13.3 – Test assumption of normality

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.

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.

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 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”

“Platykurtosis”

Negative skew, left 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 distribution and yield a “goodness of fit” test for normal distribution. We discussed the distribution in Chapter 6.9, and used the test in Chapter 9.1.

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

Then obtain the critical value of the with df = 6 – 1 = 5 (see Appendix, Table of chi-square critical values , critical  = 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).

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

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

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).

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

and the cumulative frequency plot (Fig 6).

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.

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

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

1. 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?
2. 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?
3. 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.
4. 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

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Data set used in this page

Chapter 13 contents

• Introduction
• ANOVA Assumptions
• Why tests of assumption are important
• Test assumption of normality
• Tests for Equal Variances
• References and suggested readings

13.1 – ANOVA Assumptions

Introduction

Like all parametric tests, assumptions are made about the data in order to justify and trust estimates and inferences drawn from ANOVA. While acknowledging the “experimental design” message suggested by the xkcd panel in Figure 1 — the presence of a large effect size between groups mitigates the technical necessities of inferential statistics — of course, not all of our projects will have such obvious results. Understanding the impact of assumptions on inference is a key concept in (bio)statistics: when the p-value is close to the Type I error rate, deviations from assumptions may reduce our confidence in the “statistical significance” of the result.

Figure 1. xkcd.com “Statistics,” https://xkcd.com/2400/.

ANOVA assumptions include:

1. Data come from normal distributed population. View with a histogram or Q-Q plot. Test with Shapiro-Wilks or other appropriate goodness of fit test (see Question 1 below). Normality tests are the subject of Chapter 13.3.
2. Sample size equal among groups.
   • This is an example of a potentially confounding factor — If sample sizes differ, then any difference in means could be simply because of differences in sample size! This gets us into weighted versus unweighted means.
   • You shouldn’t be surprised that modern implementations of ANOVA in software easily handle (adjust for) these known confounding factors. Depending on the program, you’ll see “Adjusted means,” “least squares means,” “Marginal means” etc. This just implies that the group means are compared after accounting for confounding factors.
   • Importantly, as long as sample sizes among the groups are roughly equivalent, normality assumption is not a big deal (low impact on risk of Type I error).
3. Independence of errors. One consequence of this assumption is that you would not view 100 repeated observations of a trait on the same subject as 100 independent data points. We’ll return to this concept more in the next two lectures. Some examples:

• • Colorimetric assay where the signal changes over time, and you measure in order (eg, samples from group 1 first, samples from group 2 second, etc.) — this confounds group with time.
    • The consequence is that you are far more likely to reject the null hypothesis, committing a Type I error.
  • Let’s say you are observing running speeds of ten mongoose. However, it turns out that five of your subjects are actually from the same family, identical quintuplets! Do you really have ten subjects?
  • Compare brain-body mass ratio among different species; this is a classic comparative method problem (Fig 3). Since 1985 (Felsenstein 1985), it was recognized that the hierarchical evolutionary relationships among the species must be accounted for to control for lack of independence among the taxa tested. See Phylogenetically independent contrasts, Chapter 20.12.

4. Equal variances (homogeneity) among groups. See Chapter 13.4  for how to test this for multiple groups.

Impact of assumptions

R — and pretty much all statistics packages — will calculate the ANOVA table and the p-value, but it is up to you to recognize that the P-value is accurate only if the assumptions are met. Violation of the assumption of normality can lead to Type I errors occurring more often than the 5% level. What to do if the assumptions are violated?

If the violation is due to only a handful of the data, you might proceed anyway. But following a significant test for normality, we could avoid the ANOVA in favor of nonparametric alternatives (Chapter 15), or, we might try to transform the data.

Consider a histogram of brains weight measures in grams for a variety of mammals (Fig 2).

Figure 2. Histogram of body mass (g) for 24 mammals (data from Boddy et al 2012).

We will introduce a variety of statistical tests of the assumption of normality in Chapter 13.3, but looking at a histogram as part of our data exploration, we clearly see the data are right-skewed (Fig 2). Is this an example of normal distributed sample of observations? Clearly not. If we proceed with statistical tests on the raw data set, then we are more likely to commit a Type I error (ie, we will reject the null hypothesis more often than we should).

