Two girls talking

(There may even be a part two, if I get around to it.)

Let me ask you a couple of questions:

1. Do you have more than just one dependent variable and one independent variable?

2. If you said, yes, do you have a CATEGORICAL or ORDINAL dependent variable? If so, use logistic regression. I have written several posts on it. You can find a list of them here. Some involve Euclid, marriage, SAS and SPSS. Alas, none involve a naked mole rat. I shall have to remedy that.

3. You said yes to #1, multiple variables, but no to number 2, so I am assuming you have multiple variables in your design and your dependent variable is interval or continuous, something like sales for the month of December, average annual temperature or IQ. The next question is do you have only ONE dependent variable and is it measured only ONCE per observation? For example, you have measured average annual temperature of each city in 2013 or sales in December , 2012. In this case, you would do either Analysis of Variance or multiple regression. It doesn’t matter much which you do if you code it correctly. Both are specific cases of the general linear model and will give you the same result. You may also want to do a general linear MIXED model, where you have city as a random effect and something else, say, whether the administration was Democratic or Republican as a fixed effect. In this case I assume that you have sales as your dependent variable because contrary to the beliefs of some extremists, political parties do not determine the weather. Generally, whether you use a mixed model or an Ordinary Least Squares (OLS) plain vanilla ANOVA or regression will not have a dramatic impact on your results unless the result is a grade in a course where the professor REALLY wants you to show that you know that school is a random effect when comparing curricula.

4. Still here? I’m guessing you have one of two other common designs. That is, you have measured the same subjects, stores, cities, whatever, more than once. Most commonly, it is the good old pretest posttest design and you have an experimental and control group. You want to know if it works. If you have only tested your people twice, you are perfectly fine with a repeated measures ANOVA. If you have tested them more than twice, you are very likely to have grossly violated the assumption of compound symmetry and I would recommend a mixed model.

5. All righty then, you DO have multiple variables, they are NOT categorical or ordinal, your dependent variable is NOT repeated, so you must have multiple dependent variables. In that case, you would do a multivariate Analysis of Variance.

Some might argue that logistic regression is not a multivariate design. Other people would argue with them that, assuming your data are multinomial, you need multiple logit functions so that really is a type of multivariate design. A third group of people would say it is multivariate in the ordinal or multinomial case because there are multiple possible outcomes.

Personally, I wonder about all of those types of people. I wonder about the amount of time in higher education spent in forcing students to learn answers to questions that have no real use or purpose as far as I can see.

On the other hand, while knowing whether something falls in the multivariate category or not probably won’t impact your life or analyses, if you treat time as an independent variable and analyze your repeated measures ANOVA with experiment and condition as a 2 x 2 ANOVA, you’re screwed.

Know your research designs.

What if you wanted to turn your PROC MIXED into a repeated measures ANOVA using PROC GLM. Why would you want to do this? Well, I don’t know why you would want to do it but I wanted to do it because I wanted to demonstrate for my class that both give you the same fixed effects F value and significance.

I started out with the Statin dataset from the Cody and Smith textbook. In this data set, each subject has three records,one each for drugs A, B and C. To do a mixed model with subject as a random effect and drug as a fixed effect, you would code it as so. Remember to include both the subject variable and your fixed effect in the CLASS statement.

Proc mixed data = statin ;
class subj drug ;
model ldl = drug ;
random subj ;

To do a repeated measures ANOVA with PROC GLM you need three variables for each subject, not three records.

First, create three data sets for Drug A, Drug B and Drug C.

Data one two three ;
set statin ;
if drug = ‘A’ then output one ;
else if drug = ‘B’ then output two ;
else if drug = ‘C’ then output three ;

Second, sort these datasets and as you read in each one, rename LDL to a new name so that when you merge the datasets you have three different names. Yes, I really only needed to rename two of them, but I figured it was just neater this way.

proc sort data = one (rename= (ldl =ldla)) ;
by subj ;

proc sort data= two (rename = (ldl = ldlb)) ;
by subj ;
proc sort data=three (rename =(ldl = ldlc)) ;
by subj ;

Third, merge the three datasets by subject.

data mrg ;
merge one two three ;
by subj ;

Fourth, run your repeated measures ANOVA .

