This is a blog about psychological assessment. However, just for fun, I am posting a short children’s story I wrote a while back.
The text version is here.
This is a blog about psychological assessment. However, just for fun, I am posting a short children’s story I wrote a while back.
The text version is here.
On 10/24/2014, I attended the Richard Woodcock Institute Event hosted at University of Texas at Austin. The three speakers had strikingly different presentation styles but were equally excellent.
Dick Woodcock gave the opening remarks. I loved hearing about the twists and turns of his career and how he made the most of unplanned opportunities. It was rather remarkable how diverse his contributions are (including an electronic Braille typewriter). Then he stressed the importance of communicating test results in ways that non-specialists can understand. He speculated on what psychological testing will look like in the future, focusing on integrative software that will guide test selection and interpretation in more sophisticated ways than has hitherto been possible. Given that he has been creating the future of assessment for decades now, I am betting that he is likely to be right. Later he graciously answered my questions about the WJ77 and how he came up with what I consider to be among the most ingenious test paradigms we have.
After a short break, Kevin McGrew gave us a wild ride of a talk about advances in CHC Theory. Actually it was more like a romp than a ride. I tried to keep track of all the interesting ideas for future research he presented but there were so many I quickly lost count. The visuals were stunning and his energy was infectious. He offered a quick overview of new research from diverse fields about the overlooked importance of auditory processing (beyond the usual focus on phonemic awareness). Later he talked about his evolving conceptualization of the memory factors in CHC theory and role of complexity in psychometric tests. My favorite part of the talk was a masterful presentation of information processing theory, judiciously supplemented with very clever animations.
After lunch, Cathy Fiorello gave one of the most thoughtful presentations I have ever heard. Instead of contrasting nomothetic and idiographic approaches to psychological assessment, Cathy stressed their integration. Most of the time, nomothetic interpretations are good first approximations and often are sufficient. However, there are certain test behaviors and other indicators that a more nuanced interpretation of the underlying processes of performance is warranted. Cathy asserted (and I agree) that well trained and highly experienced practitioners can get very good at spotting unusual patterns of test performance that completely alter our interpretations of test scores. She called on her fellow scholars to develop and refine methods of assessing these patterns so that practitioners do not require many years of experience to develop their expertise. She was not merely balanced in her remarks—lip service to a sort of bland pluralism is an easy and rather lazy trick to seem wise. Instead, she offered fresh insight and nuance in her balanced and integrative approach to cognitive and neuropsychological assessment. That is, she did the hard work of offering clear guidelines of how to integrate nomothetic and idiographic methods, all the while frankly acknowledging the limits of what can be known.
I have many reservations about this model of cognitive predictors of academic abilities that I presented at APA today (along with co-presenters Lee Affrunti, Renée Tobin, and Kimberley Collins) but I think that it illustrates an important point: prediction and explanation of cognitive and academic abilities is so complex that it is impossible to do in one’s head. Eyeballing scores and making pronouncements is not likely to be accurate and will result in misinterpretations. We need good software that can manage the complex calculations for us. We can still think creatively in the diagnostic process but the creativity must be grounded in realistic probabilities.
The images from the poster are from a single exploratory model based on a clinical sample of 865 college students. The model was so big and complex I had to split the path diagram into two images:
Because cognitive abilities are positively correlated, there is an assumption that cognitive abilities should be evenly developed. When psychologists examine cognitive profiles, they often describe any features that deviate from the expected flat profile.
It is true, mathematically, that the expected profile IS flat. However, this does not mean that flat profiles are common. There is a very large set of possible profiles and only a tiny fraction are perfectly flat. Profiles that are nearly flat are not particularly common, either. Variability is the norm.
Sometimes it helps to get a sense of just how uneven cognitive profiles typically are. That is, it is good to fine-tune our intuitions about the typical profile with many exemplars. Otherwise it is easy to convince ourselves that the reason that we see so many interesting profiles is that we only assess people with interesting problems.