A comment on normality and biology. It is very possible that data may not be normally distributed or have equal variances on the original scale, but a simple mathematical manipulation may take care of that. In fact, in many case in biology that involve growth, many types of variables are expected to not be normal on the original scale. For example, while the relationship between body mass, , and metabolic rate, , in many groups of organisms is allometric and increases positively, the relationship

is not directly proportional (linear) on the original scale. By taking the logarithm of both body mass and metabolic rate, however, the relationship is linear

In fact, taking the logarithm (base 10, base 2, or base e) is often a common solution to both non-normal data (Fig 2) and unequal variances.

Figure 3. Histogram of log₁₀-transformed body mass observations from Figure 2.

Other common transformations include taking the square root or the inverse of the square-root for skewed or kurtotic sample distributions, and the arcsine for frequencies (since frequencies can only be from 0 to 1 — need to “stretch the data” to make frequencies fit procedures like ANOVA). There are many issues about data transformation, but keep in mind three points. After completing the transformation, you should check the assumptions (normality, equal variances) again.

You may need to recode the data before applying a transform. For example, you cannot take the square root or logarithm of negative numbers. If you do not recode the data, then you will lose these observations in your subsequent analyses. In many cases, this problem is easily solved by adding 1 to all data points before making the transform. I prefer to make the minimum value 1 and go from there. The justification for data transformation is basically to realize that there is no necessity to use the common arithmetic or linear scale: many relationships are multiplicative or nonadditive (eg, rates in biology and medicine).

Statistical outlier

Another topic we should at least mention here is the concept of outliers. While most observations tend to cluster around the mean or middle of the sample distribution, occasionally one or more observations may differ substantially from the others. Such values are called outliers, and we note that there are two possible explanations for an outlier

1. the value could be an error.
2. it is a true value.

We encountered a clear outlier in the BMI homework. If the reason is (1), then we go back and either fix the error or delete it from the worksheet. If (2), however, then we have no objective reason to exclude the point from our analyses.

We worry if the outlier influences our conclusions — so it is a good idea to run your analyses with and without the outlier. If your conclusions remain the same, then no worries. If your conclusions change based on one observation, then this is problematic. For the most part you are then obligated to include the outlier and the more conservative interpretation of your statistical tests.

ANOVA is robust to modest deviations from assumptions

A comment about ANOVA assumptions ANOVA turns out to be robust to violations of item (1) or (2). That means unless the data are really skewed or the group sizes are very different, ANOVA will perform well (Type I error rate stays close to the specified 5% level). We worry about this however when p-value is very close to alpha!!

The third assumption is more important in ANOVA.

Like the t-test, ANOVA makes the assumption of equal variances among the groups, so it will be helpful to review why this assumption is important to both the t-test and ANOVA. In the two-sample independent t-test, the pooled sample variance, , is taken as an estimate of the population variance, . If you recall,

where refers to the sum of squares for the first group and refers to the second group sum of squares (see our discussion on measures of dispersion) and refers to the degrees of freedom for the first group and refers to the second group degrees of freedom. We make a similar assumption in ANOVA. We assume that the variances for each sample are the same and therefore that they all estimate the population variance . To say it in another way, we are assuming that all of our samples have identical variability

Once we make this assumption, we may pool (or combine) all of the SSs and DFs for all groups as our best estimate of the population variance, . The trick to understanding ANOVA is to realize that there can be two types of variability: there is variability due to being part of a group (eg, even though ten human subjects receive the same calorie-restricted diet, not all ten will loose the same amount of weight) and there is variability among or between groups (eg, on average, all subjects who received the calorie-restricted diet lost more weight than did those subjects who were on the non-restricted diet).

Example

The encephalization index (or encephalization quotient) is defined as the ratio of size the brain compared to body size. While there is a well-recognized increase in brain size given increased body size, encephalization describes a shift of function to cortex (frontal, occipital, parietal, temporal) from noncortical parts of the brain (cerebellum, brainstem). Increased cortex is associated with increased complexity of brain function; for some researchers, the index is taken as a crude estimate of intelligence. Figure 4A shows plot of brain mass in grams versus body size (grams) for 24 mammal species (data sampled from Boddy et al 2012); Figure 4B shows the same data, but following log₁₀-transform of both variables.

Figure 4. Plot of brain and body weights (A) and log₁₀-log₁₀ transform (B) for a variety of species (data from Boddy et al 2012). The ratio is called encephalization index.

