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

8.5 – One sample t-test

Introduction.

We’re now talking about the traditional, classical two group comparison involving continuous data types. Thus begins your introduction to parametric statistics. One sample tests involve questions like, how many — what proportion of — people would we expect are shorter or taller than two standard deviations from the mean? This type of question assumes a population and we use properties of the normal distribution and, hence, these are called parametric tests because the assumption is that the data has been sampled from a particular probability distribution.

However, when we start asking questions about a sample statistic (eg, the sample mean), we cannot use the normal distribution directly, ie, we cannot use Z and the normal table as we did before (Chapter 6.7). This is because we do not know the population standard deviation and therefore must use an estimate of the variation (s) to calculate the standard error of the mean.

With the introduction of the t-statistic, we’re now into full inferential statistics-mode. What we do have are estimates of these parameters. The t-test — aka Student’s t-test — was developed for the purpose of testing sample means when the true population parameters are not known.

The equation of the one sample t-test. Note the resemblance in form with the Z-score!

where is the sample standard error of the sample mean (SEM).

For example, weight change of mice given a hormone (leptin) or placebo. The , but under the null hypothesis, the mean change is “really” zero (). How unlikely is our value of 5 g?

Speaking of null hypotheses, can you say (or write) the null and alternative hypotheses in this example? How about in symbolic form?

We want to know if our sample mean could have been obtained by chance alone from a population where the true change in weight was zero.

$s = 3, n = 20$

and

$s_{\overline{X}} = \frac{s}{\sqrt{n}} = \frac{3}{\sqrt{20}} = 0.6708$

and we take these values and plug them into our equation of the t-test

$t = \frac{\left(5 – 0\right)}{0.67} = 7.45$

Then recall that Degrees of Freedom are DF = n – 1 so we have DF = 20 – 1 = 19 for the one sample t-test. And the Critical Value is found in the appropriate table of critical values for the t distribution (Fig 1)

Figure 1. Table of a portion of the Critical values of the t distribution. Red selections highlight critical value for t-test at α = 5% and df = 19.

Or, and better, use R

qt(c(0.025), df=19, lower.tail=FALSE)

where qt() is function call to find t-score of the p^th percentile (cf 3.3 – Measures of dispersion) of the Student t distribution. For a two tailed test, we recall that 0.025 is lower tail and 0.025 is upper tail.

In this example we would be willing to reject the Null Hypothesis if there was a positive OR a negative change in weight.

This was an example of a “two-tailed test” which is “2-tail” or α₍₂₎ in Table of critical values of the t distribution.

Critical Value for α₍₂₎ = 0.05, df = 19, = 2.093

Do we accept or reject the Null Hypothesis?

A typical inference workflow.

Note the general form of how the statistical test is processed, a form which actually applies to any statistical inference test.

1. Identify the type of data
2. State the null hypothesis (2 tailed? 1 tailed?)
3. Select the test statistic (t-test) and determine its properties
4. Calculate the test statistic (the value of the result of the t-test)
5. Find degrees of freedom
6. For the DF, get the critical value
7. Compare critical value to test statistic
8. Do we accept or reject the null hypothesis?

And then we ask, given the results of the test of inference, What is the biological interpretation? Statistical significance is not necessarily evidence of biological importance. In addition to statistical significance, the magnitude of the difference — the effect size — is important as part of interpreting results from an experiment. Statistical significance is at least in part because of sample size — the large the sample size, the smaller the standard error of the mean, therefore even small differences may be statistically significant, yet biologically unimportant. Effect size is discussed in Ch9.1 – Chi-square test: Goodness of fit, Ch11.4 – Two sample effect size and Ch12.5 – Effect size for ANOVA.

R Code.

Let’s try a one-sample t-test. Consider the following data set: body mass of four geckos and four Anoles lizards (Dohm unpublished data).

For starters, let’s say that you have reason to believe that the true mean for all small lizards is 5 grams (g).

Geckos: 3.186, 2.427, 4.031, 1.995
Anoles: 5.515, 5.659, 6.739, 3.184

Get the data into R (Rcmdr)

By now you should be able to load this data in one of several ways. If you haven’t already entered the data, check out Part 07. Working with your own data in Mike’s Workbook for Biostatistics.

Once we have our data.frame, proceed to carry out the statistical test.

To get the one-sample t-test in Rcmdr, click on Statistics → Means → Single-sample t-test… Because there is only one numerical variable, Body.mass, that is the only one that shows up in the Variable (pick one) window (Fig 2).

Figure 2. Screenshot Rcmdr single-sample t-test menu.

Type in the value 5.0 in the Null hypothesis: m = u box.

Question 1: Quick! Can you write, in plain old English, the statistical null hypothesis???

Answer 1: For example: No difference between gecko and Anolis lizard mean body mass.