If we use the correlation matrix from the WAIS-IV to randomly simulate multivariate normal profiles, we can see that even in the general population, flat, “plain-vanilla” profiles are relatively rare. There are features that draw the eye in most profiles.
If cognitive abilities were uncorrelated, profiles would be much more uneven than they are. But even with moderately strong positive correlations, there is still room for quite a bit of within-person variability.
Let’s see what happens when we look at profiles that have the exact same Full Scale IQ (80, in this case). The conditional distributions of the remaining scores are seen in the “violin” plots. There is still considerable diversity of profile shape even though the Full Scale IQ is held constant.
In my introductory statistics class, I feel uneasy when I have explain what variance explained means. The term has two things I don’t like. First, I don’t like variance very much. I feel much more comfortable with standard deviations. I understand that at a deep level variance is a more fundamental concept than the standard deviation. However, variance is a poor descriptive statistic because there is no direct visual analog for variance in a probability distribution plot. In contrast, the standard deviation illustrates very clearly how much scores typically deviate from the mean. So, variance explained is hard to grasp in part because variance is hard to visualize.
The second thing I don’t like about variance explained is the whole “explained” business. As I mentioned in my last post, variance explained does not actually mean that we have explained anything, at least in a causal sense. That is, it does not imply that we know what is going on. It simply means that we can use one or more variables to predict things more accurately than before.
In many models, if X is correlated with Y, X can be said to “explain” variance in Y even though X does not really cause Y. However, in some situations the term variance explained is accurate in every sense:
In the model above, the arrow means that X really is a partial cause of Y. Why does Y vary? Because of variability in X, at least in part. In this example, 80% of Y’s variance is due to X, with the remaining variance due to something else (somewhat misleadingly termed error). It is not an “error” in that something is wrong or that someone is making a mistake. It is merely that which causes our predictions of Y to be off. Prediction error is probably not a single variable. It it likely to be the sum total of many influences.
Because X and error are uncorrelated z-scores in this example, the path coefficients are equal to the correlations with Y. Squaring the correlation coefficients yields the variance explained. The coefficients for X and error are actually the square roots of .8 and .2, respectively. Squaring the coefficients tells us that X explains 80% of the variance in Y and error explains the rest.
Okay, if X predicts Y, then the variance explained is equal to the correlation coefficient squared. Unfortunately, this is merely a formula. It does not help us understand what it means. Perhaps this visualization will help:
If you need to guess every value of Y but you know nothing about Y except that it has a mean of zero, then you should guess zero every time. You’ll be wrong most of the time, but pursuing other strategies will result in even larger errors. The variance of your prediction errors will be equal to the variance of Y. In the picture above, this corresponds to a regression line that passes through the mean of Y and has a slope of zero. No matter what X is, you guess that Y is zero. The squared vertical distance from Y to the line is represented by the translucent squares. The average area of the squares is the variance of Y.
If you happen to know the value of X each time you need to guess what Y will be, then you can use a regression equation to make a better guess. Your prediction of Y is called Y-hat (Ŷ):
When X and Y have the same variance, the slope of the regression line is equal to the correlation coefficient, 0.89. The distance from Ŷ (the predicted value of Y) to the actual value of Y is the prediction error. In the picture above, the variance of the prediction errors (0.2) is the average size of the squares when the slope is equal to the correlation coefficient.
Thus, when X is not used to predict Y, our prediction errors have a variance of 1. When we do use X to predict Y, the average size of the prediction errors shrinks from 1 to 0.2, an 80% reduction. This is what is meant when we say that “X explains 80% of the variance in Y.” It is the proportion by which the variance of the prediction errors shrinks.