Looking at the two figures the linear relationship between the two variables is obvious in Figure 4B, less so for Figure 4A. Thus, one biological justification for transformation of the raw data is exemplified with the brain-body mass dataset: the association is allometric not additive. The other reason to apply a transform is statistical; the log₁₀-transform improves the normality of the variables. Take a look at the Q-Q plot for the raw data (Fig 5) and for the log10-transformed data (Fig 6).

Figure 5. Q-Q plot, body mass, raw data. Compare to Figure 2.

Figure 5 is a good test of your observation skills — what are you supposed to look for in a Q-Q plot? Follow the dots — the data don’t fall on the straight line. A few even fall outside of the confidence interval (the curved dashed lines), which suggests the data were not normally distributed (see histogram, Fig 2). And for the transformed data, the Q-Q plot is shown in Figure 6.

Figure 6. Q-Q plot same data, log₁₀-transformed, compare to Figure 2.

Compared to the raw data, the transformed data now fall on the line and none are outside of the confidence interval. We would conclude that the transformed data are more normal, thus, better meeting the assumptions of our parametric tests.

Lack of independence among data

Species comparisons are common in evolutionary biology and related fields. As noted earlier, comparative data should not be treated as independent data points (discussed in 20.12 – Phylogenetically independent contrasts). For our 24 species, I plotted the estimate of the phylogeny (timetree.org), represented in Figure 7. Some species are closely related (eg the two elephant shrews, Rhynchocyon genus). Thus, because of the hierarchical, nested nature of evolution — first articulated by Charles Darwin — instead of 24 independent data points, we have eight clusters of points (see Question 4).

Figure 7. Phylogenetic tree of 24 species used in this report.

The conclusion? We don’t have 24 data points, more like 8 points. Because the species are more or less related, there are fewer than 24 independent data points.  Statistically, this would mean that the errors are correlated. Various approaches to account for this lack of independence have been developed; perhaps the most common approach is to apply phylogenetically independent contrasts, a topic discussed in Chapter 20.12. (Boddy et al 2012 used this approach.)

Questions

1. Make a histogram for the brain mass data set.
2. Shapiro-Wilks is one test of normality.
   1. Can you recall the name of the other normality tests we named?
   2. Run tests of normality on both body mass and brain mass data. Interpret the results, frame your response with keywords: “central tendency,” “outliers,” “shape,” and “symmetry.”
3. Calculate moment statistics for the body mass and brain mass data set, both raw and log₁₀-transformed data. Interpret your results, taking care to frame your answer with the following keywords: “shape,” “symmetry,” and “peakedness” (aka “tailedness”).
4. We introduced the lack of independence for comparative data attributed to the evolutionary process. Statistically, the impact is on degrees of freedom; for example, if we ignore the lack of independence in the data set, df = 22 for the slope coefficient in a linear regression. However, given the degrees of freedom would be closer to six! What are the effects on Type I error rate of lack of independence?

Quiz Chapter 13.1

ANOVA assumptions

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Data set and R code used in this page

species, order, body, brain
'Herpestes ichneumon', Carnivora, 1764, 24.1
'Potos flavus', Carnivora, 2620, 31.05
'Vulpes vulpes', Carnivora, 3080, 49.5
'Madoqua kirkii', Cetartiodactyla, 4570, 37
'Sus scrofa', Cetartiodactyla, 1000, 47.7
'Tragulus napu', Cetartiodactyla, 2510, 18.5
'Casinycteris argynnis', Chiroptera, 40.5, 0.92
'Cynopterus brachyotis', Chiroptera, 29, 0.88
'Potorous platyops', Chiroptera, 718, 6.5
'Cercartetus nanus', Diprotodontia, 12, 0.44548
'Dactylopsila palpator', Diprotodontia, 474, 7.15876
'Vombatus ursinus', Diprotodontia, 1902, 11.396
'Crocidura fuscomurina', Eulipotyphla, 5.6, 0.13
'Scalopus aquaticus', Eulipotyphla, 39.6, 1.48
'Sorex hoyi', Eulipotyphla, 2.6, 0.107
'Elephantulus fuscipes', Macroscelidea, 57, 1.33
'Rhynchocyon cirnei', Macroscelidea, 490, 6.1
'Rhynchocyon petersi', Macroscelidea, 717.3, 4.46
'Aotus trivirgatus', Primates, 480, 15.5
'Cebus capucinus', Primates, 590, 53.28
'Leontopithecus rosalia', Primates, 502.5, 11.7
'Dipodomys agilis', Rodentia, 61.4, 1.34
'Peromyscus melanocarpus', Rodentia, 58.8, 1.03
'Sciurus carolinensis', Rodentia, 367, 6.49

The Newick code for the tree in Figure 6.