Your three times measuring LDL are the dependent . It seems weird to not have an independent on the other side of the equation, but that’s the way it is. In your REPEATED statement you give a name for the repeated variable and the number of levels. I used “drug” here to be consistent but actually, this could be any name at all. I could have used “frog” or “rutabaga” instead and it would have worked just as well.

proc glm data = mrg ;
model ldla ldlb ldlc = /nouni ;
repeated drug 3 (1 2 3) ;
run ;

Compare the results and you will see that both give you the numerator and denominator degrees of freedom, F-statistic and p-value for the fixed effect of drug.

Now you can be happy.

Any time you learn anything new it can be intimidating. That is true of programming as well as anything else. It may be even more true of using statistical software because you combine the uneasiness many people have about learning statistics with learning a new language.

To a statistician, this error message makes perfect sense:

ERROR: Variable BP_Status in list does not match type prescribed for this list.

but to someone new to both statistics and SAS it may be clear as mud.

Here is your problem.

The procedure you are using, PROC UNIVARIATE , PROC MEANS is designed ONLY for numeric variables. You have tried to use it for a categorical variable.

This error means you’ve used a categorical variable in a list where only numeric variables are expected. For example, bp_status is “High”, “Normal” and “Optimal”

You cannot find the mean or standard deviation of words, so your procedure has an error.

So … what do you do if you need descriptive statistics?

Go back to your PROC UNIVARIATE or PROC MEANS and delete the offending variables. Re-run it with only numeric variables.

For your categorical variables, use a PROC FREQ  for a frequency distribution  and/ or PROC GCHART.

Problem solved.

You’re welcome.

Computing confidence intervals is one of the areas where beginning statistics students have the most trouble. It is not as difficult if you break it down into steps, and if you use SAS or other statistical software.

Here are the steps:

1. Compute the statistic of interest– that is mean, proportion, difference between means
2. Compute the standard error of the statistic
3. Obtain critical value. Do you have 30 or more in your sample and are you interested in the 95% confidence interval?

  • If yes, multiply standard error by 1.96
  • If no (fewer people), look up t-value for your sample size for .95
  • If no (different alpha level) look up z-value for your alpha level
  • If no (different alpha level  AND less than 30) look up the t-value for your alpha level.

4. Multiply the critical value you obtained in step #3 by the standard error you obtained in #2

5. Subtract the result  you obtained in step #4 from the statistic you obtained in #1 . That is your lower confidence limit.

6. Add the result you obtained in step #4 to the statistic you obtained in #1. That is your upper confidence limit.
Simplifying it with SAS

Here is a homework problem:

The following data are collected as part of a study of coffee consumption among undergraduate students. The following reflect cups per day consumed:

3          4          6          8          2          1          0          2

 

A. Compute the sample mean.

B. Compute the sample standard deviation.

C. Construct a 95% confidence interval

I did this in SAS as so

data coffee ;
input cups ;
datalines ;
3
4
6
8
2
1
0
2
;
proc means mean std stderr;
var cups ;

I get the follow results.

Analysis Variable : cups
Mean Std Dev Std Error
3.2500000 2.6592158 0.9401748

These results give me A and B. Now, all I need to do to compute C is find the correct critical value.  I look it up and find that it is 2.365

3.25   – 2.365 * .94 = 1.03

3.25 + 2.365 * .94 = 5.47

That is my confidence interval (1.03, 5.47)

=========================

If you want to verify it, or just don’t want to do any computations at all, you can do this

Proc means clm mean stddev ;
var cups ;

You will end up with the same confidence intervals.