Suppose that we flip 50 coins and record how many heads there are. We do this over and over. The values we record constitute the variable Y. The number of heads we get each time we flip a coin happens to have a binomial distribution. The mean of a binomial distribution is determined by the probability p of an event occurring on a single trial (i.e., getting a head on a single toss) and the number of events k (i.e., the number of coins thrown). As k increases, the binomial distribution begins to resemble the normal distribution. The probability p of getting a head on any one coin toss is 0.5 and the number of coins k is 50. The mean number of heads over the long run is:
The variance of the binomial distribution:
Before we toss the coins, we should guess that we will toss an average number of heads, 25. We will be wrong much of the time but our prediction errors will be as small as they can be, over the long run. The variance of our prediction errors is equal to the variance of Y, 12.5.
Now suppose that after tossing 80% of our coins (i.e., 40 coins), we count the number of heads. This value is recorded as variable X. The remaining 20% of the coins (10 coins) are then tossed and the total number of heads is counted from all 50 coins. We can use a regression equation to predict Y from X. The intercept will be the mean number of heads from the remaining 10 coins:
In the diagram below, each peg represents a coin toss. If the outcome is heads, the dot moves right. If the outcome is tails, the dot moves left. The purple line represents the probability distribution of Y before any coin has been tossed.
When the dot gets to the red line (after 40 tosses or 80% of the total), we can make a new guess as to what Y is going to be. This conditional distribution is represented by a blue line. The variance of the conditional distribution has a mean equal to Ŷ, with a variance of 2.5 (the variance of the 10 remaining coins).
The variability in Y is caused by the outcomes of 50 coin tosses. If 80% of those coins are the variable X, then X explains 80% of the variance in Y. The remaining 10 coins represent the variability of Y that is not determined by X (i.e., the error term). They determine 20% of the variance in Y.
If X represented only the first 20 of 50 coins, then X would explain 40% of the variance in Y.
I like most of the technical terms we use in statistics. However, there are a few of them that I wish were easier to teach and remember. Many others have opined on such matters. This is my list of complaints:
I am sure that there are many more!
I’m afraid that the question posed by the title does not have a single answer. It depends on how we define and measure academic performance.
Let’s sidestep some difficult questions about what exactly an “academic deficit” is and for the sake of convenience pretend that it is a score at least 1 standard deviation below the mean on a well normed test administered by a competent psychologist with good clinical skills.
Suppose that we start with the 9 core WJ III achievement tests (the answers will not be all that different with the new WJ IV):
|Applications||Passage Comprehension||Writing Samples||Applied Problems|
|Fluency||Reading Fluency||Writing Fluency||Math Fluency|
What is the percentage of the population that does not have any score below 85? If we can assume that the scores are multivariate normal, the answer can be found using data simulation or via the cumulative density function of the multivariate normal distribution. I gave examples of both methods in the previous post. If we use the correlation matrix for the 6 to 9 age group of the WJ III NU, about 47% of the population has no academic scores below 85.
Using the same methods we can estimate what percent of the population has no academic scores below various thresholds. Subtracting these numbers from 100%, we can see that fairly large proportions have at least one low score.
|Threshold||% with no scores below the threshold||% with at least one score below the threshold|
The numbers in the table above include people with very low cognitive ability. It would be more informative if we could control for a person’s measured cognitive abilities.
Suppose that an individual has index scores of exactly 100 for all 14 subtests that are used to calculate the WJ III GIA Extended. We can calculate the means and the covariance matrix of the achievement tests for all people with this particular cognitive profile. We will make use of the conditional multivariate normal distribution. As explained here (or here), we partition the academic tests and the cognitive predictor tests like so:
If the cognitive variables have the vector of particular values , then the conditional mean vector of the academic variables is:
The conditional covariance matrix:
If we can assume multivariate normality, we can use these equations, to estimate the proportion of people with no scores below any threshold on any set of scores conditioned on any set of predictor scores. In this example, about 51% of people with scores of exactly 100 on all 14 cognitive predictors have no scores below 85 on the 9 academic tests. About 96% of people with this cognitive profile have no scores below 70.
Because there is an extremely large number of possible cognitive profiles, I cannot show what would happen with all of them. Instead, I will show what happens with all of the perfectly flat profiles from all 14 cognitive scores equal to 70 to all 14 cognitive scores equal to 130.
