16.4 – Spearman and other correlations

Introduction
Spearman rank correlation
R code
Other correlations
Questions
Quiz
Chapter 16 contents

Introduction

Pearson product moment correlation is used to describe the level of linear association between two variables. There are many types of correlation estimators in addition to the familiar Product Moment Correlation, r.

Spearman rank correlation

If you take the ranks for X₁ and the ranks for X₂, the correlation of ranks is called Spearman rank correlation, r_s. Spearman correlation is a nonparametric statistic. Like the product moment correlation, it can take values between -1 and +1.

For variables X₁ and X₂, the rank order correlation may be calculated on the ranks as

$\begin{align*} \rho_{s} = 1 - \frac{6 \sum d_{i}^2}{n \left (n^2 - 1 \right )} \end{align*}$

where d_i is the difference between the ranks of X₁ and X₂ for each experimental unit. This formula assumes that there are no tied ranks; if there are, use the equation for the product moment correlation instead (but on the ranks).

R commander has an option to calculate the Spearman rank correlation simply by selecting the check box in the correlation sub menu. However, if the data set is small, it may be easier to just run the correlation in the script window.

Our example for the product moment correlation was between Drosophila fly wing length and wing area (Table 1).

Table 1. Fly wing lengths and area, units mm and mm², respectively (Dohm pers obs.)

Obs	Student	Length	Area
1	S01	1.524	0.446
2	S01	2.202	0.876
3	S01	1.52	0.39
4	S01	1.62	0.51
5	S01	1.71	0.736
6	S03	1.551	0.453
7	S03	2.228	0.882
8	S03	1.46	0.394
9	S03	1.659	0.503
10	S03	1.719	0.535
11	S05	1.534	0.441
12	S05	2.223	0.889
13	S05	1.49	0.391
14	S05	1.633	0.514
15	S05	1.704	0.546
16	S08	1.551	0.453
17	S08	2.228	0.882
18	S08	1.468	0.394
19	S08	1.659	0.503
20	S08	1.719	0.535

Repeated observations by image analysis (ImageJ) were collected by four students (S01, S03, So5, S08) from fixed wings to glass slides.

Here’s the scatterplot of the ranks of fly wing length and fly wing area (Fig 1).

fly wings, area by length
Figure 1. Drosophila wing area (mm²) by wing length (mm).

A nonparametric alternative to the product moment correlation, the Spearman Rank correlation can be obtained directly. The Spearman correlation involves ranking the data, ie, converting data types, from ratio scale data to ordinal scale, then applying the same formula used for the Product moment correlation to the ranked data. The Spearman correlation would be the choice for testing linear association between two ordinal type variables. It is also appropriate in lieu of the parametric product moment correlation when the statistical assumptions are not met, eg, normality assumption.

R code

For the Spearman rank correlation, at the R prompt type

cor.test(Area, Length, alternative="two.sided", method="spearman")

R returns with
    Spearman's rank correlation rho

data:  Area and Length
S = 58.699, p-value = 5.139e-11
alternative hypothesis: true rho is not equal to 0
sample estimates:
      rho 
0.9558658

Alternatively, to calculate either correlation, use R Commander.

Rcmdr: Statistics → Summaries→ Correlation test

Example

BM=c(29,29,29,32,32,35,36,38,38,38,40)
Matings=c(0,2,4,4,2,6,3,3,5,8,6)

cor.test(BM,Matings, method="spearman")
Warning in cor.test.default(BM, Matings, method = "spearman") :
  Cannot compute exact p-value with ties

        Spearman's rank correlation rho

data:  BM and Matings
S = 77.7888, p-value = 0.03163
alternative hypothesis: true rho is not equal to 0
sample estimates:
      rho 
0.6464143

cor.test(BM,Matings, method="pearson")

        Pearson's product-moment correlation

data:  BM and Matings
t = 2.6728, df = 9, p-value = 0.02551
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
 0.1087245 0.9042515
sample estimates:
      cor 
0.6652136

