6.8 – Moments

Introduction

We care about the shape of data distribution because it can impact statistical inference. If the shape of the observed distribution differs from the standard, eg, normal, then one option is to transform the data (see Chapter 13 – Assumptions of parametric tests).

Moments are used to describe the shape of a distribution. For those of you who remember your calculus, moments were discussed as a method to find the center of mass, or balancing point (Herman and Strang 2018). For distributions, the center and shape moments follow from the expected value of the probability function.

Four moments apply for describing the shape of a distribution. The 1st moment describes the middle, the 2nd describes the spread from the middle, the 3rd describes symmetry about the middle, and the 4th describes the shape, whether peaked and sharp (leptokurtic), or broad and flattened (platykurtic).

Equations for the moments

Over the years, several equations  have been proposed to estimate skewness and kurtosis. The above formulas are just one example from the list (Joanes and Gill 1998).

Pearson’s standardized moments:

where E is expected value of random variable. The expected value concept follows from rules of probability — basically, the average of large number, n, of X.

Four moments can be used to describe the shape of a distribution.

1^st moment, μ (mean): population mean, 3.1 – Measures of Central Tendency

2^nd moment, σ² (variance): population variance, 3.2 – Measures of dispersion

3^rd moment, (skew):

4^th moment, (kurtosis):

Estimating moments in R and R Commander

Consider results for more than a thousand runners who completed a 5K timed run (Fig 1).

Figure 1. Histogram finishing times in minutes for 1307 runners at 2016 Banana 5K.

The task is to describe the moments for the 5K data set.

Recall: The first moment is the mean. The second moment is the variance. The third moment is skewness. And the fourth moment is kurtosis.

In R Commander, we select Statistics → Summaries → Numerical summaries…, which brings up a popup menu. First, select the variable, in this case Minutes, from the Data tab (not shown). Next, click on Statistics tab to choose options (Fig 2).

Figure 2. Rcmdr Numerical summaries Statistics tab.

For estimates of the moments, check Mean, Standard Deviation, Skewness, and Kurtosis.

Large sample size, n = 1307

Type 1

    mean       sd  skewness    kurtosis    n
34.42999 10.31437 0.6159258 -0.01593882 1307

Type 2

    mean       sd  skewness    kurtosis    n
34.42999 10.31437 0.6166337 -0.01139521 1307

Type 3

    mean       sd skewness    kurtosis    n
34.42999 10.31437 0.615219 -0.02050335 1307

Small sample size

To test the claim about sample size and the moment statistics, draw a random sample of 30 from the larger data set. Sample without replacement

sample.banana <- data.frame(sample(banana5K$Minutes, 30, replace = FALSE))

I forgot to specify a new variable name, so R used the whole command as the variable name. I could go back and fix my function call, or simply rename the variable as follows

names(sample.banana)[c(1)] <- c("Minutes")

The random sample yielded a distribution (Fig 3).

Figure 3. Histogram finishing times in minutes for random sample of 30 drawn from 1307 runners at 2016 Banana 5K.

Repeat Numerical summaries on small data set, n = 30

Type 1

    mean       sd  skewness  kurtosis  n
33.16667 10.00373 0.5538637 0.5024438 30

Type 2

    mean       sd  skewness  kurtosis  n
33.16667 10.00373 0.5834511 0.8276415 30

Type 3

    mean       sd  skewness  kurtosis  n
33.16667 10.00373 0.5264025 0.2728392 30

Conclusion: We can compare consistency of the estimators by calculating coefficient of variation. The three types of skewness estimators differed by only 1% and 5% for large and small sample size, respectively. In contrast, the three types of kurtosis estimators differed by 29% and 52% for large and small sample size, respectively.

Questions

1. Explore the consistency of skewness and kurtosis estimates by calculating and comparing coefficient of variation estimates. R Commander provide a nice way to draw randomly from various defined distributions. Draw two data sets of 15 (small) and 1000 (large), from the chi-square distribution (1 degree of freedom) and a minimum of one other continuous distribution.

Example, draw random sample of 1000 from chi-square distribution. Rcmdr: Distributions → Continuous distributions → Chi-squared distribution → Sample from chi-squared distribution... (Fig 4).

Enter name for the variable, enter degrees of freedom (eg, 1), number of samples (eg, 1000), and number of observations (variables, columns). Leave Sample means checked under data sets.

Figure 4. Screenshot Rcmdr menu: Sample from Chisquare distribution.

This results in a new data set. Get “moments” from Numerical summaries and calculate coefficient of variations. Which moments have the most consistency regardless of the kind of distribution.

2. Make histograms for each of your created data sets. Describe what you see about the shape of the plotted distributions.

Quiz Chapter 6.8

Moments

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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

Install R

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Download pdf of this page

Introduction

The first time installing R can seem intimidating. To start, be clear about the overall goal of the procedure: providing the student with an accessible environment for solving statistics problems.

In brief, this page explains how to get R set up on your computer. First, you need to download the R installer from the official CRAN website. When you run the installer,in general, accept the default choices. However, for Windows users, it’s important to right-click the file and choose “Run as administrator.” This step ensures that R has the proper permissions to install correctly and avoids problems with user access later on. Once installed, you can open R and test it by typing a simple command like `2 + 2` in the console to confirm everything is working.

The flow chart presented in Fig 1 suggests a way to orient the student to solving the task. First, understand which type of computing environment you have available. Second, if a macOS, then additional software (XQuartz) is required to provide full function to the R software. Third, download and install the base R software appropriate for the computer system.

Figure 1 suggests an R installation flow chart.

Figure 1. Suggested flow chart for R installation. 

It’s possible to follow the steps in Figure 1, accepting all default options presented along the way, to end up with a working R environment. As with many software processes, there are choices beyond defaults that can be made to improve the software use.

This page presents a detailed guide about how to install R onto your computer — this is referred to as building a local development environment or LDE. Additional install R help was provided in Chapter 1.1 – A quick look at R and R Commander.

Instructions for RStudio are also provided (optional for BI311 students). A guide to install R Commander is provided in Install R Commander.

Instructions for how to run R via a “cloud computing” (serverless) option — a remote development environment — are also provided, Use R in the Cloud.

For help upgrading installed packages after upgrading new R version, see R packages.

Run R on your computer (LDE install)

CoLab, skip this step: Instead, go to Use R in the cloud.

So why in this day in age should you install and build R on your own computer? The remote options to run R in the cloud are a wonderful option, convenient: you can access anywhere you have internet, from any device that connects to the internet. It’s easy to share and work together on projects, particularly those based on Jupyter Notebooks.

I think the main benefits to a local installation is it’s a more efficient environment to work in — you have control of everything and, provided your PC has power, a working R install on your computer will always be available to you. Since you can control the update cycle for your computer, you won’t run into times you cannot access the remote server to work on your project. Testing code is faster on a local install, feedback — think error messages — apply to your installed version. And, while remote R servers may come with low initial costs to students, any significant use will quickly require paid accounts. As a reminder, the good folks at the R-project continue to offer R as free software. All you need to do is work through the install process.

Start at the R-project homepage, r-project.org. To download software, first click on CRAN link,  located on left hand side of the screen (here, highlighted by green arrow, Fig 2).

Figure 2. Screenshot homepage for R-project.org. 

Figure 3 shows a screenshot of the CRAN mirror page. The idea is to select the mirror site closest to your location. In Hawaiʻi, that’s likely to be any of the sites in California. However, I recommend selecting the first in the list, 0-cloud, at cloud.r-project.org (highlighted by green arrow, Fig 3).

Figure 3. Screenshot of portion of R-Project CRAN mirror page.

After selecting the mirror site, the download page is presented (Fig 4). Click on the link that corresponds to your computer system (Linux, macOS, or Windows).

Figure 4. Screenshot of portion of base R download page.

Once the installation file is located onto your computer, proceed to install base R.

Detailed instructions

For screenshots of installation steps on WinPC, see Win10/11 setup, screenshots

1. Windows PCs, download the base application from selected CRAN Mirror site, select Download R for Windows, and install the R software as you would any other software. All of you are likely to have the 64-bit version of Windows 11, so install the 64-bit version of R. Follow the instructions as they are presented. Screenshots of the install process are available at the end of this page (click here or scroll down to Win11 setup, Screenshots).
   • Current versions of Microsoft Windows come in several flavors, the simplest distinction is between home and pro. R runs perfectly well on both.
     • Windows 10 is reaching end of life cycle.
     • Some inexpensive Microsoft Windows PCs are built on ARM64, not Intel or AMD64 CPU. Thus, installing R and or RStudio may prove problematic.
     • Also note: Windows in S mode only run applications from the Microsoft store. To install R, you first may need to switch out of S mode (see Microsoft FAQ about S mode).
   • You should install R with Administrator privileges. Highlight the install file, right-click the file, and select “Run as administrator” from the popup menu.
   • When you first try to run R you may get a popup screen “Windows protected your PC,” locate and click on the “More info” link and select “Run anyway.”
     • This in no way will harm your computer — provided you have downloaded from official sites. R is a verified program. Microsoft has taken an aggressive line on developers and favors apps that are part of their app store.
     • It is advisable to confirm for yourself: check the md5sum against the fingerprint on the CRAN server
   • When prompted, I recommend that you change the install directory to root folder, e.g., C:\R\R-4.5.1. This will allow for installation of packages to the common library as opposed to a personal library.
     • I recommend this change because of how Windows assigns home folders. During initial setup Windows 10 prompted you to choose a username and whether you wanted your work stored locally or in your OneDrive folder. A worse case scenario? You select a username with spaces, e.g.,”Mike Dohm,” and you selected OneDrive. Both will cause challenges later for running and or installing packages for R.
       • If you install R anywhere but the default Program files folder on your Win10/11 PC, chances are you will need to add the folder containing the executable, r.exe, to you path.
         • Search: “env”
         • Open “Edit the system environment variables” in the Control Panel
         • Click Advanced tab, then click on Environment Variables… button (lower right of panel)
         • Under System variables, scroll to and select Path.
         • Click Edit… button, then click New button.
         • Type in the path to the folder containing R.exe. That’s likely to be C:\R\R-4.5.1, assuming R-4.5.1 is the latest version of R installed on your computer.
     • I made a video for you. Video is about 26 minutes long; at 22 minute mark, video includes how to install R Commander (instructions provided Install R Commander).
       • Apologies for the production quality — videos are not my thing.

https://youtu.be/upjmBieh3bM

For screenshots of installation steps on WinPC, see MacOS setup, screenshots

1. macOS PCs, first you must download and install XQuartz from https://www.xquartz.org. Best to restart your mac after installing XQuartz then proceed to install R.
   After installing XQuartz, then return to https://cran.r-project.org, select Download for Mac(OS) X, and run the installer. Screenshots of the install process are available at the end of this page (click here or scroll down to Macos setup, Screenshots).
   • As of August 2025, be advised that there are two distinct R versions for your MacBook or iMac.
     • For MacBook or iMac with Apple’s M1 or M2 ARM chip sets, download and install R-4.5.1-arm64.pkg.
       • If you recently purchased a new MacBook or iMac (2020 to present), then you probably have the M2 or M3 chipset (check by clicking the Apple icon, then selecting About this Mac or System Information (/Applications/Utilities/System Information.app)).
       • XQuartz version 2.8.5 works on macs with either the M1 or Intel chipsets.
     • For older MacBook or iMacs with Intel processors, download R-4.5.1-x86_64.pkg.
     • Depreciated 8/4/2021: Be advised that these instructions are for Intel-based macs. At the time of writing these instructions (April 2021), the installation of XQuartz and R should work on new M1-based macs. At the time of this writing (April 2021), however, R will not run natively on your M1 mac. It will run using Rosetta 2, an emulator that is included with your M1 mac. The R folks are busy working on a version that will run natively, which may be ready within a few months.
   • At the completion of the install process, don’t forget to drag the R app to your Applications folder.
   • When you first try to run R, you may get a popup screen which provides no option to start the app, and perhaps even a rather ominous option to move the app to Trash. Just close the warning message and right-click on the R app. A new screen pops up, which looks very much like the previous warning, but now you will see and option to open the app. Click on open to start R.
   • Like the message to Windows PC users, bypassing Apple’s Gatekeeper to run R in no way will harm your computer — provided you have downloaded from official sites — R is a verified program. Apple has taken an aggressive line on developers and favors apps that are part of their app store.
2. LINUX distros. If your PC platform is Linux, then you should be comfortable with installation and updating of software. R base is already included in Debian distributions (e.g., Mint, Ubuntu). See https://cloud.r-project.org/ for additional instructions.
   • For Chromebook users, if you can install a Linux subsystem, then you can also install and run R. For instructions to install R see Levi’s excellent writeup at levente.littvay.hu/chromebook/.

Test R

For both macOS and Windows PCs, successful installation of R on your computer installs base R programming language and a simple graphical user interface. Test your install by running code in the terminal (one line at a time) or via script:

1. Rgui.exe (Windows PC)
2. File → New script

Enter code in script editor, e.g.,

myX <- c(1,2,3,4)
myY <- c(5,10,15,20)
plot(myY,myX)

3. Run code: Ctrl+R

Figure 5 shows screenshot of R running on WinPC. RGui.exe (1), script editor (2), and results of plot() (3) on WinPC.

Figure 5. Screenshot of RGui.exe (1), script editor (2), and results of plot() (3) on WinPC.

