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Ex. for 15 Nov. Lecture
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19.8 | |
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# ANOVA | |
# Clean up the workspace | |
rm(list=ls()) | |
# Import the data | |
# We have life length for rats, subject to different class of Poison: | |
# I, II, III | |
# and Treatment | |
# A, B, C, D | |
# The final argument of our table is tha length of the life. We can | |
# directly import data from the internet: | |
df <- read.table("https://goo.gl/W9CXQV", header=T) | |
# Understand the problem with boxplots and interaction plot | |
boxplot(Life ~ Poison, d=df) | |
boxplot(Life ~ Treat, d=df) | |
# In this case we are not sure about residual analysis | |
# thus it is better to perform ana analysis of residuals | |
interaction.plot(df$Poison, df$Treat, df$Life) | |
# Let's make a first model for this process. | |
# We consider that the the Life process as a dependecies | |
# from the Treat, the Poison and the combination of Treat and Poison | |
# Converted in formula it means | |
df.formula <- Life ~ Treat + Poison + Treat:Poison | |
# And we can create a model | |
df.lm <- lm(df.formula, data=df) | |
# And we can make an analysis of residuals (qualitative) to understand | |
# if we are making a good assumption of the model | |
qqnorm(rstandard(df.lm)) | |
qqline(rstandard(df.lm)) | |
# residual pplot | |
plot(rstandard(df.lm)) | |
# Let's try to simplify this model | |
df.formula.2 <- Life ~ Treat + Poison | |
# And we can create a model | |
df.lm.2 <- lm(df.formula.2, data=df) | |
# And we can make an analysis of residuals (qualitative) to understand | |
# if we are making a good assumption of the model | |
qqnorm(rstandard(df.lm.2)) | |
qqline(rstandard(df.lm.2)) | |
# residual pplot | |
plot(rstandard(df.lm.2)) | |
# Let's try to improve the model | |
boxplot(1/Life ~ Poison, d=df) | |
boxplot(1/Life ~ Treat, d=df) | |
df.formula.3 <- 1/Life ~ Treat + Poison | |
# And we can create a model | |
df.lm.3 <- lm(df.formula.3, data=df) | |
# And we can make an analysis of residuals (qualitative) to understand | |
# if we are making a good assumption of the model | |
qqnorm(rstandard(df.lm.3)) | |
qqline(rstandard(df.lm.3)) | |
# residual pplot | |
plot(rstandard(df.lm.3)) | |
# Let's try with anova analisys | |
anova(df.lm.3) | |
anova(df.lm) |
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# GOODNESS OF FIT | |
# This example represent bservations of vehicle per minute | |
# in a street. Evaluate if this pcess can be modeled with | |
# a Poisson distribution. | |
rm(list=ls()) | |
# Data are available at https://goo.gl/v91n7L | |
df <- read.table("https://goo.gl/v91n7L")$V1 | |
df.n <- length(df) | |
df.m <- mean(df) | |
df.s <- sd(df) | |
df.l <- 46 | |
# Create the bins | |
# We put freq=F to get a density distribution instead of a pure | |
# frequency distribution | |
(df.h <- hist(df, freq=F)) | |
# Create the expected distribution | |
df.b <- df.h$breaks # x-values in hist plot | |
#df.b[1] = -Inf | |