Click OK

The results go to the Output Window.

t.test(lizards$Body.mass, alternative='two.sided', mu=5.0, conf.level=.95)
One Sample t-test
data: lizards$Body.mass
t = -1.5079, df = 7, p-value = 0.1753
alternative hypothesis: true mean is not equal to 5
95 percent confidence interval:
2.668108 5.515892
sample estimates:
mean of x 4.092

end of R output

Let’s identify the parts of the R output from the one sample t-test. R reports the name of the test and identifies

1. The dataset$variable used (lizards$Body.mass). The data set was called “lizards” and the variable was “Body.mass”. R uses the dollar sign ($) to denote the dataset and variable within the data set.
2. The value of the t test statistic was (t = -1.5079). It is negative because the sample mean was less than the population mean — you should be able to verify this!
3. The degrees of freedom, df = 7
4. The p-value = 0.1753
5. 95% confidence interval of the population mean; lower limit = 2.668108, upper limit = 5.515892
6. The sample mean = 4.092

Take a step back and review.

Let’s make sure we “get” the logic of the hypothesis testing we have just completed.

Consider the one-sample t-test.

Step 1. Define H_O and H_A. The null hypothesis might be that a sample mean, , is equal to μ = 5.

$H_0 : \mu = 5$

The alternate is that the sample mean is not equal to 20.

$H_A : \mu \ne 5$

Where did the value 5 come from? It could be a value from the literature (does the new sample differ from values obtained in another lab?). The point is that the value is known in advance, before the experiment is conducted, and that makes it a one-sample t-test.

One tailed hypothesis or two?

We introduced you to the idea of “tails of a test” (Ch08.4). As you should recall, a null/alternative hypothesis for a two-tailed test may be written as

Null hypothesis

versus the alternative hypothesis

where  is the sample mean and is the population mean.

Alternatively, we can write one-tailed tests of null/alternative hypothesis

for the null hypothesis versus the alternative hypothesis

Question 2: Are all possible outcomes of the one-tailed test covered by these two hypotheses?

Answer 2: Yes

Question 3: What was the SEM for this problem?

Answer 3: It would be the sample standard deviation divided by the square root of the sample size.

Step 2. Decide how certain you wish to be (with what probability) that the sample mean is different from μ. As stated previously, in biology, we say that we are willing to be incorrect 5% of the time (Cowles and Davis 1982; Cohen 1994). This means we are likely to correctly reject the null hypothesis 100% – 5% = 95% of the time, which is the definition of statistical power. We do this by setting the Type I error to be 5% (alpha, α = 0.05). The Type I error is the chance that we will reject a null hypothesis, but the true condition in the population we sampled was actually “no difference.”

Step 3. Carry out the calculation of the test statistic. In other words, get the value of t from the equation above by hand, or, if using R (yes!) simply identify the test statistic value from the R output after conducting the one sample t test.

Step 4. Evaluate the result of the test. If the value of the test statistic is greater than the critical value for the test, then you conclude that the chance (the P-value) that the result could be from that population is not likely and you therefore reject the null hypothesis.

Question 4: What is the critical value for a one-sample t-test with df = 7?

Answer 4: From R, we get + 2.365 for the two-tailed test. R code was qt(c(.025), df=7, lower.tail=FALSE)

Hint; you need the table or better, use R

Rcmdr: Distributions → Continuous distributions → t distributions → t quantiles

You also need to know three additional things to answer this question.

1. You need to know alpha (α), which we have said generally is set at 5%.
2. You also need to know the degrees of freedom (DF) for the test. For a one sample t-test, DF = n – 1, where n is the sample size.
3. You also must know whether your test is one or two-tailed.
4. You then use the t-distribution (the tables of the t-distribution at the back of your book) to obtain the critical value. Note that if you use R, the actual p-value is returned.

Why learn the equations when I can just do this in R?

Rcmdr does this for you as soon as you click OK. Rcmdr returns the value of the test statistic and the p-value. R does not show you the critical value, but instead returns the probability that your test statistic is as large as it is AND the null hypothesis is true. From our one-sample t-test example, the Rcmdr output. The simple answer is that in order to understand the R output properly you need to know where each item of the output for a particual test comes from and how to interpret it. Thus, the best way is to have the equations available and to understand the algorithmic approach to statistical inference.

And, this is as good of time as any to show you how to skip the RCmdr GUI and go straight to R.

First, create your variables. At the R prompt enter the first variable

liz <- c("G","G","G","G","A","A","A","A")

and then create the second variable

bm <- c(3.186,2.427,4.031,1.995,5.515,5.659,6.739,3.184)

Next, create a data frame. Think of a data frame as another word for worksheet.

lizz <- data.frame(liz,bm)

Verify that entries are correct. At the R prompt type “lizz” wthout the quotes and you should see

lizz
 liz bm
1 G 3.186
2 G 2.427
3 G 4.031
4 G 1.995
5 A 5.515
6 A 5.659
7 A 6.739
8 A 3.184

End of R output

Carry out the t-test by typing at the R prompt the following

t.test(lizz$bm, alternative='two.sided', mu=5, conf.level=.95)

And, like the Rcmdr output we have for the one-sample t-test the following R output

One Sample t-test
data: lizards$Body.mass
t = -1.5079, df = 7, p-value = 0.1753 alternative hypothesis: true mean is not equal to 5
95 percent confidence interval:
2.668108 5.515892
sample estimates:
mean of x
4.092

End of R output

which, as you probably guessed, is the same as what we got from RCmdr.