1. R.app (macOS): run code in the terminal or via script
2. File → New Document

Enter code in script editor, e.g.,

myX <- c(1,2,3,4)
myY <- c(5,10,15,20)
plot(myY,myX)

3. Run code: Cmd+Enter

Figure 6 shows screenshot of R running on macOS. R.app (1), script editor (2), and results of plot() (3).

Figure 6. Screenshot of R.app (1), script editor (2), and results of plot() (3) on macOS.

Many of you would like a video. Do a little search and you’ll find plenty, although most are also showing how to install RStudio in addition to base R.

For BI311, we also use R Commander package

R Commander is a package that adds function to R; it provides a familiar point-and-click interface to R, which allows the user to access functions via a drop-down menu system (Fox 2017).

Go to Install R Commander guide.

Personalize R startup

At startup, R app looks to a series of environment files to instruct how it will provide a working session environment. One important file to know about is the .Rprofile file, which will reside in your home directory on your computer. This file generally won’t be present after the first installation, but can be easily added.

First, verify the home directory where R will first look for an .Rprofile file.

Identify home directory

At the R prompt, verify your home directory, type and run

path.expand("~")

On a macOS, this should return something like

"~Users/USERNAME"

On a Windows 11 machine, this may return

"C:/Users/USERNAME/OneDrive/Documents"

if default instructions were accepted during initial Windows setup.

How to modify .Rprofile

Good practice suggests setting a project folder to contain work done for a project. For biostatistics students, minimally, this might be the course name, BIOSTATS. Within this folder we’d expect to find all course data files, homework, R coding scripts and related for the class. While it is easy enough to point R to a data file (see Part 07. Working with your own data, Mike’s Workbook for Biostatistics), it’s often easier to tell R in advance where to expect to find these files by setting a working folder at the beginning of a session using the setwd() command.

Alternatively, you can edit .Rprofile so that R always starts in the folder you want.

To open (or create if the first time) .Rprofile, at the R prompt type and submit

file.edit("~/.Rprofile")

options(test_success = "Success! .Rprofile is loaded.")

getOption("test_success")

Windows 11 Setup

• Open your local home documents folder (usually C:\Users\YOUR_NAME\Documents).
• Create or open a file named exactly .Rprofile (Ensure it does not end in .txt).

Copy this code for Windows:

.First <- function() {
# 1. Grab the literal, local hard drive user folder
local_root <- Sys.getenv("USERPROFILE")

# 2. Open a native Windows browse window for folder selection
message("Please select your local project folder...")
flush.console()

chosen_path <- choose.dir(
default = local_root,
caption = "Select your local project folder"
)

# 3. Set the directory to what they clicked on
if (!is.na(chosen_path)) {
setwd(chosen_path)
message("Working directory successfully set to: ", chosen_path)
} else {
warning("No folder selected. Working directory unchanged.")
}
}

macOS Setup

• Ensure R.app has been granted Full Disk Access in your Mac’s System Settings > Privacy & Security.
• Open Terminal and run open -a TextEdit ~/.Rprofile to open your profile configuration.

Copy this code for Mac:

.First <- function() {
# 1. Open a macOS Finder folder selection window
message("Please select your local project folder...")
flush.console()

# Calls native macOS AppleScript to open a secure folder picker
chosen_path <- tryCatch({
system("osascript -e 'POSIX path of (choose folder with prompt \"Select your local project folder:\")'", intern = TRUE)
}, error = function(e) {
return(NA)
})

# 2. Set the directory to what they clicked on
if (!is.na(chosen_path) && length(chosen_path) > 0) {
setwd(chosen_path)
message("Working directory successfully set to: ", chosen_path)
} else {
warning("No folder selected. Working directory unchanged. Keep in mind R.app needs 'Full Disk Access' in macOS System Settings.")
}
}

Set your CRAN mirror site

To add function to R, you’ll download many R packages. These, too, are stored at the mirror sites. Each time you run install.packages(), R will ask for the mirror site. The R app on Windows pops up a long selection box; macOS R.app is a bit more friendly, but regardless, it’s best to choose a mirror site at the beginning of a session. At the R prompt, type and submit the command

options(repos = c(CRAN = "https://cloud.r-project.org"))

this will remain in effect for the current R session.

To make the change permanent, add to .Rprofile in your personal home directory

local({
r <- getOption("repos")
r["CRAN"] <- "https://cloud.r-project.org"
options(repos = r)
})

If you use textedit or notepad, make sure to return after the }); the file requires an empty new line to run.

Restart R, then enter at the prompt

file.exists("~/.Rprofile")

If you’ve created the Rprofile file correctly, R should repeat.

Run R from the terminal

Whether yours is a mac or win11 pc, you have a powerful computing environment lurking beneath the glossy graphic user interface. It’s called the terminal. The terminal is a place where text-based instructions may be written and submitted to command your computer to do something.

On win11 pc, the modern terminal is the Power Shell. Search “PowerShell,” then open Windows PowerShell. Alternatively, shortcut + X, then I key. This opens terminal in your home folder.

• You’ll want to navigate to your working folder, e.g., BI311 on the Desktop. Setting your working folder can be done in R of course, but at the terminal type the command “cd \users\default\desktop\bi311“.

On macOS, Spotlight (search) “Terminal.app“. Alternatively, right-click on your working folder icon and select New terminal at Folder from the popup menu.

On Ubuntu Linux, shortcut keys + alt, then t key.

Open the included script editor

Use RStudio

Rstudio is a very popular data science IDE, specifically written for R programming. Therefore, if your goal is to become adept at R programming, go with RStudio. Like R Commander, Rstudio provides an environment to write an manage code, generate markdown reports, and importantly, manage files in R projects. Figure 7 shows a screenshot of RStudio: Clockwise, four panes: Source, Environment, Output, Console on WinPC. BI311 students, please note that I recommend use of R Commander and Google CoLab.

Figure 7. Screenshot of RStudio IDE.

Installation of RStudio desktop is straightforward.

Go to https://posit.co/download/rstudio-desktop/

Install R first. Since at this point you’ve likely already installed R, go directly to 2 (green box, Fig 8).

Figure 8. Screenshot RStudio desktop download page, current as of September 2025.

After downloading the installation software, double click the file to begin. Follow on screen instructions for your computer system.

Run R in the “Cloud”

If you do not wish to install R, or, if you have a Chromebook and, therefore cannot gracefully install R, then there are alternatives; Run R in the Cloud. I’ll list three ways to run R in the cloud for free. Go to Use R in the Cloud guide.

MacOS setup, Screenshots

Download R install package from R-project.org, then select the R install package from your Download folder (Fig 9).

Figure 9. Screenshot — Find the R install file in your download folder.

First screen, R install for macOS (Fig 10). Select continue.

Figure 10. Screenshot: first instructions pop-up window.

Second screen, R install on macOS (Fig 11). Select continue.

Figure 11. Screenshot: second instructions pop-up window.

Third screen, R install on macOS (Fig 12). Select continue.

Figure 12. Screenshot: third instructions pop-up window.

Fourth screen, R install on macOS (Fig 13). Agree to continue.

Figure 13. Screenshot: fourth instructions pop-up window.

Fifth screen, R install on macOS (Fig 14). Select Install.

Figure 14. Screenshot: fifth instructions pop-up window.

Sixth screen, R install on macOS (Fig 15). Enter your username and password for your computer, then select Install Software.

Figure 15. Screenshot: sixth instructions pop-up window.

Seventh screen, R install on macOS (Fig 16). Several screens will popup, reporting progress. Be patient!

Figure 16. Screenshot: seventh instructions pop-up window.

Eighth and final screen, R install on macOS (Fig 17). Select close.

Figure 17. Screenshot: eighth and final instructions pop-up window.

Optional — Keep or discard the install file (Fig 18). I keep and then do manual delete after I’ve confirmed the installation.

Figure 18. Screenshot: MacOS will prompt with option to delete the installation file. This has no effect on the installation of R.

From Applications folder, start R.app. You should see the R Console (Fig 19).

Figure 19. Screenshot: R Console in the R.app on macOS.

Wind10/11 setup, screenshots

Download from R-project.org, then right-click the R install package from your Download folder. Run as administrator (Fig 20).

Figure 20. WinPC  screenshot: Pop-up menu after right-click on the R installation file in windows Explorer.

First screen, select language (Fig 21). Select OK to continue

Figure 21. WinPC screenshot: first instructions pop-up window.

Second screen (Fig 22), click Next to continue

Figure 22. WinPC screenshot: second instructions pop-up window.

Third screen (Fig 23). Change the default location (show in the screenshot) to root folder, e.g., C:\R\R-4.5.1 (current version as of August 2025).

Figure 23. WinPC screenshot: third instructions pop-up window. Note — R-4.5.1 (current version as of August 2025).

Fourth screen (Fig 24). Change startup options. Select Yes (customized startup) to continue.

Figure 24. WinPC screenshot: fourth instructions pop-up window.

Fifth screen (Fig 25), select SDI — single document interface — not the default MDI — multiple document interface, then Next to continue.

Figure 25. WinPC screenshot: fifth instructions pop-up window. Recommend setting to SDI.

Sixth screen (Fig 26), select HTML help, then Next to continue.

Figure 26. WinPC screenshot: sixth instructions pop-up window.

Seventh screen (Fig 27), leave start menu folder as is (R), then Next to continue.

Figure 27. WinPC screenshot: seventh instructions pop-up window.

Eighth screen (Fig 28), check all boxes, then Next to continue.

Figure 28. WinPC screenshot: eighth instructions pop-up window.

Ninth screen (Fig 29), a series of status updates during the installation.

Figure 29. WinPC screenshot: ninth instructions pop-up window.

Final screen (Fig 30), successful install.

Figure 30. WinPC screenshot: final instructions pop-up window — successful installation.

/MD

16.6 – Similarity and Distance

Introduction

A measure of dependence between two random variables. Unlike Pearson Product Moment correlation, distance correlation measures strength of association between the variables whether or not the relationship is linear. Distance correlation is a recent addition to the literature, first reported by Gábor J. Székely (e.g., Székely et al. 2007). The package correlation (Makowski et al 2019) offers distance correlation and significance test.

Example, fly wing dataset introduced 16.1 – Product moment correlation

library(correlation)
Area <- c(0.446, 0.876, 0.390, 0.510, 0.736, 0.453, 0.882, 0.394, 0.503, 0.535, 0.441, 0.889, 0.391, 0.514, 0.546, 0.453, 0.882, 0.394, 0.503, 0.535)
Length <- c(1.524, 2.202, 1.520, 1.620, 1.710, 1.551, 2.228, 1.460, 1.659, 1.719, 1.534, 2.223, 1.490, 1.633, 1.704, 1.551, 2.228, 1.468, 1.659, 1.719)
FlyWings <- data.frame(Area, Length)
correlation(FlyWings,method="distance")

Output from R

# Correlation Matrix (distance-method)
Parameter1 | Parameter2 |    r |       95% CI | t(169) |      p
------------------------------------------------------------------
      Area |     Length | 0.92 | [0.80, 0.97] |  30.47 | < .001***

p-value adjustment method: Holm (1979)
Observations: 20

The product-moment correlation was 0.97 with 95% confidence interval (0.92, 0.99). The note about “p-value adjustment method: Holm (1979)” refers to the algorithm used to mitigate the multicomparison problem, which we first introduced in Chapter 12.1. The correction is necessary in this context because of how the algorithm conducts the test of the distance correlation. Please see Székely and Rizzo (2013) for more details.

Which should you report? For cases where it makes sense to test for a linear association, then the Product-moment correlation is the one to use. For other cases where no inference of linearity is expected, then the distance correlation makes sense.

Similarity and Distance

Similarity and distance are related mathematically. When two things are similar, the distance between them is small; When two things are dissimilar, the distance between them is great. Whether similarity (sometimes dissimilarity) or distance, the estimate is a statistic. The difference between the two is that the typical distance measures one sees in biostatistics all obey the triangle inequality rule while similarity (dissimilarity) indices do not necessarily obey the triangle inequality rule.

Distance measures

Distance is a way to talk about how far (or how close) two objects are from each other (Fig. 1). The distance may be relate to physical distance (map distance), or in mathematics, distance is a metric or statistic. Euclidean distance is the distance between two points in either the X-Y plane or 3-dimensional space measures the length of a segment connecting the two points (e.g., x₁,y₁ = 1, 4 and x₂,y₂ = 1,4).

Figure 1. Cartesian plot of two points, the first at x₁ = 1 and y₁ = 1 and the second at x₂ = 4 and y₂ = 4.

For two points (x₁, y₁ and x₂, y₂) described in two dimensions (e.g., an xy plane), the distance d is given by

For two points described in three (e.g., an xyz-space), or more dimensions, the distance d is given by

Distances of this form are Euclidean distances and can be directly obtained by use of the Pythagorean Theorem. The triangle inequality rule then applies (i.e., the sum of any two sides must be less than the length of the remaining side). Euclidean distance measures also include

• Manhattan distance: the sum of absolute difference between the measures in all dimensions of two points.

• Chebyshev distance: also called the maximum value distance, the distance between two points is the greatest of their differences along any coordinate dimension.

Note: We first met Chebyshev in Chapter 3.5.

Example

There are a number of distance measures. Let’s begin discussion of distance with geographic distance as an example. Consider the distances between cities (Table 1).

Table 1. Distances (miles) among cities.

