Reducing Skewness

Do you want to see me quickly reduce skewness?

skewness

More seriously, here are some comments after looking at your homework (the grades are posted and you did generally well on this homework).

1. Why is symmetry important?

Some of you may be wondering why it is useful to make datasets symmetric. I don't think making a dataset symmetric is as important as making datasets have equal spreads, but here are a couple of reasons why symmetry is helpful.

-- Symmetric datasets are simpler to summarize. There is an obvious "average". Also if the data is bell-shaped, one can use the empirical rule (2/3 of the data fall within one standard deviation of the mean).

-- Many statistical procedures like ANOVA assume normally distributed data. One might wish to reexpress data before applying these procedures.

2. Monotone increasing reexpressions.

-- In our definition of power transformations, recall that when p is negative, we consider

"minus data raised to a power"

We did that so that all of the transformations are monotone increasing. So when you have a right-skewed dataset and you move from p=1 to p=0 to p=-1, you should be moving toward symmetry and left-skewness.

3. Skewness in the middle and skewness in the tails.

Sometimes it is tough to make a data symmetric, since there will be different behavior in the middle half and in the tails. You can detect this by a symmetry plot. The points to the left may be close to the line and the points to the right may fall off the line. This indicates that the middle of the data is symmetric, but there is a long tail to the right (or the left).

4. When does reexpression work?

You need sufficient spread in the data as measured by HI/LO. If this ratio is not much different from 1, reexpressions won't help much.

5. Are some of you "R resistant"?

Many of you seem to prefer using Fathom or Minitab. That's okay, but the best way to get comfortable using R is to practice using it. I made the package LearnEDA to make R easier to use, but some of you aren't taking advantage of the special functions.

Last Homework and Dreamtowns

We've been talking about taking power reexpressions of data to achieve particular objectives, such as equalizing spreads between batches or making a batch symmetric.

When do these reexpressions work?  

1.  First you need to have data with sufficient spread, that we measure by the ratio HI/LO for the power transformations to have much effect.

2.  You have to be taking a "significant" power.  In the last Fathom homework, most of you tried p = .82 (from a starting value of p = 1) which would typically have little effect on the data.  Typically, we try reexpressions in steps of 0.5, so we move from raw to roots to logs, and so on.

This week, we are reexpressing data to achieve approximate symmetry.  Here's an example.

Last summer, there was an interesting article posted on bizjournals.com that ranked 140 "dreamtowns" -- these towns offer refuge from big cities and conjested traffice.  I got interested in the article since my town, Findlay, made the list.

Anyway, they collected a number of variables from each city, including the percentage of adults (25 and over) who hold college degrees.

Here's a histogram of the these percentages from the 140 towns.

This looks right-skewed with one outlier (I never knew that Bozeman, Montana had a lot of highly educated people) and this is a good candidate for reexpression.

In the notes, I give several methods (plotting the mids, using a symmetry plot, using Hinkley's method) for choosing the "right" rexpression.

I'd suggest that you should try at least 2 of these methods in your homework.

Simple Statistical Comparisons

I just finished grading your "comparing batches" homework.  You did fine in the mechanics (computing the summaries, constructing a spread versus level graph, and deciding on an appropriate reexpression), but it wasn't clear that you understand WHY we are doing all of this work.

Let's review the main points of this section.

1.  We wish to compare two batches.

2.  What does compare mean?  Well, it could mean many things.  Batch 1 has a larger spread, batch 2 has three more outliers than batch 2, and so on.  You illustrated many type of comparisons in your homework writeups.

Here we wish to compare the general or average locations of the two batches.   A simple statistical comparison is something like

"batch 2 is 10 units larger than batch 1"

Maybe we should call this a SSC (statisticians love to use acroynms.)

What this could mean is that if I added 10 units to each value in batch 1, then I would get a new dataset that resembles batch 2.

3.  Is it always appropriate to make a SSC?

No.

It won't work if the two batches have unequal spreads.  If they have unequal spreads, then adding a number to batch 1 will NOT resemble batch 2.

4.  So if the batches have different spreads, we give up?

No.