Question 5: From the R output of the one sample t-test, what was the value of the test statistic?

1. -1.5079
2. 7
3. 0.1753
4. 2.668108
5. 5.515892
6. 4.092

Answer 5: -1.5079

Why fall back on statistical tables? Think of this as a basic skill. In statistics and for some statistical tests, Rcmdr and other software may not provide the information needed to decide that your test statistic is large, and a table in a statistics book is the best way to evaluate the test.

For now, double check Rcmdr by looking up the critical value from the t-table.

Check critical value against our test statistic

Df = 8 – 1 = 7

The test is two-tailed, therefore α(2)

α = 0.05 (note that two-tailed critical value is 2.365. T was equal to 1.51 (since t-distribution is symmetrical, we can ignore the negative sign), which is smaller than 2.365 and so we would agree with Rcmdr — we cannot reject the null hypothesis.

Question 6: From the R output of the one sample t-test, what was the P-value?

1. -1.5079
2. 7
3. 0.1753
4. 2.668108
5. 5.515892

Answer 6: 0.1753

Question 7: We would reject the null hypothesis

1. False
2. True

Answer 7: False — p-value, 17.5%, is greater than Type I error of 5%.

Questions

Seven questions, with answers, were provided for you within the text in this chapter. Here’s one more, but without answers.

8. Here’s a small data set for you to try your hand at the one-sample t-test and Rcmdr. The dataset contains cell counts, five counts of the numbers of beads in a liquid with an automated cell counter (Scepter, Millipore USA). The true value is 200,000 beads per milliliter fluid; the manufacturer claims that the Scepter is accurate within 15%. Does the data conform to the expectations of the manufacturer? Write a hypothesis then test your hypothesis with the one-sample t-test. Here’s the data.

Quiz Chapter 8.5

One sample t-test

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Chapter 8 contents

• Introduction
• The null and alternative hypotheses
• The controversy over proper hypothesis testing
• Sampling distribution and hypothesis testing
• Tails of a test
• One sample t-test
• Confidence limits for the estimate of population mean
• References and suggested readings

8.3 – Sampling distribution and hypothesis testing

Introduction

Understanding the relationship between sampling distributions, probability distributions, and hypothesis testing is the crucial concept in the NHST — Null Hypothesis Significance Testing — approach to inferential statistics. is crucial, and many introductory text books are excellent here. I will add some here to their discussion, perhaps with a different approach, but the important points to take from the lecture and text are as follows.

Our motivation in conducting research often culminates in the ability (or inability) to make claims like:

1. “Total cholesterol greater than 185 mg/dl increases risk of coronary artery disease.”
2. “Average height of US men aged 20 is 70 inches (1.78 m).”
3. “Species of amphibians are disappearing at unprecedented rates.”

Lurking beneath these statements of “fact” for populations (just what IS the population for #1, for #2, and for #3?) is the understanding that not ALL members of the population were recorded.

How do we go from our sample to the population we are interested in? Put another way — How good is our sample? We’ve talked about how “biostatistics” can be generalized as sets of procedures you use to make inferences about what’s happening in populations. These procedures include:

• Have an interesting question
• Experimental design (Observational study? Experimental study?)
• Sampling from populations (Random? Haphazard?)
• Hypotheses: H_O and H_A
• Estimate parameters (characterize the population)
• Tests of hypotheses (inferences)

We have control of each of these — we choose what to study, we design experiments to test our hypotheses…We have already introduced these topics (Chapters 6 – 8).

We also obtain estimates of parameters, and inferential statistics applies to how we report our descriptive statistics (Chapter 3). Estimates of parameters like the sample mean and sample standard deviation can be assessed for accuracy and precision (eg, confidence intervals).

Sampling distribution

Imagine drawing a sample of 30 from a population, calculating the sample mean for a variable (eg, systolic blood pressure), then calculating a second sample mean after drawing a new sample of 30 from the same population. Repeat, accumulating one estimate of the mean, over and over again. What will be the shape of this distribution of sample means? The Central Limit Theorem states that the shape will be a normal distribution, regardless of whether or not the population distribution was normal, as long as the sample size is large (ie, Law of Large Numbers). We alluded to this concept when we introduced discrete and continuous distributions (Chapter 6).

It’s this result from theoretical statistics that allows us to calculate the probability of an event from a sample without actually carrying out repeated sampling or measuring the entire population.

A worked example

To demonstrate the CLT we want R to help us generate many samples from a particular distribution and calculate the same statistic on each sample. We could make a for loop, but the replicate() function provides a simpler framework.  We’ll sample from the chi-square distribution. You should extend this example to other distributions on your own, see Question 5 below.