df.b[length(df.b)] = 0 | |
# Evaluate the expected Poisson distribution | |
ei <- diff(ppois(df.b, df.l)) * df.n | |
plot(ei) | |
points(df.h$counts, col="red") | |
chi <- sum((df.h$counts - ei) ** 2 / ei) | |
# Check against pchisq | |
pchisq(chi, df=length(df.h$counts) - 1 - 1) |
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# CONFIDENCE INTERVAL | |
rm(list=ls()) | |
# An article describes t results of tensile adhesion tests on | |
# n = 22 U-700 Alloy specimens. The failure load is described in file | |
# at: https://goo.gl/6XWdME | |
df <- read.table(file="https://goo.gl/6XWdME")$V1 | |
(m <- mean(df)) | |
(s <- sd(df)) | |
(n <- length(df)) | |
# Let's check our model for a distribution | |
boxplot(df, ylab="Load at Failure") | |
qqnorm(df) | |
qqline(df) | |
# We can assume that the population is normally distributed | |
# and we want to understand which is the confidence interval | |
# on the mean of our distribution | |
# The confidence interval is the confidence interval on the mean | |
# with variance unknown for the t-distribution | |
conf <- 0.95 | |
t <- qt((1 - conf)/2, n - 1, low=F) | |
(c( | |
m - t * s / sqrt(n), | |
m + t * s / sqrt(n) | |
)) |
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Poison Treat Life | |
I A 0.31 | |
I B 0.82 | |
I C 0.43 | |
I D 0.45 | |
I A 0.45 | |
I B 1.1 | |
I C 0.45 | |
I D 0.71 | |
I A 0.46 | |
I B 0.88 | |
I C 0.63 | |
I D 0.66 | |
I A 0.43 | |
I B 0.72 | |
I C 0.76 | |
I D 0.62 | |
II A 0.36 | |
II B 0.92 | |
II C 0.44 | |
II D 0.56 | |
II A 0.29 | |
II B 0.61 | |
II C 0.35 | |
II D 1.02 | |
II A 0.4 | |
II B 0.49 | |
II C 0.31 | |
II D 0.71 | |
II A 0.23 | |
II B 1.24 | |
II C 0.4 | |
II D 0.38 | |
III A 0.22 | |
III B 0.3 | |
III C 0.23 | |
III D 0.3 | |
III A 0.21 | |
III B 0.37 | |
III C 0.25 | |
III D 0.36 | |
III A 0.18 | |
III B 0.38 | |
III C 0.24 | |
III D 0.31 | |
III A 0.23 | |
III B 0.29 | |
III C 0.22 | |
III D 0.33 |
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# T-Test example - 15 Nov 2016 | |
# Clear workspace | |
rm(list=ls()) | |
# PART 1 | |
# DATA GENERATION | |
# Define two means µ[a] and µ[b], suche that µ[a] > µ[b] | |
mu.a <- 0.1 | |
mu.b <- -0.1 | |
# Define two standard deviation σ[a] and σ[b], such that σ[a]!= σ[b] | |
sigma.a <- 0.1 | |
sigma.b <- 0.15 | |
# Create 4 groups of samples such that: | |
# * N(µa, σa) with 12 samples -> s1 | |
# * N(µa, σb) with 15 samples -> s2 | |
# * N(µb, σa) with 13 samples -> s3 | |
# * N(µb, σb) with 10 samples -> s4 | |
n.1 <- 12 | |
s.1 <- rnorm(n.1, mu.a, sigma.a) | |
n.2 <- 15 | |
s.2 <- rnorm(n.2, mu.a, sigma.b) | |
n.3 <- 13 | |
s.3 <- rnorm(n.3, mu.b, sigma.a) | |
n.4 <- 10 | |
s.4 <- rnorm(n.4, mu.b, sigma.b) | |
# PART 2 | |
# T-TEST ONE SAMPLE, TWO SIDES | |
# t.test if s4 belongs to a distribution with mean µ[b] | |
# H0: µ1 = µ2 | |
# H1: µ1 != µ2 | |
# In this case there is no need to evaluate the pooled variance, | |
# but we can directly evaluate the t0 | |
test.1.t0 <- (mean(s.4) - mu.b)/(sd(s.4) / sqrt(n.4)) | |
# And we get a p-value from a two sided t-distribution | |
( pt(abs(test.1.t0), n.4 - 1, lower.tail=FALSE) * 2 ) | |