Prediction: At least one person who reads this won’t believe me, will run the analysis and be surprised when I am right.

A recent tweet about mixed martial arts decisions set me to thinking about probability. @Fight_ghost tweeted that a TV commentator made no sense when she said that she thought a fighter should have won by split, not unanimous decision. Others on twitter agreed with him that was a stupid comment, and asked did she think judges should say the other fighter only 2/3 won or what.

I thought it did make sense in statistical terms. Think of it this way:

The “true score” of the population in this case is the mean of what an infinite number of judges would rate a fighter’s performance. Of course, there is going to be variation around that mean. Some judges may tend to weight take downs a tiny bit more. Judges vary in their definition of a significant strike. Some judges are just going to be clueless or inattentive and give a score that is far from accurate. On the average, though, these balance out and the mean of all of those infinite judges’ scores should be the true score. Let’s say our fighter, Bob, had a true score of 27. The most common score we should see a judge give him is 27, but a 26 or 28 would not be totally unexpected. Given that the standard deviation of fight scores is low, we would be surprised to see him given a score of 25 or 29 and completely floored if he received a 24 or a 30.

Let’s say we have a second fighter, Fred. His true score is 29. The most common score we should see for him is a 29, but again, a 28 or a 30 would not be unexpected because there is variation in our sample of judges.

Here is the point … when fighters are far apart in the true score of their performance, judges should very seldom have a difference of opinion in who won. Even when Bob is scored high, for him, at 28 and Fred is scored his average of 29, Fred still wins. Let’s say the standard deviation of judge’s scores is 1. I believe it is really lower than that and I do know that the winner of a round has to get 10 points, but for ease of computation, just go with me.

For Bob to win, he must be rated at least two standard deviations above his true score (which occurs 2.5% of the time) and Fred must be rated below his true score, which occurs half the time. Since the scores for Bob and Fred are independent probabilities the probability of BOTH of these events happening is .025  x .5 = .0125

The other way for Bob to win is if Fred scores two standard deviations below his true score, which will occur 2.5% of the time AND for Bob to score above his true score. Again, the combined probability is .0125.   SO …. only 2.5% of the time (.0125 + .0125)  would Bob win. Since judges’ scores are independent, the probability of any one scoring it for him, causing a split decision is .025 + .025 + .025 = 7.5%

(If all three judges scored it for Bob, that would be a very, very low probability  of .o25 * .025 * .025 because, again, the judges scores are assumed independent of one another. In only 0.063% of the cases would this occur. We should probably subtract that and the probability of two of them scoring it for Bob to be exact, but I have to finish grading papers tonight so we’ll just acknowledge that it is not exactly 7.5% and move on.)

Let’s go back to the fight that actually happened. I didn’t see it so I am going to take some people’s word that it was a close fight. They might be lying but let’s assume not.

In this case, Bob, who has a true score of 27, is not fighting Fred, but rather, Ignatz, who has a true score of 27.3 (with three judges, he’d get a 27, 27, 28 score). There is great overlap in Bob and Ignatz’s scores. To outscore Ignatz’s average score, Bob would need a score of  27.4  – well, a z-score of .4 occurs about 35% of the time. Half of the time Ignatz is going to score 27.3 or lower so the probability of him both having an average or below score AND Bob having a 27.4 or high score is .5 *.35 or .175.  So 17.5% of the time, a judge would give Bob a higher score. Since there are three judges, the probability of ONE of them giving him a higher score would be .175 + .175 + .175  = 52.5%

There is also the small probability that it could go unanimous the other way, but that’s not really pertinent to our argument.

The point is simply this … if two fighters’ true scores are close, it is much less likely that you will see a unanimous decision than if their true scores are really far apart. The closer they are, the more that statement holds. So, no, it is not a stupid comment to say that you believe someone warranted a split decision rather than a unanimous decision. It may simply mean that you think the fighters’ were so close that you were surprised there was not any variance in favor of the only slightly better fighter.