((Vombatus_ursinus:48.94499077,((Potorous_platyops:47.59556667,Cercartetus_nanus:47.59556667)'14':0.66887333,Dactylopsila_palpator:48.26444000)'13':0.68055077)'11':109.65259681,(((((Crocidura_fuscomurina:33.74066667,Sorex_hoyi:33.74066667)'10':33.03022424,Scalopus_aquaticus:66.77089091)'19':22.55292749,(((Herpestes_brachyurus:54.32144118,(Vulpes_vulpes:45.52834967,Potos_flavus:45.52834967)'9':8.79309151)'22':23.43351523,((Tragulus_napu:43.96862857,Madoqua_kirkii:43.96862857)'8':17.99735995,Sus_scrofa:61.96598852)'6':15.78896789)'30':0.77378567,(Casinycteris_argynnis:35.20000000,Cynopterus_brachyotis:35.20000000)'29':43.32874208)'27':10.79507632)'35':7.13857077,(((Peromyscus_melanocarpus:69.89837667,Dipodomys_agilis:69.89837667)'43':0.64655123,Sciurus_carolinensis:70.54492789)'42':19.27825952,((Leontopithecus_rosalia:18.38385647,Aotus_trivirgatus:18.38385647)'40':1.29720005,Cebus_capucinus:19.68105652)'48':70.14213090)'51':6.63920175)'47':8.99739162,(Elephantulus_fuscipes:39.23366667,(Rhynchocyon_cirnei:15.34500000,Rhynchocyon_petersi:15.34500000)'39':23.88866667)'56':66.22611412)'55':53.13780679);

Chapter 13 contents

• Introduction
• ANOVA Assumptions
• Why tests of assumption are important
• Test assumption of normality
• Tests for Equal Variances
• References and suggested readings

3.2 – Measures of Central Tendency

Introduction to summary statistics.

For a sample of observations we can begin the summary by identifying the “typical” value. Various statistics are used to describe the middle and collectively these are referred to as measures of central tendency. The mean, the median, and the mode are the most common measures of central tendency. In the situation in which we work with data from a population census, we would calculate population descriptive statistics — not inferential statistics; because we more often work with samples from populations, we report sample descriptive statistics.

First we review the familiar arithmetic mean and introduce the weighted mean. Next we introduce “other means,” which you may not be as familiar with.

Means.

There are several means beyond the simple arithmetic average or mean. Here we review a few. We use this topic also to start introducing standard notation we will use throughout the book.

The population mean, μ (pronounced “mu”) is given by

X_i, an observation on the i-th individual

Σ, or “sigma,” which instructs you to add them up (the X’s), from i = 1 (the first observation) to i = N (the last observation).

The sample mean, , pronounced “eks bar,” of a collection of observations is given by 

Weighted arithmetic mean.

In some cases you may have several samples from the same population. If the sample sizes are the same, you can calculate the average of averages without any fuss — just take all of the sample means and add them up, then divide by the total number of samples. If the sample sizes differ, then you needs to weight (W) each sample mean by its sample size. Simply divide each sample mean by its appropriate sample size, then add all of these up. That is the weighted average.

More generally, we can write

For example, consider a variable containing the following observations, for example, over the course of watching birds gathering at a drinking source (eg, recorded over 20 days, a single 5 minute interval per day), we can count how many times zero, one, two, etc birds are observed. 

Table 1. A sample of counts of observations, eg, how many birds

The observation “4” was observed four times; the observation “5” was observed twice, and so on for a total of 15 observations.

What is the arithmetic mean of these 15 observations? To solve this, well, you have a couple of choices. You could copy the numbers down as often as they appear and then calculate the mean in the usual way.

Other “means”.

The arithmetic average (illustrated above) is not the only way to estimate the mean.

