1. Basic statistic inference in R
Shai Meiri

2. “We expect to find differences between x and y” is a trivial saying
Everything differs!!!
“We expect to find differences between x and y” is a trivial saying
The statistician within you asks “Are the differences we found are larger that expected by chance?”
The biologist within you asks “Why the differences I found are in the direction and the level they are?”

3. Moments of central tendency
Mean
Arithmetic mean: Σxi/n
Geometric mean: (x1*x2*…*xn)1/n
Harmonic mean:

4. Moments of central tendency in R
1. Arithmetic mean: Σxi/n
Example:
Use the function “mean”
data<-c(2,3,4,5,6,7,8)
mean(data)
[1] 5
2. Geometric mean: (x1*x2*…*xn)1/n
You can also use the .csv file :
Example:
dat<-read.csv("island_type_final2.csv")
Attach(dat)
mean(lat)
[1]
data<-c(2,3,4,5,6,7,8) exp(mean(log(data)))
[1]

5. Moments of central tendency
A. mean
B. Median
C. Mode
General example:
data<-c(2,3,4,5,6,7,8)
median(data)
[1] 5
Example from the .csv:
median(mass)
[1] 0.69

6. Moments of central tendency
Mean
Variance
= Σ(xi-μ)2 / n
Is the mean is a good measurement to what is happening in the population when the variance is low?
data<-c(2,3,4,5,6,7,8)
var(data)
[1]
Var(lat)
[1]
Example:

7. Moments of central tendency
Mean
Variance
The second moment of central tendency is the measurement of how much the data is scattered around the first moment (mean)
An example for the second moment are the variance, the standard variation, standard error, coefficient of variation and the confidence interval of 90%, 95% and 99% from something

8. Moments of central tendency
#for:
data<-c(2,3,4,5,6,7,8)
Sample size:
length(data)
Variance:
var(data)
Standard deviation:
sd(data)
se<-(sd(data)/length(data)^0.5) se
[1]
Standard error:
CV<-sd(data)/mean(data) CV
[1]
coefficient of variation:

9. Moments of central tendency
Mean
Variance
Skew
Skewed distribution of frequencies is not symmetric
Do you think that the arithmetic mean is a good measurement of central tendency for a skewed frequency distribution
What is the mean salary of the student here and of Bill Gates?

10. Moments of central tendency
Skew
skew<-function(data){
m3<-sum((data-mean(data))^3)/length(data)
s3<-sqrt(var(data))^3
m3/s3}
skew(data)
The SE of skewness:
sdskew<-function(x) sqrt(6/length(x))

11. Moments of central tendency
Mean
Variance
Skew
Kurtosis

12. Moments of central tendency
Kurtosis
kurtosis<-function(x){
m4<-sum((x-mean(x))^4)/length(x)
s4<-var(x)^2
m4/s4-3 }
kurtosis(x)
sdkurtosis<-function(x) sqrt(24/length(x))
SE of kurtosis:

13. A normal distribution can get a value of mean and variance but its skewness and the kurtosis should equal to zero
Values of skew and kurtosis have their own variance – and zero should be outside of their confidence interval in order for them to be significantly different from zero

14. Residuals When doing statistics we’re creating models of the reality
One of the most simple models is the mean:
The mean height of Israeli citizens is 173 cm
The mean salary is 9271 ₪ (correct for April 2014)
The mean service in IDF is 24 months (I guess)
46,699 ₪ for a month
(excluding the bottles)
Rab. Dov Lior
Served in IDF for 1 month
m2.06

15. Residuals When doing statistics we’re creating models of the reality
We can see here that our models: 24 months, 9271 ₪ and 173 cm are not very successful
The Residual Is how much a certain value is far from the prediction of the model. Omri Caspi is far away in 32 cm from the model “Israeli = 173” and in 29 cm from the more complicated model: “Israeli man = 177, Israeli women = 168”
Residual = ₪ 37428
Residual = -23 month IDF service
Residual = 33 cm

