Potato Chips and ANOVA, Part 2: Using Analysis of Variance to Improve Sample Preparation in Analytical Chemistry

In this second article of a 2-part series on the official JMP blog, I use analysis of variance (ANOVA) to assess a sample-preparation scheme for quantifying sodium in potato chips.  I illustrate the use of the “Fit Y by X” platform in JMP to implement ANOVA, and I propose an alternative sample-preparation scheme to obtain a sample with a smaller variance.  This article is entitled “Potato Chips and ANOVA, Part 2: Using Analysis of Variance to Improve Sample Preparation in Analytical Chemistry“.

If you haven’t read my first blog post in this series on preparing the data in JMP and using the “Stack Columns” function to transpose data from wide format to long format, check it out!  I presented this topic at the last Vancouver SAS User Group (VanSUG) meeting on Wednesday, November 4, 2015.

My thanks to Arati Mejdal, Louis Valente, and Mark Bailey at JMP for their guidance in writing this 2-part series!  It is a pleasure to be a guest blogger for JMP!

 

potato-chips-and-analytical-chemistry-part-2

Potato Chips and ANOVA in Analytical Chemistry – Part 1: Formatting Data in JMP

I am very excited to write again for the official JMP blog as a guest blogger!  Today, the first article of a 2-part series has been published, and it is called “Potato Chips and ANOVA in Analytical Chemistry – Part 1: Formatting Data in JMP“.  This series of blog posts will talk about analysis of variance (ANOVA), sampling, and analytical chemistry, and it uses the quantification of sodium in potato chips as an example to illustrate these concepts.

The first part of this series discusses how to import the data into the JMP and prepare them for ANOVA.  Specifically, it illustrates how the “Stack Columns” function is used to transpose the data from wide format to long format.

I will present this at the Vancouver SAS User Group (VanSUG) meeting later today.

Stay tuned for “Part 2: Using Analysis of Variance to Improve Sample Preparation in Analytical Chemistry“!

 

potato-chips-and-analytical-chemistry-part-1

Odds and Probability: Commonly Misused Terms in Statistics – An Illustrative Example in Baseball

Yesterday, all 15 home teams in Major League Baseball won on the same day – the first such occurrence in history.  CTV News published an article written by Mike Fitzpatrick from The Associated Press that reported on this event.  The article states, “Viewing every game as a 50-50 proposition independent of all others, STATS figured the odds of a home sweep on a night with a full major league schedule was 1 in 32,768.”  (Emphases added)

odds of all 15 home teams winning on same day

Screenshot captured at 5:35 pm Vancouver time on Wednesday, August 12, 2015.

Out of curiosity, I wanted to reproduce this result.  This event is an intersection of 15 independent Bernoulli random variables, all with the probability of the home team winning being 0.5.

P[(\text{Winner}_1 = \text{Home Team}_1) \cap (\text{Winner}_2 = \text{Home Team}_2) \cap \ldots \cap (\text{Winner}_{15}= \text{Home Team}_{15})]

Since all 15 games are assumed to be mutually independent, the probability of all 15 home teams winning is just

P(\text{All 15 Home Teams Win}) = \prod_{n = 1}^{15} P(\text{Winner}_i = \text{Home Team}_i)

P(\text{All 15 Home Teams Win}) = 0.5^{15} = 0.00003051757

Now, let’s connect this probability to odds.

It is important to note that

  • odds is only applicable to Bernoulli random variables (i.e. binary events)
  • odds is the ratio of the probability of success to the probability of failure

For our example,

\text{Odds}(\text{All 15 Home Teams Win}) = P(\text{All 15 Home Teams Win}) \ \div \ P(\text{At least 1 Home Team Loses})

\text{Odds}(\text{All 15 Home Teams Win}) = 0.00003051757 \div (1 - 0.00003051757)

\text{Odds}(\text{All 15 Home Teams Win}) = 0.0000305185

The above article states that the odds is 1 in 32,768.  The fraction 1/32768 is equal to 0.00003051757, which is NOT the odds as I just calculated.  Instead, 0.00003051757 is the probability of all 15 home teams winning.  Thus, the article incorrectly states 0.00003051757 as the odds rather than the probability.