This table is a distance matrix — note that along the diagonal are “zeros,” which should make sense — the distance between an object and itself is, well zero. Above and below the diagonal you see the distance between one city and another. This is a special kind of matrix called a symmetric matrix. Enter the distance in miles (1 mile = 1.609344) between 2 cities (this is “pairwise“). There are many resources “out there,” to help you with this. For example, I found a web site called mapcrow that allowed me to enter the cities and calculate distances between them.

To get the distance matrix, use this online resource, the Geographic Distance Matrix Calculator.

Real world problem, use geodist package. Provide latitude and longitude coordinates

The dist() function in R base will return Manhattan distance and others () on matrices.

R script

HNL <- c(0, 2667.57, 5323.37, 3849.99, 3891.82)
SEA <- c(2667.57, 0, 6590.23, 4776.81, 1888.06)
MAN <- c(5323.37, 6590.23, 0, 1835.1, 8471.48)
TOK <- c(3849.99, 4776.81, 1835.1, 0, 6664.82)
HOU <- c(3891.82, 1888.06, 8471.48, 6664.82, 0)
mat.city <- rbind(HNL, SEA, MAN, TOK, HOU)
colnames(mat.city) <- c("HNL", "SEA", "MAN", "TOK", "HOU")
mat.city
dist(mat.city, method = "manhattan")

R output

        HNL       SEA     MAN      TOK
SEA 9532.58 
MAN 21163.95 25361.39 
TOK 16070.49 20267.93 8763.66 
HOU 14526.09 8769.63 27906.40 22896.60

Other distances, replace “manhattan” with

For example,

dist(mat.city, method = "euclidean")

R output

         HNL       SEA     MAN       TOK
SEA 4550.917 
MAN 9853.774 12079.347 
TOK 7345.155 9615.750 3931.736 
HOU 6980.976 3966.360 13820.922 11883.931

The Chebyshev distance doesn’t seem to be part of base R, but is available in other packages (philentropy, function distance).

R script

require(philentropy)
distance(mat.city, method = "chebyshev", diag=FALSE, upper=FALSE)

R output

     v1         v2      v3      v4      v5
v1    0.00 2667.57 5323.37 3849.99 3891.82
v2 2667.57    0.00 6590.23 4776.81 1888.06
v3 5323.37 6590.23    0.00 1835.10 8471.48
v4 3849.99 4776.81 1835.10    0.00 6664.82
v5 3891.82 1888.06 8471.48 6664.82    0.00

Distance measures used in biology

It is easy to see how the concept of genetic distance between a group of species (or populations within a species) could be used to help build a network, with genetically similar species grouped together and genetically distant species represented in a way to represent how far removed they are from each other. Here, we speak of distance as in similarity: two species (populations) that are similar are close together, and the distance between them is short. In contrast, two species (populations) that are not similar would be represented by a great distance between them. Genetic distance is the amount of divergence of species from each other. Smaller genetic distances reflects close genetic relationship. Hamming distance is a common choice, good for categorical data.

Here’s an example (Fig. 2), RAPD gel for five kinds of beans. RAPD stands for random amplified polymorphic DNA.

Figure 2. RAPD gel (simulated) five kinds of beans.

Samples were small red beans (SRB), garbanzo beans (GB), split green pea (SGP), baby lima beans (BLB), and black eye peas (BEP). RAPD primer 1 was applied to samples in lanes 1 – 6; RAPD primer 2 was applied to samples in lane 7 – 12. Lane 1 & 7 = SRB; Lane 2 & 8 = GB; Lanes 3 & 9 = SGP; Lane 4 & 10 = BLB; Lane 5 & 11 = BB; Lane 6 & 12 = BEP.

Here’s how to go from gel documentation to the information needed for genetic distance calculations (see below). I’ll use “1” to indicate presence of a band, “0” to indicate absence of a band, and “?” to indicate no information. For simplicity, I ignored the RF value, but ranked the bands by order of largest (= 1) to smallest (=8) fragment.

We need three pieces of information from the gel to calculate genetic distance.

N_A = the number of markers for taxon A

N_B = the number of markers for taxon B

N_AB = the number of markers in common between A and B (this is the pairwise part — we are comparing taxa two at a time.

First, compare the beans against the same primer. My results for primer 1 are in Table 2; results for primer 2 are in Table 3.

Table 2. Bands for Primer 1

Table 3 for Primer 2

From Table 2 and Table 3, count N_A (= N_B ) for each taxon. Results are in Table 4.

Table 4. Bands for each taxon.

To get Hamming distance between SRB and BB (from Table 2 and Table 3), R script as follows.

SRB <- c(1, 1, 1, 0, 0, 1, 0, 1)
GB <- c(0, 0, 1, 1, 0, 1, 1, 1)
# sum the inequality between the two vectors 
sum(SRB != GB)

R output

> sum(SRB != GB)
[1] 4

Hamming distance between small red beans and black beans for the markers is 4.

As you can see, there is no simple relationship among the taxa; there is no obvious combination of markers that readily group the taxa by similarity. So, I need a computer to help me. I need a measure of genetic distance, a measure that indicates how (dis)similar the different varieties are for our genetic markers. I’ll use a distance calculation that counts only the “present” markers, not the “absent” markers, which is more appropriate for RAPD. I need to get the N_AB values, the number of shared markers between pairs of taxa.

Table 5. N_AB

The equation for calculating Nei’s distance is:

where N_A = number of bands in taxon “A”, N_B = number of bands in taxon “B”, and N_AB is the number of bands in common between A and B (Nei and Li 1979). Here’s an example calculation.

Let A = SRB and B = GB, then

The R package adegenet includes Nei’s distance (and others), but the data set assumes more information than I provided here (e.g., ploidy, alleles).

Which distance measure to use?

Unfortunately, not a simple answer. A quick search of journals will likely return “Euclidean distance” the most common (Evolution = 328, Ecology = 298). The different distance (similarity) measures were developed to solve sometimes related, but often different problems, and so a review of the statistical properties would be warranted before making a choice. Helpful reviews

Questions

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

2. Review all of the different correlation estimates we introduced in Chapter 16 and construct a table to help you learn. Product moment correlation is presented as example.

3. Be able to define and distinguish between similarity measures, dissimilarity measures quantify, and distance.

4. Return to our counts of colored candies from mini bags of Skittles, M&M plain chocolate, and M&M peanut. Which two are most similar? Here’s a data sample. Calculate the Euclidean pairwise distances, then the Manhattan distances among the bags of candies. Do you get the same answer?

The data

Example R code

mySites <- data.frame(
Site <- c("A","B","C"),
Spp1 <- c(10, 5, 12),
Spp2 <- c(7, 15, 8),
Spp3 <- c(3, 2, 9)
)

# View the matrix
print(mySites)

# Exclude the "Site" column
myAbundances = mySites[,-1]

# Calculate the Euclidean distance matrix
dist(myAbundances,method="euclidean")

The output should be

          1          2
2  9.486833  
3  6.403124  12.124356

where “1” refers to site A, “2” refers to site B, and “3” refers to site C. We would conclude that sites A and C are most similar, or have similar species abundance, because they have the smallest Euclidean distance. As you can imagine, there’s more to interpreting these statistics, and in particular, Euclidean distance can lead to some confusing interpretations about abundance and species composition between sites, eg, Orlóci paradox. Interested readers should see Ricotta 2021.

Why then teach Euclidean distance? The Euclidean distance is a standard measure, it helps lay the foundation for understanding geometric concepts. Its limitations are a reminder that students need to be aware that alternative methods are sometimes needed.

Quiz Chapter 16.6

Similarity and Distance

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Chapter 16 contents

• Introduction
• Product moment correlation
• Causation and Partial correlation
• Data aggregation and correlation
• Spearman and other correlations
• Instrument reliability and validity
• Similarity and Distance
• References and suggested readings

16.4 – Spearman and other correlations

Introduction

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

Spearman rank correlation

If you take the ranks for X ₁ and the ranks for X ₂, the correlation of ranks is called Spearman rank correlation, r_s. Spearman correlation is a nonparametric statistic. Like the product moment correlation, it can take values between -1 and +1.

For variables X ₁ and X ₂, the rank order correlation may be calculated on the ranks as

where d_i is the difference between the ranks of X ₁ and X ₂ for each experimental unit. This formula assumes that there are no tied ranks; if there are, use the equation for the product moment correlation instead (but on the ranks).

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

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

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

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

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

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

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

R code

For the Spearman rank correlation, at the R prompt type

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

R returns with
    Spearman's rank correlation rho

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

Alternatively, to calculate either correlation, use R Commander.

Rcmdr: Statistics → Summaries→ Correlation test

Example

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

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

        Spearman's rank correlation rho

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

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

        Pearson's product-moment correlation

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

Other correlations

Kendall’s tau

Another nonparametric correlation is Kendall’s tau (τ). Rank the X ₁ values, then rank the X ₂ values. Count the number of X ₁, X ₂ pairs that have the same rank (concordant pairs) and the number of X ₁, X ₂ pairs that do not have the same rank (discordant pairs), Kendall’s tau is then

where n is the number of pairs.

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

Example: Judging of Science Fair posters

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

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

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

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

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

  Cannot compute exact p-value with ties

        Kendall's rank correlation tau

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

End of R output

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

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

        Pearson's product-moment correlation

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

End R output

Tetrachoric and Polychoric correlation

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

R package psych.

R code: Tetrachoric correlation

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

R code: Polychoric correlation

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

Polyserial correlation

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

R code: Polyserial correlation

polyserial(x,y)

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

R code: Polyserial correlation

biserial(x,y)

Intra-class correlation coefficient

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

Example. Data from Table 2

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

R output follows

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

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

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

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

Questions

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

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

Quiz Chapter 16.4

Spearman and other correlations

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Chapter 16 contents

• Introduction
• Product moment correlation
• Causation and Partial correlation
• Data aggregation and correlation
• Spearman and other correlations
• Instrument reliability and validity
• Similarity and Distance
• References and suggested readings

20.10 – Growth equations and dose response calculations

Introduction

In biology, growth may refer to increase in cell number, change in size of an individual across development, or increase of number of individuals in a population over time. Nonlinear, time-series, several models proposed to fit growth data, including the Gompertz, logistic, and the von Bertalanffy. These models fit many S-shaped growth curves. These models are special cases of generalized linear models, also called Richard curves.

Growth example

This page describes how to use R to analyze growth curve data sets.

Key to parameter estimates: y0 is the lag, mumax is the growth rate, and K is the asymptotic stationary growth phase. The spline function does not return an estimate for K.

R code

require(growthrates)
#Enter the data. Replace these example values with your own 
#time variable (Hours)
Hours <- c(0.000, 0.274, 0.384, 0.658, 0.986, 1.260, 1.479, 1.699, 1.973, 2.137, 2.466, 2.795, 3.123, 3.671, 4.110, 4.548, 5.151, 5.534, 5.863, 6.521, 7.068, 7.671, 8.164, 8.877, 9.644, 10.466, 11.233, 11.890, 12.493, 13.205, 14.082)
#absorbance or concentration variable (Abs)
Abs <- c(0.002207, 0.010443, 0.033688, 0.063257, 0.111848, 0.249240, 0.416236, 0.515578, 0.572632, 0.589528, 0.619091, 0.608486, 0.621136, 0.616850, 0.614689, 0.614643, 0.612465, 0.606082, 0.603933, 0.595407, 0.589006, 0.578372, 0.567749, 0.559217, 0.546451, 0.537907, 0.537826, 0.529300, 0.516551, 0.505905, 0.491013)

#Make a dataframe and check the data; If error, then check that variables have equal numbers of entries
Yeast <- data.frame(Hours,Abs); Yeast

#Obtain growth parameters from fit of a parametric growth model
#First, try some reasonable starting values
p <- c(y0 = 0.001, mumax = 0.5, K = 0.6)
model.par <- fit_growthmodel(FUN = grow_logistic, p = p, Hours, Abs, method=c("L-BFGS-B"))
summary(model.par)
coef(model.par)

#Obtain growth parameters from fit of a nonparametric smoothing spline
model.npar <- fit_spline(Hours,Abs)
summary(model.npar)
coef(spline.md)

#Make plots
par(mfrow = c(2, 1))
plot(Yeast, ylim=c(0,1), cex=1.5,pch=16, main="Parametric Nonlinear Growth Model", xlab="Hours", ylab="Abs")
lines(model.par, col="blue", lwd=2)
plot(model.npar, ylim=c(0,1), lwd=2, main="Nonparametric Spline Fit", xlab="Hours", ylab="Abs")

Results from example code

Parameters:
      Estimate Std. Error t value Pr(>|t|)    
y0    0.003765   0.001049    3.59  0.00125 ** 
mumax 3.954458   0.238810   16.56 5.37e-16 ***
K     0.579354   0.008092   71.60  < 2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.03833 on 28 degrees of freedom

Parameter correlation:
            y0   mumax        K
y0     1.00000 -0.8755  0.05847
mumax -0.87550  1.0000 -0.14623
K      0.05847 -0.1462  1.00000
         y0       mumax           K 
0.003764518 3.954457995 0.579354378 
Fitted smoothing spline:
Call:
smooth.spline(x = time, y = ylog)

Smoothing Parameter  spar= 0.4409789  lambda= 1.221762e-05 (12 iterations)
Equivalent Degrees of Freedom (Df): 14.70929
Penalized Criterion (RSS): 0.5024203
GCV: 0.05868796

Parameter values of exponential growth curve:
Maximum growth at x= 0.0004255438 , y= 0.003072228 
y0 = 0.003065586 
mumax = 5.085845 

r2 of log transformed data= 0.9906254 
         y0       mumax 
0.003065586 5.085845036

Figure 1. Top: Parametric Nonlinear Growth Model; Bottom: Nonparametric Spline Fit

LD₅₀

In toxicology, the dose of a pathogen, radiation, or toxin required to kill half the members of a tested population of animals or cells is called the lethal dose, 50%, or LD₅₀. This measure is also known as the lethal concentration, LC₅₀, or properly after a specified test duration, the LCt₅₀ indicating the lethal concentration and time of exposure. LD₅₀ figures are frequently used as a general indicator of a substance’s acute toxicity. A lower LD₅₀ is indicative of increased toxicity.