16. Residuals When doing statistics we’re creating models of the reality
dat<-read.csv("island_type_final2.csv")
model<-lm(mass~iso+area+age+lat, data=dat)
out<-model$residuals
out
write.table(out, file = "residuals.txt",sep="\t",col.names=F,row.names=F)
#note that residual values are in the order entered (i.e., not alphabetic, not by residual size – first in, first out)
Residual = ₪ 37428
Residual = 33 cm
Residual = -23 month service

17. Theoretical statistics and statistical
inference
When we have data it is best that we first describe them: plot graphs, calculate the mean and so on
In statistical inference we are testing the behavior of our data compared to a certain hypothesis
We can present our hypothesis as a statistical model
For Example:
The distribution of the heights is normal
Number of species increases with area
Number of species increases with area with a power function of 0.25

18. Frequency distribution*
How many observations are in each bin?
dat<-read.csv("island_type_final2.csv")
attach(dat)
names(dat)
Hist(mass)
Describes the distribution of all observations
*graphic form = “histogram”

19. Frequency distribution
What did we learn?
dat<-read.csv("island_type_final2.csv")
attach(dat)
Hist(mass)
There are no mass smaller than one tenth of a gram or larger than 100 kg
Lizard with mass between 1 and 10 are very common – larger or smaller lizards are rare
The distribution is unimodal and skewed to the right

20. Frequency distribution
Histograms don’t have to be so ugly
dat<-read.csv("island_type_final2.csv")
attach(dat)
hist(mass, col="purple",breaks=25,xlab="log mass (g)",main="masses of island lizards - great data by Maria",cex.axis=1.2,cex.lab=1.5)

21. Presenting a categorical predictor with a continuous response variable
dat<-read.csv("island_type_final2.csv")
attach(dat)
plot(type,brood)
Always prefer boxplot to barplot

22. Presenting a continuous variable against another continuous variable
dat<-read.csv("island_type_final2.csv")
attach(dat)
plot(mass,clutch)
plot(mass,clutch,pch=16, col=“blue”)

23. Which test should we choose?
It changes according to the nature of our response variable (=y variable), and mostly according to the nature of our predictor variables
If the response variable is “success or failure” and the null hypothesis is equality of both we’ll use a binomial test
If the response variable is counts we’ll usually use chi-square or G
In many cases our response variable will be continuous (14 species, 78 individuals, 54 heartbeats per second, 7.3 eggs, 23 degrees)

24. Which test should we choose?
What is your response variable ?
Continuous
(14 species, 78 individuals, 23 degrees, 7.3 eggs)
Counts
(frequency: 6 females, 4 males)
Success or failure
(found the cheese/idiot)
Chi-square or G (=log-likelihood)
Binomial
Soon…

25. Binomial test in R
You need to define the number of successes from the whole sample size.
For example: 19 out of 34 is not significant
19 out of 20 is significant
binom.test(19,34)
Exact binomial test data: 19 and 34 number of successes = 19, number of trials = 34
p-value = alternative hypothesis: true probability of success is not equal to percent confidence interval: sample estimates: probability of success
binom.test(19,20)
Exact binomial test data: 19 and 20 number of successes = 19, number of trials = 20,
p-value = 4.005e-05 alternative hypothesis: true probability of success is not equal to percent confidence interval: sample estimates: probability of success 0.95

26. Chi-square test in R chisq.test
Data: lizard insularity & diet:
chisq.test
habitat
diet
species#
island
carnivore
488
herbivore 43
omnivore
177
mainland
1901
101
269
M<-as.table(rbind(c(1901,101,269),c(488,43,177)))
chisq.test(M)
data: M
χ2 = 80.04, df = 2, p-value < 2.2e-16

27. Chi-square test in R chisq.test χ2 = 17.568, df = 4, p-value = 0.0015
Now lets use our dataset:
chisq.test
dat<-read.csv("island_type_final2.csv")
install.packages("reshape")
library(reshape)
cast(dat, type ~ what, length)
type
anoles
else
gecko
Continental 7 45
Land_bridge 1 30 14
Oceanic 23
110 44
M<-as.table(rbind(c(7,45,45),c(1,30,14),c(23,110,44)))
chisq.test(M)
data: M
χ2 = , df = 4, p-value =

