Correlation between numerical variables

• Reading Assignments:
  • Chapter 16: Correlation
  • Chapter 17: Regression
• Non-Homework Posted (Chapters 16-17)
• PLEASE COMPLETE COURSE EVALUATIONS!

Variables: For a correlation, our data consist of two numerical variables (continuous or discrete).

Definition: The (linear) correlation coefficient \( \rho \) measures the strength and direction of the association between two numerical variables in a population.

The linear (Pearson) correlation coefficient measures the tendency of two numerical variables to co-vary in a linear way.

The symbol \( r \) denotes a sample estimate of \( \rho \).

\[ r = \frac{\sum_{i}(X_{i}-\bar{X})(Y_{i}-\bar{Y})}{\sqrt{\sum_{i}(X_{i}-\bar{X})^2}\sqrt{\sum_{i}(Y_{i}-\bar{Y})^2}} \]

\[ -1 \leq r \leq 1 \]

\[ r = \frac{\frac{1}{n-1}\sum_{i}(X_{i}-\bar{X})(Y_{i}-\bar{Y})}{\sqrt{\frac{1}{n-1}\sum_{i}(X_{i}-\bar{X})^2}\sqrt{\frac{1}{n-1}\sum_{i}(Y_{i}-\bar{Y})^2}} \]

\[ \begin{eqnarray*} r & = & \frac{\frac{1}{n-1}\sum_{i}(X_{i}-\bar{X})(Y_{i}-\bar{Y})}{\sqrt{\frac{1}{n-1}\sum_{i}(X_{i}-\bar{X})^2}\sqrt{\frac{1}{n-1}\sum_{i}(Y_{i}-\bar{Y})^2}} \\ & = & \frac{\mathrm{Covariance}(X,Y)}{s_{X}s_{Y}} \end{eqnarray*} \]

http://guessthecorrelation.com/

Practice Problem #1

In their study of hyena laughter, or “giggling”, Mathevon et al. (2010) asked whether sound spectral properties of hyenas' giggles are associated with age. The accompanying figure and data show the giggle fundamental frequency (in hertz) and the age (in years) of 16 hyenas.

Read in the data and take a peek

hyenaData <- read.csv("http://whitlockschluter.zoology.ubc.ca/wp-content/data/chapter16/chap16q01Hyena%20GigglesAndAge.csv")
str(hyenaData)

'data.frame':   16 obs. of  2 variables:
 $ Age_years               : int  2 2 2 6 9 10 13 10 14 14 ...
 $ Fundamental_frequency_Hz: int  840 670 580 470 540 660 510 520 500 480 ...

Plot in a scatter plot

plot(Fundamental_frequency_Hz ~ Age_years, data=hyenaData, cex =1.5, cex.lab=1.5)

Compute correlation coefficient

hyenaCorr <- cor.test(hyenaData$Age_years,
                      hyenaData$Fundamental_frequency_Hz)
(r <- hyenaCorr$estimate)

      cor 
-0.601798 

How do we quantify the uncertainty of this estimate?

We need the standard error and confidence intervals.

\[ \mathrm{SE}_{r} = \sqrt{\frac{1-r^2}{n-2}} \]

n <- nrow(hyenaData)
sderr <- sqrt((1-r^2)/(n-2))
unname(sderr)

[1] 0.2134478

Note: Sampling distributions for \( r \) are not normally distributed, so we shouldn't use the standard error to compute a confidence interval.

Fisher (of course) discovered an approximate confidence interval:

• Compute Fisher's \( z \)-transformation:   \( z = 0.5\ln\left(\frac{1+r}{1-r}\right) \)
• Compute standard error for \( z \):   \( \sigma_{z} = \sqrt{\frac{1}{n-3}} \)
• Compute approximate 95% confidence interval: \[ z - 1.96\sigma_{z} < \zeta < z + 1.96\sigma_{z} \]
• Transform lower and upper bounds back: \[ r = \frac{e^{2z}-1}{e^{2z}+1} \]