Here is what happens with the same procedure when the threshold is 70 for the academic scores:
Here is the R code I used to perform the calculations. You can adapt it to other situations fairly easily (different tests, thresholds, and profiles).
library(mvtnorm) WJ <- matrix(c( 1,0.49,0.31,0.46,0.57,0.28,0.37,0.77,0.36,0.15,0.24,0.49,0.25,0.39,0.61,0.6,0.53,0.53,0.5,0.41,0.43,0.57,0.28, #Verbal Comprehension 0.49,1,0.27,0.32,0.47,0.26,0.32,0.42,0.25,0.21,0.2,0.41,0.21,0.28,0.38,0.43,0.31,0.36,0.33,0.25,0.29,0.4,0.18, #Visual-Auditory Learning 0.31,0.27,1,0.25,0.33,0.18,0.21,0.28,0.13,0.16,0.1,0.33,0.13,0.17,0.25,0.22,0.18,0.21,0.19,0.13,0.25,0.31,0.11, #Spatial Relations 0.46,0.32,0.25,1,0.36,0.17,0.26,0.44,0.19,0.13,0.26,0.31,0.18,0.36,0.4,0.36,0.32,0.29,0.31,0.27,0.22,0.33,0.2, #Sound Blending 0.57,0.47,0.33,0.36,1,0.29,0.37,0.49,0.28,0.16,0.23,0.57,0.24,0.35,0.4,0.44,0.36,0.38,0.4,0.34,0.39,0.53,0.27, #Concept Formation 0.28,0.26,0.18,0.17,0.29,1,0.35,0.25,0.36,0.17,0.27,0.29,0.53,0.22,0.37,0.32,0.52,0.42,0.32,0.49,0.42,0.37,0.61, #Visual Matching 0.37,0.32,0.21,0.26,0.37,0.35,1,0.3,0.24,0.13,0.22,0.33,0.21,0.35,0.39,0.34,0.38,0.38,0.36,0.33,0.38,0.43,0.36, #Numbers Reversed 0.77,0.42,0.28,0.44,0.49,0.25,0.3,1,0.37,0.15,0.23,0.43,0.23,0.37,0.56,0.55,0.51,0.47,0.47,0.39,0.36,0.51,0.26, #General Information 0.36,0.25,0.13,0.19,0.28,0.36,0.24,0.37,1,0.1,0.22,0.21,0.38,0.26,0.26,0.33,0.4,0.28,0.27,0.39,0.21,0.25,0.32, #Retrieval Fluency 0.15,0.21,0.16,0.13,0.16,0.17,0.13,0.15,0.1,1,0.06,0.16,0.17,0.09,0.11,0.09,0.13,0.1,0.12,0.13,0.07,0.12,0.07, #Picture Recognition 0.24,0.2,0.1,0.26,0.23,0.27,0.22,0.23,0.22,0.06,1,0.22,0.35,0.2,0.16,0.22,0.25,0.21,0.19,0.26,0.17,0.19,0.21, #Auditory Attention 0.49,0.41,0.33,0.31,0.57,0.29,0.33,0.43,0.21,0.16,0.22,1,0.2,0.3,0.33,0.38,0.29,0.31,0.3,0.25,0.42,0.47,0.25, #Analysis-Synthesis 0.25,0.21,0.13,0.18,0.24,0.53,0.21,0.23,0.38,0.17,0.35,0.2,1,0.15,0.19,0.22,0.37,0.21,0.2,0.4,0.23,0.19,0.37, #Decision Speed 0.39,0.28,0.17,0.36,0.35,0.22,0.35,0.37,0.26,0.09,0.2,0.3,0.15,1,0.39,0.36,0.32,0.3,0.3,0.3,0.25,0.33,0.23, #Memory for Words 0.61,0.38,0.25,0.4,0.4,0.37,0.39,0.56,0.26,0.11,0.16,0.33,0.19,0.39,1,0.58,0.59,0.64,0.5,0.48,0.46,0.52,0.42, #Letter-Word Identification 0.6,0.43,0.22,0.36,0.44,0.32,0.34,0.55,0.33,0.09,0.22,0.38,0.22,0.36,0.58,1,0.52,0.52,0.47,0.42,0.43,0.49,0.36, #Passage Comprehension 