28. Which test should we choose?
If our response variable is continuous then we’ll choose our test based on the predictor variables
If our predictor variable is categorical (Area 1, Area 2, Area 3 or species A, species B, species C)
We’ll use ANOVA
If our predictor variable is continuous (temperature, body mass, height)
We’ll use REGRESSION

29. t-test in R t.test(x,y) dimorphism<-read.csv("ssd.csv",header=T)
Sex
size
female
79.7
male 85
120
133.0
118
126.0
105.8
112
106
121.0 95
111.0 86
93.0 65
75.0
230
240.0
t.test(x,y)
dimorphism<-read.csv("ssd.csv",header=T)
attach(dimorphism)
names(dimorphism)
males<-size[Sex=="male"]
females<-size[Sex=="female"]
t.test(females,males)
Welch Two Sample t-test data: females and males t = , df = , p-value = alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: sample estimates: mean of x mean of y

30. t-test in R (2) lm(x~y) Estimate standard error t p value (Intercept)
Sex
size
female
79.7
male 85
120
133.0
118
126.0
105.8
112
106
121.0 95
111.0 86
93.0 65
75.0
230
240.0
lm(x~y)
dimorphism<-read.csv("ssd.csv",header=T)
attach(dimorphism)
names(dimorphism)
model<-lm(size~Sex,data=dimorphism)
summary(model)
Estimate
standard error t
p value
(Intercept)
88.17
1.291
68.32
<2e-16
***
Sexmale
3.932
1.825
2.154
0.031 *

31. Paired t-test in R t.test(x,y,paired=TRUE) female male 88.17 92.10
Species
size
Sex
Xenagama_zonura
79.7
female 85
male
Xenosaurus_grandis
120
133.0
Xenosaurus_newmanorum
118
126.0
Xenosaurus_penai
105.8
112
Xenosaurus_platyceps
106
121.0
Xenosaurus_rectocollaris 95
111.0
Zonosaurus_anelanelany 86
93.0
Zootoca_vivipara 65
75.0
Zygaspis_nigra
230
240.0
Zygaspis_quadrifrons
195
227.0
t.test(x,y,paired=TRUE)
dimorphism<-read.csv("ssd.csv",header=T)
attach(dimorphism)
names(dimorphism)
males<-size[Sex=="male"]
females<-size[Sex=="female"]
t.test(females,males, paired=TRUE)
Paired t-test data:
females and males t = , df = 3503, p-value < 2.2e-16 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: sample estimates: mean of the differences
female
male
88.17
92.10
tapply(size,Sex,mean)

32. ANOVA in R aov model<-aov(x~y) Df Sum sq Mean sq F value Pr(>F)
species
type
clutch
Trachylepis_sechellensis
Continental
0.6
Trachylepis_wrightii
0.65
Tropidoscincus_boreus
0.4
Tropidoscincus_variabilis
0.45
Urocotyledon_inexpectata
0.3
Varanus_beccarii
0.58
Algyroides_fitzingeri
Land_bridge
Anolis_wattsi
Archaeolacerta_bedriagae
Cnemaspis_affinis
Cnemaspis_limi
0.18
Cnemaspis_monachorum
Amblyrhynchus_cristatus
Oceanic
0.35
Ameiva_erythrocephala
Ameiva_fuscata
Ameiva_plei
0.41
Anolis_acutus
Anolis_aeneus
Anolis_agassizi
Anolis_bimaculatus
Anolis_bonairensis
ANOVA in R
aov
model<-aov(x~y)
island<-read.csv("island_type_final2.csv",header=T)
names(island)
[1] "species" "what" "family" "insular" "Archipelago" "largest_island" [7] "area" "type" "age" "iso" "lat" "mass"
[13] "clutch" "brood" "hatchling" "productivity“
model<-aov(clutch~type,data=island)
summary(model) Df
Sum sq
Mean sq
F value
Pr(>F)
type 2
0.466
2.784
0.0635 .
Residuals
289
24.184

