects analysis of the relationship between height and treatment of trees, as studied over a period of time. As fixed effects, we entered time and treatment  (with an interaction term) into the model. As random effects, we had intercepts for plotnr (plot numbers). Visual inspection of residual plots did not reveal any obvious deviations from homoscedasticity or normality. P-values were obtained by likelihood ratio tests via anova of the full
model with the effect in question against the model without the effect in question.



. . ..  . and it’s a wrap,folks!

Thanks again for stopping by, and again, I’d be happy to hear your comments and suggestions. Help me make this content better! 🙂




Stepwise Regression – What’s not to like ?



Plenty, apparently.

Besides encouraging you not to think , it doesn’t exactly do a great job at what it claims to do. Given a set of predictors, there is no guarantee that stepwise regression will find the optimal combination. Many of my statisticians buddies , whom I consult from time to time,  have a  gripe with it because it’s not  sensitive to the context of the research. Seems fair.


I built an interactive Shiny app to evaluate results from Stepwise regression (direction = “backward) when applied to different predictors and datasets. What I observed during model building and cross-validation was that the model performed better on the data at hand but performs much worse when subjected to cross-validation.After a lot of different random selections and testing, I eventually did find a model that worked well on both the fitted dataset and the cross-validation set, but it performed poorly when applied to new data.Therein lies at least most of the problem.


Shiny app can be found here.

The initial model was built to predict the ‘Life Expectancy’ and it does, to a certain extent, do it’s job . But when generalized, it pretty much turned out to be a bit of an uncertainty ridden damp squib. For example, predictions for the variables from the same dataset , such as, ‘Population’ ,’Frost’, ‘Area’ are nowhere close to the observed values. At the same time, the model did okay for variables such as ‘Illiteracy’ and ‘HS.Grad’.

Given all these drawbacks ( and more! ), people do find the motivation to use stepwise regression to produce a simpler model in terms of number of coefficients. It does not necessarily find the optimal model, but it does give a hunch of the possible combination of predictors.

While no one would conclude a statistical study based on stepwise results or publish a paper with it, some might find uses for it, say, to  verify models already created by software systems. Or as an easy-to-use tool for initial exploratory data analysis (with all the necessary caveats in place !) .

You win some, you lose some.

What do you think ? Leave a comment!

p.s. You can find the (needs-to-be-cleaned-up) code for the Shiny app here.




Multiple Regression (sans interactions) : A case study.



state.x77 – Standard built-in dataset with 50 rows and 8 columns giving the following statistics in the respective columns.

Populationpopulation estimate as of July 1, 1975

Incomeper capita income (1974)

Illiteracy: illiteracy (1970, percent of population)

Life Exp: life expectancy in years (1969–71)

Murder: murder and non-negligent manslaughter rate per 100,000 population (1976)

HS Grad: percent high-school graduates (1970)

Frost: mean number of days with minimum temperature below freezing (1931–1960) in capital or large city

Area: land area in square miles




Conduct multiple regression to arrive at an optimal model which is able to predict the life expectancy .



We begin this exercise by running a few standard diagnostics on the data including by not limited to  head(), tail() , class(), str(), summary() ,names() etc.

The input dataset turns out to be a matrix which will need to be coerced into a data frame using

Next,we’ll do a quick exploratory analysis on our data to examine the variables for outliers and distribution before proceeding. One way to do this is to plot qqplots for all the variables in the dataset.

for(i in 1:ncol(st)){



It’s a good idea to check the correlation between the variables to get a sense of the dependencies . We do this here using the cor(dataframe) function.


We can see that life expectancy appears to be moderately to strongly related to “Income” (positively), “Illiteracy” (negatively), “Murder” (negatively), “HS.Grad” (positively), and “Frost” (no. of days per year with frost, positively).


We can also view this by plotting our correlation matrix using pairs(dataframe)  which corroborates the cor.matrix  as seen above.


As a starting point for multiple regression would be to build a “full” model which will have Life Expectancy as the response variable and all other variables as explanatory variables. The summary of that linear model will be our “square one” and we will proceed to whittle it down until we reach our optimal model.