Really, I think most people would find that a reasonable statement.

Extra credit points:

Give one reason why the Central Limit Theorem does not apply in the above scenario.

Answer this question:

Does the fact that the distribution of errors is necessarily non-symmetric in Fred’s case (cannot score above 30) negate the application of the Central Limit Theorem?

Last year, I went from teaching in classrooms in a pretty building with a library on the ground floor to teaching on-line. I also went from the semester system to teaching the same content in four weeks. This has led curious friends of mine, used to teaching in the traditional format, to ask ,

How does that work? Does it work?

Initially, I was skeptical myself. I thought if students were really serious they would make the sacrifices to take the class in a “regular” setting. Interestingly, I had to take a class on a new system and had the option to sign up for a session held on a local campus or on-line. After looking at my schedule, I chose the on-line option. No one has ever accused me of being a slacker – in fact, it may be the only negative thing I’ve not been called. Still, I thought it was possible I might have conflicts those days, whether meeting with clients, employees or investors. The option of taking the course in smaller bits – an hour here or there – was a lot more convenient for me than several hours at a time. To be truthful, too, I didn’t really want to spend hours hanging out with people with whom I didn’t expect I would have that much in common. It wasn’t like a class on statistics that I was really interested in.

So … if we are willing to accept that students who sign up for on-line, limited-term classes might be just as motivated and hard-working as anyone else, do they work? I think the better question is how they work or for what type of students they work.

National University, where I teach, offers courses in a one course one month format. Students are not supposed to take more than one course at a time and , although exceptions can be made, I advise against it. The courses work for those students (and faculty) who can block off a month, and then, during that month DEVOTE A LOT OF TIME TO THE COURSE.  Personally, I give two-hour lectures twice a week. If a student cannot attend – and some are in time zones where it is 2 a.m. when I’m teaching – the lectures are recorded and they can listen to them at their leisure. Time so far – 16 hours in the month. Normally, a graduate course I teach will require 50-100 pages of reading per week. Depending on your reading speed that could take you from one to four hours.

I just asked our Project Manager, Jessica, how long she thought it took the average person to read 75 pages of technical material she said,

“Whatever it is, I’m sure it’s a lot more than you are thinking!”

Talking it over, we agreed it probably took around 3-5 minutes per page, because even if some pages you get right away, others you have to read two or three times to figure out wait, that -1 next to a capital letter in bold means to take the inverse of a matrix while the single quote next to it means to transpose the matrix. These are things that are not second nature to you when you are just learning a field. Discussing this made me think I want to reduce the required reading in my multivariate statistics course. Let’s say on the low end, then it takes five hours to read the assigned material and review it for a test or just for your own clarity. Now we are up to 20 hours a month + 16 =  36 hours.

I give homework assignments because I am a big believer in distributed practice. We have all had classes we crammed for in college that we can’t remember a damn thing about. Okay, well, I have, any way. So, I give homework assignments every week, usually several problems like, “What is the cumulative incidence rate given the data in Table 2?” as well s assignments that require you to write a program, run it and interpret the results. I estimate these take students 4-5 hours per week.  Let’s go on the low end and say 16 + 20 + 16 =  52 hours

There is also a final paper, a final exam and two quizzes. The final and quizzes are given 5 hours total and it is timed so students can’t go over. I think, based on simply page length, programs required and how often they call me, the average student spends 14 hours on the paper. Total hours for the course  52 hours plus another 19 =  71 hours in four weeks.

IF students put in that amount of time, they definitely pass the course with a respectable grade and probably learn enough that they will retain a useful amount of it. The kiss of death in a course like this is to put off the work. It is impossible to finish in a week.

My personal bias is that I require students actually DO things with the information they learn. It is not just memorizing formula and a lot of calculations because  I really do think students will forget that after a few weeks. However, if they have to post a question that is a serious personal interest and then conduct a study to answer that question, the whole time posting progress and discussions on line with their classmates , then I think they WILL retain more of the material.