It is possible that can can reexpress the batches to a new scale, so that the new batches have approximate new scales.

5.  So the plan is to (1) try to find a suitable reexpression and (2) do a SSC on the reexpressed data.

The snowfall data example using Fathom is one example where our strategy works.  But generally, you did a poor job in making a SCC on the reexpressed data.

A Statistical Salute to the Steelers

I just finished watching the Super Bowl and I'm happy that the Steelers won.  The winning quarterback Ben Roethlisberger is from my home town (Findlay) and it was very exciting seeing him direct his team to the winning touchdown at the end of the game.

As I was watching this game, I wondered: 

"Which team has the better offense?"

To answer this question, I collected the yards gained for every single play of the two teams, the Steelers and the Cardinals, for this Superbowl.  I created a datafile with two variables, Yards and Team, and entered this data into R.

Using the R function

boxplot(Yards~Team,horizontal=T,xlab="Yards Gained",ylab="Team",col="gold")

I created the following boxplot.

What do we see in this boxplot display?

• Both batches of yards gained look a bit right-skewed.  
• There are four outliers for the Steelers and two for the Cardinals.  These correspond to big plays for the teams that gained a lot of yards.

Why did I draw this boxplot display?  Several comments:

• We wish to make a comparison between the two batches.
• What is a comparison?  Well, we could say that one batch tends to be larger than the second batch.  For this example, if it were true, then I would say that the Cardinals gained more yards (per play) than the Steelers.
• But that isn't really saying much.   When I say "make a comparison", this means that I want to say that one batch is a particular number larger or smaller than the second batch.
• When can we say 
  "batch 1 is 10 larger than batch 2"?
• As you'll read in the notes, we can only make this type of comparison "batch 1 is 10 larger than batch 2" when the two batches have similar spreads.

Returning to my football example, it doesn't appear that the two teams were very different with respect to yards gained per play.  Both teams had approximately the same median.

Letter values by group

In the next topic, we are talking about comparing groups.  The first step is to construct some graph (such as a stemplot or boxplot or dotplot) across groups and the next step is to compute summaries (such as letter values) for each group.  Here are some comments about using R to do this stuff.

Let's return to the boston marathon data where there are two variables, time and age.  We are interested in comparing the completion times for different ages.

data(boston.marathon)

attach(boston.marathon)

By the way, this data frame is organized with two variables, response (here time) and

group (here age).

1.  GRAPHS

There is no simple way to construct parallel stemplots in R for different groups.  Parallel boxplots are easy to construct by typing, say, 

boxplot(time~age,ylab="Time",xlab="Group")

Also, it is easy to construct parallel dotplots by use of the stripchart function.

stripchart(time~age,ylab="Group")

Personally, I prefer dots that are solid black and are stacked:

stripchart(time~age,method="stack",pch=19,ylab="Group")

2.  SUMMARIES BY GROUP

If you apply the boxplot option with the plot = FALSE option, it will give you five number summaries (almost) for each group.  But the output isn't very descriptive, so I wrote a simple "wrapper" function where the output is easier to follow.

lval.by.group=function(response,group)

{

B=boxplot(response~group, plot=FALSE, range=0)

S=as.matrix(B$stats)

dimnames(S)[[1]]=c("LO","QL","M","QH","HI")

dimnames(S)[[2]]=B$names

S

}

To use this in R, you just type the function into the R Console window.  Or if you store this function in a file called lval.by.group.R, then you read the function into R by typing

source("lval.by.group.R")

Anyway, let me apply this function for the marathon data.

lval.by.group(time,age)

    20  30  40  50  60

LO 150 194 163 222 219

QL 222 213 224 251 264

M  231 235 239 262 274

QH 240 259 262 281 279

HI 274 330 346 349 338

attr(,"class")

       20 

"integer" 

I think the output is pretty clear.