This is an example of a common confusion between probability and odds that the media and the general public often make.  Probability and odds are two different concepts and are calculated differently, and my calculations above illustrate their differences.  Thus, exercise caution when reading statements about probability and odds, and make sure that the communicator of such statements knows exactly how they are calculated and which one is more applicable.

Using PROC SQL to Find Uncommon Observations Between 2 Data Sets in SAS

A common task in data analysis is to compare 2 data sets and determine the uncommon rows between them.  By “uncommon rows”, I mean rows whose identifier value exists in one data set but not the other. In this tutorial, I will demonstrate how to do so using PROC SQL.

Let’s create 2 data sets.

data dataset1;
      input id $ group $ gender $ age;
      cards;
      111 A Male 11
      111 B Male 11
      222 D Male 12
      333 E Female 13
      666 G Female 14
      999 A Male 15
      999 B Male 15
      999 C Male 15
      ;
run;
data dataset2;
      input id $ group $ gender $ age;
      cards;
      111 A Male 11
      999 C Male 15
      ;
run;

First, let’s identify the observations in dataset1 whose ID variable values don’t exist in dataset2.  I will export this set of observations into a data set called mismatches1, and I will print it for your viewing.  The logic of the code is simple – find the IDs in dataset1 that are not in the IDs in dataset2.  The code “select *” ensures that all columns from dataset1 are used to create the data set in mismatches1.

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The advantages of using count() to get N-way frequency tables as data frames in R

Introduction

I recently introduced how to use the count() function in the “plyr” package in R to produce 1-way frequency tables in R.  Several commenters provided alternative ways of doing so, and they are all appreciated.  Today, I want to extend that tutorial by demonstrating how count() can be used to produce N-way frequency tables in the list format – this will magnify the superiority of this function over other functions like table() and xtabs().

 

2-Way Frequencies: The Cross-Tabulated Format vs. The List-Format

To get a 2-way frequency table (i.e. a frequency table of the counts of a data set as divided by 2 categorical variables), you can display it in a cross-tabulated format or in a list format.

In R, the xtabs() function is good for cross-tabulation.  Let’s use the “mtcars” data set again; recall that it is a built-in data set in Base R.

> y = xtabs(~ cyl + gear, mtcars)
> y
          gear
 cyl      3     4     5
 4        1     8     2
 6        2     4     1
 8        12    0     2

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How to Get the Frequency Table of a Categorical Variable as a Data Frame in R

Introduction

One feature that I like about R is the ability to access and manipulate the outputs of many functions.  For example, you can extract the kernel density estimates from density() and scale them to ensure that the resulting density integrates to 1 over its support set.

I recently needed to get a frequency table of a categorical variable in R, and I wanted the output as a data table that I can access and manipulate.  This is a fairly simple and common task in statistics and data analysis, so I thought that there must be a function in Base R that can easily generate this.  Sadly, I could not find such a function.  In this post, I will explain why the seemingly obvious table() function does not work, and I will demonstrate how the count() function in the ‘plyr’ package can achieve this goal.

The Example Data Set – mtcars

Let’s use the mtcars data set that is built into R as an example.  The categorical variable that I want to explore is “gear” – this denotes the number of forward gears in the car – so let’s view the first 6 observations of just the car model and the gear.  We can use the subset() function to restrict the data set to show just the row names and “gear”.

> head(subset(mtcars, select = 'gear'))
                     gear
Mazda RX4            4
Mazda RX4 Wag        4
Datsun 710           4
Hornet 4 Drive       3
Hornet Sportabout    3
Valiant              3

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Using PROC SGPLOT to Produce Box Plots with Contrasting Colours in SAS

I previously explained the statistics behind box plots and how to produce them in R in a very detailed tutorial.  I also illustrated how to produce side-by-side box plots with contrasting patterns in R.