The point at which 50% response of studied organisms to range of doses of a substance (e.g., agonist, antagonist, inhibitor, etc.) to any response, from change in behavior or life history characteristics up to and including death can be described by the methods described in this chapter. The procedures outlined below assume that there is but one inflection point, i.e., an “s-shaped” curve, either up or down; if there are more than one inflection points, then the logistic equations described will not fit the data well and other choices need to be made (see Di Veroli et al 2015). We will use the drc package (Ritz et al 2015).

Example

First we’ll work through use of R. We’ll follow up with how to use Solver in Microsoft Excel.

After starting R, load the drc library.

library(drc)

Consider some hypothetical 24-hour survival data for yeast exposed to salt solutions. Let resp equal the variable for frequency of survival (e.g., calculated from OD₆₆₀ readings) and NaCl equal the millimolar (mm) salt concentrations or doses.

At the R prompt type

resp <- c(1,1,1,.9,.7,.3,.4,.2,0,0,0)
NaCl=seq(0,1000,100) 
#Confirm sequence was correctly created; alternatively, enter the values.
NaCl 
[1] 0 100 200 300 400 500 600 700 800 900 1000 
#Make a plot
plot(NaCl,resp,pch=19,cex=1.2,col="blue",xlab="NaCl [mm]",ylab="Survival frequency")

And here is the plot (Fig. 2).

Figure 2. Hypothetical data set, survival of yeast in different salt concentrations.

Note the sigmoidal “S” shape — we’ll need an logistic equation to describe the relationship between survival of yeast and NaCl doses.

The equation for the four parameter logistic curve, also called the Hill-Slope model, is

where c is the parameter for the lower limit of the response, d is the parameter for the upper limit of the response, e is the relative EC₅₀, the dose fitted half-way between the limits c and d, and b is the relative slope around the EC₅₀. The slope, b, is also known as the Hill slope. Because this experiment included a dose of zero, a three parameter logistic curve would be appropriate. The equation simplifies to

EC₅₀ from 4 parameter model

Let’s first make a data frame

dose <- data.frame(NaCl, resp)

Then call up a function, drm, from the drc library and specify the model as the four parameter logistic equation, specified as LL.4(). We follow with a call to the summary command to retrieve output from the drm function. Note that the four-parameter logistic equation

model.dose1 = drm(dose,fct=LL.4())
summary(model.dose1)

And here is the R output.

Model fitted: Log-logistic (ED50 as parameter) (4 parms) 
Parameter estimates: 
                   Estimate  Std. Error    t-value  p-value 
b:(Intercept)      3.753415    1.074050   3.494636   0.0101 
c:(Intercept)     -0.084487    0.127962  -0.660251   0.5302 
d:(Intercept)      1.017592    0.052460  19.397441   0.0000 
e:(Intercept)    492.645128   47.679765  10.332373   0.0000 

Residual standard error: 
0.0845254 (7 degrees of freedom)

The EC₅₀, technically the LD₅₀ because the data were for survival, is the value of e: 492.65 mM NaCl.

You should always plot the predicted line from your model against the real data and inspect the fit.

At the R prompt type

plot(model.dose1, log="",pch=19,cex=1.2,col="blue",xlab="NaCl [mm]",ylab="Survival frequency")

As long as the plot you made in earlier steps is still available, R will add the line specified in the lines command. Here is the plot with the predicted logistic line displayed (Fig. 3).

Figure 3. Logistic curve added to Figure 1 plot.

While there are additional steps we can take to decide is the fit of the logistic curve was good to the data, visual inspection suggests that indeed the curve fits the data reasonably well.

More work to do

Because the EC₅₀ calculations are estimates, we should also obtain confidence intervals. The drc library provides a function called ED which will accomplish this. We can also ask what the survival was at 10% and 90% in addition to 50%, along with the confidence intervals for each.

At the R prompt type

ED(model.dose1,c(10,50,90), interval="delta")

And the output is shown below.

Estimated effective doses
(Delta method-based confidence interval(s))
     Estimate Std. Error   Lower   Upper
1:10  274.348     38.291 183.803  364.89
1:50  492.645     47.680 379.900  605.39
1:90  884.642    208.171 392.395 1376.89

Thus, the 95% confidence interval for the EC₅₀ calculated from the four parameter logistic curve was between the lower limit of 379.9 and an upper limit of 605.39 mm NaCl.

EC₅₀ from 3 parameter model

Looking at the summary output from the four parameter logistic function we see that the value for c was -0.085 and p-value was 0.53, which suggests that the lower limit was not statistically different from zero. We would expect this given that the experiment had included a control of zero mm added salt. Thus, we can explore by how much the EC₅₀ estimate changes when the additional parameter c is no longer estimated by calculating a three parameter model with LL.3().

model.dose2 = drm(dose,fct=LL.3())
summary(model.dose2)

R output follows.

Model fitted: Log-logistic (ED50 as parameter) with lower limit at 0 (3 parms)
Parameter estimates:
               Estimate Std. Error   t-value p-value
b:(Intercept)   4.46194    0.76880   5.80378   4e-04
d:(Intercept)   1.00982    0.04866  20.75272   0e+00
e:(Intercept) 467.87842   25.24633  18.53253   0e+00
Residual standard error:
 0.08267671 (8 degrees of freedom)

The EC₅₀ is the value of e: 467.88 mM NaCl.

How do the four and three parameter models compare? We can rephrase this as as statistical test of fit; which model fits the data better, a three parameter or a four parameter model?

At the R prompt type

anova(model.dose1, model.dose2)

The R output follows.

1st mode
fct:      LL.3()
2nd model
fct:      LL.4()
ANOVA table
          ModelDf      RSS Df F value p value
2nd model       8 0.054684                   
1st model       7 0.050012  1  0.6539  0.4453

Because the p-value is much greater than 5% we may conclude that the fit of the four parameter model was not significantly better than the fit of the three parameter model. Thus, based on our criteria we established in discussions of model fit in Chapter  16 – 18, we would conclude that the three parameter model is the preferred model.

The plot below now includes the fit of the four parameter model (red line) and the three parameter model (green line) to the data (Fig. 4).

Figure 4. Four parameter (red) and three parameter (green) logistic models fitted to data.

The R command to make this addition to our active plot was

lines(dose,predict(model.dose2, data.frame(x=dose)),col="green")

We continue with our analysis of the three parameter model and produce the confidence intervals for the EC₅₀ (modify the ED() statement above for model.dose2 in place of model.dose1).

Estimated effective doses
(Delta method-based confidence interval(s))
     Estimate Std. Error   Lower  Upper
1:10  285.937     33.154 209.483 362.39
1:50  467.878     25.246 409.660 526.10
1:90  765.589     63.026 620.251 910.93

Thus, the 95% confidence interval for the EC₅₀ calculated from the three parameter logistic curve was between the lower limit of 409.7 and an upper limit of 526.1 mm NaCl. The difference between upper and lower limits was 116.4 mm NaCl, a smaller difference than the interval calculated for the 95% confidence intervals from the four parameter model (225.5 mm NaCl). This demonstrates the estimation trade-off: more parameters to estimate reduces the confidence in any one parameter estimate.

Additional notes of EC₅₀ calculations

Care must be taken that the model fits the data well. What if we did not have observations throughout the range of the sigmoidal shape? We can explore this by taking a subset of the data.

dd = dose[1:6,]

Here, all values after dose 500 were dropped

dd
  resp dose
1  1.0    0
2  1.0  100
3  1.0  200
4  0.9  300
5  0.7  400
6  0.3  500

and the plot does not show an obvious sigmoidal shape (Fig. 5).

Figure 5. Plot of reduced data set.

We run the three parameter model again, this time on the subset of the data.

model.dosedd = drm(dd,fct=LL.3())
summary(model.dosedd)

Output from the results are

Model fitted: Log-logistic (ED50 as parameter) with lower limit at 0 (3 parms)
Parameter estimates:
                Estimate Std. Error    t-value p-value
b:(Intercept)   6.989842   0.760112   9.195801  0.0027
d:(Intercept)   0.993391   0.014793  67.153883  0.0000
e:(Intercept) 446.882542   5.905728  75.669344  0.0000
Residual standard error:
 0.02574154 (3 degrees of freedom)

Conclusion? The estimate is different, but only just so, 447 vs. 468 mm NaCl.

Thus, within reason, the drc function performs well for the calculation of EC₅₀. Not all tools available to the student will do as well.

NLopt and nloptr

draft

Free open source library for nonlinear optimization.

Steven G. Johnson, The NLopt nonlinear-optimization package, http://github.com/stevengj/nlopt

https://cran.r-project.org/web/packages/nloptr/vignettes/nloptr.html

Alternatives to R

What about online tools? There are several online sites that will allow students to perform these kinds of calculations. Students familiar with MatLab know that it can be used to solve nonlinear equation problems. An open source alternative to MatLab is called GNU Octave, which can be installed on a personal computer or run online at http://octave-online.net. Students also may be aware of other sites, e.g., mycurvefit.com and IC50.tk.

Both of these free services performed well on the full dataset (results not shown), but fared poorly on the reduced subset: mycurvefit returned a value of 747.64 and IC50.tk returned an EC₅₀ estimate of 103.6 (Michael Dohm, pers. obs.).

Simple inspection of the plotted values shows that these values are unreasonable.

EC₅₀ calculations with Microsoft Excel

Most versions of Microsoft Excel include an add-in called Solver, which will permit mathematical modeling. The add-in is not installed as part of the default installation, but can be installed via the Options tab in the File menu for a local installation or via Insert for the Microsoft Office online version of Excel (you need a free Microsoft account). The following instructions are for a local installation of Microsoft Office 365 and were modified from information provided by sharpstatistics.co.uk.

After opening Excel, set up worksheet as follows. Note that in order to use my formulas your spreadsheet values need to be set up exactly as I describe.

• Dose values in column A, beginning with row 2
• Response values in column B, beginning with row 2
• In column F type b in row 2, c in row 3, d in row 4, and e in row 5.
• In cells G2 – G5, enter the starting values. For this example, I used b = 1, c = 0, d = 1, e = 400.
  • For your own data you will have to explore use of different starting values.
• Enter headers in row 1.
  • Column A Dose
  • Column B Response
  • Column C Predicted
  • Column D Squared difference
  • Column F Constants
• In cell C14 enter sum squares
• In cell D14 enter =sum(D2:D12)

Here is an image of the worksheet (Fig. 6), with equations entered and values updated, but before Solver run.

Figure 6. Screenshot Microsoft Excel worksheet containing our data set (col A & B), with formulas added and calculated. Starting values for constants in column G, rows 2 – 4.

Next, enter the functions.

• In cell C2 enter the four parameter logistic formula — type everything between the double quotes: “=$G$3+(($G$4-$G$3)/(1+(A2/$G$5)^$G$2))“. Next, copy or drag the formula to cell C12 to complete the predictions.
  • Note: for a three parameter model, replace the above formula with “=(($G$4)/(1+(A2/$G$5)^$G$2))“.
• In cell D2 type everything between the double quotes: “=(B2-C2)^2“.
• Next, copy or drag the formula to cell D12 to complete the predictions.
• Now you are ready to run solver to estimate the values for b, c, d, and e.
  • Reminder: starting values must be in cells G2:G5
• Select cell D14 and start solver by clicking on Data and looking to the right in the ribbon.

Solver is not normally installed as part of the default installation of Microsoft Excel 365. If the add-in has been installed you will see Solver in the Analyze box (Fig. 7).

Figure 7. Screenshot Microsoft Excel, Solver add-in available.

If Solver is not installed, go to File → Options → Add-ins (Fig. 8).

Figure 8. Screenshot Microsoft Excel, Solver add-in available and ready for use.

Go to Microsoft support for assistance with add-ins.

With the spreadsheet completed and Solver available, click on Solver in the ribbon (Fig. 7) to begin. The screen shot from the first screen of Solver is shown below (Fig. 9).

Figure 9. Screenshot Microsoft Excel solver menu.

Setup Solver

• Set Objective, enter the absolute cell reference to the sum squares value, \$D$14
• Set To: Min.
• For By Changing Variable Cells, enter the range of cells for the four parameters, $G$2:$G$5
• Uncheck the box by “Make Unconstrained Variables Non-Negative.”
  • Where constraints refers to any system of equalities or inequalities equations imposed on the algorithm.
• Select a Solving method, choose “GRG Nonlinear,” the nonlinear programming solver option (not shown in Fig. 7, select by clicking on the down arrow).
• Click Solve button to proceed.