# Lets test it versus built in function | |
( test.1.tt <- t.test(s.4, alternative="two.sided", mu=mu.b, paired=F) ) | |
# PART 3 | |
# INTERLUDE: Functions | |
# Write that evaluates t-test for sigma equal and sigma different | |
my.t.test <- function(s1, s2, sigma.equal=TRUE) { | |
n1 <- length(s1) | |
n2 <- length(s2) | |
t0 <- 0 | |
if (sigma.equal) { | |
n <- n1 + n2 - 2 | |
sp <- ((n1-1)*var(s1) + (n2-1)*var(s2)) / n | |
return((mean(s1)-mean(s2))/(sqrt(sp*(1/n1+1/n2)))) | |
} else { | |
return((mean(s1) - mean(s2))/sqrt(var(s1)/n1 + var(s2)/n2)) | |
} | |
} | |
# PART 4 | |
# T-TEST TWO SAMPLES, TWO SIDES | |
# t.test between s1 and s2 with | |
# H0: µ1 = µ2 | |
# H1: µ1 != µ2 | |
test.2.t0 <- my.t.test(s.1, s.2) | |
( pt(abs(test.2.t0), (n.1 + n.2 - 2), lower.tail=FALSE) * 2 ) | |
( test.2.tt <- t.test(s.1, y=s.2, alternative="two.sided", var.equal=T) ) | |
# PART 5 | |
# T-TEST TWO SAMPLES, ONE SIDE | |
# t.test between s1 and s3 with | |
# H0: µ1 = µ2 | |
# H1: µ1 < µ2 | |
test.3.t0 <- my.t.test(s.1, s.3) | |
( pt(test.3.t0, n.1 + n.3 - 2, low=T) ) | |
( test.3.tt <- t.test(s.1, y=s.3, alt="less", var.equal=T)) | |
# PART 6 | |
# T-TEST TWO SAMPLES, ONE SIDE | |
# t.test between s2 and s4 with | |
# H0: µ1 = µ2 | |
# H1: µ1 > µ2 | |
test.4.t0 <- my.t.test(s.2, s.4) | |
( pt(test.4.t0, n.2 + n.4 - 2, low=F) ) | |
( test.4.tt <- t.test(s.2, y=s.4, paired=F, alt="greater", var.equal=T)) | |
# HOMEWORK | |
# Imagine that you now know a priori: | |
# | |
# * s.1 deviation differs from s.2 deviation | |
# * s.1 deviation is the same of s.3 deviation | |
# * s.2 deviation is the same of s.4 deviation | |
# | |
# Modify the script to make a "good" comparison, using | |
# var.equal=FALSE where it is necessary. | |
# Write a function to evaluate the degree of freedom | |
# that is necessary in that case. | |
# hint: Welch-Satterthwaite equation | |
# PART 7 | |
# Evaluate the confidence interval for each of the previous | |
# two-samples comparison | |
pool.deviation <- function(s1, s2) { | |
n1 <- length(s1) | |
n2 <- length(s2) | |
n <- n1 + n2 - 2 | |
sp <- (n1-1)*var(s1) + (n2-1)*var(s2) | |
return(sqrt(sp/n)) | |
} | |
scaling.coefficient <- function(s1, s2, var.equal=T) { | |
n1 <- length(s1) | |
n2 <- length(s2) | |
if(var.equal) { | |
sp <- pool.deviation(s1, s2) | |
sn <- sqrt((1/n1) + (1/n2)) | |
return(sp*sn) | |
} else { | |
return(sqrt((var(s1)/n1) + (var(s2)/n2))) | |
} | |
} | |
confidence.interval <- function(s1, s2, conf=0.95, var.equal=T) { | |
dmu <- mean(s1) - mean(s2) | |
n <- 0 | |
if (var.equal) { | |
n <- length(s1) + length(s2) - 2 | |
} else { | |
# To be completed with a paricular equation... | |
# this means... HOMEWORK!! yup! | |
n <- 0 # ? | |
} | |
cf <- scaling.coefficient(s1, s2, var.equal=var.equal) | |
tp <- qt((1 - conf)/2, n, low=F) | |
i_min <- dmu - tp * cf | |
i_max <- dmu + tp * cf | |
return(c(i_min, i_max)) | |
} | |
( test.2.tt$conf.int ) | |
( confidence.interval(s.1, s.2) ) | |
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