So, yes, students can learn online and they can learn in a compressed term. It IS harder, though, I think, both for the students and the instructor, and takes a lot of commitment on the part of both, which is why I don’t teach very many courses a year.

 

 

 

 

 

 

 

 

 

Am I missing something here? All of the macros I have seen for the parallel analysis criterion for factor analysis look pretty complicated, but, unless I am missing something, it is a simple deal.

The presumption is this:

There isn’t a number like a t-value or F-value to use to test if an eigenvalue is significant. However, it makes sense that the eigenvalue should be larger than if you factor analyzed a set of random data.

Random data is, well, random, so it’s possible you might have gotten a really large or really small eigenvalue the one time you analyzed the random data. So, what you want to do is analyze a set of random data with the same number of variables and the same number of observations a whole bunch of times.

Horn, back in 1965, was proposing that the eigenvalue should be higher than the average of when you analyzed a set of random data. Now, people are suggesting it should be higher than 95% of the time you analyzed random data (which kind of makes sense to me).

Either way, it seems simple. Here is what I did and it seems right so I am not clear why other macros I see are much more complicated. Please chime in if you see what I’m missing.

  1. Randomly generate a set of random data with N variables and Y observations.
  2. Keep the eigenvalues.
  3. Repeat 500 times.
  4. Combine the 500 datasets  (each will only have 1 record with N variables)
  5. Find the 95th percentile

%macro para(numvars,numreps) ;
%DO k = 1 %TO 500 ;
data A;
array nums {&numvars} a1- a&numvars ;
do i = 1 to &numreps;
do j = 1 to &numvars ;
nums{j} = rand(“Normal”) ;
if j < 2 then nums{j} = round(100*nums{j}) ;
else nums{j} = round(nums{j}) ;
end ;
drop i j ;
output;
end;

proc factor data= a outstat = a&k  noprint;
var a1 – a&numvars ;
data a&k ;
set  a&k  ;
if trim(_type_) = “EIGENVAL” ;

%END ;
%mend ;

%para(30,1000) ;

data all ;
set a1-a500 ;

proc univariate data= all noprint ;
var a1 – a30 ;
output out = eigvals pctlpts =  95 pctlpre = pa1 – pa30;

*** You don’t need the transpose but I just find it easier to read ;
proc transpose data= eigvals out=eigsig ;
Title “95th Percentile of Eigenvalues ” ;
proc print data = eigsig ;
run ;

It runs fine and I have puzzled and puzzled over why a more complicated program would be necessary. I ran it 500 times with 1,000 observations and 30 variables and it took less than a minute on a remote desktop with 4GB RAM. Yes, I do see the possibility that if you had a much larger data set that you would want to optimize the speed in some way. Other than that, though, I can’t see why it needs to be any more complicated than this.

If you wanted to change the percentile, say, to 50, you would just change the 95 above. If you wanted to change the method from say, Principal Components Analysis (the default, with commonality of 1) to saying else, you could just do that in the PROC FACTOR step above.

The above assumes a normal distribution of your variables, but if that was not the case, you could change that in the RAND function above.

As I said, I am puzzled. Suggestions to my puzzlement welcome.

 

I’m about to tear my hair out. I’ve been reading this statistics textbook which shall remain nameless, ostensibly a book for graduate students in computer science, engineering and similar subjects. The presumption is that at the end of the course students will be able to compute and interpret a factor analysis, MANOVA and other multivariate statistics. The text spends 90% of the space discussing the mathematics in computing the results, 10% discussing the code to get these statistics and 0% discussing decisions one makes in selection of communality estimates, rotation, post hoc tests or anything else.