The trimmed mean, also called the truncated mean, is a useful approach when data is widely dispersed — data spread away from the middle (see Chapter 3.3). Thus, the trimmed mean will be less influenced compared to the arithmetic mean by outlier data points, ie, data far from other data points in the set.

You would use the trimmed mean to describe the middle of a data set in which a plot shows most of the values are clumped together around a middle – and yet you see a few values that are much smaller or much greater. A specified percentage of the smallest and largest values are removed from the data set and then the simple arithmetic mean is calculated for the trimmed data set. For example, given a data set of daily rainfall for different cities, you might wish to remove the driest 5% and wettest 5% of the days in order to better compare the rainfall trends for the cities.

Calculating the trimmed mean is straight-forward in R: use the same built-in function, mean(), but add some options.

Note 2: This is a good time to remind you how to get help with R commands. Do you recall how to get help in R? 

At the R prompt type

help(mean) 

or

?mean

The R Documentation page for mean() will pop up (assuming you allowed R to install help pages as html). Figure 1 shows a screenshot of a portion of the help page for mean()

Figure 1. A portion of the R help page about the function mean.

From the help page (Fig 1) we can see that we can specify a trimmed mean by adding options to the mean(x, ...) command. For our x variable defined above, get the trimmed mean after 25% of the data are removed.

mean(x, trim=0.25)

[1] 6

From the help page we see that the only required option you need to feed the mean command is the name of the variable, in this case, “x” (it can be, of course, any name provided the data are attached). In this case we removed 25% of the values – 12.5% of the smallest values and 12.5% of the largest values – that’s also called the interquartile mean. We use a dot plot to show these trends (Fig 2).

Figure 2. Dot plot of our x variable with locations of the mean (blue) and the trimmed mean (red).  The Dotplot(x) function in package RcmdrMisc was used in Rcmdr to make this graphic. Arrows were added by hand. Dotplot() example code presented in Chapter 3.4.

If we recalculate a trimmed mean after dropping 10% of the points, or even 40% of the points, we get the same mean value of 6. The trimmed mean is an example of a robust estimator; it’s resistant to the influence of outliers.

Another useful descriptor of the middle is the geometric mean. The geometric mean is useful for calculating the average of ratios.  Geometric mean would be used when you want to compare central tendency for different variables, each differing in scale. For example, gene expression results, reported as fold-changes, for different genes often shows tremendous differences among genes and would be best described by logarithmic scale, not arithmetic scale. Geometric mean expression values would be better choice for central tendency. Other examples are found in economics: for example, calculating compound interest or interest. The geometric mean applies whenever the scale is multiplicative and not additive

The geometric mean is given by the equation

The geometric mean (gm) is equivalent to log-transforming your data, then calculating the arithmetic mean, and transforming the result back (with the antilog exponent.) As you recall, for our simple data set the arithmetic mean was 6.2. The geometric mean for this data was 5.977. Taking the natural log for each of the values from our simple data set, then calculating the arithmetic mean we have 1.788. 

The antilog of this value is

exp(1.788)

[1] 5.977

Another frequently encountered mean is the harmonic mean, which is defined by the equation

Harmonic mean is appropriate for averaging rates. For example, what is the average speed traveled if you travel 30 miles per hour (mph) between point A and B, then on the return trip, your speed was 40 mph? If you think (30 + 40)/2 = 35 mph, then this would be incorrect — after all, the distance covered has not changed, just the time. The harmonic mean returns 34.2 mph. Let

y = c(30,40)

The harmonic mean returns 34.2 mph ( see below “How to calculate these other means”)

Both harmonic and geometric means apply for values greater than zero.

How to calculate these “other” means.

In Microsoft Excel, calculate geometric mean via the function GEOMEAN(); calculate harmonic mean via the function HARMEAN().

The base R (and Rcmdr) doesn’t have built in functions for these, although you could download and install some R packages which do (eg, package psych, geometric.mean(variable), harmonic.mean(variable)). It is quicker to just to calculate these by submitting a snippet of code into the script window

For geometric mean of variable “x” at the R prompt type

exp(mean(log(x)))

For harmonic mean of variable “x” at the R prompt type

1/mean(1/x)

where is the base of the natural logarithm, Euler’s number, and log is the natural logarithm (in R, to get log to other bases you can use log10 for base 10 logarithm or log2 for base 2 logarithm, or  log(x, base = n) for any base n of the variable x, and variable is the name of the variable you wish to do the calculations on.