Generate x replicate samples (eg, x = 10, 100, 1000, one million) of 30 each from chi-square distribution with one degree of freedom, test the distribution against null hypothesis (assume normal distributed, eg, Shapiro-Wilk test, see Chapter 13.3), then make a histogram (Chapter 4.2), like Figure 1 or Figure 2.

x.10 <- replicate(10, {
my.mean <- rchisq(30, 1)
mean(my.mean)
})

normalityTest(~x.10, test="shapiro.test")
hist(x.10, col="orange")

Result from R

Shapiro-Wilk normality test

data: x.10
W = 0.87016, p-value = 0.1004

Figure 1. means of ten replicate samples drawn at random from chi-square distribution, df = 1.

Modify the code to draw 100 samples, we get Fig 2.

Figure 2. means of 100 replicate samples drawn at random from chi-square distribution, df = 1. Results from Shapiro-Wilks test: W = 0.97426, p-value = 0.04721.

And finally, modify the code to draw one million samples, we get Figure 3.

Figure 3. means of one million replicate samples drawn at random from chi-square distribution, df = 1. Normality test will fail to run, sample size of 5000 limit.

How to apply sampling distribution to hypothesis testing

First, a reminder of some definitions.

Estimate = we will always (almost) concern ourselves with how good our sample mean (such values are called estimates) is relative to the population mean, the thing we really want, but can only hope to get an estimate of.

Accuracy = how close to the true value is our measure?

Precision = how repeatable is our measure?

How can we tell if we have a good estimate? We want an estimate with an evaluation for accuracy and for precision. The sampling error provides an assessment of precision, whereas the confidence interval provides a statement of accuracy. We need an estimate of the sampling error for the statistic,

Sample standard error of the mean

We introduced sample error of the mean in section 3.4 of Chapter 3. Everything we measure can have a corresponding statement about how accurate (sampling error) is our estimate! First, we begin by asking, “how accurate is the mean that we estimate from a sample of a population?” How do we answer this? We could prove it in the mathematical sense of proof (and people have and do) OR we can use the computer to help. We’ll try this approach in a minute.

What we will show relates to the standard error of the population mean (SEM) or

, whose equation is shown below.

or equivalently, from the standard deviation we have

Note that the SEM takes the variance and divides through by the sample size. In general, then, the larger the sample size, the smaller the “error” around the mean. As we work through the different statistical tests, t-tests, analysis of variance, and related, you will notice that the test statistic is calculated as a ratio between a difference or comparison divided by some form of an error measurement. This is to remind you that “everything is variable.”

A note on standard deviation (SD) and standard error of the mean (SEM): SD estimates the variability of a sample of X_i‘s whereas SEM estimates the variability of a sample of means.

Let’s return to our thought problem and see how to demonstrate a solution. First, what is the population? Second, can we get the true population mean?

One way, a direct (but impossible?) approach would be to measure it — get all of the individuals in a population and measure them, then calculate the population mean. Then, we could compare our original sample mean against the true mean and see how close it was. This can be accomplished in some limited cases. For example, the USA conducts a census of her population every ten years, a procedure which costs billions of dollars. We can then compare samples from the different states or counties to the USA mean. And these statistics are indeed available via the census.gov website. But even the census uses sampling — individuals are randomly selected to answer more questions and from this sample trends in the population are inferred.

So, sampling from populations is the way to go for most questions we will encounter. The procedures we will use to show how a sample mean relates to the population mean are general and may be used to show how any estimate of a variable (sample mean and sample standard deviation, etc.), relates to properties of a parameter. We’ll get to the other issues, but for now, think about sample size.

Sampling from populations is necessary and inevitable, and, to a certain extent, under your control. But how many individuals do we need? The quick answer is for me to direct your attention to the equation for the SEM. Can you see in that ratio the secret to obtaining more precise estimates? There are many ways to approach this question, but let’s use the tools from last time, those based on properties of a normal distribution.

If we can view the sampling as having come from a population at least approximately normally distributed for our variable, then we can now examine empirically the effect of different sample sizes on the estimate of the mean.

A hint: variability is important!

From one population we obtain two samples, A and B. Sample sizes are

Assume for now that we know the true mean (μ) and standard deviation (σ) for the population. Note. This is one of the points of why we use computer simulation so much to teach statistics — it allows us to specify what the truth is, then see how our statistical tools work or how our assumptions affect our statistically based conclusions.

Confidence intervals

Reliability is another word for precision. We define a confidence interval as a statistic to report the reliability of our estimated statistic. We introduced confidence interval in Chapter 3.4. At least in principle, confidence intervals can be calculated for all statistics (mean, variance, etc.,) and for all data types. Confidence intervals define a lower limit, L, and an upper limit, U, and that you are making a statement that you are “95% certain that the true value (parameter value) is between these two limits.