If all goes well, this next screen (Fig. 10) will appear which shows the message, “Solver has converged to the current solution. All constraints are satisfied.”

Figure 10. Screenshot solver completed run.

Click OK and note that the values of the parameters have been updated (Table 1).

Table 1. Four parameter logistic model, results from Solver 

Note the values obtained by Solver are virtually identical to the values obtained in the drc R package. The differences are probably because of the solver algorithm.

More interestingly, how did Solver do on the subset data set? Here are the results from a three parameter logistic model (Table 2).

Table 2. Three parameter logistic model, results from Solver

The results are again very close to results from the drc R package.

Thus, we would conclude that Solver and Microsoft Excel would be a reasonable choice for EC₅₀ calculations, and much better than IC50.tk and mycurvefit.com. The advantages of R over Microsoft Excel is that the model building is more straightforward than entering formulas in the cell reference format required by Excel.

Questions

[pending]

References and suggested readings

Beck B, Chen YF, Dere W, et al. Assay Operations for SAR Support. 2012 May 1 [Updated 2012 Oct 1]. In: Sittampalam GS, Coussens NP, Nelson H, et al., editors. Assay Guidance Manual [Internet]. Bethesda (MD): Eli Lilly & Company and the National Center for Advancing Translational Sciences; 2004-. Available from: www.ncbi.nlm.nih.gov/books/NBK91994/

Di Veroli G. Y., Fornari C., Goldlust I., Mills G., Koh S. B., Bramhall J. L., Richards, F. M., Jodrell D. I. (2015) An automated fitting procedure and software for dose-response curves with multiphasic features. Scientific Reports 5: 14701.

Ritz, C., Baty, F., Streibig, J. C., Gerhard, D. (2015) Dose-Response Analysis Using R PLOS ONE, 10(12), e0146021

Tjørve, K. M., & Tjørve, E. (2017). The use of Gompertz models in growth analyses, and new Gompertz-model approach: An addition to the Unified-Richards family. PloS one, 12(6), e0178691.

Chapter 20 contents

• Additional topics
• Area under the curve
• Peak detection
• Baseline correction
• Surveys
• Time series
• Dimensional analysis
• Estimating population size
• Diversity indexes
• Survival analysis
• Growth equations and dose response calculations
• Plot a Newick tree
• Phylogenetically independent contrasts
• How to get the distances from a distance tree
• Binary classification
• Meta-analysis

/MD

12.3 – Fixed effects, random effects, and agreement

Introduction

Within discussions of one-way ANOVA models the distinction between two general classes of models needs to be made clear by the researcher. The distinction lies in how the levels of the factor are selected. If the researcher selects the levels, then the model is a Fixed Effects Model, also called a Model I ANOVA. On the other hand, if the levels of the factor were selected by random sampling from all possible levels of the factor, then the model is a Random Effects Model, also called a Model II ANOVA. This page is about such models and I’ll introduce the intraclass correlation coefficient, abbreviated as ICC, as a way to illustrate applications of Model II ANOVA.

Here’s an example to help the distinction. Consider an experiment to see if over the counter painkillers are as good as prescription pain relievers at reducing numbers of migraines over a six week period. The researcher selects Tylenol^®, Advil^®, Bayer^® Aspirin, and Sumatriptan (Imitrex^®), the latter an example of a medicine only available by prescription. This is clearly an example of fixed effects; the researcher selected the particular medicines for use.

Random effects, in contrast, implies that the researcher draws up a list of all over the counter pain relievers and draws at random three medicines; the researcher would also randomly select from a list of all available prescription medicines.

Fixed effects are probably the more common experimental approaches. To be complete, there is a third class of ANOVA called a Mixed Model or Model III ANOVA, but this type of model only applies to multidimensional ANOVA (eg, two-way ANOVA or higher), and we reserve our discussion of the Model III until we discuss multidimensional ANOVA (Table 1).

Table 1. ANOVA models

Although the calculations for the one-way ANOVA under Model I or Model II are the same, the interpretation of the statistical significance is different between the two.

In Model I ANOVA, any statistical difference applies to the differences among the levels selected, but cannot be generalized back to the population. In contrast, statistical significance of the Factor variable in Model II ANOVA cannot be interpreted as specific differences among the levels of the treatment factor, but instead, apply to the population of levels of the factor. In short, Model I ANOVA results apply only to the study, whereas Model II ANOVA results may be interpreted as general effects, applicable to the population.

This distinction between fixed effects and random effects can be confusing, but it has broad implications for how we interpret our results in the short-term. This conceptual distinction between how the levels of the factor are selected also has general implications for our ability to acquire generalizable knowledge by meta-analysis techniques (Hunter and Schmidt 2000). Often we wish to generalize our results: we can do so only if the levels of the factor were randomly selected in the first place from all possible levels of the factor. In reality, this may not often be the case. It is not difficult to find examples in the published literature in which the experimental design is clearly fixed effects (ie, the researcher selected the treatment levels for a reason), and yet in the discussion of the statistical results, the researcher will lapse into generalizations.

Random Effects Models and agreement

Model II ANOVA is common in settings in which individuals are measured more than once. For example, in behavioral science or in sports science, subjects are typically measured for the response variable more than once over a course of several trials. Another common setting of Model II ANOVA is where more than one raters are judging an event or even a science project. In all of these cases what we are asking is about whether or not the subjects are consistent, in other words, we are asking about the precision of the instrument or measure.

In the assessment of learning by students, for example, different approaches may be tried and the instructor may wish to investigate whether the interventions can explain changes in test scores. To assess instrument validity — an item score from a grading rubric (Goertzen and Klaus 2023), tumor budding as biomarker for colorectal cancer (Bokhorst et al 2020) — agreement or concordance among two or more raters or judges (inter-rater reliability) needs to be established (same citations and references therein). For binomial data, we would be tempted to use the phi coefficient, Is it actually reliable Cohen’s kappa (two judges) or Fleiss Kappa (two or more judges) is a better tool. Phi coefficient and both assessment scores were introduced in Chapter 9.2.

For ratio scale data, agreement among scores of  two or more scores from judges There is an enormous number of articles on reliability measures in the social sciences and you should be aware of a classical paper on reliability by Shrout and Fleiss (1979) (see also McGraw and Wong, 1996). Both the ICC and the Pearson product moment correlation, r, which we will introduce in Chapter 16, are measures of strength of linear association between two ratio scale variables (Jinyuan et al 2016). But ICC is more appropriate for association between repeat measures of the same thing, eg, repeat measures of running speed or agreement between judges. In contrast, the product moment correlation can be used to describe association between any two variables, eg, between repeat measures of running speed, but also between say running speed and maximum jumping height. The concept of repeatability of individual behavior or other characteristics is also a common theme in genetics and so you should not be surprised to learn that the concept actually traces to RA Fisher and his invention of ANOVA. Just as in the case of the sociology literature, there are many papers on the use and interpretation of repeatability in the evolutionary biology literature (eg, Lessels and Boag 1987; Boake 1989; Falconer and Mackay 1996; Dohm 2002; Wolak et al 2012).

Agreement and repeatability are Model II (random effects) because the subjects, items, or raters being studied are treated as a random sample from a larger population rather than specific, fixed levels of interest. In reliability studies, subjects are not chosen for their specific characteristics but as representatives of the population variability. The goal is to infer whether a measurement tool works for any future subject or rater, not just the ones tested in the study. These factors are designed to estimate variance components and generalize results beyond the specific sample used. Put another way, the goal of Model II is often to estimate how much variance is due to differences between subjects versus within subjects (repeatability), which requires treating the subject effect as random.

There are many ways to analyze these kinds of data but a good way is to treat this problem as a one-way ANOVA with Random Effects. Thus, the Random Effects model permits the partitioning of the variation in the study into two portions: the amount that is due to differences among the subjects or judges or intervention versus the amount that is due to variation within the subjects themselves. The Factor is the Subjects and the levels of the factor are how ever many subjects are measured twice or more for the response variable.

If the subjects performance is repeatable, then the Mean Square Between (Among) Subjects, , component will be greater than the Mean Square Error component, , of the model. There are many measures of repeatability or reliability, but the intraclass correlation coefficient or ICC is one of the most common. The ICC may be calculated from the Mean Squares gathered from a Random Effects one-way ANOVA. ICC can take any value between zero and one.

where

and

B and W refers to the among group (between or among groups mean square) and the within group components of variation (error mean square), respectively, from the ANOVA, MS refers to the Mean Squares, and k is the number of repeat measures for each experimental unit. In this formulation k is assumed to be the same for each subject.

By example, when a collection of sprinters run a race, if they ran it again, would the outcome be the same, or at least predictable? If the race is run over and over again and the runners cross the finish lines at different times each race, then much of the variation in performance times will be due to race differences largely independent of any performance abilities of the runners themselves and the Mean Square Error term will be large and the Between subjects Mean Square will be small. In contrast, if the race order is preserved race after race: Jenny is 1st, Ellen is 2nd, Michael is third, and so on, race after race, then differences in performance are largely due to individual differences. In this case, the Between-subjects Mean Square will be large as will the ICC, whereas the Mean Square for Error will be small, and so will the ICC.

Can the intraclass correlation be negative?

In theory, no. Values for ICC range between zero and one. The familiar Pearson product moment correlation, Chapter 16, takes any value between -1 and +1. However, in practice, negative values for ICC will result if .

In other words, if the within group variability is greater than the among group variability, then negative ICC are possible. The question is whether negative estimates are ever biologically relevant or, simply result of limitations of the estimation procedure. Small ICC values and few repeats increases the risk of negative ICC estimates. Thus, a negative ICC would be “simply a(n) “unfortunate” estimate (Liljequist et al 2019). However, there may be situations in which negative ICC is more than unfortunate.  For example, repeatability, a quantitative genetics concept (Falconer and Mackay 1996), can be negative (Nakagawa and Schielzeth 2010).

Repeatability reflects the consistency of individual traits over time (or spatially, eg, scales on on a snake), is defined as the ratio of within individual differences and among individuals differences. Thus, within individual variance is attributed to general environmental effects, , whereas variance among individuals is both genetic and environmental. Scenarios for which repeated measures on the same individual may occur under environmental conditions, termed specific environmental effects or by Falconer and Mackay (1996), that negatively affect performance are not hard to imagine — it remains to be determined how common these conditions are and whether or not estimates of repeatability are negative.

ICC Example

I extracted 15 data points from a figure about nitrogen metabolism in kidney patients following treatment with antibiotics (Figure 1, Mitch et al 1977). I used a web application called WebPlot Digitizer (https://apps.automeris.io/wpd/), but you can also accomplish this task within R via the digitize package. I was concerned about how steady my hand was using my laptop’s small touch screen, a problem that very much can be answered by thinking statistically, and taking advantage of the ICC. So, rather than taking just one estimate of each point I repeated the protocol for extracting the points from the figure three times generating a total of three points for each of the 15 data points or 45 points in all. How consistent was I?

Let’s look at the results just for three points, #1, 2, and 3.

In the R script window enter

points = c(1,2,3,1,2,3,1,2,3)

Change points to character so that the ANOVA command will treat the numbers as factor levels.

points = as.character(points)
extracts = c(2.0478, 12.2555, 16.0489, 2.0478, 11.9637, 16.0489, 2.0478, 12.2555, 16.0489)

Make a data frame, assign to an object, eg, “digitizer”

digitizer = data.frame(points, extracts)

The dataset “digitizer” should now be attached and available to you within Rcmdr. Select digitizer data set and proceed with the one-way ANOVA.

Output from oneway ANOVA command

Model.1 <- aov(extracts ~ points, data=digitizer)

 summary(Model.1)
            Df    Sum Sq   Mean Sq  F value     Pr(>F)    
points       2 313.38895 156.69448 16562.49 5.9395e-12 ***
Residuals    6   0.05676   0.00946                        
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

numSummary(extracts , groups = points, 
+   statistics=c("mean", "sd"))
         mean           sd data:n
1  2.04780000 0.0000000000      3
2 12.15823333 0.1684708085      3
3 16.04890000 0.0000000000      3

End R output. We need to calculate the ICC

I’d say that’s pretty repeatable and highly precise measurement!

But is it accurate? You should be able to disentangle accuracy from precision based on our previous discussion (Chapter 3.5), but now in the context of a practical way to quantify precision.

ICC calculations in R

We could continue to calculate the ICC by hand, but better to have a function. Here’s a crack at the function to calculate ICC along with a 95% confidence interval.

myICC <- function(m, k, dfN, dfD) {
 testMe <- anova(m) 
 MSB <- testMe$"Mean Sq"[1]
 MSE <- testMe$"Mean Sq"[2]
 varB <- MSB - MSE/k
 ICC <- varB/(varB+MSE)
 fval <- qf(c(.025), df1=dfN, df2=dfD, lower.tail=TRUE)
 CI = (k*MSE*fval)/(MSB+MSE*(k-1)*fval)
 LCIR = ICC-CI
 UCIR = ICC+CI
 myList = c(ICC, LCIR, UCIR)
 return(myList)
}

The user supplies the ANOVA model object (eg, Model.1 from our example), k, which is the number of repeats per unit, and degrees of freedom for the among groups comparison (2 in this example), and the error mean square (6 in this case). Our example, run the function

m2ICC = myICC(Model.1, 3, 2,6); m2ICC

and R returns

[1] 0.9999396 0.9999350 0.9999442

with the ICC reported first, 0.9999396, followed by the lower limit (0.9999350) and the upper limit (0.9999442) of the 95% confidence interval.