R code: Do try on your own!

Here’s some numbers to try your hand. For example, create a variable containing a few numbers, any numbers. and write it to the variable named z

myZ = c(3,4,6,7,9,11,4)

Now, calculate the arithmetic mean, the geometric mean, and the harmonic mean for the variable myZ. You should get Table 4.2)}

Table 2. Comparison of different means for myZ.

Try three more. In R (or R Commander script window), create three new variables.

varA = c(3,3,3,3)

varB = c(1,2,3,12)

varC = c(-3,0,1,3)

Now, calculate the arithmetic mean, geometric mean, and harmonic mean for each variable.

For the simple arithmetic mean

mean(varA)

For the geometric mean, use the formula above

exp(mean(log(varA)))

For the harmonic mean, use the formula above

1/mean(1/varA)

What did you get?

Other measures of central tendency

Median 

The median, which lacks an accepted notation — we’ll go with , divides a set of observed numbers into two equal halves. Half the observations are above the median, half of the observations are below the median. Arrange data from lowest to highest, take the middle measurement:

odd number of measurements minus the middle value minus even number of measurements minus average of the 2 middle values

Or, more succinctly, we have

To get the median in R type at the R prompt

median(variable)

and of course, replace variable with the name of the variable containing the numbers. For our x variable created earlier, the function median returns in R

[1] 6

[Note that the median for x was the same as the trimmed mean for x, which is consistent with with our view that the trimmed mean is a robust estimator of the middle of a data set.

Mode

Mode is another way to express the middle and it refers to the most frequent occurring measurement. Use of mode makes most sense for discrete or countable numbers. For a normal distribution, the mean, median and mode will be the same value. Note that a data set may have more than one mode. For example, what is the mode for the variable  we created earlier?

myX = c(4,4,4,4,5,5,6,6,6,7,7,8,8,8,15)

For this small data set we see that “4” is the most frequent with a count of four occurrences in the set.

Mode would seem like a straightforward function in R. However, it turns out there is not a mode function in the base package.

A little explanation is in order. In R, typing mode at the R prompt like so

mode(myX)

returns

[1] "numeric" 

Not the answer we were expecting. In R, mode command is used to tell you what the mode (ie, way or manner in which some task is accomplished) of storage is for the variables.

In order to get the statistical mode we want, we either hunt down a package that contains mode estimation (eg, install the package DescTools use the Mode function — see below), or we can write a little code.

A quick Google search found a number of answers at stackoverflow.com (eg, question 2547402). The simplest response was to use names and max commands like so

temp = table(as.vector(myX))

names (temp)[temp==max(temp)]

Why the median may be a better middle than the mean

Comparing the two measures of central tendency can tell you without plotting how your data are distributed about the middle. Sample distributions are discussed in Chapter 6. 

• When the distribution of the data is symmetric or normally distributed (discussed in Chapter 6.7) then the mean and the median will be about the same value
• When data are right-skewed (a few large values), then the mean will be greater than the median.
• When the data are left-skewed (a few small values), then the mean will be less than the median.

Here’s an illustration (Fig 3). I sampled 100 points from a random normal distribution with mean zero and standard deviation 0ne and another 100 points from a log-normal distribution also with mean zero and standard deviation one. In Figure 3, the histograms (see Chapter 4.2 – Histograms), along with summary statistics (see Note 4). Means are indicated with red arrows and medians are indicated with blue arrows.

Figure 3. Normal and lognormal distributions with mean (red) and median (blue) noted for comparison.

So the median is a better descriptor or the central tendency of a sample distribution when the distribution is NOT normally distributed.

Scaling and transformation of data

Sometimes it is useful to standardize your data so that the variables all have the same scale. One algorithm for standardization is called normalization. Normalization implies that you correct the data so that data has a mean, μ, of zero, and a standard deviation, σ, of 1 (unit variance). There are several ways to standardize, each with strengths and limitations. To normalize we use the Z-score equation (see Chapter 6.7 for other uses of Z score).

where X_i is each observation in your data set.

Normalization will make outliers, the few points in a data set that are noticeably different from the central tendency of the rest of the data, smaller and less influential. When you normalize multiple sets of data, then each will have the same mean (0) and variance (unit variance), but the ranges will differ. An example of this is the simple product moment correlation — by standardizing you change the variances for the different variables to have the same unit variance.