We previously reported how to calculate an approximate confidence intervals for proportions and for NNT; simply multiple standard error estimate by 2. Here we introduce an improved approximate calculation of the 95% confidence interval for the sample mean

where Z is something you would look up from the table of the normal distribution. For a 95% confidence interval, 100% – 95% = 5% and divide 5% by two: the lower limit corresponds to 2.5% and the upper limit corresponds to 2.5% on our normal distribution. We look up the table and we find that Z for 0.025 is 1.96 and that is the value we would plug into our equation above. For large sample sizes, you can get a pretty decent estimate of the confidence interval by replacing 1.96 with “2.”

Questions

1. What is the probability of having a sample mean greater than 50 (mean > 50) for a sample of n = 9 ?

We’ll use a slight modification of the Z-score equation we introduced in Chapter 6.6 — the modification here is that previously we referred to the distribution of Xi‘s and how likely a particular observation would be. Instead, we can use the Z score with the standard normal distribution (aka Z-distribution), approach to solving how likely an estimated sample mean is given the population parameters μ and σ. Recall the Z score

We have everything we need except the SEM, which we can calculate by dividing the standard deviation by squared root of sample size.

For

, σ = 12.0 (given above), and μ = 47, n = 9, plug in the values:

Therefore, after applying the equation for Z score, . This corresponds to how far away from the standard mean of zero.

Look up from the table of normal distribution. The answer is , which corresponds to that is EQUAL to or GREATER than 0.75, which is what we wanted. Translated, this implies that, given the level of variability in the sample, 22.66% of your sample means would be greater than 50! We write: .

Some care needs to be taken when reading these tables — make sure you understand how the direction (less than, greater than) away from the mean is tabulated.

2. Instead of greater, how would you get the probability less than 50?

Total area under the curve is 1 (100%), so subtract .

I recommend that you do these by hand first, then check your answers. You’ll need to be able to do this for exams.

Here’s how to use Rcmdr to do these kind of problems.

Rcmdr: Distributions → Continuous distributions → Normal distribution → Normal probabilities …

Figure 5. Screenshot Rcmdr menu to get normal probability.

Here’s the answer from Rcmdr

pnorm(c(50), mean=47, sd=12, lower.tail=TRUE)

[1] 0.5987063

3. Now, try a larger sample size. For , what is the probability of having a sample mean greater than 50 ()?

, μ = 47, σ = 12, n = 50 and

Therefore, after applying the equation for Z score, . Look up (Normal table, subtract answer from 1) and we get .

Or 3.84% of your sample means would be greater than 50! We write: .

Said another way: If you have a sample size of 50 () and you obtain a mean greater than 50 then there is only a 3.84% chance that the TRUE MEAN IS 47.

4. What happens if the variability is smaller? Chance σ from 12 to 6 then repeat questions 1 and 4.

5. Repeat the demonstration of Central Limit Theorem and Law of Large Numbers for discrete  distributions

1. binomial distribution. Replace rchisq() with rbinom(n, size, prob) in the replicate() function example. See Chapter 6.5
2. poisson distribution. Replace rchisq() with rpois(n, lambda) in the replicate() function example. See Chapter 6.5

Quiz Chapter 8.3

Sampling distribution and hypothesis testing

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Chapter 8 contents

• Introduction
• The null and alternative hypotheses
• The controversy over proper hypothesis testing
• Sampling distribution and hypothesis testing
• Tails of a test
• One sample t-test
• Confidence limits for the estimate of population mean
• References and suggested readings

6.10 – t distribution

Introduction

Student’s t distribution is a sampling distribution where values are sampled from a normal distributed population, but σ the standard deviation and μ the mean of the population are not known. When sample size is large and we know σ the standard deviation, we would use the Z-score to evaluate probabilities of the sample mean. The t-distribution applies when σ is not known and sample size is small (eg, less than 30, per rule of thirty).

Because of its thicker tails compared to the normal distribution, the t-distribution accounts for the uncertainty in estimating the population variability. This is important because we would require a more extreme value to be considered “statistically significant” compared to the normal distribution, making your conclusions more conservative.

The equation of the t-test is

where the difference between “X bar,” the sample mean, and μ, the population mean, is divided by the standard error of the mean, , defined in Chapter 3.2 and again in Chapter 3.3. This formulation of the t-test is called the one sample t-test (Chapter 8.5).  We call the result of this calculation the test statistic for t. We evaluate how often that value or greater of a test statistic will occur by applying the t distribution function.

There are many t-distributions, actually, one for every degree of freedom. Like the normal distribution, the t distribution is always symmetrical about the mean. But it is stacked up (leptokurtic) around the middle at low degrees of freedom. As degrees of freedom increase, the t distribution spreads and becomes increasingly like the normal distribution.

Relationship between t distribution and standard normal curve.

First, here is our standard normal plot, mean = 0, standard deviation = 1 (Fig 1).

Figure 1. Density plot of standard normal distribution

Next, here’s the t-distribution for five degrees of freedom (Fig 2).

Figure 2. Density plot of t-distribution for five degrees of freedom.

Lets see what happens to the shape of the t-distribution as we increase the degrees of freedom from df = 5, 10, 20, 50, 1000, 10000 (Fig 3). The last graphic in the series is the standard normal curve again (Fig 3).