> model1 = lm(Life.Exp ~ ., data=st)
> summary(model1)

lm(formula = Life.Exp ~ ., data = st)

 Min         1Q     Median   3Q       Max 
-1.47514 -0.45887 -0.06352 0.59362 1.21823 

             Estimate Std. Error t value Pr(>|t|) 
(Intercept) 6.995e+01  1.843e+00  37.956  < 2e-16 ***
Population  6.480e-05  3.001e-05   2.159   0.0367 * 
Income      2.701e-04  3.087e-04   0.875   0.3867 
Illiteracy  3.029e-01  4.024e-01   0.753   0.4559 
Murder     -3.286e-01  4.941e-02  -6.652   5.12e-08 ***
HS.Grad     4.291e-02  2.332e-02    1.840  0.0730 . 
Frost      -4.580e-03  3.189e-03   -1.436  0.1585 
Area       -1.558e-06  1.914e-06   -0.814  0.4205 
Density    -1.105e-03  7.312e-04   -1.511  0.1385 
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.7337 on 41 degrees of freedom
Multiple R-squared: 0.7501, Adjusted R-squared: 0.7013 
F-statistic: 15.38 on 8 and 41 DF, p-value: 3.787e-10

We see that this 70% of the variation in Life Expectancy is explained by all the variables together (Adjusted R-squared).

Let’s see if this can be made better. We’ll try to do that by removing one predictor at a time from model1, starting with “Illiteracy” (p-value = 0.4559 i.e. least significance) , until we have a model with all significant predictors.

> model2 = update(model1, .~. -Illiteracy)
> summary(model2)

lm(formula = Life.Exp ~ Population + Income + Murder + HS.Grad + 
 Frost + Area + Density, data = st)

 Min 1Q Median 3Q Max 
-1.47618 -0.38592 -0.05728 0.58817 1.42334 

              Estimate Std. Error t value Pr(>|t|) 
(Intercept)  7.098e+01  1.228e+00   57.825   < 2e-16 ***
Population   5.675e-05  2.789e-05     2.034    0.0483 * 
Income       1.901e-04  2.883e-04     0.659    0.5133 
Murder       -3.122e-01 4.409e-02     -7.081    1.11e-08 ***
HS.Grad       3.652e-02 2.161e-02      1.690    0.0984 . 
Frost        -6.059e-03 2.499e-03      -2.425    0.0197 * 
Area         -8.638e-07 1.669e-06      -0.518     0.6075 
Density       -8.612e-04 6.523e-04       -1.320   0.1939 
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.7299 on 42 degrees of freedom
Multiple R-squared: 0.7466, Adjusted R-squared: 0.7044 
F-statistic: 17.68 on 7 and 42 DF, p-value: 1.12e-10


Next, Area must go (p-value = 0.6075!)


> model3 = update(model2, .~. -Area)
> summary(model3)

lm(formula = Life.Exp ~ Population + Income + Murder + HS.Grad + 
 Frost + Density, data = st)

 Min 1Q Median 3Q Max 
-1.49555 -0.41246 -0.05336 0.58399 1.50535 

            Estimate Std. Error t value Pr(>|t|) 
(Intercept)  7.131e+01 1.042e+00 68.420 < 2e-16 ***
Population   5.811e-05 2.753e-05 2.110 0.0407 * 
Income       1.324e-04 2.636e-04 0.502 0.6181 
Murder      -3.208e-01 4.054e-02 -7.912 6.32e-10 ***
HS.Grad      3.499e-02 2.122e-02 1.649 0.1065 
Frost       -6.191e-03 2.465e-03 -2.512 0.0158 * 
Density     -7.324e-04 5.978e-04 -1.225 0.2272 
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.7236 on 43 degrees of freedom
Multiple R-squared: 0.745, Adjusted R-squared: 0.7094 
F-statistic: 20.94 on 6 and 43 DF, p-value: 2.632e-11

We will continue to throw out the predictors in this fashion . . . . .