Instead of writing your own function, several R packages can calculate the intraclass correlation coefficient (ICC) and its variants. These packages include: irr, psy, psych, and rptR.  For complex experiments involving multiple predictor variables, these packages are helpful for obtaining the correct ICC calculation (cf Shrout and Fleiss 1979; McGraw and Wong 1996). For the one-way ANOVA it is easier to just extract the information you need from the ANOVA table and run the calculation directly. We do so for a couple of examples.

Example: Are marathon runners consistent more consistent than my commute times?

Figure 1.Honolulu Marathon 2024 participant. Image credit: Pdubs.94, licensed under CC BY 4.0 (Creative Commons Attribution 4.0 International License), via Wikipedia.

A marathon is 26 miles, 385 yards (42.195 kilometers) long (World Athletics.org). And yet, tens of thousands of people choose to run in these events. For many, running a marathon is a one-off, the culmination of a personal fitness goal. For others, it’s a passion and a few are simply extraordinary, elite runners who can complete the courses in 2 to 3 hours (Table 3). That’s about 12.5 miles per hour. For comparison, my 20 mile commute on the H1 freeway on Oahu typically takes about 40 minutes to complete, or 27 miles per hour (Table 2, yes, I keep track of my commute times, per Galton’s famous maxim: “Whenever you can, count”).

Figure 2. View west along Interstate H-1 (Lunalilo Freeway) at 6:30AM on a weekday, Honolulu, Oahu, Hawaii. Image used with permission, credit: S. Dohm.

Table 2. A sampling of average commute speeds, miles per hour (mph),  on the H1 freeway during DrD’s morning commute (distance about 19.3 miles).

Try on your own.

Calculate the ICC for my commute speeds.

Run the one-way ANOVA to get the necessary mean squares and input the values into our ICC function. We have

require(psych)
m2ICC = myICC(AnovaModel.1, 4, 4,11); m2ICC
[1] 0.7390535 0.6061784 0.8719286

Repeatability, as estimated by the ICC, was 0.74 (95% CI 0.606, 0.872), for repeat measures of commute times.

We can ask the same about marathon runners — how consistent from race to race are these runners? The following data are race times drawn from a sample of runners who completed the Honolulu Marathon in both 2016 and 2017 in 2 to 3 hours (times recorded in minutes). In other words, are elite runners consistent?

Table 3. Honolulu marathon running times (min) for eleven repeat, elite runners.

Stacked worksheet, P refers to person, therefore P1 is person ID 1, P2 is person ID 2, and so on.

After running a one-way ANOVA, here are results for the marathon runners

m2ICC = myICC(Model.1, 2, 10,11); m2ICC
[1] 0.9780464 0.9660059 0.9900868

Repeatability, as estimated by the ICC, was 0.98 (95% CI 0.966, 0.990), for repeat measures of marathon performance. Put more simply, knowing what a runner did in 2016 I would be able to predict their 2017 race performance with high confidence, 98%!

And now, we compare: the runners are more consistent!

Clearly this is an apples-to-oranges comparison, but it gives us a chance to think about how we might make such comparisons. The ICC will change because of differences among individuals. For example, if individuals are not variable, if everyone finishes the race at the same time, then the ICC will be low (or even zero).  Because there is no variance to be attributed to individual differences, the proportion of total variance explained by those differences will be being minimal.

Visualizing repeated measures.

Waterfall plot.

If just two measures, then a profile plot will do. We introduced profile plots in the paired t-test chapter (Ch10.3). For more than two repeated measures, a line graph will do (see question 9 below). Another graphic increasingly used is the waterfall plot (waterfall chart), particularly for before and after measures or other comparisons against a baseline. For example, it’s a natural question for our eleven marathon runners — how much did they improve on the second attempt of the same course? Figure 3 shows my effort to provide a visual.

Figure 3. Simple waterfall plot of race improvement for Table 3 data. Dashed horizontal line at zero.

R code

Data wrangling: Create a new variable, eg, race21, by subtracting the first time from the second time. Multiply by -1 so that improvement (faster time in second race) is positive. Finally, apply descending sort so that improvement is on the left and increasingly poorer times are on the right.

df <- data.frame(
ID = rep(paste0("P", 1:11), each = 2),
Time = c(179.9,192.0,
129.9,130.8,
128.5,129.6,
179.4,179.7,
174.3,181.7,
177.2,176.2,
169.0,173.4,
174.1,175.2,
175.1,174.2,
163.9,175.9,
179.3,179.8)
)

# Convert to wide format
df_wide <- reshape(df, idvar = "ID", timevar = "ID",
direction = "wide")

# Extract first and second times manually
time_matrix <- matrix(df$Time, ncol = 2, byrow = TRUE)
colnames(time_matrix) <- c("Time1", "Time2")

# Calculate improvement (second - first)
base21 <- time_matrix[,2] - time_matrix[,1]

# Sort for waterfall display
base21 <- sort(base21, decreasing = TRUE)

# Waterfall plot
barplot(base21, col="blue", border="blue", space=0.5,
ylim=c(-20,20),
ylab="Improvement from first race (min)")
abline(h=0, lty=2)

Interpretation.

From the graph (Fig 3) we see that few of the racers improved their times the second time around — a few had decreased times.

Spaghetti plots.

Spaghetti plots, which display individual trajectories over time or multiple measurements, is a specialized line plot effective for visualizing the consistency of performance. The vertical thickness or spread of a single subject’s line across measurements indicates their level of repeatability. A “flat” line indicates high repeatability, while a volatile, zig-zagging line indicates low repeatability. Spaghetti plots are useful to visualize growth, decline, or stability over time for a single performance measure.

Figure 4. A spaghetti plot of average commute speeds from Table 2 data. 

R code

# 1. Enter the data from Table 2.
commute <- data.frame(
Week = 1:4,
Monday = c(28.5, 25.8, 26.2, 23.3),
Tuesday = c(23.8, 22.4, 22.6, 26.9),
Wednesday = c(28.5, 29.3, 24.9, 31.3),
Thursday = c(30.2, 26.2, 24.2, 26.2),
Friday = c(26.9, 27.7, 34.3, 30.2)
)

head(commute)

# 2. Set the X-axis points
days <- 1:5
day_names <- c("Mon", "Tue", "Wed", "Thu", "Fri")

# 3. Create empty plot
# Find range of Y to ensure all data fits
ymin <- min(commute[, 2:6]) - 2
ymax <- max(commute[, 2:6]) + 2

plot(NA, xlim = c(1, 5), ylim = c(ymin, ymax),
xlab = "Day of Week", ylab = "Average Speed (mph)",
main = "Weekly Commute Speeds", xaxt = "n")

# Add X-axis labels
axis(1, at = 1:5, labels = day_names)

# 4. Add lines for each week (spaghetti)
colors <- c("red", "blue", "darkgreen", "purple")
for (i in 1:nrow(commute)) {
# Plot lines
lines(days, commute[i, 2:6], col = colors[i], lwd = 2, type = "b", pch = 19)
}

# 5. Add legend
legend("topleft", legend = paste("Week", 1:4), col = colors, lwd = 2, pch = 19)

Parallel coordinates plots.

Parallel coordinates plot allows each entity to be represented by a unique, continuous line traced across several vertical axes (each representing a different measurement or variable). High repeatability is indicated when a line remains roughly horizontal across all axes, while poor repeatability is visualized as lines that zig-zag or change direction between measurements.

Figure 5. A parallel coordinates plot of average commute speeds from Table 2 data.

Both spaghetti plots and parallel coordinate plots can be used to visualize consistency of performance over multiple measures, but they differ significantly in their handling of data, scale, and context. Spaghetti plots are best for tracking within-subject variability over time, but quickly become a mess — like pasta tossed and sticking to a wall — when subject count increases. Parallel coordinate plots are a multivariate generalization — each line represents an individual’s profile across different tests, eg, accuracy, speed, VO2max.

R code.

# 1. Create the data frame
# see above

# 2. Extract only the numerical commute data for plotting
commute_data <- commute[, 2:6]

# 3. Setup the plot area
# Define range for y-axis
y_range <- range(commute_data)
# Define number of days (5)
num_days <- ncol(commute_data)

# Create empty plot
plot(NA, xlim = c(1, num_days), ylim = y_range, 
xaxt = "n", xlab = "Day of Week", ylab = "Average Speed (mph)",
main = "Average Commute Speed: Weekly Comparison")

# Add x-axis labels
axis(1, at = 1:num_days, labels = colnames(commute_data))

# 4. Draw vertical lines for each day
for (i in 1:num_days) {
lines(c(i, i), y_range, col = "grey", lty = 2)
}

# 5. Add lines for each week (repeated measures)
colors <- c("red", "blue", "darkgreen", "orange")
for (i in 1:nrow(commute_data)) {
lines(1:num_days, commute_data[i, ], col = colors[i], lwd = 2, type = "b", pch = 19)
}

# 6. Add legend
legend("topleft", legend = paste("Week", commute$Week), col = colors, lwd = 2, pch = 19)

Heat map.

Heat map quickly identify patterns, such as an individual’s performance improving, degrading, or remaining stable over time. For example, uniform, solid colors across rows (individuals) or columns (time/trials) indicate high consistency, while varied colors indicate poor repeatability.

Figure 6. A heat map of the commute speeds data set.

R code.

# 1. Create the data
# Week is excluded from matrix to only include numerical commute times
commute_data <- matrix(c(
28.5, 23.8, 28.5, 30.2, 26.9,
25.8, 22.4, 29.3, 26.2, 27.7,
26.2, 22.6, 24.9, 24.2, 34.3,
23.3, 26.9, 31.3, 26.2, 30.2
), nrow = 4, byrow = TRUE)

# 2. Add row and column names
rownames(commute_data) <- c("Week 1", "Week 2", "Week 3", "Week 4")
colnames(commute_data) <- c("Mon", "Tue", "Wed", "Thu", "Fri")

# 3. Create the heatmap
# Use t() to transpose so weeks are rows and days are columns
# Use rev() on rows to have Week 1 at the top
image_data <- t(apply(commute_data, 2, rev))

# Set up the plotting area (margins for labels)
par(mar = c(5, 6, 4, 2))

# Plot the matrix
# Option 1
image(1:nrow(image_data), 1:ncol(image_data), image_data,
col = heat.colors(12),
axes = FALSE, xlab = "Days", ylab = "",
main = "Commute Times Heatmap")

# Add axes labels
axis(1, at = 1:5, labels = colnames(commute_data))
axis(2, at = 1:4, labels = rev(rownames(commute_data)), las = 1)
box()

Interpretation.

The graphs (Fig 4 – 6) show a lack of trends in consistency of drive speeds. The heat map suggests more consistency on Monday — low speeds (Fig 4 & Fig 5), but that’s a stretch.

An example for you to work, from our Measurement Day

If you recall, we had you calculate length and width measures on shells from samples of gastropod and bivalve species. In the table are repeated measures of shell length, by caliper in mm, for a sample of Conus shells (Fig 7 and Table 4).

Figure 7. Conus shells, image by M. Dohm.

Repeated measures of the same 12 Conus shells.

Table 4. Unstacked dataset of repeated length measures on 12 shells

Maximum length from from anterior end to apex with calipers. Repeated measures were conducted blind to shell id, sampling was random over three separate time periods.

Questions

1. Consider data in Table 2, Table 3, and Table 4. True or False: The arithmetic mean is an appropriate measure of central tendency. Explain your answer.
2. Enter the shell data into R; Best to copy and stack the data in your spreadsheet, then import into R or R Commander. Once imported, don’t forget to change Sample to character, otherwise R will treat Sample as ratio scale data type. Run your one-way ANOVA and calculate the intraclass correlation (ICC) for the dataset. Is the shell length measure repeatable?
3. The marathon results in Table 2 are paired data, results of two races run by the same individuals. Make a profile plot for the data (see Ch10.3 for help).
4. A profile plot is used to show paired data; for two or more repeat measures, use a line graph instead.
   A. Produce a line graph for data in Table 4. The line graph is an option in R Commander (Rcmdr: Graphs Line graph…), and can be generated with the following simple R code:

lineplot(Sample, Measure1, Measure2, Measure3)

B. What trends, if any, do you see? Report your observations.
C. Make another line plot for data in Table 2. Report any trend and include your observations.

Quiz Chapter 12.3

Fixed effects, random effects, and agreement

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Chapter 12 contents

• Introduction
• The need for ANOVA
• One way ANOVA
• Fixed effects, random effects, and agreement
• ANOVA from “sufficient statistics”
• Effect size for ANOVA
• ANOVA posthoc tests
• Many tests one model
• References and suggested readings

12.2 – One way ANOVA

Introduction

In ANalysis Of VAriance, or ANOVA for short, we likewise have tools to test the null hypothesis of no difference between between categorical independent variables — often called factors when there’s just a few levels to keep track of — and a single, dependent response variable. But now, the response variable is quantitative, not qualitative like the tests.

Analysis of variance, ANOVA, is such an important statistical test in biology that we will take the time to “build it” from scratch. We begin by reminding you of where you’ve been with the material.