Figure 3. Animated GIF of density plot t distribution, from df = 5 to 10,000 plus standard normal curve.

By convention in the Null Hypothesis Significance Testing protocol (NHST), we compare the test statistic to a critical value. The critical value is defined as the value of the test statistic that occurs at the Type I error rate, which is typically set to 5%. We introduced logic of NHST approach in Chapter 6.9 with the chi-square distribution. Again ,this is just an introduction; we teach it now as a sort of mechanical understanding to develop. The justification for this approach to testing of statistical significance is developed in Chapter 8.

Table of Critical values of the t distribution for df  1 – 5, one tail (upper)

See Appendix 20.3 for a complete table of t-distribution.

Questions

Be able to answer these questions using the t table, Appendix 20.3, or using Rcmdr

1. For probability α = 5%, what is the critical value of the t distribution (upper tail) for 1 degree of freedom? For 5 df? For 20 df? For 30 df?
2. The value of the t test statistic is given as 12. With 3 degrees of freedom, what is the approximate probability of this value, or greater from the t distribution?

Quiz Chapter 6.10

t distribution

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Chapter 6 contents

• Introduction
• Some preliminaries
• Ratios and proportions
• Combinations and permutations
• Types of probability
• Discrete probability distributions
• Continuous distributions
• Normal distribution and the normal deviate (Z)
• Moments
• Chi-square distribution
• t distribution
• F distribution
• References and suggested readings

6.9 – Chi-square distribution

Introduction

As noted earlier, the normal deviate or Z score can be viewed as randomly sampled from the standard normal distribution. The chi-square distribution describes the probability distribution of the squared standardized normal deviates with degrees of freedom, df, equal to the number of samples taken. (The number of independent pieces of information needed to calculate the estimate, see Ch. 8.) We will use the chi-square distribution to test statistical significance of categorical variables in goodness of fit tests and contingency table problems.
The equation of the chi-square is

where k is the number of groups or categories, from 1 to k, and f_i is the observed frequency and f_i “hat” is the expected frequency for the k^th category. We call the result of this calculation the chi-square test statistic. We evaluate how often that value or greater of a test statistic will occur by applying the chi-square distribution function. Graphs to show chi-square distribution for degrees of freedom equal to 1 – 5, 10, and 20 (Fig 1).

Figure 1. Animated GIF of plots of chi-square distribution over range of degrees of freedom.

Note that the distribution is asymmetric, particularly at low degrees of freedom. Thus tests using the chi-square are one-tailed (Fig 2).

Figure 2. The test of the chi-square is typically one-tailed. In this case, probability of values greater than the critical value.

By convention in the Null Hypothesis Significance Testing protocol (NHST), we compare the test statistic to a critical value. The critical value is defined as the value of the test statistic — the cutoff boundary between statistical significance and insignificance — that occurs at the Type I error rate, which is typically set to 5%. The interpretation of the result is as follows: after calculating a test statistic, we can judge significance of the results relative to the null hypothesis expectation. If our test statistics is greater than the critical value, then the p-value of our results are less than 5% (R will report an exact p-value for the test statistic).  You are not expected to be able to follow this logic just yet — rather, we teach it now as a sort of mechanical understanding to develop in the NHST tradition. The justification for this approach to testing of statistical significance is developed in Chapter 8. A portion of the critical values of the chi-square distribution are shown in Figure 3.

Figure 3. Portion of the table of some critical values of chi-square distribution, one tailed (right-tailed or “upper” portion of distribution).

See Appendix for a complete chi-square table.

The chi-squared test is a one-tailed test.

The result of the calculation of the test statistic is always non-negative because it is calculated by squaring the differences between the observed and expected frequencies.  As a consequence, the test is always “right-tailed.”

The chi-square test does not provide information about the direction of an association, nor the strength of the association, only whether or not the variables are independent or associated. We use “correlation,” (eg, Phi coefficient or Cramer’s V, see Chapter 9.1 – Chi-square test: Goodness of fit and Chapter 9.2 – Chi-square contingency tables), to quantify the strength and direction of the association. While statistical significance is important, examining the direction and magnitude, the effect size, provides actionable insight and context for making informed decisions. For example, a new drug tested in a clinical trial may show an association between subject improvement (statistical significance), but yet have a small effect.

Example

Professor Hermon Bumpus of Brown University in Providence, Rhode Island, received 136 House Sparrows (Passer domesticus) after a severe winter storm 1 February 1898. The birds were collected from the ground; 72 of the birds survived, 64 did not (Table 1). Bumpus made several measures of morphology on the birds and the data set has served as a classical example of Natural Selection (Chicago Field Museum). We’ll look at this data set when we introduce Linear Regression.

Table 1. Survival statistics of Bumpus House sparrows

Was there a survival difference between male and female House Sparrows? This is a classic contingency table analysis, something we will at length in Chapter 9. For now, we report the Chi-square test statistic for this test was 3.1264 and the test had one degree of freedom. What is the critical value of the chi-square distribution at 5% and one degree of freedom?. Typically we would simply use R to look this up

qchisq(c(0.05), df=1, lower.tail=FALSE)

But we can also get this from the table of critical values (Fig 4). Simply select the row based on the degrees of freedom for the test then scan to the column with the appropriate significance level, again, typically 5% (0.05).