Here is an example of how to make box plots in SAS using the VBOX statement in PROC SGPLOT.  I modified the built-in data set SASHELP.CLASS to generate one that suits my needs.

The PROC TEMPLATE statement specifies the contrasting colours to be used for different classes.  I also include code for exportingthe result into a PDF file using ODS PDF.  (I used varying shades of grey to allow the contrast to be shown when printed in black and white.)

boxplots

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Getting a List of the Variable Names of a SAS Data Set

Getting a list of the variable names of a data set is a fairly common and useful task in data analysis and manipulation, but there is actually no procedure or function that will do this directly in SAS.  After some diligent searching on the Internet, I found a few tricks within PROC CONTENTS do accomplish this task.

Here is an example involving the built-in data set SASHELP.CLASS.  The ultimate goal is to create a new data set called “variable_names” that contains the variable names in one column.

The results of PROC CONTENTS can be exported into a new data set.  I will call this data set “data_info”, and it contains just 2 variables that we need – “name” and “varnum“.

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Getting All Duplicates of a SAS Data Set

Introduction

A common task in data manipulation is to obtain all observations that appear multiple times in a data set – in other words, to obtain the duplicates.  It turns out that there is no procedure or function that will directly provide the duplicates of a data set in SAS*.

*Update: As Fareeza Khurshed kindly commented, the NOUNIQUEKEY option in PROC SORT is available in SAS 9.3+ to directly obtain duplicates and unique observations.  I have written a new blog post to illustrate her solution.

The Wrong Way to Obtain Duplicates in SAS

You may think that PROC SORT can accomplish this task with the nodupkey and the dupout options.  However, the output data set from such a procedure does not have the first of each set of duplicates.  Here is an example.

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Performing Logistic Regression in R and SAS

Introduction

My statistics education focused a lot on normal linear least-squares regression, and I was even told by a professor in an introductory statistics class that 95% of statistical consulting can be done with knowledge learned up to and including a course in linear regression.  Unfortunately, that advice has turned out to vastly underestimate the variety and depth of problems that I have encountered in statistical consulting, and the emphasis on linear regression has not paid dividends in my statistics career so far.  Wisdom from veteran statisticians and my own experience combine to suggest that logistic regression is actually much more commonly used in industry than linear regression.  I have already started a series of short lessons on binary classification in my Statistics Lesson of the Day and Machine Learning Lesson of the Day.    In this post, I will show how to perform logistic regression in both R and SAS.  I will discuss how to interpret the results in a later post.

The Data Set

The data set that I will use is slightly modified from Michael Brannick’s web page that explains logistic regression.  I copied and pasted the data from his web page into Excel, modified the data to create a new data set, then saved it as an Excel spreadsheet called heart attack.xlsx.

This data set has 3 variables (I have renamed them for convenience in my R programming).

  1. ha2  – Whether or not a patient had a second heart attack.  If ha2 = 1, then the patient had a second heart attack; otherwise, if ha2 = 0, then the patient did not have a second heart attack.  This is the response variable.
  2. treatment – Whether or not the patient completed an anger control treatment program.
  3. anxiety – A continuous variable that scores the patient’s anxiety level.  A higher score denotes higher anxiety.

Read the rest of this post to get the full scripts and view the full outputs of this logistic regression model in both R and SAS!

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Extracting the Postal Codes from Addresses of Hospitals in British Columbia – An Exercise in SAS Text Processing

Introduction

In my job as a Biostatistical Analyst at the British Columbia (BC) Cancer Agency in Vancouver, I recently needed to get the postal codes for the hospitals in BC.  I found a data table of the hospitals with their addresses, but I needed to extract the postal codes from the addresses.  In this tutorial, I will show you some text processing techniques in SAS that I used to extract the postal codes from that raw data file.

* This blog post contains information licensed under the Open Government License – British Columbia.