As we will see later in class it is also useful to expand or contract the variability of the data or to change the shape of the distribution (if the data is not normally distributed). For example, if you compare individuals of a population for many morphological traits (eg body size, growth rate), the spread of points (called a distribution) will look more like a Poisson distribution (not symmetrical about the mean, a few individuals may be much larger…). This is partly due to the way in which morphological traits are measured. We normally measure body size on a linear scale (inches or centimeters). However, body size is affected by physiological processes that are more related to volume. Therefore, the more appropriate scale of measurement is on a log scale. We can transform the data measured on a linear scale to a log scale. For morphological traits this can produce a distribution that is normally distributed (bell shape). There are many more statistical procedures for data that is normally distributed than there are statistical procedures for Poisson distributions or any other type of distribution. Additional discussion about data transformation is introduced in Chapter 13.3.

You can always uncode or unstandardize your data after performing the statistical procedures and return to the original scale. In fact, when reporting descriptive statistics you should report the untransformed, uncoded data. Moreover, you will find it useful to report means adjusted for other variables (eg, from ANOVA or regression); if the ANOVA or regression equations are performed on transformed or coded data you would want to back calculate to the original scale after applying the ANOVA or regression adjustments. This advise will make more sense after we’ve discussed ANOVA (Chapter 12) and linear regression (Chapter 17).

R operators

The names command can be used to retrieve the names contained in the variable (if text types) or to set the names of the observations, which is what we are using it for here. We set the numbers to text names “4”, “5”, etc. then find the maximum count of named items in the temp table. The double equals operator (==) is used to tell R to find the object that is “equal to” something we specify, in this case, the max value (R Language Definition 2014). Table 3 shows common operators in R.

Table 3. Common arithmetic* and comparison** operators

help(Arithmetic)

help(Comparison)

The R package modeest has a number of algorithms for calculating the mode, depending on the kind of data you are working with. After installing the package and its dependencies, type at the R prompt 

require(modeest)
mfv(myX)
[1] 4

Creating objects in R

Everything in R is an object (Chambers 2008). We provide many worked examples in Part 07. Working with your own data – Mike’s Workbook for Biostatistics. In brief, create a variable in R by assigning the vector x, either directly at the R prompt or in a script window (Rcmdr, RStudio), like so

myX = c(4,4,4,4,5,5,6,6,6,7,7,8,8,8,15)

The function c(), which stands for combine  is used to combine the set of numbers into the object, myX. For small sets like this you may find it convenient to enter the values one by one and let R store it into the vector for you. For R installed on your computer, use the scan() function and your keyboard — scan() doesn’t work consistently in serverless R instances (eg., Google CoLab). Careful! Make sure that you remember to assign the results from scan to a vector.

I’ll create the object tryScan just to distinguish it from x, although I will enter the same values. Until R receives an interrupt signal from you, it will prompt you to enter numbers one row at a time. When you’ve reached the end, use the keyboard shortcut Ctrl + q (Windows) or Cmd + q (macOS) to interrupt keyboard input. 

tryScan = scan()
1: 4
2: 4
3: 4
4: 4
5: 5
6: 5
7: 6
8: 6
9: 6
10: 7
11: 7
12: 8
13: 8
14: 8
15: 15
Read 15 items

tryScan
 [1] 4 4 4 4 5 5 6 6 6 7 7 8 8 8 15

The function tryScan is a very useful command, with many options, and can be used for more than keyboard entry. For example, you can paste from your computer’s clipboard a column of numbers from your spreadsheet.

R code calculate central tendency

Once we have the vectormyX, calculate the mean by entering at the R prompt

mean(myX)

and you should get the answer of 6.466667

And of course, you don’t type in the R prompt >, right?

Or, for the better option, create two variables, one containing the list of observed numbers and the second that contains the frequency for each observed number in the series. You would then use the command for weighted mean.

myY = c(4,5,6,7,8,15)

myW = c(4/15, 1/15, 3/15, 2/15, 4/15, 1/15)

Note — you can check that the frequencies sum to 1 by using the sum command like so

sum(myW)

For the weighted mean, the command is

weighted.mean(myY,myW)

and, the answer returned is 6.466667, the same as before.

There’s an R package for that.