> model4 = update(model3, .~. -Income)
> summary(model4)
> model5 = update(model4, .~. -Density)
> summary(model5)

. . . .  until we reach this point

> summary(model5)

lm(formula = Life.Exp ~ Population + Murder + HS.Grad + Frost, 
 data = st)

 Min 1Q Median 3Q Max 
-1.47095 -0.53464 -0.03701 0.57621 1.50683 

              Estimate Std. Error t value Pr(>|t|) 
(Intercept)  7.103e+01 9.529e-01 74.542   < 2e-16 ***
Population   5.014e-05 2.512e-05 1.996     0.05201 . 
Murder      -3.001e-01 3.661e-02 -8.199    1.77e-10 ***
HS.Grad      4.658e-02 1.483e-02 3.142     0.00297 ** 
Frost       -5.943e-03 2.421e-03 -2.455    0.01802 * 
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.7197 on 45 degrees of freedom
Multiple R-squared: 0.736, Adjusted R-squared: 0.7126 
F-statistic: 31.37 on 4 and 45 DF, p-value: 1.696e-12

Now, here we have “Population” with a borderline p-value of 0.052 . Do we throw it out ? Do we keep it ? One way to know is to test for interactions.

> interact= lm(Life.Exp ~ Population * Murder * HS.Grad * Frost,data=st)
> anova(model5,interact)
Analysis of Variance Table

Model 1: Life.Exp ~ Population + Murder + HS.Grad + Frost
Model 2: Life.Exp ~ Population * Murder * HS.Grad * Frost
 Res.Df RSS Df Sum of Sq F Pr(>F) 
1 45 23.308 
2 34 14.235 11 9.0727 1.9699 0.06422 .
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

P-value of 0.06422 indicates that there is no significant interaction between the terms so we can drop “Population” as it doesn’t really contribute to anything at this point.BUT, if we drop “Population”, the adj R-squared drops a little as well . So as a trade-off, we’ll just keep it. 

So there you have it,folks.Our minimal optimal model with statistically significant slopes.


Run model diagnostics

par(mfrow=c(2,2)) # to view all 4 plots at once


1. Residuals vs Fitted

We find equally spread residuals around a horizontal line without distinct patterns so that is a that is a good indication we don’t have non-linear relationships.

2. Normal Q-Q plot

#This plot shows if residuals are normally distributed. Our residuals more or less follow a straight line well, so that’s an encouraging sign.

3. Scale-location
This plot shows if residuals are spread equally along the ranges of predictors. This is how we can check the assumption of equal variance (homoscedasticity). It’s good that we can see a horizontal(-ish) line with equally (randomly) spread points.

4. Residuals vs Leverage

This plot helps us locate the influential cases that have an effect on the regression line.We look for subjects outside the dashed line , the Cook’s distance.The regression results will be altered if we exclude those cases.
‘Full’ model VS ‘Optimal’ model

> anova(model1,model5)
Analysis of Variance Table

Model 1: Life.Exp ~ Population + Income + Illiteracy + Murder + HS.Grad + 
 Frost + Area + Density
Model 2: Life.Exp ~ Population + Murder + HS.Grad + Frost
 Res.Df RSS Df Sum of Sq F Pr(>F)
1 41 22.068 
2 45 23.308 -4 -1.2397 0.5758 0.6818

The p-value of 0.6818 suggests that we haven’t lost any significance from when we started, so we’re good.

Onto some cross-validation

We will perform a 10-fold cross validation (m=10 in CVlm function).

A comparison of the cross- validation plots shows that model5 performs the best out of all other models.The lines of best fit are relatively parallel(-ish) and most big symbols are close to the lines indicating a decent accuracy of prediction.Also, the mean squared error is small , 0.6 .



cvResults <- suppressWarnings(CVlm(data=st, form.lm=model5, m=10, dots=FALSE, seed=29, legend.pos="topleft", printit=FALSE));

mean_squared_error <- attr(cvResults, 'ms')



The relative importance of predictors

The relaimpo package provides measures of relative importance for each of the predictors in the model.


 # Bootstrap Measures of Relative Importance (1000 samples) 
 boot <- boot.relimp(model5, b = 1000, type = c("lmg", 
 "last", "first", "pratt"), rank = TRUE, 
 diff = TRUE, rela = TRUE)

# print result
 plot(booteval.relimp(boot,sort=TRUE)) # plot result


Yup. ‘Murder’ is the most important predictor of Life Expectancy, followed by ‘HS.Grad’ and ‘Frost. ‘Population’  sits smug at the other end of the spectrum with marginal effect and importance.