Figure 4. Portion of the chi-square distribution which shows how to find critical value of the chi-square distribution. 

For 1 degree of freedom at 5% significance, the critical value is 3.841. Back to our hypothesis: Did male and female survival differ in the Bumpus data set? Following the NHST logic, if the test statistic value (eg, 3.1264) is greater than the critical value (3.841), then we would reject the null hypothesis. For this example, we would conclude no statistical difference between male and female survival because the test statistic was smaller than the critical value. How likely are these results due to chance? That’s where the p-value comes in. Our test statistic value falls between 5% and 10% (2.706 < 3.1264 < 3.841). In order to get the actual p-value of our test statistic we would need to use R.

R code

Given a chi-square test statistic you can use R to calculate the probability of that value against the null hypothesis. At the R prompt

pchisq(c(3.1264), df=1, lower.tail=FALSE)

And R output

[1] 0.07703368

Because we are using R Commander, simply select the command by following the menu options.

Rcmdr: Distributions → Continuous distributions → Chi-squared distribution → Chi-squared probabilities …

Enter the chi-square value and degrees of freedom (Fig 5).

Figure 5. Screenshot of input box in Rcmdr for Chi-square probability values.

Questions

1. What happens to the shape of the chi-square distribution as degrees of freedom are increased from 1 to 5 to 20 to 100?

Be able to answer these questions using the Chi-square table, Appendix 20.2, or using Rcmdr

2. For probability α = 5%, what is the critical value of the chi-square distribution (upper tail)?
3. The value of the chi-square test statistic is given as 12. With 3 degrees of freedom, what is the approximate probability of this value, or greater from the chi-square distribution?

Quiz Chapter 6.9

Chi-square distribution

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Chapter 6 contents

• Introduction
• Some preliminaries
• Ratios and proportions
• Combinations and permutations
• Types of probability
• Discrete probability distributions
• Continuous distributions
• Normal distribution and the normal deviate (Z)
• Moments
• Chi-square (Χ²) distribution
• t distribution
• F distribution
• References and suggested readings

6.7 – Normal distribution and the normal deviate (Z)

Introduction

In chapter 3.3 we introduced the normal distribution and the Z score, aka normal deviate, as part of a discussion about how some knowledge about characteristics of the dispersion of our data sampled from a population could be used to calculate how many samples we need (the empirical rule). We introduced Chebyshev’s inequality as a general approach to this problem, where little is known about the distribution of the population and contrasted it with the Z score, for cases where the distribution is known to be Gaussian or the normal distribution. The normal distribution is one of the most important distributions in classical statistics. All normal distributions are bell-shaped and symmetric about the mean. To describe a normal distribution only two parameters are needed: the population mean, μ, and the population standard deviation, σ. The normal distribution with mean equal to zero and standard deviation equal to one is called the standard normal, or Z distribution. With use of the Z score any normal distribution can be quickly converted to the standard normal distribution.

Proportions of a Normal Distribution

This concept will become increasingly important for the many statistical tests we will learn over the next few weeks. What is the proportion of the populations that is greater than some specific value? Below (Fig 1), again, I have generated a large data set, now with population mean µ = 5 and σ = 2. The red line corresponds to the equation of the normal curve using our values of µ = 5 and σ = 2.

Figure 1. Frequency of observations expected to be greater than 7 (red bars) from a large population with mean µ = 5 and σ = 2.

R code for Fig 1.

set.seed(123)
myData <- rnorm(1000, mean = 5, sd = 2)
h <- hist(myData, breaks = 20, plot = FALSE) 
threshold <- 7
# Bars with midpoints greater than threshold will be "red", 
# and others will be "white."
bar_colors <- ifelse(h$mids > threshold, "red", "white")
plot(h, col = bar_colors, xlab = "Value", ylab = "Frequency")

Note that this is a crucial step! We assume that our sample distribution is really a sample from a population density (= “area under the curve”) function (= “an equation”) for a normal random (= “population”) variable.

Once I (you) make this assumption, then we have powerful and easy to use tools at our command to answer questions like

Question: What proportion of the population is greater than 7 (Fig 1, colored in red)?

This gets to the heart of the often-asked question, How many samples should I measure? If we know something about the mean and the variability, then we can predict how many samples will be of a particular kind. Let’s solve the problem.

The Z score

We could use the formula for the normal curve (and a lot of repetitions), but fortunately, some folks have provided tables that short-cut this procedure. R and other programs also can find these numbers because the formulas are “built in” to the base packages. First, let’s introduce a simple formula that lets us standardize our population numbers so that we can use established tables of probabilities for the normal distribution.

Below, we will see how to use Rcmdr for these kinds of problems.