33. Post-hoc test for ANOVA in R
TukeyHSD(model)
Tukey multiple comparisons of means
95% family-wise confidence level
Fit: aov(formula = clutch ~ type, data = island)
$type
diff
lwr
upr
p adj
Land_bridge-Continental
0.124
0.2505
0.0561
Oceanic-Continental
0.0218
0.1108
0.8318
Oceanic-Land_bridge
-0.102
0.0163
0.1066
The difference is not significant. Notice that zero is always in the confidence interval. The difference between Land bridge islands and Continental islands is very close to significance (p = 0.056)

34. correlation in R cor.test(x,y)
mass 5
1.21
0.83 4
1.84 18
1.39
0.42
0.29 20
0.45
1.54
0.36
0.27
0.04
0.01 21
0.95
0.51 22
0.74
0.92
island<-read.csv("island_type_final2.csv",header=T)
names(island)
[1] "species" "what" "family" "insular" "Archipelago" "largest_island"
[7] "area" "type" "age" "iso" "lat" "mass"
[13] "clutch" "brood" "hatchling" "productivity“
attach(island)
cor.test(mass,lat)
Pearson's product-moment correlation
data: mass and lat
t = , df = 317, p-value = 0.256
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
sample estimates: cor
The variable “cor” is the correlation coefficient r

35. Same data as in the previous example
regression in R
Same data as in the previous example
lm (=“linear model”):
lm (y~x)
model<-lm(mass~lat,data=island)
summary(model)
Call: lm(formula = mass ~ lat, data = island)
Residuals:
Min 1Q Median 3Q Max
Coefficients:
Estimate
Std. Error
t value
Pr(>|t|)
(Intercept)
9.934
<2e-16
***
lat
-1.138
0.256
Residual standard error: on 317 degrees of freedom
Multiple R-squared: , Adjusted R-squared:
F-statistic: on 1 and 317 DF, p-value: 0.256

36. lm vs. aov We can also use ‘lm’ with data that fits ANOVA
In this case we’ll receive all the data that ‘summary’ gives for ‘lm’ function for regression including parameter estimates, SE, difference between factors and p-values for contrasts between categories of our predictor variable

37. aov vs. lm aov results lm results More later on
We can use ‘lm’ also on data that fits ANOVA
In this case we’ll receive all the data that ‘summary’ gives for ‘lm’ function for regression including parameter estimates, SE, difference between factors and p-values for contrasts between category pairs of our predictor variable
island<-read.csv("island_type_final2.csv",header=T)
model<-aov(clutch~type,data=island)
model2<-lm(clutch~type,data=island)
summary(model)
summary(model2) Df
Sum sq
Mean sq
F value
Pr(>F)
type 2
0.466
2.784
0.0635 .
Residuals
289
24.184
aov results
Estimate
Std. Error
t value
Pr(>|t|)
(Intercept)
11.11
<2e-16
***
typeLand_bridge
2.309
0.0216 *
typeOceanic
0.578
0.5635
lm results
Residual standard error: on 289 degrees of freedom
(27 observations deleted due to missingness)
Multiple R-squared: , Adjusted R-squared:
F-statistic: on 2 and 289 DF, p-value:
More later on

38. Assumptions of statistical tests (all statistical tests)
A non-random, non-independent sample of Israeli people
Random sampling (assumption of all tests not only parametric)
Independence (spatial, phylogenetic etc.)

39. Assumptions of parametric test. A. ANOVA
In addition to the assumptions of all tests
Homoscedasticity
Normal distribution of the residuals
"Comments on earlier drafts of this manuscript made it clear that for many readers who analyze data but who are not particularly interested in statistical questions, any discussion of statistical methods becomes uncomfortable when the term ‘‘error variance’’ is introduced.“
Smith, R. J Use and misuse of the reduced major axis for line-fitting. American Journal of Physical Anthropology 140:
Richard Smith & 3 friends
Reading material:
Sokal & Rohlf Biometry. 3rd edition. Pages (especially for normality)

40. Always look at your data
Don’t just rely on the statistics!
Anscombe's quartet Summary statistics are the same for all four data sets:
n = 11
means of x & y (9, 7.5),
standard deviation (4.12)
regression & residual SS
R2 = (0.816)
regression line (y = 3 + 0.5x)
Anscombe Graphs in statistical analysis. The American Statistician 27: 17–21.