Well, looks like this is as good as it is going to get with this model.

I will leave you here, folks. Please feel free to work through various datasets (as I will) to get a firm grasp on things.

Did you find this article helpful?  Can it be improved upon ? Please leave me a comment !

Until next time!











An Investigation for Determining the Optimum Length of Chopsticks : A case for Single-Factor Repeated Measures ANOVA.


In the pursuit to determine the optimum length of chopsticks, two laboratory studies were conducted, using a randomised complete block design, to evaluate the effects of the length of the chopsticks on the food-serving performance of adults and children. Thirty-one male junior college students and 21 primary school pupils served as subjects for the experiment.The response was recorded in terms of food-pinching efficiency (number of peanuts picked, and placed in a cup).

Data source

The data was read in using read.csv() and allotted appropriate column names.

stick<-read.csv(file.choose(),header = TRUE,stringsAsFactors = FALSE)


Next, we’ll convert  the ‘Length’ and ‘Individual’ to factors and the dependent variable ‘Efficiency’ to numeric to carry out our repeated measures ANOVA analysis.

chop$Length<- factor(chop$Length)

Let’s use the tapply() to find out the mean pinching efficiency grouped by the chopstick length and plot the result.


plot(group_mean,type="p",xlab="Chopstick Length",ylab="Mean Efficiency",
 main="Average Food Pinching Efficiency \n by Chopstick Length",col="green",pch=16)



Visually, we can see that the efficiency grows steadily from 180 mm through to the 240  and then falling sharply at 270 mm and despite a slight increase at 300,it continues to fall at 330.

Before we can perform the repeated measures ANOVA, we will need to check for the Sphericity which assumes the homogeneity of variance among differences between all possible pairs of groups. This is done by the Mauchly’s test.

chop1<- cbind(chop$Efficiency[chop$Length==180],

mauchly.test (lm (chop1 ~ 1), X = ~1)

> mauchly.test (lm (chop1 ~ 1), X = ~1)

           Mauchly's test of sphericity
           Contrasts orthogonal to

data: SSD matrix from lm(formula = chop1 ~ 1)
W = 0.43975, p-value = 0.05969

Since, (thankfully!) the p-value is >0.05, the homogeneity of variance assumption holds and no further adjustment is necessary.

We can now get on with the repeated measures ANOVA.

aov.chop = aov(Efficiency~Length + Error(Individual/Length),data=chop)

Error: Individual
 Df Sum Sq Mean Sq F value Pr(>F)
Residuals 30 2278 75.92 

Error: Individual:Length
         Df Sum Sq Mean Sq F value Pr(>F) 
Length    5   106.9 21.372   5.051 0.000262 ***
Residuals 150 634.6 4.231 
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

The F-ratio is 5.051 and p-value is 0.000262 which means there is significant effect of the length of the chopstick on the eating efficiency.

Since, the F-ratio is significant , we can now process the post-hoc analysis to perform pair-wise comparisons.

1. Holm’s adjustment


> with(chop,pairwise.t.test(Efficiency,Length,paired=T))

 Pairwise comparisons using paired t tests 

data: Efficiency and Length 

      180 210   240   270    300 
210 1.000 - - - - 
240 0.323 1.000 - - - 
270 1.000 0.034 0.035 - - 
300 1.000 1.000 0.198 1.000 - 
330 0.630 0.048 8.1e-05 1.000 0.435

P value adjustment method: holm

2. Bonferroni adjustment


> with(chop,pairwise.t.test(Efficiency,Length,paired=T,p.adjust.method="bonferroni"))

 Pairwise comparisons using paired t tests 

data: Efficiency and Length 

       180 210 240 270 300 
210 1.000 - - - - 
240 0.485 1.000 - - - 
270 1.000 0.036 0.040 - - 
300 1.000 1.000 0.269 1.000 - 
330 1.000 0.060 8.1e-05 1.000 0.725

P value adjustment method: bonferroni

As we can see, the Bonferroni adjustment inflates the p-values as compared to Holm’s adjustment.