Other correlations

Kendall’s tau

Another nonparametric correlation is Kendall’s tau (τ). Rank the X₁ values, then rank the X₂ values. Count the number of X₁, X₂ pairs that have the same rank (concordant pairs) and the number of X₁, X₂ pairs that do not have the same rank (discordant pairs), Kendall’s tau is then

$\begin{align*} \tau = \frac{\left ( no. \ of \ concordant \ pairs \right ) - \left ( no. \ of \ discordant \ pairs \right ) }{\frac{1}{2} \left ( n - 1 \right )} \end{align*}$

where n is the number of pairs.

Note 1: The denominator is our familiar number of pairwise comparisons if we take k = n

$\begin{align*} C = \frac{k \left ( k - 1 \right )}{2} \end{align*}$

We introduced concordant and discordant pairs when we presented McNemar’s test and cross-classified experimental design in Chapter 9.6.

Example: Judging of Science Fair posters

What is the agreement between two judges, A and B, who evaluated the same science fair posters? Posters were evaluated on if the student’s project was hypothesis-based and judges used a Likert-like scale Strongly disagree (1), Somewhat disagree (2), Neutral (3), Somewhat agree (4), Strongly agree (5).

Table 2. Two judges evaluated six posters for evidence of hypothesis-based project

Poster	Judge.A	Judge.B
1	5	4
2	2	3
3	4	2
4	3	1
5	2	1
6	4	3

A concordant pair represents a poster ranked higher by both judges, while a discordant pair is a poster ranked high by one judge but low by another judge. Poster 1 and poster 5 were concordant pairs,

In R, it is simple to get this correlation directly by invoking the cor.test function and specifying the method equal to kendall. The cor.test assumes that the data are in a matrix, so use the cbind function to bind two vectors together – note the vectors need to have the same number of observations. If the data set is small, it is easier to just enter the data directly in the script window of R commander.

A = c(2,2,3,4,4,5)
B = c(1,3,1,2,3,4)
m = cbind(A,B)
cor.test(A,B, method="kendall")

  Cannot compute exact p-value with ties

        Kendall's rank correlation tau

data:  A and B
z = 1.4113, p-value = 0.1581
alternative hypothesis: true tau is not equal to 0
sample estimates:
      tau 
0.5384615

End of R output

There were no ties in this data set, but we can run the product moment correlation just for comparison

cor.test(A,B, method="pearson")

        Pearson's product-moment correlation

data:  A and B
t = 1.4649, df = 4, p-value = 0.2168
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
 -0.4239715  0.9478976
sample estimates:
      cor 
0.5909091

End R output

Tetrachoric and Polychoric correlation

Tetrachoric correlations used for binomial outcomes (yes, no), polychoric correlation used for ordinal categorical data like the Likert scale. Introduced by Karl Pearson, commonly applied correlation estimate for Item Analysis in psychometric research. Pyschometrics, a sub-discipline within psychology and now a significant part of education research, is about evaluating assessment tools.

R package psych.

R code: Tetrachoric correlation

tetrachoric(x,y=NULL,correct=.5,smooth=TRUE,global=TRUE,weight=NULL,na.rm=TRUE,
     delete=TRUE)

R code: Polychoric correlation

polychoric(x,smooth=TRUE,global=TRUE,polycor=FALSE,ML=FALSE, std.err=FALSE, 
     weight=NULL,correct=.5,progress=TRUE,na.rm=TRUE,  delete=TRUE)

Polyserial correlation

R package polychor. Used to estimate linear association between a ratio scale variable and an ordinal variable.

R code: Polyserial correlation

polyserial(x,y)

Biserial correlation would be a special case of the polyserial correlation, where ordinal variable is replaced by a dichotomous (binomial) variable.