Read the rest of this post to get the SAS code for extracting the postal codes and the final spreadsheet that contains the postal codes of the hospitals in British Columbia!

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Calculating the sum or mean of a numeric (continuous) variable by a group (categorical) variable in SAS

Introduction

A common task in data analysis and statistics is to calculate the sum or mean of a continuous variable.  If that variable can be categorized into 2 or more classes, you may want to get the sum or mean for each class.

This sounds like a simple task, yet I took a surprisingly long time to learn how to do this in SAS and get exactly what I want – a new data with with each category as the identifier and the calculated sum/mean as the value of a second variable.  Here is an example to show you how to do it using PROC MEANS.

Read more to see an example data set and get the SAS code to calculate the sum or mean of a continuous variable by a categorical variable!

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Online index of plots and corresponding R scripts

Dear Readers of The Chemical Statistician,

Joanna Zhao, an undergraduate researcher in the Department of Statistics at the University of British Columbia, produced a visual index of over 100 plots using ggplot2, the R package written by Hadley Wickham.

An example of a plot and its source R code on Joanna Zhao's catalogue.

An example of a plot and its source R code on Joanna Zhao’s catalog.

Click on a thumbnail of any picture in this catalog – you will see the figure AND all of the necessary code to reproduce it.  These plots are from Naomi Robbins‘ book “Creating More Effective Graphs”.

If you

  • want to produce an effective plot in R
  • roughly know what the plot should look like
  • but could really use an example to get started,

then this is a great resource for you!  A related GitHub repository has the code for ALL figures and the infrastructure for Joanna’s Shiny app.

I learned about this resource while working in my job at the British Columbia Cancer Agency; I am fortunate to attend a wonderful seminar series on statistics at the British Columbia Centre for Disease Control, and a colleague from this seminar told me about it.  By sharing this with you, I hope that it will immensely help you with your data visualization needs!

The Chi-Squared Test of Independence – An Example in Both R and SAS

Introduction

The chi-squared test of independence is one of the most basic and common hypothesis tests in the statistical analysis of categorical data.  Given 2 categorical random variables, X and Y, the chi-squared test of independence determines whether or not there exists a statistical dependence between them.  Formally, it is a hypothesis test with the following null and alternative hypotheses:

H_0: X \perp Y \ \ \ \ \ \text{vs.} \ \ \ \ \ H_a: X \not \perp Y

If you’re not familiar with probabilistic independence and how it manifests in categorical random variables, watch my video on calculating expected counts in contingency tables using joint and marginal probabilities.  For your convenience, here is another video that gives a gentler and more practical understanding of calculating expected counts using marginal proportions and marginal totals.

Today, I will continue from those 2 videos and illustrate how the chi-squared test of independence can be implemented in both R and SAS with the same example.

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Video Tutorial – Calculating Expected Counts in a Contingency Table Using Joint Probabilities

In an earlier video, I showed how to calculate expected counts in a contingency table using marginal proportions and totals.  (Recall that expected counts are needed to conduct hypothesis tests of independence between categorical random variables.)  Today, I want to share a second video of calculating expected counts – this time, using joint probabilities.  This method uses the definition of independence between 2 random variables to form estimators of the joint probabilities for each cell in the contingency table.  Once the joint probabilities are estimated, the expected counts are simply the joint probabilities multipled by the grand total of the entire sample.  This method gives a more direct and deeper connection between the null hypothesis of a test of independence and the calculation of expected counts.

I encourage you to watch both of my videos on expected counts in my Youtube channel to gain a deeper understanding of how and why they can be calculated.  Please note that the expected counts are slightly different in the 2 videos due to round-off error; if you want to be convinced about this, I encourage you to do the calculations in the 2 different orders as I presented in the 2 videos – you will eventually see where the differences arise.

You can also watch the video below the fold!