3. Treatment-by-Subjects

Another method which can be employed is the Treatment-by-Subjects i.e. as a 2 factor design without interaction , without replication. The advantage of this approach is that it allows us to employ the TukeyHSD() post-hoc .


aov.chop1 = aov(Efficiency ~ Length + Individual, data=chop)

> summary(aov.chop1)
         Df   Sum Sq Mean Sq     F value Pr(>F) 
Length     5   106.9       21.37 5.051    0.000262 ***
Individual 30 2277.5       75.92 17.944   < 2e-16 ***
Residuals 150 634.6 4.23 
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


> tukey
 Tukey multiple comparisons of means
 95% family-wise confidence level

Fit: aov(formula = Efficiency ~ Length + Individual, data = chop)

           diff        lwr       upr        p adj
210-180 0.54870968 -0.9595748 2.05699418 0.8999148
240-180 1.38774194 -0.1205426 2.89602644 0.0904885
270-180 -0.61129032 -2.1195748 0.89699418 0.8503866
300-180 0.03290323 -1.4753813 1.54118773 0.9999999
330-180 -0.93548387 -2.4437684 0.57280063 0.4749602
240-210 0.83903226 -0.6692522 2.34731676 0.5959492
270-210 -1.16000000 -2.6682845 0.34828450 0.2346843
300-210 -0.51580645 -2.0240910 0.99247805 0.9213891
330-210 -1.48419355 -2.9924781 0.02409096 0.0565555
270-240 -1.99903226 -3.5073168 -0.49074775 0.0025803
300-240 -1.35483871 -2.8631232 0.15344579 0.1053005
330-240 -2.32322581 -3.8315103 -0.81494130 0.0002412
300-270 0.64419355 -0.8640910 2.15247805 0.8199855
330-270 -0.32419355 -1.8324781 1.18409096 0.9893780
330-300 -0.96838710 -2.4766716 0.53989741 0.4349561

By studying the difference between the means and the Tukey adjusted p-value, the most significant diff is between the groups 240-330 , with a mean difference of 2.32.



The results show that the food-pinching performance is significantly affected by the length of the chopsticks, and that chopsticks of about 240 mm long performed the best.



Beauty is talent: A case for 2-way ANOVA.





Landy and Sigall (1974). “Beauty is Talent: Task Evaluation as a Function of the Performer’s Physical Attraction,” Journal of Personality and Social Psychology, 29:299-304.

60 male undergraduates read an essay supposedly written by a female college freshman. They then evaluated the quality of the essay and the ability of its writer on several dimensions. By means of a photo attached to the essay, 20 evaluators were led to believe that the writer was physically attractive and 20 that she was unattractive. The remaining evaluators read the essay without any information about the writer’s appearance. 30 evaluators read a version of the essay that was well written while the other evaluators read a version that was poorly written. Significant main effects for essay quality and writer attractiveness were predicted and obtained.


Two Factor experiment to assess if appearance effects  judgment of student’s essay.             Appearance (Attractive/Control/Unattractive) and Essay Quality (Good/Poor) are factors (Pictures given with essay, except in control).


Grade on essay. Data simulated to match their means and standard deviations.

Essay Quality 8 /* 1=Good, 2=Poor */
Student Attactiveness 16 /* 1= Attractive, 2=Control, 3=Unattractive */
Score 18-24


When we import halo.dat using read.delim() ,we get a single  column V1 containing all the data values. The separation of V1 into 3 distinct columns and other subsequent transformations are done using functions from the very handy dplyr package, as shown below.

hal<-read.delim(file.choose(),header = FALSE,stringsAsFactors = FALSE)
halo<- select(halo,2:4)

Next we’ll convert the 2 independent variables (Essay Quality, Attractiveness) to factors and the dependent variable (Score) to numeric to carry out our ANOVA analysis.

halo$attractiveness <- factor(halo$attractiveness)

We can test 3 hypotheses:

1. Are scores higher for more attractive students ?
2. Are scores higher for higher quality essay?
3. Whether student attractiveness and essay quality interact resulting in higher scores?

Let’s start by look at the means of our groups. Here, we will use the tapply() function to generate a table of means and use the results to plot a graph.