R code: Polyserial correlation

biserial(x,y)

Intra-class correlation coefficient

Both the ICC and the product moment correlation, r, which we introduced in Chapter 16.1, are measures of strength of linear association between two ratio scale variables (Jinyuan et al 2016). But ICC is more appropriate for association between repeat measures of the same thing, eg, repeat measures of running speed. In contrast, the product moment correlation can be used to describe association between any two variables, eg, between repeat measures of running speed, but also between say running speed and maximum jumping height. ICC is used when quantitative measures are organized into paired groups, eg, before and after on same subjects, or cross-classified designs. ICC was introduced in Chapter 12.3 as part of discussion of repeated measures and random effects. ICC is used extensively to assess reliability of a measurement instrument (Shrout and Fleiss 1979; McGraw and Wong 1996).

Example. Data from Table 2

library(psych)
ICC(myJudge, lmer=FALSE)

R output follows

Intraclass correlation coefficients 
                         type  ICC   F df1 df2      p lower bound upper bound
Single_raters_absolute   ICC1 0.40 2.3   5   6  0.166      -0.306        0.84
Single_random_raters     ICC2 0.46 3.9   5   5  0.081      -0.093        0.85
Single_fixed_raters      ICC3 0.59 3.9   5   5  0.081      -0.130        0.90
Average_raters_absolute ICC1k 0.57 2.3   5   6  0.166      -0.880        0.91
Average_random_raters   ICC2k 0.63 3.9   5   5  0.081      -0.205        0.92
Average_fixed_raters    ICC3k 0.74 3.9   5   5  0.081      -0.299        0.95

Number of subjects = 6 Number of Judges = 2
See the help file for a discussion of the other 4 McGraw and Wong estimates

Lots of output, lots of “ICC”. However, rather than explaining each entry, reflect on the type and review the data. Were the posters evaluated repeatedly? Posters were evaluated twice, but only once per judge, so there is a repeated design with respect to the posters. Had the judges been randomly selected from a population of all possible judges? No evidence was provided to suggest this, so judges were a fixed factor (see Chapter 12.3 and Chapter 14.3).

The six ICC estimates reported by R follow discussion in Shrout and Fliess (1979), and our description fits their Case 3: “Each target is rated by each of the same k judges, who are the only judges of interest (p. 421)” Thus, we find ICC for single fixed rater, ICC = 0.59. Note that we would fail to reject the hypothesis that the judges evaluations were associated.

Questions

1. Write up three learning outcomes for this page. Hint: Point your favorite generative AI to this page and ask for help.

2. See Homework 10, Mike’s Workbook for biostatistics

Quiz Chapter 16.4

Spearman and other correlations

Chapter 16 contents

13.3 – Test assumption of normality

Introduction
Goodness of fit tests of normality hypothesis
Example
Which test of normality?
Questions
Quiz
Data set
Chapter 13 contents

Introduction

I’ve commented numerous times that your conclusions from statistical inference are only as good as the validity of making the and applying the correct procedures. This implies that we know the assumptions that go into the various statistical tests, and where possible, we critically test the assumptions. From time to time then I will provide you with “tests of assumptions.”

Here’s one. The assumption of normality, that your data were sampled from a population with a normal distribution for the variable of interest, is key and there are a number of ways to test this assumption. Hopefully as you read through the next section you can extend the logic to any distribution; if the data are presumed to come from a binomial, or a Poisson, or a uniform distribution, then the logic of goodness of fit tests would apply.

How to test normality assumption

It’s not a statistical test per se, but the best is to simply plot (histogram) the data. You can do these graphics by hand, or, install the data mining package rattle which will generate nice plots useful for diagnosing normality issues.

Note 1: rattle (R Analytic Tool To Learn Easily) is a great “data-mining” package. We introduced the package in Chapter 3 — Exploring data.

The rattle histogram plot (Fig 1) superimposes a normal curve over the data, which allows you to “eyeball” the data.