41. Assumptions of parametric tests. B. Regression
1. Homoscedasticity
Smith, R. J Use and misuse of the reduced major axis for line-fitting. American Journal of Physical Anthropology 140:

42. Assumptions of parametric tests. B. Regression
Homoscedasticity
The explanatory variable was sampled without error
Smith, R. J Use and misuse of the reduced major axis for line-fitting. American Journal of Physical Anthropology 140:

43. Assumptions of parametric tests. B. Regression
Homoscedasticity
The explanatory variable was sampled without error
Normal distribution of the residuals of each response variable
Smith, R. J Use and misuse of the reduced major axis for line-fitting. American Journal of Physical Anthropology 140:

44. Assumptions of parametric tests. B. Regression
Homoscedasticity
The explanatory variable was sampled without error
Normal distribution of the residuals of each response variable
Equality of variance between the values of the explanatory variables
Smith, R. J Use and misuse of the reduced major axis for line-fitting. American Journal of Physical Anthropology 140:

45. Assumptions of parametric tests. B. Regression
Homoscedasticity
The explanatory variable was sampled without error
Normal distribution of the residuals of each response variable
Equality of variance between the values of the explanatory variables
Linear relationship between the response and the predictor
Smith, R. J Use and misuse of the reduced major axis for line-fitting. American Journal of Physical Anthropology 140:

46. How will we test if our model follows the assumptions?
R has a very useful model diagnostic functions which allows us to evaluate in a graphical matter how much our model follows the model assumption (especially in regression)
ראו גם:

47. What can we do when our data doesn’t follow the assumptions?
We can ignore it and hope that our test is robust enough to break the assumptions: this is not as unreasonable as it sounds
Use non-parametric tests
Use generalized linear models (glm); which means:
Transformation (in glm it means changing the link functions)
Change error distribution in glm) to non-normal distribution)
Use non-linear tests
Use randomization (more about it in Roi’s lessons)

48. I think it is really wrong to have a presentation without any animal pictures in it
Non-parametric test
Non-parametric test do not assume equality of variance or normal distribution. They are based on Ranks
Disadvantages:
There are no test for models with multiple predictors
Many times their statistical power is very low compared to a equivalent parametric test
They do not give you parameter estimation (slopes, intercepts)

49. נראה לי ממש לא בסדר שבמצגת שלמה אין לי תמונות של חיות
Non-parametric tests
Non-parametric test do not assume equality of variance or normal distribution. They are based on Ranks
חסרונות:
לא קיימים מבחנים למודלים מרובי predictors
לעיתים קרובות ה-statistical power שלהם נמוך משל מבחן פרמטרי מקביל
לא מאפשרים הערכת פרמטרים (שיפועים ונקודות חיתוך)

50. The photographed is not related to the lectures
A few useful non-parametric tests
Orycteropus afer
The photographed is not related to the lectures
Chi-square test is a non-parametric test
Kolmogorov-Smirnov is a non-parametric test used to compare to frequency distributions (or to compare “our” distribution to a known distribution. For example: a normal distribution
Mann-Whitney U = Wilcoxon rank sum Is a non-parametric test equivalent to students t-test
Wilcoxon two-sample (=Wilcoxon signed-rank) test replaces paired-t-test
Kruskal-Wallis replaces one-way ANOVA
Spearman test Kendall’s-tau test replaces correlation tests

51. Non-parametric tests in R
Kolmogorov-Smirnov is a non-parametric test used to compare to frequency distributions (or to compare “our” distribution to a known distribution. For example: a normal distribution
Orycteropus afer
The photographed is not related to the lectures
We need to define in R the grouping variable and the response: lets say we want to compare between the frequency distribution of lizard body mass on oceanic and land bridge islands
island<-read.csv("island_type_final2.csv",header=T)
attach(island)
levels(type)
[1] "Continental" "Land_bridge" "Oceanic“
Land_bridge<-mass[type=="Land_bridge"]
Oceanic <-mass[type==" Oceanic"]
ks.test(Land_bridge, Oceanic)
Two-sample Kolmogorov-Smirnov test
 data: Land_bridge and Oceanic
D = , p-value =
alternative hypothesis: two-sided