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Video Tutorial – Allelic Frequencies Remain Constant From Generation to Generation Under the Hardy-Weinberg Equilibrium

The Hardy-Weinberg law is a fundamental principle in statistical genetics.  If its 7 assumptions are fulfilled, then it predicts that the allelic frequency of a genetic trait will remain constant from generation to generation.  In this new video tutorial in my Youtube channel, I explain the math behind the Hardy-Weinberg theorem.  In particular, I clarify the origin of the connection between allelic frequencies and genotyopic frequencies in the second generation – I have not found a single textbook or web site on this topic that explains this calculation, so I hope that my explanation is helpful to you.

You can also watch the video below the fold!

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Video Tutorial – Calculating Expected Counts in Contingency Tables Using Marginal Proportions and Marginal Totals

A common task in statistics and biostatistics is performing hypothesis tests of independence between 2 categorical random variables.  The data for such tests are best organized in contingency tables, which allow expected counts to be calculated easily.  In this video tutorial in my Youtube channel, I demonstrate how to calculate expected counts using marginal proportions and marginal totals.  In a later video, I will introduce a second method for calculating expected counts using joint probabilities and marginal probabilities.

In a later tutorial, I will illustrate how to implement the chi-squared test of independence on the same data set in R and SAS – stay tuned!

You can also watch the video below the fold!

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Useful Options For Every SAS Program – Lessons and Resources from Dr. Jerry Brunner

Introduction

Today, I want to share some useful options to put at the beginning of every SAS program that I write.  These options will make the practicality of using SAS much easier.  My applied statistics professor from the University of Toronto, Jerry Brunner*, taught me some of these options when I first learned SAS in our class, and I’m grateful for that.  In later instances of using SAS in team projects, I have met SAS programmers who were delightfully surprised by the existence of these options and desperately wished that they had learned them earlier.  I hope that they will help you with your SAS programming.  I have also learned some useful options by posting questions on the SAS Support Communities online forum.

 

Clearing Output

After running your SAS program many times to test and debug, you will have accumulated numerous pages of old and useless output and log.  Scrolling through and searching for the desired portion to read in either file can be tedious and difficult.  Thus, it’s really helpful to have the option of clearing all of the output and the log whenever you run your script.  I put the following commands on top of every one of my SAS scripts.

/* 
Useful Options For Every SAS Program 
- With Some Tips Learned From Dr. Jerry Brunner
by Eric Cai - The Chemical Statistician
*/

dm 'cle log; cle out;';
ods html closeods html;

dm 'odsresults; clear';
ods listing close;
ods listing;

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Determining chemical concentration with standard addition: An application of linear regression in JMP – A Guest Blog Post for the JMP Blog

I am very excited to announce that I have been invited by JMP to be a guest blogger for its official blog!  My thanks to Arati Mejdal, Global Social Media Manager for the JMP Division of SAS, for welcoming me into the JMP blogging community with so much support and encouragement, and I am pleased to publish my first post on the JMP Blog!  Mark Bailey and Byron Wingerd from JMP provided some valuable feedback to this blog post, and I am fortunate to get the chance to work with and learn from them!

Following the tradition of The Chemical Statistician, this post combines my passions for statistics and chemistry by illustrating how simple linear regression can be used for the method of standard addition in analytical chemistry.  In particular, I highlight the useful capability of the “Inverse Prediction” function under “Fit Model” platform in JMP to estimate the predictor given an observed response value (i.e. estimate the value of x_i given y_i).  Check it out!

JMP blog post - standard addition

Video Tutorial – Useful Relationships Between Any Pair of h(t), f(t) and S(t)

I first started my video tutorial series on survival analysis by defining the hazard function.  I then explained how this definition leads to the elegant relationship of

h(t) = f(t) \div S(t).

In my new video, I derive 6 useful mathematical relationships that exist between any 2 of the 3 quantities in the above equation.  Each relationship allows one quantity to be written as a function of the other.

I am excited to continue adding to my Youtube channel‘s collection of video tutorials.  Please stay tuned for more!

You can also watch this new video below the fold!

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