First, the eye test. I used the R-package rattle for this on a data set of comet tail lengths of rat lung cells exposed to different amounts of copper in growth media (scroll to bottom of page or click here to get the data).

In addition to the histogram (Fig 1 top image), I plotted the cumulative function (Fig 1 bottom image). In short, if the data set were normal, then the cumulative frequency plot should look like a straight line.

rattle normal

cumulative frequency

Figure 1. Rattle descriptive graphics on Comet Copper dataset. Dotted line (top image) and red line (bottom image) follow the combined observations regardless of treatment.

So, just looking at the data set, we don’t see clear evidence for a normal-like data set. The top image (Fig 1) looks stacked to the left and the cumulative plot (bottom image) is bumped in the middle, not falling on a straight line. We’ll need to investigate the assumption of normality more for this data set. We’ll begin by discussing some hypothetical points first, then return to the data set.

Goodness of fit tests of normality hypothesis

While graphics and the “eyeball test” are very helpful, you should understand that whether or not your data fits a normal distribution, that’s a testable hypothesis. The null hypothesis is that your sample distribution fits are normal distribution. In general terms, these “fit” hypotheses are viewed as “goodness of fit” tests. Often times, the test is some variation of a $\chi^{2}$ problem: you have your observed (sample distribution) and you compare it to some theoretical expected value (eg, in this situation, the normal distribution). If your data fit a normal curve, then the test statistics will be close to zero.

We have discussed before that the data should be from a normal distributed population. To the extent this is true, then we can trust our statistical tests are performing the way they are expected to do. We can test the assumption of normality by comparing our sample distribution against a theoretical distribution, the normal curve. I’ve shown several graphs in the past that “looked normal”. What are the alternatives for unimodal (single peak) distributions (Fig 2)?

Kurtosis describes the shape of the distribution, whether it is stacked up in the middle (leptokurtosis), or more spread out and flattened (platykurtosis).

Skewness describes differences from symmetry about the middle. Left skew the tail of the distribution extends to the left, ie, smaller values.

“Leptokurtosis”

Leptokurtosis

“Platykurtosis”

Platykurtosis

Negative skew, left skewed

Negative skew, left skewed

Positive skew, right skewed

Positive skew, right skewed

Figure 2. Graphs describing different distributions. From top to bottom: Leptokurtosis, platykurtosis, negative skew, positive skew.

The easiest procedures for goodness of fit tests of normality are based on the $\chi^{2}$ distribution and yield a “goodness of fit” test for normal distribution. We discussed the $\chi^{2}$ distribution in Chapter 6.9, and used the test in Chapter 9.1.

$\begin{align*} \chi^2 = \sum_{i = 1}^{k}\frac{\left ( O_{i}-E{i} \right )^2}{E_{i}} \end{align*}$

where O refers to the observed data (what we’ve got) and E refers to the expected (eg, data from a normal curve with same mean and standard deviation as our sample).

To illustrate, I simulated a data set in R.

Rcmdr: Distributions → Continuous distributions → Normal distribution → Sample from normal distribution

I created 100 values, stored them in column 1, and set the population mean = 125 and population standard deviation = 10. And therefore the population standard error of the mean was 1.0.

The resulting MIN/MAX was from 99.558 to 146.16; the sample mean was 124.59 with a sample standard deviation of 9.9164. And therefore the sample standard error of the mean was 0.9916.

Question: After completing the steps above, your data will be slightly different from mine… Why?

But getting back to the main concept, Does our data agree with a normal curve? We have discussed how to construct histograms and frequency distributions.

Let’s try six categories (why six? we discussed this when we talked about histograms).

All chi-square tests are based on categorical data, so we use the counts per category to get our data. Group the data, then count the number of OBSERVED in each group. To get the EXPECTED values, use the Z-score (normal deviate) with population mean and standard deviation as entered above.

Table 1. Tabulated values for test of normality.