However, it’s one of the basic tasks in statistics that you should be able to do by hand. We’ll use the Z score as a way to take advantage of known properties of the standard normal curve.

Z = 1 (with the mean and SD). Z (say “Z-score”) is called the normal deviate (aka “standard normal score”; it is also called the “Z-score”); it gives us a shortcut for finding the proportion of data greater than 7 in this case).

We use the normal deviate to do a couple of things; one use is to standardize a sample of observations so that they have a mean of zero and a standard deviation of one (the Z distribution). The data would then said to have been normalized.

The second use is to make predictions about how often a particular observation is likely to be encountered. As you can imagine, this last use is very helpful for designing an experiment — if we need to see a specified difference, we can conduct a pilot study (or refer to the literature) to determine a mean and level of variability for our observation of choice, plug these back into the normal equation and predict how likely we can expect to see a particular difference. In other words, this is one way to answer that question — how many observations need I make for my experiment to be valid?

Table of normal distribution

A portion of the table of the normal curve is provided at our web site and in your workbook. For our discussions, here’s another copy to look at (Fig 2).

Figure 2. Portion of the table of the normal distribution. Only values equal to or greater than Z = 0 are visible.

See Table 1 in the Appendix for a full version of the normal table.

We read values of Z from the first column and the first row. For Z = 0.23 we would scan the top row, scoot over to fourth column, then trace to where the row and column intersect (Fig 3); the frequency of occurrence of values at Z = 0.23 is 0.409046 or 40.9% (Fig 3).

Figure 3. Highlight Z = 0.23, frequency is 0.409046.

Z on the standard normal table is going to range between -4 and +4, with Z = 0 corresponding to 0.500. The Normal table values are symmetrical about the mean of zero.

What to make of the values of Z, from -4, -3, … +2, +3, up to +4 and beyond?  These are the standard deviations! Recall that using the Z score you corrected to mean of zero (got it!), and a standard deviation of one! A Z of 2 is twice the standard deviation; a Z = 3 is therefore three times the standard deviation, and so forth. The distribution is symmetrical: you get the same frequency for negative as for positive values. So on the “X” axis on a standard normal distribution, we have units of standard deviation plus (greater) or minus (less) than the mean. In Figure 4, area less than -1 standard deviations is highlighted (Fig 4).

Figure 4. Plot of standard normal distribution; area less than -1 σ.

R code for Fig 4.

set.seed(123)
myData2 <- rnorm(100000, mean = 0, sd = 1)
threshold <- -1
bar_colors <- ifelse(h$mids < threshold, "red", "white")
# Additional code used, same as Figure 1.

Question. How many multiples of standard deviations would you have for a Z score of Z = 1.75?

Answer = 1.75 times

Examples

See Table 1 in the Appendix for a full version of the normal table as you read this section.

What proportion of the data set will have values greater than 7? After applying our Z score equation, I get Z = 1.0 and for a Z of 1.0 = 0.1587 or 15.87% of the observations are greater than 7.

What proportion of the data set will have values less than -7? After applying our Z score equation, I get Z = -1.0. Taking advantage of symmetry argument, I just take my Z = -1.0 and make it positive — instead of values smaller than -1.0 we now have values greater than +1.0. And for a Z of 1.0 = 0.1587 or 15.87% of the observations are greater than 7, which means that 15.87% will be -7 or smaller.

What proportion of the data set will have values greater than 8? Again, apply the Z score equation I get Z of 1.5 = 0.0668 or 6.68% of the observations are greater than Z.

What proportion of the population is between 5 and 7 ? (colored in red). Draw the problem (example shown in Fig 5).

Figure 5. Proportion of the population is between 5 and 7.

R code for Fig 5.

# Used myData from Fig 1 code.

highlight_start <- 5
highlight_end <- 7
bar_colors <- rep("white", length(h$counts)) 
bar_colors[h$breaks[-length(h$breaks)] >= highlight_start & h$breaks[-1] <= highlight_end] <- "red"
# Additional code used, same as Figure 1.

Worked problem

1 – (proportion beyond 7) – (proportion less than 5)

1 – (0.1587) – (proportion less than 5)

And the proportion less than 5?

Use Z-score equation again. Now we find that

Z = 0

And look up Z of 0 from the table  = 0.5 proportion or 50.0%

Therefore, the proportion between 5 & 7 equals

1 -0.1587 – 0.50 = 0.3413

Answer = 34.13 % of the observations are between 5 and 7 when the mean µ = 5 and s = 2.

Questions

Several questions were asked in the text.

1. Proportion of observations greater than 7  from a large population with mean µ = 5 and σ = 2.

2. How many multiples of standard deviations would you have for a Z score of Z = 1.75?

3.

Quiz Chapter 6.7

Normal distribution questions

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Chapter 6 contents

• Introduction
• Some preliminaries
• Ratios and probabilities
• Combinations and permutations
• Types of probability
• Discrete probability distributions
• Continuous distributions
• Normal distribution and the normal deviate (Z)
• Moments
• Chi-square (Χ²) distribution
• t distribution
• F distribution
• 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