We already saw an example of this null hypothesis. When there’s only one factor (but with two or more levels), we call the analysis of means and “one-way ANOVA.” In the independent sample t-test, we tested whether two groups had the same mean. We made the connection between the confidence interval of the difference between the sample means and whether or not it includes zero (ie, no difference between the means). In ANOVA, we extend this idea to a test of whether two or more groups have the same mean. In fact, if you perform an ANOVA on only two groups, you will get the exact same answer as the independent two-sample t-test, although they use different distributions of critical values (t for the t-test, F for the ANOVA — a nice little fact for you square the t-test statistic, you’ll get the F-test statistic: ).

Let’s say we have an experiment where we’ve looked at the effect of different three different calorie levels on weight change in middle-aged men.

I’ve created a simulated dataset which we will use in our ANOVA discussions. The data set is available at the end of this page (scroll down or click here).

We might graph the mean weight changes ( SEM); Below are two possible outcomes of our experiment (Fig 1).

Figure 1. Hypothetical results of an experiment, box plots. Left, no difference among groups; Right, large differences among groups.

As statisticians and scientists, we would first calculate an overall or grand mean for the entire sample of observations; we know this as the sample mean whose symbol is . But this overall mean is made up of the average of the sample means. If the null hypothesis is true, then the all of the sample means all estimate the overall mean. Put another way, being a member of a treatment group doesn’t matter, ie, there is no systematic effect, all differences among subjects are due to random chance.

The hypotheses among are three groups or treatment levels then are:

Null that there are no differences among the group means. And the alternative hypotheses include any (or all) of the following possibilities

or maybe

or… have we covered all possible alternate outcomes among three groups?

In either case, we could use one-way ANOVA to test for “statistically significant differences.”

Three important terms you’ll need for one-way ANOVA

FACTOR: We have one factor of interest. For example, a factor of interest might be

• Diet fed to hypertensive subjects (men and women)
• Distribution of coral reef sea cucumber species in archipelagos
• Antibiotic drug therapy for adolescents with Acne vulgaris (click see Webster 2002 for review).

Gender <- c("Gender non-conforming", "Men", "Non-binary", "Transgender", "Women", "Declined"); Gender
[1] "Gender non-conforming" "Men" "Non-binary" "Transgender" 
[5] "Women" "Declined"

is.factor(Gender)
[1] FALSE

LEVELS: We can have multiple levels (2 or more) within the single factor. Some examples of levels for the Factors listed:

• Three diets (DASH), diet rich in fruits & vegetables, control diet)
• Five archipelagos (Hawaiian Islands, Line Islands, Marshal Islands, Bonin Islands, and Ryukyu Islands)
• Five antibiotics (ciprofloxacin, cotrimoxazole, erythromycin, doxycycline, minocycline).

RESPONSE: There is one outcome or measurement variable. This variable must be quantitative (ie, on the ratio or interval scale). Continuing our examples then

• Reduction in systolic pressure
• Numbers of individual sea cucumbers in a plot
• number of microcomedo.

The response variable can be just about anything we can measure, but because ANOVA is a parametric test, the response variable must be normally distributed!

Note on experimental design

As we discuss ANOVA, keep in mind we are talking about analyzing results from an experiment. Understanding statistics, and in particular ANOVA, informs how to plan an experiment. The basic experimental design is called a completely randomized experimental design, or CDR, where treatments are assigned to experimental units at random.

In this experimental design, subjects (experimental units) must be randomly assigned to each of these levels of the factor. That is, each individual should have had the same probability of being found in any one of the levels of the factor. The design is complete because randomization is conducted for all levels, all factors.

Thinking about how you would describe an experiment with three levels of some treatment, we would have

Table 1. Summary statistics of three levels for some ratio scale response variable.

ANOVA sources of variation

ANOVA works by partitioning total variability about the means (the grand mean, the group means). We will discuss the multiple samples and how the ANOVA works in terms of the sources of variation. There are two “sources” of variation that can occur:

• Within Group Variation
• Among Groups Variation

So let’s look first at the variability within groups, also called the Within Group Variation.

Consider and experiment to see if DASH diet reduces systolic blood pressure in USA middle-aged men and women with hypertension (Moore et al 2001). After eight weeks we have

We get the corrected sum of squares, SS, for within group

and the degrees of freedom, DF, for within groups

where i is the identity of the groups,  is the individual observations within group i, is the group i mean, n_i is the sample size within each group, N is the total sample size of all subjects for all groups, and k is the number of groups.

Importantly, this value is also referred to as “error sums of squares” in ANOVA. It’s importance is as follows — In our example, the within group variability would be zero if and only if all subjects within the same diet lost exactly the same amount of weight. This is hardly ever the case of course in a real experiment. Because there are almost always some other unknown factors or measurement error that affect the response variable, there will be some unknown variation among individuals who received the same treatment (within the same group). Thus, the error variance will generally be larger than zero.

The first point to consider: your ANOVA will never result in statistical differences among the groups if the error variance is greater than the second type of variability, the variability between groups.

The second type of variability in ANOVA is that due to the groups or treatments. Individuals given a calorie-restricted diet will lose some weight; individuals allowed to eat a calorie-rich diet likely will gain weight, therefore there will be variability (a difference) due to the treatment. So we can calculate the variability among groups. We get the corrected sum of squares for among groups

and the degrees of freedom for among groups

where i is the identity of the groups, is the grand mean as defined in Measures of Central Tendency (Chapter 3.1), is the group i mean, n_i is the sample size within each group, N is the total sample size of all subjects for all groups, and k is the number of groups.

The sums of squares here is simply subtracting the mean of each population from the overall mean.

• If the Factor is not important in explaining the variation among individuals then all the population means will be similar and the sums of squares among populations would be small.
• If the Factor is important in explaining some of the variation among the individuals then all the population means will NOT be the same and the sums of squares among populations would be large.

Finally, we can identify the total variation in the entire experiment. We have the total sum of squares.

Thus, the insight of ANOVA is that variation in the dataset may be attributed to a portion explained by differences among the groups and differences among individual observations within each group. The inference comes from recognizing that if the among group effect is greater than the within group effect, then there will be a difference due to the treatment levels.

Mean squares

To decide whether the variation associated with the among group differences are greater than the within group variation, we calculate ratios of the sums of squares. These are called Mean Squares or MS for short. The ratio of the Mean Squares is called F, the test statistic for ANOVA.

For the one-way ANOVA we will have two Mean Squares  and one F, tested with degrees of freedom for both the numerator , and denominator ).

The Mean Square for (among) groups is

The Mean Square for error is

And finally, the value for F, the test statistic for ANOVA, is

Worked example with R

A factor with three levels, A, B, and C

group <- c("A", "A", "A", "B", "B", "B", "C", "C", "C")

and their responses, simulated

response <- c(10.8, 11.8, 12.8, 6.5, 7, 8, 3.8, 2.8, 3)

We create a data frame

all <- data.frame(group, response)

Of course, you could place the data into a worksheet and then import the worksheet into R. Regardless, we now have our dataset.

Now, call the ANOVA function, aov, and assign the results to an object (eg, Model.1)

Model.1 <- aov(response ~ group, data=all)

Now, visualize the ANOVA table

summary(Model.1)

and the output from R, the ANOVA table, is shown below

Table 2. Output from aov() command, the ANOVA table, for the “Difference” outcome variable.

           Df   Sum Sq    Mean Sq   F value       Pr(>F) 
group       2   111.16      55.58     89.49    0.0000341 ***
Residuals   6     3.73       0.62
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Let’s take the ANOVA table one row at a time

• The first row has subject headers defining the columns.
• The second row of the table “groups” contains statistics due to group, and provides the comparisons among groups.
• The third row, “Residuals” is the error, or the differences within groups.

Moving across the columns, then, for each row, we have in turn,

• the degrees of freedom (there were 3 groups, therefore 2 DF for group),
• the Sums of Squares, the Mean Squares,
• the value of F, and finally,
• the P-value.

R provides a helpful guide on the last line of the ANOVA summary table, the “Signif[icance] codes,” which highlights the magnitude of the P-value.

What to report? ANOVA problems can be much more complicated than the simple one-way ANOVA introduced here. For complex ANOVA problems, report the ANOVA table itself! But for the one-way ANOVA it would be sufficient to report the test statistic, the degrees of freedom, and the p-value, as we have in previous chapters (eg, t-test, chi-square, etc.). Thus, we would report:

F = 89.49, df = 2 and 6, p = 0.0000341

where F = 89.49 is the test statistic, df = 2 (degrees of freedom for the among group mean square) and 6 (degrees of freedom for the within group mean square), and p = 0.0000341 is the p-value.

In Rcmdr, the appropriate command for the one-way ANOVA is simply

Rcmdr: Statistics → Means

Figure 2. Screenshot Rcmdr select one-way ANOVA.

which brings up a simple dialog. R Commander anticipates factor (Groups) and Response variable. Optional, choose Pairwise comparisons of means for post-hoc test (Tukey’s) and, if you do not want to assume equal variances (see Chapter 13), select Welch F-test.

Figure 3. Screenshot Rcmdr select one-way ANOVA options.

Questions

1. Review the example ANOVA Table and confirm the following

• How many levels of the treatment were there?
• How many sampling units were there?
• Confirm the calculation of and  using the formulas contained in the text.
• Confirm the calculation of F using the formula contained in the text.
• The degrees of freedom for the F statistic in this example were 2 and 6 (F_2,6). Assuming a two-tailed test with Type I error rate of 5%, What is the critical value of the F distribution (see Appendix 20.4)

2. Repeat the one-way ANOVA using the simulated data, but this time, calculate the ANOVA problem for the “No.difference” response variable.

3. Leaf lengths from three strains of Arabidopsis thaliana plants grown in common garden are shown in Fig 2. Data are provided for you in the following R script.

arabid <- c("wt","wt","wt","AS1","AS1","AS1","AS2","AS2","AS2")
leaf <- c(4.909,5.736,5.108,6.956,5.809,6.888,4.768,4.209,4.065)
leaves <- data.frame(arabid,leaf)

• Write out a null and alternative hypotheses
• Conduct a test of the null hypothesis by one-way ANOVA
• Report the value of the test statistic, the degrees of freedom, and the P-value
• Do you accept or reject the null hypothesis? Explain your choice.

Figure 4. Box plot of lengths of leaves of a 10-day old plant from on of three strains of Arabidopsis thaliana.

4. Return to your answer to question 7 from Chapter 12.1 and review your answer and modify as appropriate to correct your language to that presented here about factors and levels.

5. Conduct the one-way ANOVA test on the Comet assay data presented question 7 from Chapter 12.1. Data are copied for your convenience to end of this page. Obtain the ANOVA table and report the value of the test statistics, degrees of freedom, and p-value.
a. Based on the ANOVA results, do you accept or reject the null hypothesis? Explain your choice.

Quiz Chapter 12.2

One-way ANOVA

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Data used in this page

Difference, no difference

Table 12.2 Difference or No Difference

Comet assay, antioxidant properties of tea

Data presented in Chapter 12.1

Chapter 12 contents

• Introduction
• The need for ANOVA
• One way ANOVA
• Fixed effects, random effects, and agreement
• ANOVA from “sufficient statistics”
• Effect size for ANOVA
• ANOVA posthoc tests
• Many tests one model
• References and suggested readings

10.1 – Compare two independent sample means

Introduction

We introduced the concept of comparing a sample statistic (mean) against a population parameter (Chapter 6.7, Normal deviate) or one-sample t-test against a specified mean (eg, from published data or from theory, Chapter 8.5).

Consider now a basic experimental design, the randomized control trial, or RCT (Fig 1), introduced in Chapter 2.4.

Figure 1. A two group Randomized Control Trial.

Subjects randomly selected from population of interest, then again — random assignment — once recruited into one of two treatment groups. Importantly, subjects belong to one treatment arm only: no subject simultaneously receives the treatment and the control. This is in contrast to the paired design, in which subjects receive both treatments (see Chapter 10.3).

In inferential statistics about an experiment, we are more likely trying to test if sample means are different. For example

• two species grown in common garden, do they differ in growth rate?
• human subjects given a new treatment have better outcomes compare to those receiving a control treatment (eg, placebo).

The equivalent null hypothesis is that two samples are pulled from the same population. We write the null hypothesis as

and the corresponding alternative hypothesis, H_A, then must be

Question: Is this a one tailed or two-tailed hypothesis?

Answer: Two-tailed (review Chapter 8.4)

Worked example

We introduce the two-sample t-test, or better, the independent sample t-test.

where the numerator is the difference between the two sample means and the denominator is the standard error of the differences between the two groups standard errors. The formula for

The choice of independent sample over two-sample is best because it emphasizes that the two groups (the two samples), must be comprise of independent sampling units. This is a pretty straight-forward requirement; you have randomly assigned twenty individuals to two groups, a control group (n = 10) and a treatment group (n = 10). Individuals are either in the control group or they are in the treatment group — they cannot simultaneously appear in both groups.

We will work our way through this test by example. For starters, let’s use the same lizard dataset (see Example data set, below), four body mass recordings (grams) each for house geckos (Hemidactylus frenatus, Fig 2) and the Carolina anole (Anolis carolensis, Fig 3), two of many lizard species introduced to Hawaiʻi.

Figure 2. Male Hemidactylus frenatus, central Oahu, M. Dohm.

Figure 3. Male Anolis carolinensis, ʻAkaka Falls, Big Island of Hawaiʻi, M. Dohm.

Example data set

Geckos: 3.186, 2.427, 4.031, 1.995

Anoles: 5.515, 5.659, 6.739, 3.184

Question: How would you go about creating a data frame with the values in long form (stacked worksheet), including a label variable and the body mass?