We list code examples for all of the statistics described on this page, but as you can imagine, there are packages that include these and other useful functions, or “wrapper functions” that make writing code easier in R. A good example is the DescTools package. One advantage of using a package like DescTools is that it cuts through the need to remember specific commands for basic data exploration. For example, just call the Desc() function to yield several measures of central tendency and dispersion (see Chapter 3.3). 

The R code and output follows, along with a plot (Fig 4). The mean and median are highlighted in orange.

Desc(myX)
──────────────────────────────────────────────────────────────────────────────── 
myX (numeric)

length      n   NAs unique   0s  mean meanCI'
    15     15     0      6    0  6.47    4.92
       100.0%  0.0%        0.0%          8.02

   .05    .10   .25 median  .75   .90     .95
  4.00   4.00  4.50   6.00 7.50  8.00   10.10

range      sd vcoef    mad  IQR  skew    kurt
11.00    2.80  0.43   2.97 3.00  1.72    2.94

  value freq   perc cumfreq cumperc
1     4    4  26.7%       4   26.7%
2     5    2  13.3%       6   40.0%
3     6    3  20.0%       9   60.0%
4     7    2  13.3%      11   73.3%
5     8    3  20.0%      14   93.3%
6    15    1   6.7%      15  100.0%

' 95%-CI (classic)

Figure 4. Histogram, box plot, and cumulative distribution function plot generated by default by Desc() call.

Lots of output! This is a good example of how to explore data — describe the middle (eg, mean and median), but also the variability about the middle (eg, range, standard deviation — sd, mean absolute deviation — mad, coefficient of variation — vcoef), and distribution shape (quantiles — 0.05, 0.25 etc, kurt, skew). The plots — histogram, box plot, and cumulative distribution plots — are explained in Chapter 4.

Use aggregate() to calculate summary statistics by group.

So far, all code presumes need to calculate descriptive statistics one vector at a time. In general, we need summary statistics for more than one group or strata. For example, we may ask, what are the average body temperature readings measured by IR device on different body areas from several subjects (question 8a, Homework 2A: Descriptive statistics in Mike’s Workbook for Biostatistics).

The aggregate() function in R simplifies calculating means by groups by providing a concise and efficient way to apply a summary function (like mean()) to a variable, grouped by one or more other variables. The format for aggregate is

 aggregate(x = variable_to_aggregate ~ grouping_variable, FUN = mean, data = your_dataframe)

Questio8a had two levels of body regions, “Forehead” and “Throat.” Assuming the variable is named bodyTemp and the grouping variable is called bodyGroup and stored in the attached data.frame question08a, then the following R code provides the requested summary statistics by groups.

aggregate(bodyTemp ~ bodyGroup, data = question08a, 
function(x) c(mean = mean(x)))

Additional statistics can be requested by adding one after the other following mean(x), eg, median = median(x).

The advantage of use of R Commander, it’s easy to get multiple statistics by groups by following the drop down menus, beginning with Statistics > Summaries > Table of statistics… (Fig 5).

Figure 5. Screenshot R Commander, summary statistics by group.

Questions

1. Find the help page in R for the median function. How does the function handle missing values?
2. For a simple data set like the following myY <- c(1,1,3,6) you should now be able to calculate, by hand, the
   • mean • median • mode
3. If the observations for a ratio scale variable are normally (symmetrically) distributed, which statistic of central tendency is best (eg, less sensitive to outlier values)?
4. In the names() command, what do you think the result will be if you replace max in the command with min?
5. If data are right skewed, what will be the order of the mean, median, and mode?
6. Body mass of Rhinella marina (formerly Bufo marinus, Fig 6),
   bufo <- c(71.3, 71.4, 74.1, 85.4, 85.4, 86.6, 97.4, 99.6, 107, 115.7, 135.7, 156.2)

Figure 6. Female Rhinella marina (formerly Bufo marinus), Chaminade University campus. Body length 23.5 cm. Click image to view full sized image.

7. Under what circumstances is it appropriate to transform raw data?
8. What does “standardize” or “normalize” imply as actions to take with data?

Quiz Chapter 3.2

Measures of Central Tendency

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Chapter 3 contents

• Exploring data
• Data types
• Measures of Central Tendency
• Measures of dispersion
• Estimating parameters
• Statistics of error
• References and suggested readings