Long way

z <- 0.5*log((1+r)/(1-r)) # Compute z
zsderr <- sqrt(1/(n-3))   # Std err of z
zlb <- z - 1.96*zsderr    # Lower bound for z
zub <- z + 1.96*zsderr    # Upper bound for z
rlb <- (exp(2*zlb) - 1)/(exp(2*zlb) + 1)
rub <- (exp(2*zub) - 1)/(exp(2*zub) + 1)
CI <- c(rlb, rub)
names(CI) <- c("lower", "upper")
CI

     lower      upper 
-0.8453322 -0.1511871 

Short way

hyenaCorr$conf.int

[1] -0.8453293 -0.1511968
attr(,"conf.level")
[1] 0.95

CI

     lower      upper 
-0.8453322 -0.1511871 

Testing for zero correlation

\[ H_{0}: \rho = 0\\ H_{A}: \rho \neq 0 \]

Test statistic \[ t = \frac{r}{\mathrm{SE}_{r}} \]

\( t \)-distributed with \( df = n-2 \) (estimated two parameters from data: \( \bar{X} \) and \( \bar{Y} \))

Long way

(tstat <- hyenaCorr$estimate/sderr)

      cor 
-2.819416 

pval <- 2*pt(abs(tstat), df=n-2, lower.tail=FALSE)
unname(pval)

[1] 0.01364843

Short way

cor.test(hyenaData$Age_years, 
         hyenaData$Fundamental_frequency_Hz)

Short way

    Pearson's product-moment correlation

data:  hyenaData$Age_years and hyenaData$Fundamental_frequency_Hz
t = -2.8194, df = 14, p-value = 0.01365
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
 -0.8453293 -0.1511968
sample estimates:
      cor 
-0.601798 

As always, we assume the sample of individuals is a random sample from the population.

The measurements are assumed to come from a joint probability distribution \( p(X,Y) \) called a bivariate normal distribution (function of two variables).

• Relationship between \( X \) and \( Y \) is linear.
• Cloud of points in a scatter plot has a circular or elliptical shape.
• Frequency distributions for \( X \) and \( Y \) (called marginals) separately are normal.

Bivariate normal distribution with \( \rho = 0.7 \)

Bivariate normal distribution with \( \rho = 0.7 \)

• Use a transformation on \( X \) alone, \( Y \) alone, or both.
• Non-parametric techniques (Spearman rank correlation)

Definition: The Spearman’s correlation coefficient is the linear correlation coefficient computed on the ranks of the data.

Practice Problem #3

As human populations became more urban from prehistory to the present, disease transmission between people likely increased. Over time, this might have led to the evolution of enhanced resistance to certain diseases in setted human populations. For example, a mutation in the SLC11A1 gene in humans causes resistance to tuberculosis. Barnes et al. (2011) examined the frequency of the resistant allele in different towns and villages in Europe and Asia and compared it to how long humans had been settled in the site (“duration of settlement”). If settlement led to the evolution of greater resistance to tuberculosis, there should be a positive association between the frequency of the resistant allele and the duration of settlement.

Load the data

'data.frame':   17 obs. of  4 variables:
 $ settlement     : chr  "Catal Hoyuk" "Susa-- others" "Harappa" "Sanxingdui" ...
 $ date           : chr  "6000 BC" "3250 BC" "2725 BC" "2000 BC" ...
 $ duration       : int  8010 5260 4735 4010 3710 2810 2730 2310 2110 1955 ...
 $ alleleFrequency: num  0.99 1 0.948 0.885 0.947 0.854 1 0.769 0.956 0.979 ...

Plot the data

Compute Spearman rank correlation coefficient

cor.test(TBData$duration, 
         TBData$alleleFrequency, 
         method="spearman")

    Spearman's rank correlation rho

data:  TBData$duration and TBData$alleleFrequency
S = 232.64, p-value = 0.001258
alternative hypothesis: true rho is not equal to 0
sample estimates:
     rho 
0.714899 

Definition: A bias called attenuation occurs with measurement error, whereby \( r \) tends to underestimate the magnitude of \( \rho \) (closer to zero on average than the true correlation).

Left: No measurement; Middle: Error in \( X \); Right: Error in both