# Plot a bar-plot
barplot(halo_mean, beside = TRUE, 
 col = c("orange","blue","green"), 
 main = "Mean Scores", 
 xlab = "Essay Quality", 
 ylab = "Score")

# Add a legend
legend("topright", fill = c("orange","blue","green"), 
 title = "Attractiveness",c("1","2","3"),

> halo_mean
    1       2
1 17.900 14.899
2 17.900 13.400
3 15.499 8.701


Visually, we can see that for a good quality essay,the students were evaluated most favorably when they were attractive or when their appearance was unknown , least when they were unattractive.For a poor grade essay,the students were evaluated most favorably when they were attractive, least when they were unattractive, and intermediately when their appearance was unknown.

Let’s explore and further refine our study by performing a 2-way ANOVA. But before we do that, we need to test the homogeneity of the variance assumption.The assumption must hold for the results of an ANOVA analysis to be valid.


The result gives us a p-value of 0.2989 which means that the assumption of homogeneity of variance holds. There is not a significant difference in the variances across the groups.

And now for the ANOVA.


> summary(aov_halo)
                                    Df Sum Sq Mean Sq F value Pr(>F) 
halo$quality                         1   340.8  340.8 17.231   0.000118 ***
halo$attractiveness                  2   211.0 105.5  5.335    0.007687 ** 
halo$quality:halo$attractiveness     2    36.6  18.3  0.925    0.402832 
Residuals                           54  1067.9  19.8 
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

The very low p-values of quality and attractiveness indicate high significance of these 2 factors independently to drive the higher scores. But the high p-value of the interaction shows that both together do not drive higher scores.Since,we do not have a significant interaction effect , we do not need to follow it up to see exactly where the interaction is coming from.


We can now proceed with Post-hoc analysis to test the main effect pairwise comparisons via the following  methods:

  1. TukeyHSD
  2. pairwise.t.test()
  3. Bonferroni Adjustment
  4. Holm’s adjustment


Pair -wise comparison using TukeyHSD

> TukeyHSD(aov_halo,"halo$quality")
  Tukey multiple comparisons of means
    95% family-wise confidence level
Fit: aov(formula = halo$scoretotal ~ halo$quality * halo$attractiveness)
      diff      lwr        upr     p adj
2-1 -4.766333 -7.068384 -2.464282 0.0001182

By studying the difference between the means and the Tukey adjusted p-value, ,GoodQ_mean-PoorQ_mean = 4.77 | p-value = 0.0001182
The mean score is higher for more good quality essays than for poor quality essays.

> TukeyHSD(aov_halo,"halo$attractiveness")
 Tukey multiple comparisons of means
 95% family-wise confidence level
Fit: aov(formula = halo$scoretotal ~ halo$quality * halo$attractiveness)
      diff      lwr      upr      p adj
2-1 -0.7495 -4.138616 2.6396163 0.8555096
3-1 -4.2995 -7.688616 -0.9103837 0.0095632
3-2 -3.5500 -6.939116 -0.1608837 0.0381079

By studying the difference between the means and the Tukey adjusted p-value, ,
Attractive_mean – Control_mean = 0.75 | p-value = 0.8555096
Attractive_mean – Unattractive_mean = 4.99 | p-value = 0.0095632
Control_mean – Unattractive_mean = 3.55 | p-value = 0.0381079

The mean score is higher for more attractive people than for unattractive people (The halo affect).


Pairwise comparison using pairwise.t.test()

pairwise.t.test(halo$scoretotal, halo$attractiveness, p.adj = "none")
> pairwise.t.test(halo$scoretotal, halo$attractiveness, p.adj = "none")

 Pairwise comparisons using t tests with pooled SD 

data: halo$scoretotal and halo$attractiveness 

   1       2 
2 0.6397 - 
3 0.0091 0.0297

P value adjustment method: none 

With no adjustments, the Attractive-Unattractive (1-3) and Control-Unattractive (2-3) comparison are statistically significant,whereas, the Attractive-Control (1-2) comparison is not.This suggests that both the attractive and control groups scored superior to the Unattractive group, but that there is insufficient statistical support to distinguish between the Attractive and Control groups.


pairwise.t.test(halo$scoretotal, halo$quality, p.adj = "none")
> pairwise.t.test(halo$scoretotal, halo$quality, p.adj = "none")

 Pairwise comparisons using t tests with pooled SD 

data: halo$scoretotal and halo$quality 

2 0.00027

P value adjustment method: none 

With no adjustments, the Good Quality-Poor Quality (1-2) comparison is statistically significant. This suggests that good quality essays scored superior as compared to the poor quality essay group.