Number of observations	Weight	Normal deviate (Z)	Expected Proportion	Expected number	(Obs – Exp)² / Exp
105	3	less than or equal to -2	0.0228	2.28	0.227368421
105 < 115	17	between -1 & -2 = 0.1587 – 0.0228	0.1359	13.59	0.855636497
115 < 125	34	between -0 & -1 = 0.5 – 0.1587	0.3413	34.13	0.000495166
125 < 135	30	between +0 & +1 = 0.5 – 0.1587	0.3413	34.13	0.499762672
135 < 145	15	between +1 & +2 = 0.9772 – 0.8413	0.1359	13.59	0.146291391
> 145	1	greater than or equal to +2 = 1 – 0.9772	0.0228	2.28	0.718596491
					$\chi^2 = 2.44815064$

Then obtain the critical value of the $\chi^{2}$ with df = 6 – 1 = 5 (see Appendix, Table of chi-square critical values , critical $\chi^{2}$ = 11.1 with df = 5).

Thus, we would not reject the null hypothesis and would proceed with the assumption that our data could have come from a normally distributed population.

This would be an OK test, but different approaches, although based on a chi-square-like goodness of fit, have been introduced and are generally preferred. We have just shown how one could use the chi-square goodness of fit approach to testing whether your data fit a normal distribution. A number of modified tests based on this procedure have been proposed; we have already introduced and used one (Wilks-Shapiro), which is easily accessed in R.

Rcmdr: Summaries → Wilks-Shapiro

Like Wilks-Shapiro, another common “goodness-of-fit” test of normality is the Anderson-Darling test. This test is now included with Rcmdr, but it’s also available in the package nortest. After the package is installed and you run the library (you should by now be able to do this!), then at the R prompt (>) type:

ad.test(dataset$variable)

replacing “dataset$variable” with the name of your data set and variable name.

In the context of goodness of fit, a perfect fit means the data are exactly distributed as a normal curve, and the test statistics would be zero. Differences away from normality increase the value of the test statistic.

How do these tests perform on data?

The histogram of our simulated normal data of 100 observations with mean = 125 and standard deviation = 10 (Fig 3).

Histogram of simulated normal dataset, μ = 125, σ = 10.

Figure 3. Histogram of simulated normal dataset, μ = 125, σ = 10.

and here’s the cumulative plot (Fig 4).

Cumulated frequency of simulated normal dataset, μ = 125, σ = 10

Figure 4. Cumulated frequency of simulated normal dataset, μ = 125, σ = 10.

Results from Anderson-Darling test were A = 0.2491, p-value = 0.7412, where A is the Anderson-Darling test statistic.

Results of the Shapiro-Wilks test on the same data. W = 0.9927, p-value = 0.8716, where W is the Shapiro-Wilks test statistic. We would not reject the null hypothesis in either case because p > 0.05.

Example

Histogram of a data set, highly skewed to the right. 90 observations in three groups of 30 each, with mean = 0 and standard deviation = 1 (Fig 5).

Histogram of simulated normal dataset, μ = 0, σ = 1.

Figure 5. Histogram of simulated normal dataset, μ = 0, σ = 1.

and the cumulative frequency plot (Fig 6).

Cumulated frequency of simulated normal dataset, μ = 0, σ = 1.

Figure 6. Cumulated frequency of simulated normal dataset, μ = 0, σ = 1.

Results from Anderson-Darling test were A = 9.0662, p-value < 2.2e^-16. Results of the Shapiro-Wilks test on the same data. W = 0.6192, p-value = 6.248e^-14. Therefore, we would reject the null hypothesis because p < 0.05.

The Shapiro-Wilk test in Rcmdr

Let’s go back to our data set and try tests of normality on the entire data set, ie, not by treatment groups.