In short, the book is entirely devoted to explaining the part that the computer does for you that students will never need to do and 10% or less on the decisions and interpretation that they will spend their careers doing. One might argue that it is good to understand what is going on “under the hood” and I’m certainly not going to argue against that but there is a limit on how much can be taught in any one course and I would argue very strenuously that there needs to be a much greater emphasis on the part the computer cannot do for you.

There was an interesting article in Wired a few years ago on The End of Theory, saying that we now have immediate access to so much data that we can use “brute force”. We can throw the data into computers and “find patterns where science cannot.”

Um. Maybe not.

Let’s take an example I was working on today, from the California Health Interview Survey. There are 47,000+ subjects but it wouldn’t matter if there were 47 million. There are also over 500 variables measured on these 47,000 people. That’s over 23,000,000 pieces of data. Not huge by some standards, but not exactly chicken feed, either.

Let’s say that I want to do a factor analysis, which I do. By some theory – or whatever that word is we’re using instead of theory – I could just dump all of  the variables into an analysis and magically factors would come out, if I did it often enough. So, I did that and came up with results that meant absolutely nothing because the whole premise was so stupid.

Here are a couple of problems

1. The CHIS includes a lot of different types of variables, sample weights, coding for race and ethnic categories, dozens of items on treatment of asthma, diabetes or heart disease, dozens more items on access to health care. Theoretically (or computationally, I guess the new word is), one could run an analysis and we would get factors of asthma treatment, health care access, etc. Well, except I don’t really see that the variables that are not on a numeric scale are going to be anything but noise. What the heck does racesex coded as 1=  “Latin male”, 10 = “African American male” etc. ever load on as a factor?

2. LOTS of the variables are coded with -1 as inapplicable. For example, “Have you had an asthma attack in the last 12 months?”
-1 = Inapplicable
1 = Yes
2 = N0

While this may not be theory, these two problems do suggest that some knowledge of your data is essential.

Once you get results, how do you interpret them? Using the default minimum eigenvalue of 1 criterion (which if all you learned in school was how to factor analyze a matrix using a pencil and a pad of paper, I guess you’d use the defaults), you get 89 factors. Here is my scree plot.

Scree plot showing 89 eigenvalues  I also got another 400+ pages of output that I won’t inflict on you.

What exactly is one supposed to do with 500 variables that load on 89 factors? Should we then factor analyze these factors to further reduce the dimensions? It would certainly be possible. All you’d need to do is output the factor scores on the 89 factors, and then do a factor analysis on that.

I would argue, though,  and I would be right, that before you do any of that you need to actually put some thought into the selection of your variables and how they are coded.

Also, you should perhaps understand some of the implications of having variables measured on vastly different scales. As this handy page on item analysis points out,

“Bernstein (1988) states that the following simple examination should be mandatory: “When you have identified the salient items (variables) defining factors, compute the means and standard deviations of the items on each factor.  If you find large differences in means, e.g., if you find one factor includes mostly items with high response levels, another with intermediate response levels, and a third with low response levels, there is strong reason to attribute the factors to statistical rather than to substantive bases” (p. 398).”

And hold that thought, because our analysis of the 517 or so variables provided a great example …. or would it be using some kind of theory to point that out? Stay tuned.

 

I’ve written here before about visual literacy and Cook’s D is just my latest example.

Most people intuitively understand that any sample can have outliers, say, an 80-year-old man who is the father of a six-year-old child, the new college graduate who is making $150,000 a year. We understand that those people may throw off our predictions and perhaps we want to exclude those outliers from our models.

What if you have multiple variables, though? It’s possible that each individual value may not be very extreme but the combination is. Take this data set below that I totally made up, with mom’s age, dad’s age and child’s age.