3.  WHAT IF YOUR DATA IS ORGANIZED AS (GROUP1, GROUP2, ...)?

Sometimes, it is convenient to read data in a matrix format, where the different groups are in different columns.  An example of this is the population densities dataset.

data(pop.densities.1920.2000)

pop.densities.1920.2000[1:4,]

  STATE y1920 y1960 y1980 y1990 y2000

1    AL  45.8  64.2  76.6  79.6  87.6

2    AK   0.1   0.4   0.7   1.0   1.1

3    AZ   2.9  11.5  23.9  32.3  45.2

4    AR  33.4  34.2  43.9  45.1  51.3

You see that the 1920 densities are in the 2nd column, the 1960 densities in the second column, etc.

Anyway, you want to put this in the

[Response, Group]

format.  You can do this by the stack command.  The input is the matrix of data (with the first column removed) and the output is what you want.

d=stack(pop.densities.1920.2000[,-1])

d[1:5,]

  values   ind

1   45.8 y1920

2    0.1 y1920

3    2.9 y1920

4   33.4 y1920

5   22.0 y1920

Now we can compute the letter values by group (year) by typing

lval.by.group(d$values,d$ind)

     y1920   y1960   y1980   y1990   y2000

LO    0.10     0.4     0.7    1.00    1.10

QL   17.30    22.5    28.4   32.05   41.40

M    39.90    67.2    80.8   79.60   88.60

QH   62.35   114.6   157.7  181.65  202.85

HI 7292.90 12523.9 10132.3 9882.80 9378.00

Note that since I didn't attach the data frame d, I'm referring to the response variable as d$values and the grouping variable by d$ind.

How to Bin?

One of you asked me a question about the Fathom assignment on constructing histograms.  Let me describe the issues by considering a histogram for a new dataset.

Today I was thinking about snow and wondering if 6 inches is really a lot of snowfall.  So I thought I would look for some data showing levels of snowfall for different locations in the United States.

At the National Climatic Data Center site http://www.ncdc.noaa.gov/ussc/USSCAppController?action=map, one can download a "1.1 MB ASCII text file containing the maximum 1-day, 2-day, and 3-day snowfall for all available stations in the Lower 48 States and Alaska based on data through December 2006."  This sounded interesting so I downloaded this data, created a text datafile, and then read this into R.  (I had to do some data cleaning -- a little tedious.)

data=read.table("snowclim-fema-sep2006.txt",header=T,sep="\t")

This dataset contains the maximum 1-day snowfall, the maximum 2-day snowfall, the maximum 3-day snowfall for 9087 weather stations in 49 states in the U.S.

Let's focus on the max 1-day snowfall for all of the stations in Ohio -- remember, I wanted to figure out the extreme nature of 6 inches.

Using the subset command, I create a new data frame ohio that contains the data for the Ohio stations.

ohio=subset(data,data$State=="OH")

The variable of interest is called Max.1.Day.

Let's say that I wish to construct a histogram by hand -- that is, choose the bins by hand. 

First, I use the summary command to figure out the extremes.  So the variable falls betweeen 6.5 and 21.

  Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 

   6.50   11.25   13.00   13.28   15.00   21.00 

Let's try using 7 bins (why not?).  The range of the data is 21-6.50 = 14.5.  If I divide this range by 7, I get 14,5/7 = 2.07 which is approximately 2 (a prettier number than 2.07).

So let's try using the bins

(6, 8),  (8, 10), (10, 12), (12, 14), (14, 16), (16, 18), (18, 20), (20, 22)

(Actually this is 8 bins, but that's ok -- it is close to 7 bins.)

In R, I create a vector that gives these bin endpoints -- we call these my.breaks.

my.breaks=seq(6,22,2)

I use the hist command with the breaks option.  Also I want to show "6 inches" on the graph.  I do this by adding a vertical line to the graph (abline command) and label it by the text command. 

hist(ohio$Max.1.Day,breaks=my.breaks)

abline(v=6,lwd=3,col="red")

text(8,45,"6 inches",col="red")

Is this a good looking histogram?  In other words, did I make a reasonable choice of bins?  

This is a pretty smooth and I have a clear idea of the shape of the data (a little right-skewed).  But maybe I need more bins to see finer structure in the histogram.