Pairwise comparison using Bonferroni adjustment

The Bonferroni adjustment simply divides the Type I error rate (.05) by the number of tests (in this case, three for attractiveness and 2 for essay quality). Hence, this method is often considered overly conservative. The Bonferroni adjustment can be made using p.adj = “bonferroni” in the pairwise.t.test() function.

pairwise.t.test(halo$scoretotal,halo$attractiveness,p.adj = "bonferroni")
> pairwise.t.test(halo$scoretotal,halo$attractiveness,p.adj = "bonferroni")

 Pairwise comparisons using t tests with pooled SD 

data: halo$scoretotal and halo$attractiveness 

 1       2 
2 1.000 - 
3 0.027 0.089

P value adjustment method: bonferroni 

Using the Bonferroni adjustment, only the Attractive-Unattractive (1-3) group comparison is statistically significant. This suggests that the attractive group is treated superior to the Unattractive group, but that there is insufficient statistical support to distinguish between the Control-Unattractive (2-3) and the the Attractive-Control (1-2) group comparisons.
Notice that these results are more conservative than with no adjustment.

pairwise.t.test(halo$scoretotal,halo$quality,p.adj = "bonferroni")
> pairwise.t.test(halo$scoretotal,halo$quality,p.adj = "bonferroni")

 Pairwise comparisons using t tests with pooled SD 

data: halo$scoretotal and halo$quality 

2 0.00027

P value adjustment method: bonferroni 

Using the Bonferroni adjustment, the Good Quality-Poor Quality (1-2) comparison is statistically significant.This suggests that good quality essays scored superior as compared to the poor quality essay group.


Pairwise comparison using Holm’s adjustment

The Holm adjustment sequentially compares the lowest p-value with a Type I error rate that is reducedfor each consecutive test. In our case of attractiveness, this means that our first p-value is tested at the .05/3 level (.017),second at the .05/2 level (.025), and third at the .05/1 level (.05).This method is generally considered superior to the Bonferroni adjustment and can be employed using p.adj = “holm” in the pairwise.t.test() function.

pairwise.t.test(halo$scoretotal,halo$attractiveness,p.adj = "holm")
> pairwise.t.test(halo$scoretotal,halo$attractiveness,p.adj = "holm")

 Pairwise comparisons using t tests with pooled SD 

data: halo$scoretotal and halo$attractiveness 

   1     2 
2 0.640 - 
3 0.027 0.059

P value adjustment method: holm 

Using the Holm procedure, our results are almost identical to using no adjustment.



Based on our statistical tests, we can conclude that

1. The students were evaluated most favorably when they were attractive , least when they were unattractive.
2. Good quality essays scored higher than poor quality ones.
3. There is no statistical proof to show that student attractiveness and essay quality interact resulting in higher scores.

Chick Weight vs Diet – A case for one-way ANOVA.


The chick1 dataset is a data frame consisting  of 578 rows and 4 columns  “weight” “Time” “Chick” & “Diet”  which represents the progression of weight of several chicks. The little chicklings are each given a specific diet. There are four types of diet and the farmer wants to know which one fattens the chicks the fastest.

It’s time to do some exploratory statistics on the data frame!

Once the data has been read in using read.csv, we perform a few operations as shown below to make the data ggplot2 ready. The data by itself is pretty clean and doesn’t require much work .


chick1$Time<- as.numeric(chick1$Time)



chick1<-chick1[-579,] #remove the one NA

And now for the ggplot-ing .


Use ggplot() to map Time to x and Weight to y within the aes() function.Add col=diet to highlight the diet fed to the chicks.