0.53,0.31,0.18,0.32,0.36,0.52,0.38,0.51,0.4,0.13,0.25,0.29,0.37,0.32,0.59,0.52,1,0.58,0.48,0.65,0.42,0.43,0.59, #Reading Fluency 0.53,0.36,0.21,0.29,0.38,0.42,0.38,0.47,0.28,0.1,0.21,0.31,0.21,0.3,0.64,0.52,0.58,1,0.5,0.49,0.46,0.47,0.49, #Spelling 0.5,0.33,0.19,0.31,0.4,0.32,0.36,0.47,0.27,0.12,0.19,0.3,0.2,0.3,0.5,0.47,0.48,0.5,1,0.44,0.41,0.46,0.36, #Writing Samples 0.41,0.25,0.13,0.27,0.34,0.49,0.33,0.39,0.39,0.13,0.26,0.25,0.4,0.3,0.48,0.42,0.65,0.49,0.44,1,0.38,0.37,0.55, #Writing Fluency 0.43,0.29,0.25,0.22,0.39,0.42,0.38,0.36,0.21,0.07,0.17,0.42,0.23,0.25,0.46,0.43,0.42,0.46,0.41,0.38,1,0.57,0.51, #Calculation 0.57,0.4,0.31,0.33,0.53,0.37,0.43,0.51,0.25,0.12,0.19,0.47,0.19,0.33,0.52,0.49,0.43,0.47,0.46,0.37,0.57,1,0.46, #Applied Problems 0.28,0.18,0.11,0.2,0.27,0.61,0.36,0.26,0.32,0.07,0.21,0.25,0.37,0.23,0.42,0.36,0.59,0.49,0.36,0.55,0.51,0.46,1), nrow= 23, byrow=TRUE) #Math Fluency WJNames <- c("Verbal Comprehension", "Visual-Auditory Learning", "Spatial Relations", "Sound Blending", "Concept Formation", "Visual Matching", "Numbers Reversed", "General Information", "Retrieval Fluency", "Picture Recognition", "Auditory Attention", "Analysis-Synthesis", "Decision Speed", "Memory for Words", "Letter-Word Identification", "Passage Comprehension", "Reading Fluency", "Spelling", "Writing Samples", "Writing Fluency", "Calculation", "Applied Problems", "Math Fluency") rownames(WJ) <- colnames(WJ) <- WJNames #Number of tests k<-length(WJNames) #Means and standard deviations of tests mu<-rep(100,k) sd<-rep(15,k) #Covariance matrix sigma<-diag(sd)%*%WJ%*%diag(sd) colnames(sigma)<-rownames(sigma)<-WJNames #Vector identifying predictors (WJ Cog) p<-seq(1,14) #Threshold for low scores Threshold<-85 #Proportion of population who have no scores below the threshold pmvnorm(lower=rep(Threshold,length(WJNames[-p])),upper=rep(Inf,length(WJNames[-p])),sigma=sigma[-p,-p],mean=mu[-p]) #Predictor test scores for an individual x<-rep(100,length(p)) names(x)<-WJNames[p] #Condition means and covariance matrix condMu<-c(mu[-p] + sigma[-p,p] %*% solve(sigma[p,p]) %*% (x-mu[p])) condSigma<-sigma[-p,-p] - sigma[-p,p] %*% solve(sigma[p,p]) %*% sigma[p,-p] #Proportion of people with the same predictor scores as this individual who have no scores below the threshold pmvnorm(lower=rep(Threshold,length(WJNames[-p])),upper=rep(Inf,length(WJNames[-p])),sigma=condSigma,mean=condMu)