Answer: At the R prompt, type

Geckos <- c(3.186, 2.427, 4.031, 1.995); Anoles = c(5.515, 5.659, 6.739, 3.184) #create two vectors
bmass <- c(Geckos, Anoles)    #combine the two vectors into a single vector holding all of the body mass records
species <- c("gecko", "gecko", "gecko", "gecko", "anole", "anole", "anole", "anole")   #create your label variable
lizards <- data.frame(species, bmass)  #create your data frame
lizards   #print your data frame  species bmass
1   gecko 3.186
2   gecko 2.427
3   gecko 4.031
4   gecko 1.995
5   anole 5.515
6   anole 5.659
7   anole 6.739
8   anole 3.184

Note also that you can enter data into the Data editor by creating the data frame first then adding values. To edit the data frame “lizards” type fix(lizards) at the R prompt, then close the data frame when you have added or changed values as needed.

As always, begin with an exploration of the data, including a graph (Fig 4).

Figure 4. Box plot of lizard body mass.

We can see already that there’s greater spread of data for the Anoles compared to the Geckos, but the median values differ. Small sample sizes can be a problem for analyses as we can only have reduced confidence in our conclusions. However, we press on for the sake of demonstration.

Let’s test the null hypothesis, H₀, ie, the two species of lizards have the same mean body mass.

Rcmdr: Statistics → Means → Independent-samples t-test…

In this next image I posted the Rcmdr menu popup for the Independent Samples t-test (Fig 5). Later versions of Rcmdr split the settings for this command into two tabs; the first tab allows for the selection of the variables and setting the hypotheses whereas the second tab, labeled Options, permits additional choices. The default selections need your attention: to actually conduct the t-test you need to answer “No” to the question, “Assume equal variance?” (Fig 6).

Figure 5. Rcmdr menu for Independent sample t-test.

Select the Options tab (Fig 6) to select null hypothesis and to select the t-test and not the Welch-test (which is the default, ie, No to the prompt “Assume equal variances?”).

Figure 6. Rcmdr Options menu for Independent sample t-test.

Let’s look at the results and break down the parts of the test.

t.test(Body.mass~Lizard, alternative='two.sided', conf.level=.95, 
+ var.equal=TRUE, data=lizards)

Two Sample t-test

data: Body.mass by Lizard 
t = 2.7117, df = 6, p-value = 0.03503
alternative hypothesis: true difference in means is not equal to 0 
95 percent confidence interval:
0.2308685 4.4981315 
sample estimates:
mean in group Anolis mean in group Gecko 
5.27425 2.90975

Consider the R session output above and answer the following questions.

Questions for the worked example

1. Which lizard group had the greater mean value, Anolis or Gecko?
2. What are the assumptions necessary for you to use the independent sample t-test?
3. What does “two-sided” mean?
4. What was the null hypothesis?
5. Was this a one-tailed or two-tailed test of the null hypothesis?
6. What is the value of the test statistic?
7. How many degrees of freedom?
8. What is the critical value for this test?
9. What is the value of the lower limit of the 95% confidence interval?
10. What is the value of the lower limit of the 99% confidence interval?
11. True or False. If the null hypothesis is accepted, then zero is a value included in the 95% confidence interval.
12. Do you accept the null hypothesis? Explain your selection.

Try another example

DNA damage, changes in the chemical structure of nucleotide bases or breakage of the DNA chains, occurs in cells under many circumstances. The comet assay, or single-cell gel electrophoresis, is one method for visualizing and measuring DNA strand breaks in cells. Exposed cells are mixed with a low-melting temperature agarose and placed onto a microscope slide. The cells are then lysed with an alkaline detergent and high salts. When current is applied across the slide, undamaged DNA remains in the nucleus, whereas damaged DNA extends towards the anode to form a comet-like tail, with imaging assisted by including a fluorescent dye like Sybr-Green. Examples of comets are shown below (Fig 7).

Figure 7. Comet examples. A Intact cell, no DNA damage, B Cell with some DNA damage, a slight tail to the right is evident, C  Cell with significant DNA damage, a large tail is evident. M. Dohm.

In an experiment, immortalized lung epithelial cells were exposed to dilute copper solutions for 30 minutes then washed with PBS. The comet assay was applied to these cells and for comparison, to cells without copper exposure but otherwise treated the same way (controls). The data are available at the bottom of this page (scroll down or click here).

Again, you should begin all analyses with an exploration of the data, including a graph (Fig 8).

Figure 8. Boxplot of comet tail lengths for cells with and without (control) exposure to copper in the cell medium for 30 minutes.

Let’s look at the R output for the t-test analysis.

    Two Sample t-test

data:  CometTail by Treatment
t = -5.8502, df = 38, p-value = 9.139e-07
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -22.39865 -10.88213
sample estimates:
mean in group Control  mean in group Copper 
             11.14533              27.78571

Consider the R session output above and answer the following questions.

Questions for Comet assay data set

1. Which cell group had the greater mean value, Copper-exposed or Control-exposed cells?
2. What are the assumptions necessary for you to use the independent sample t-test?
3. What does “two-sided” mean?
4. What was the null hypothesis?
5. Was this a one-tailed or two-tailed test of the null hypothesis?
6. What is the value of the test statistic?
7. How many degrees of freedom?
8. What is the critical value for this test?
9. What is the value of the lower limit of the 95% confidence interval?
10. What is the value of the lower limit of the 99% confidence interval?
11. True or False. If the null hypothesis is accepted, then zero is a value included in the 95% confidence interval.
12. Do you accept the null hypothesis? Explain your selection.

T test from summary statistics

In some cases you may only have to summary statistics for data, eg, the means and the standard deviations. We can use the equations of the t test to write a simple formula, where the user provides the known means, standard deviations, and sample size. For example, create a simple function with readline for user input.

myTtest <- function() {
   mnx <- as.numeric(readline(prompt="Enter mean of x: "))
   stdevx <- as.numeric(readline(prompt="Enter sd of x: "))
   nx <- as.numeric(readline(prompt="Enter n of x: "))
   mny <- as.numeric(readline(prompt="Enter mean of y: "))
   stdevy <- as.numeric(readline("Enter sd of y: "))
   ny <- as.numeric(readline(prompt="Enter n of y: "))
   myTvalue <- abs(((mnx-mny)-0)/sqrt(((stdevx^2)/nx)+(stdevy^2)/ny))
   myDF <- as.integer(nx+ny-2)
   myPvalue <- pt(myTvalue,myDF,lower.tail=FALSE)*2
   myResults <- c(myTvalue, myDF, myPvalue)
   report <- c("T-test: ", "df: ", "two-tailed p-value: ")
   cat(sprintf("%s %3.3f, ", report, myResults))
}

then run the function by typing myTest() at the R prompt and entering the means, standard deviations, and sample size when prompted.

myTtest()
Enter mean of x: 2.91
Enter sd of x: .895
Enter n of x: 4
Enter mean of y: 5.27
Enter sd of y: 1.497
Enter n of y: 4
T-test: 2.706, df: 6.000, two-tailed p-value: 0.035

P-values from confidence intervals

While we expect certain reporting criteria for published, it is not uncommon that one or more elements are missing. For example, while means plus or minus standard deviations for two groups may be reported, the confidence interval of the difference may not be reported, or, an inexact p-value is reported like “< 0.05,” but we “need” an exact p-value for our meta-analysis.  A little math, and these missing statistics can be calculated.

We can use the lizard study as an example; while all of the expected elements were reported, including the p-value (0.03503), let’s say the p value was reported as “statistically significant at p < 0.05.”

We need

lower and upper confidence intervals: 0.2309 and 4.498, respectively.

critical value from t distribution, df = 6 (2 tailed, therefore 0.05/2). A little R code we have:

the standard error, which can be calculated as

and then the p-value,

which is pretty close to the result from R (p = 0.03503). The difference is likely due to the small sample size.

Questions

1. Don’t forget to work through the Questions for the Comet tail data set (scroll up or click here).
2. Microsoft Excel, LibreOffice Calc, and Google sheets spreadsheet software all include t-test functions and return the p-value. Consider two variables big (100, 110, 120, 100, 110, 210, 200) and small (0,1,1,2,0,1,0). (Note — these two groups are obviously very different, calculating a t-test on their difference is silly, just for this question.)  If formatting is set to the default two decimal places for Number cell category, the p-value will return as “0.00.” How should you report the p-value in this case?

Quiz Chapter 10.1

Compare two independent sample means

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Comet assay data set

Chapter 10 contents

• Introduction
• Compare two independent sample means
• Digging deeper into t-test Plus the Welch test
• Paired t-test
• References and suggested reading

9.3 – Yates continuity correction

Introduction

Yates continuity correction: Most statistical textbooks at this point will note that critical values in their table (or any chi-square table for that matter) are approximate, but don’t say why. We’ll make the same directive — you may need to make a correction to the for low sample numbers. It’s not a secret, so here’s why.

We need to address a quirk of the test: the chi-square distribution is a continuous function (if you plotted it, all possible values between, say, 4 and 3 are possible). But the calculated statistics we get are discrete. In our HIV-HG co-infection problem from the previous subchapter, we got what appears to be an exact answer for P, but it is actually an approximation.

We’re really not evaluating our test statistic at the alpha levels we set out. This limitation of the goodness of fit statistic can be of some consequence — increases our chance to commit a Type I error — unless we make a slight correction for this discontinuity. The good news is that the does just fine for most df, but we do get concerned with it’s performance at df = 1 and for small samples.

Therefore, the advice is to use a correction if your calculated is close to the critical value for rejection/acceptance of the null hypothesis and you have only one df. Use the Yate’s continuity correction to standard calculation, .

For the 2×2 table (Table 1), we can rewrite Yates correction

Our concern is this: without the correction, Pearson’s test statistic will be biased (eg, the test statistic will be larger than it “really” is), so we’ll end up rejecting the null hypothesis when we shouldn’t (that’s a Type I error). This becomes an issue for us when p-value is close to 5%, the nominal rejection level: what if p-value is 0.051? Or 0.049? How confident are we in concluding accept or reject null hypothesis, respectively?

I gave you three examples of goodness of fit and one contingency table example. You should be able to tell me which of these analyses it would be appropriate to apply to correction.

More about continuity corrections

Yates suggested his correction to Pearson’s back in 1934. Unsurprisingly, new approaches have been suggested (eg, discussion in Agresti 2001). For example, Nicola Serra (Serra 2018; Serra et al 2019) introduced

Serra reported favorable performance when sample size was small and the expected value in any cell was 5 or less.

R code

When you submit a 2X2 table with one or more cells less than five, you could elect to use a Fisher exact test, briefly introduced here (see section 9.5 for additional development), or, you may apply the Yates correction. Here’s some code to make this work in R.

Let’s say the problem looks like Table 1.

Table 1. Example 2X2 table with one cell with low frequency.

At the R prompt type the following

library(abind, pos=15)
#abind allows you to combine matrices into single arrays
.Table <- matrix(c(8,12,3,22), 2, 2, byrow=TRUE)
rownames(.Table) <- c('A', 'B')
colnames(.Table) <- c('Yes', 'No') # when you submit, R replies with the following table
.Table  # Counts
 Yes No
A  8 12
B  3 22

Here’s the command; the default is no Yates correction (ie, correct=FALSE); to apply the Yates correction, set correct=TRUE.

.Test <- chisq.test(.Table, correct=TRUE)

Output from R follows

.Test
    Pearson's Chi-squared test with Yates' continuity correction
data:  .Table
X-squared = 3.3224, df = 1, p-value = 0.06834

Compare without the Yates correction

.Test <- chisq.test(.Table, correct=FALSE)
.Test
    Pearson's Chi-squared test
data:  .Table
X-squared = 4.7166, df = 1, p-value = 0.02987

Note that we would reach different conclusions! If we ignored the potential bias of the un-corrected we would be tempted to reject the null hypothesis, when in fact, the better answer is not to reject because Yates-corrected p-value is greater than 5%.

Just to complete the work, what does the Fisher Exact test results look like (see section 9.5)?

fisher.test(.Table)
    Fisher's Exact Test for Count Data
data:  .Table
p-value = 0.04086
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
0.9130455 32.8057866
sample estimates:
odds ratio 
  4.708908

Which to use? The Fisher exact test is just that, an exact test of the hypothesis. All possible outcomes are evaluated and we interpret the results as likely as p=0.04086 if there is actually no association between the treatment (A vs B) and the outcome (Yes/No) (see section 9.5).

Questions

1. With respect to interpreting results from a test for small samples, why use the Yates continuity correction?
2. Try your hand at the following four contingency tables (a – d). Calculate the test, with and without the Yates correction. Make note of the p-value from each and note any trends.

(a)

(b)

(c)

(d)

3. Ch09.1, Question 1 provided an example of a count from a small bag of M&Ms. Apply the Yates correction to obtain a better estimate of p-value for the problem. The data were four blue, two brown, one green, three orange, four red, and two yellow candies.

• Construct a table and compare p-values obtained with and without the Yates correction. Note any trend in p-value

Quiz Chapter 9.3

Yates continuity correction

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Chapter 9 contents

• Introduction
• Chi-square test: Goodness of fit
• Chi-square contingency tables
• Yates continuity correction
• Heterogeneity chi-square tests
• Fisher exact test
• McNemar’s test
• 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