Let's try using bins of length 1.

my.breaks=seq(6,22,1)

hist(ohio$Max.1.Day,breaks=my.breaks)

abline(v=6,lwd=3,col="red")

text(8,30,"6 inches",col="red")

Well, I think we went too far.  The histogram is less smooth -- I see some bumps in the histogram that may be a byproduct of random noise.  I definitely prefer the first histogram to this one.

I am guessing that an "optimal" bin width is between 1 and 2.

If you use the hist command without a breaks option, it will use the "Sturges" method to determine the optimal bin width for this particular sample size.  This method is similar to the one described in the Fathom lab.

I tried it.  Suprisingly (really), it turns out that R used the same breaks as I did in the first histogram.  

Wrap-up:  How does this discussion relate to the Fathom homework?  In Fathom, it is very easy to play with the bin width and see the effect of the choice of bins on the histograms.  In the famous kid's tale "Goldilocks and the Three Bears", remember that Goldilocks tried several bowls of porridge -- one bowl was "too hot", a second bowl was "too cold", and a third bowl was "just right".  Similarly, you'll get bad histograms when you choose too few bins or too many bins, and there will be a middle number of bins that will be "just right".  By playing with different histograms, you will decide (by eye) what choice of bin width is "just right", and then you will see if your choice of best bin width corresponds to the bin choice using the optimal histogram rule.

By the way, how does the "optimal rule" 

  bin width = 2 (fourth spread) / n^(1/3)

perform here?  Here the quartiles are 11.25 and 15, the fourth spread is 15-11.25 = 3.75, so the optimal bin width is 2 (3.75)/179^(1/3) = 1,33 which is a bit smaller than the bin width of 2 used in the hist command.

Final thought -- "6 inches" is really not a lot of snow, although it seemed to cripple the city where I live.

President Ages

As you know, yesterday was a big day in the history of the US as Barack Obama became our 44th president.  In honor of that event, I collected the ages at inauguration and lifespans of all 44 presidents and read this dataset into R.

data=read.table("president.ages.txt",header=T,sep="\t")

I attach the dataframe to make the variables visible.

attach(data)

Then using the stem.leaf function in the aplstats package, I construct a stemplot of the president ages:

> stem.leaf(Age)

1 | 2: represents 12

 leaf unit: 1

            n: 44

   2     t | 23

         f | 

   6     s | 6677

   9    4. | 899

  15    5* | 011111

  17     t | 22

  (9)    f | 444445555

  18     s | 6667777

  11    5. | 8

  10    6* | 0111

   6     t | 2

   5     f | 445

         s | 

   2    6. | 89

What do I see?  Here are some things I notice:

• The ages at inauguration seem pretty symmetric shaped about the values of 54 or 55.
• President ages range from 42 to 69.  Actually, one has to be a particular age to be president, so I believe that 42 is close to the rule.
• Barrack Obama is one of the youngest presidents in history at 48

I didn't play with the options for stemplot -- I just used the default settings in the stem.leaf function.  Could I produce a better stemplot?

Let's break between the tens and ones with five leaves per stem.

> stem.leaf(Age,unit=1,m=2)

1 | 2: represents 12

 leaf unit: 1

            n: 44

    2    4* | 23

    9    4. | 6677899

   22    5* | 0111112244444

  (12)   5. | 555566677778

   10    6* | 0111244

    3    6. | 589

> 

Or I could go the other way and split between the ones and tenths with 10 leaves per stem.

> stem.leaf(Age,unit=.1,m=1)

1 | 2: represents 1.2

 leaf unit: 0.1

            n: 44

   1    42 | 0

   2    43 | 0

        44 | 

        45 | 

   4    46 | 00

   6    47 | 00

   7    48 | 0

   9    49 | 00

  10    50 | 0

  15    51 | 00000

  17    52 | 00

        53 | 

  22    54 | 00000

  (4)   55 | 0000

  18    56 | 000

  15    57 | 0000

  11    58 | 0

        59 | 

  10    60 | 0

   9    61 | 000

   6    62 | 0

        63 | 

   5    64 | 00

   3    65 | 0

HI: 68 69

> 

I think it is pretty obvious that the first stemplot is the best.   If I have too few lines, then I lose some of the structure of the distribution; with too many lines, I don't see any structure at all.