Add geom_line() at the end to draw the lines.To draw one line per chick, add group=chick to the aes() of geom_line()

Finally, add the geom_smooth  to add a smoothing curve that shows the general trend of the data.The gray area around the curve is a confidence interval, suggesting how much uncertainty there is in this smoothing curve.

ggplot(chick1,aes(x=Time,y=Weight,col=Diet)) + geom_line(aes(group=Chick)) + 


If we want to turn off the confidence interval, we can add an option to the geom_smooth later; specifically “se=FALSE”, where “s.e.” stands for “standard error.”

ggplot(chick1,aes(x=Time,y=Weight,col=Diet)) + geom_line(aes(group=Chick)) + 



The visual clearly shows that Diet 3 fattens up the chicks the fastest.

Now we have 4 groups of the independent variable i.e. Diet and one dependent variable. This is a good case to be analysed by ANOVA (Analysis of Variance). ANOVA is a commonly used statistical technique for investigating data by comparing the means of subsets of the data. In one-way ANOVA the data is sub-divided into groups based on a single classification factor.

Our assumptions here are :
# Dependent variable:Weight
# Independent variable : Diet 1,2,3,4  (treatment)
# Null hypothesis H0: all diets lead to equal weight gain
# Alternative hypothesis H1: All diets do not lead to equal weight gain

When we conduct an analysis of variance, the null hypothesis considered is that there is no difference in treatments mean, so once rejected the null hypothesis, the question is what treatment differs

ggplot(chick1,aes(x=Time,y=Weight,col=Diet)) + geom_line(aes(group=Chick)) + 

Next, we need to test whether or not the assumption of homogeneity of variance holds true . The latter must hold true for the results of ANOVA to be valid. This is done using the Levene’s test.


The test gives a high F-value of 9.6 and a low P value ,this means the data does not show homogeneity of variance i.e. we have violated principles of ANOVA and need to take steps to rectify this.We do this by logarithmic adjustment.


With the log transformation , it gives a p value of 0.049 ~ 0.05 .A value of p = .05 is probably good enough to consider the homogeneity of variance assumption to be met . Now we can perform ANOVA knowing that the results would hold true.


The results show the f-value as 8.744 and p value as 1.12e-05  showing significant effect . As a result we have 8.744 times between group variance as within group variance. i.e.the variation of weight means among different diets is much larger than the variation of weight within each diet, and our p-value is less than 0.05 (as suggested by normal scientific standard). Hence we can conclude that for our confidence interval we accept the alternative hypothesis H1 that there is a significant relationship between diets and weight gain.



The F-test showed a significant effect somewhere among the groups. However, it did not tell us which pairwise comparisons are significant. This is where post-hoc
tests come into play. They help you to find out which groups differ significantly from one other and which do not. More formally, post-hoc tests allow for multiple pairwise comparisons without inflating the type I error (i.e. rejecting the type-1 error when it is valid).

In R you can use Tukey’s procedure via the TukeyHSD() function. Input to TukeyHSD() is anova.

> tukey
 Tukey multiple comparisons of means
 95% family-wise confidence level

Fit: aov(formula = chick2$Weight ~ chick2$Diet)

           diff      lwr      upr     p adj
2-1 0.15278600 -0.01463502 0.3202070 0.0879198
3-1 0.28030756 0.11288653 0.4477286 0.0001109
4-1 0.26747943 0.09914285 0.4358160 0.0002824
3-2 0.12752155 -0.06293542 0.3179785 0.3114949
4-2 0.11469342 -0.07656886 0.3059557 0.4112309
4-3 -0.01282813 -0.20409041 0.1784342 0.9981647

From the table above (looking at “diff” and “p adj” columns) we can see which diets have significant differences in weight from others. For example we can conclude that:

  • There is no significant difference in weight between groups 3-2 (p=0.311) ,groups 4-2 (p=0.411) , groups 4-3 (p=0.998)  & groups 2-1 (p=0.0878).
  • There is a significant difference in weight between groups  3-1 ( p=0.0001109) and 4-1 (p=0.0002824).

Finally, the results obtained above can also be visualized  by plotting the the “tukey” object in R. Significant differences are the ones which not cross the zero value.

plot(tukey) # plot the confidence intervals.


Each line represents the mean difference between both groups with the according confidence interval. Whenever the confidence interval doesn’t include zero (the vertical line), the difference between both groups is significant.

I hope you found it helpful and as always, comments are welcome!

Thanks for stopping by!