Mom Dad Child

30 32 6
20 27 5
31 33 8
29 28 6
40 42 20
44 44 21
37 39 14
25 29 7
30 32 6
20 27 5
31 33 8
29 28 6
39 42 19
43 44 20
37 39 13
25 28 6
40 29 15

Look at our last record. The mother has an age of 40, the father an age of 29 and the child an age of 15. None of these individually are extreme scores. These aren’t even the minimum or maximum for any of the variables. There are mothers older (and younger) than 40, fathers younger (and older) than 29; 15 isn’t that extreme an age in our sample of children.  The COMBINATION, however, of a 40-year-old mother, 29-year-old father and 15-year-old child is an extreme case.

Enter Cook’s distance, a.k.a. Cook’s D, which measures the effect of deleting an observation. The larger the distance, the more influential that point is on the results. Take a look at my graph below.

Cook's D showing one very high pointIt is pretty clear that the last observation is very influential. Now, you might have guessed that if you had thought to look at the data. However, if you had 11 variables and 100 observations it wouldn’t be so easy to see by looking at the data and you might be really happy you had Cook around to help you out.

Let’s look at the data re-analyzed without that last observation. Here is what our plot of Cook’s D looks like now.

Plot with 2 points moderately high but not vastly differentThis gives you a very different picture. While a couple of points are higher than the others, it is certainly not the extreme case we saw before.

In fact, dropping out that one point changed our explained variance from 89%  to 93%.

So … knowing how to use Cook’s D for regression diagnostics is our latest lesson in visual literacy.

You’re welcome.

Ever wonder why with goodness of fit tests non- significance is what you want?

Why is that sometimes when you have a significant p-value it means your hypothesis is correct, there is a relationship between the price of honey and the number of bees, and in other cases, significance means your model is rejected? Well, if you are reading this blog, it’s possible you already know all of this, but I can guarantee you that students who start off in statistics learning that a significant p-value is a good thing often are confused to learn that with model fit statistics, non-significance is (usually) what you want.

You are hoping that you find non-significance when you are looking at model fit statistics  because the hypothesis you are testing is that the full model – one that has as many parameters as there are observations  – is different than this model you have postulated.

To understand model fit statistics, you should think about three models.

The null model, and contains only one parameter, the mean. Think of it this way, if all of your explanatory variables are useless then your best prediction for the dependent variable is the mean. If you knew nothing about the next woman likely to walk into the room, your best prediction of her height would be 5’4″ , if you live in the U.S., because that is the average height.

The full model  has one parameter per observation. With this model, you can predict the data perfectly. Wouldn’t that be great? No, it would be useless. Using the full model is a bad idea because it is non- replicable

Here is an example data set where I predict IQ using gender, number of M & M’s in your pocket and hair color.

EXAMPLE

Male 10 redhead   100

Female 0 blonde. 70

Male 10 blonde 60

Female 30 brunette 100

50 + MMx1  + female x 20 + redhead x 40

Is that replicable at all? If you selected another random sample of 4 people from the population do you think you could predict their scores perfectly using this equation?

No.

Also, I do not know why that woman has so many M & M’s in her pocket.

The M and M store

In between these two useless models is your model. The hypothesis you are testing is that your model, whatever it is, is non-significantly different from the full model. If you throw out one of your parameters, your new model won’t be as good as the full model – that one extra parameter may explain one case – but the question is, does the model without that parameter differ significantly from the full model. If it doesn’t then we can conclude that the parameters we have excluded from the model were unimportant.

We have a more parsimonious model and we are happy.

But WHY do more parsimonious models make us happy? Well, because that is kind of the whole point of model building. If you need a parameter for each person, why not just examine each person individually?  The whole point of a model is dimension reduction, that is, reducing the number of dimensions you need to measure while still adequately explaining the data.

If, instead of needing 2,000 parameters to explain the data gathered from 2,000 people you can do just as well with 47 parameters, then you would have made some strides forward in understanding how the world works.

Coincidentally, I discussed dimension reduction on this blog almost exactly a year ago, in a post with the title “What’s all that factor analysis crap mean, anyway?”

(Prediction: At least one person who follows this link will be surprised at the title of the post.)

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