Rcmdr: Statistics → Summaries → Test of normality…

normalityTest(~CometTail, test="shapiro.test", data=CometCopper)

 Shapiro-Wilk normality test

data: CometCopper$CometTail
W = 0.91662, p-value = 0.006038

End of R output

Another test, built on the basic idea of a chi-square goodness of fit is the Anderson Darling test. Some statisticians prefer this test and it is one built-in to some commercial statistical packages (eg, Minitab). To obtain the Anderson-Darling test in R, you need to install a package. After installing nortest package, run the AD test at the command prompt.

require(nortest)
ad.test(CometCopper$CometTail)

Anderson-Darling normality test

data: CometCopper$CometTail
A = 1.0833, p-value = 0.006787

End of R output.

Note 2: Because you’ve installed Rcmdr the Anderson-Darling test is available (since version 2.6) without installing the nortest package. You can call Anderson-Darling via the menu system, Statistics → Summaries → Test of normality … or type in the script the following command

normalityTest(~CometTail, test="ad.test", data=CometCopper)

In contrast, Shapiro-Wilks test is part of the base R package, so can be called directly

shapiro.test(CometCopper$CometTail)

P-values are both much less than 0.05, so we would reject the assumption of normality for this data set.

Which test of normality?

Why show you two tests for normality, the Shapiro-Wilks and Anderson-Darling? The simple answer is that both are good as general tests of normality, both are widely used in scientific papers, so just pick one and go with it as your general test of normality.

The more complicated answer is that each is designed to be sensitive to different kinds of departure from normality. By some measures, the Shapiro-Wilks test is somewhat better (ie, more statistical power to test the null hypothesis), than other tests, but this is not something you want to get into as a beginner. So, I show both of them to you so that you are at least introduced to the concept that there are often more than one way to test a hypothesis. The bottom-line is that plots may be best!

Questions

Work describe in this chapter involves statistical tests of the assumption of normality. It is just as important, maybe more so, to also apply graphics to take advantage of our built-in pattern recognition functions. What graphic techniques, besides histogram, should be used to view the distribution of the data?
In R, what command would you use so that you can call the variable name, CometTail, directly instead of having to refer to the variable as CometCopper$CometTail?
Why are Anderson-Darling, Shapiro-Wilks and other related tests referred to as “goodness of fit” tests? You may wish to review discussion in Chapter 9.1.
The example tests presented for the Comet Copper data set were conducted on the whole set, not by treatment groups. Re-run tests of normality via Rcmdr, but this time, select the By groups option and select Treatment.

Quiz Chapter 13.3

Test assumption of normality

Data set used in this page

Treatment	CometTail
Control	17.86
Control	16.52
Control	14.93
Control	14.03
Control	13.33
Control	8.81
Control	14.70
Control	9.26
Control	21.78
Control	6.18
Control	9.20
Control	5.54
Control	6.72
Control	2.63
Control	7.19
Control	5.39
Control	11.29
Control	15.44
Control	17.86
Contro	l4.25
Copper	53.21
Copper	38.93
Copper	18.93
Copper	30.00
Copper	28.93
Copper	15.36
Copper	17.86
Copper	17.50
Copper	21.07
Copper	29.29
Copper	28.21
Copper	16.79
Copper	21.07
Copper	37.50
Copper	38.22
Copper	17.86
Copper	29.64
Copper	11.07
Copper	35.00
Copper	49.29

16.4 – Spearman and other correlations

Introduction

Spearman rank correlation

R code

Example

Other correlations

Kendall’s tau

Example: Judging of Science Fair posters

Tetrachoric and Polychoric correlation

R code: Tetrachoric correlation

R code: Polychoric correlation

Polyserial correlation

R code: Polyserial correlation

R code: Polyserial correlation

Intra-class correlation coefficient

Questions

Chapter 16 contents

13.3 – Test assumption of normality

Introduction

How to test normality assumption

Goodness of fit tests of normality hypothesis

Example

The Shapiro-Wilk test in Rcmdr

Which test of normality?

Questions

Data set used in this page

Chapter 13 contents