Interpretation of a linear regression

I have some problems in understanding the results of two linear regressions on the same data.

I have 2 variables and I want to know if there is a linear correlation between the two. Let's say velocity of a car and perceived velocity of the conductor. Let's say I have 4 velocities and 2 evaluations of the perceived velocities performed in 2 trials by 15 participants, on a 7 points Likert scale where 1 is very slow and 7 = very fast.

Now if I take the AVERAGE values of the evaluations given by the 15 participants for the 4 velocities I get a clear linear relationship with $R^2$ = 0.7.

However if I perform the analysis on the raw data (i.e. 15 participants * 2 trials = 30 values for each velocity) I get a $R^2$ = 0.2. Plotting the graph of the velocity data vs perceived velocity I get a mess of points very widely distributed.

Why do I get this behavior? Which of the two analyses should I use? The first or the second?

How can I interpret my values?

Please let me know

• You are getting the conclusions that different individuals often describe their speed differently, but that overall they often notice when their speed changes. Commented Feb 22, 2012 at 2:07
• Hi, could you clarify a few things. Which is your response variable (on the vertical axis) - perceived velocity or actual velocity? What is the scale of actual velocity (is it the same seven point scale as perceived, or is it a four point scale - one interpretation of what you mean by "I have 4 velocities"). What do you mean by "I have 4 velocities" - is this 4 possible velocities? In the first regression, do you have 60 points or 15? And in the second are there 120 or 30? Commented Feb 22, 2012 at 2:27
• Hi, can you show us the scatterplot of your two velocity variables? Commented Feb 22, 2012 at 2:52
• Hello everybody thanks for your reply. On the vertical axis there is the perceived velocity. On the horizontal axis I have the velocities, on a 4 point scale. Just velocity 1, 2 , 3 and 4. It is a general case, I don't care about the differences between the velocities (Please note: this velocity example is just a generalization of the real problem that I am trying to solve...). In the first regression I just have 4 points (the average perceived velocity for the 4 velocities). In the second regression I have 120 points (15 participants * 2 trials * 4 estimate of the velocities).
– L_T
Commented Feb 22, 2012 at 14:37
• It would help if you edited your question to show the first couple of rows of your data, and put up a scatterplot. Commented Feb 22, 2012 at 16:46

I agree with @PeterEllis & @Michelle that it is hard to follow your question exactly, and more information (such as they request) would probably be very helpful in getting an answer that is useful for you. I can make a guess at what is going on here based on what I can glean, however: I suspect what you are seeing is just the resolving power of the mean. The strength of correlations are influenced by the amount of noise present. Take a variable that has a correlation, $\rho$, with another variable and add additional variability to it, and you will decrease the estimated value, $r$. You don't have to change the underlying trend; that is, the random noise you add can be centered on 0. Looking at the formula for the product-moment correlation makes this more apparent:
$$r=\frac{{cov}(x,y)}{s_xs_y}$$
As the SD for one of the variables gets larger, the product will get larger, leading to a larger denominator, and a smaller $r$. On the other hand, using means instead of raw data will shrink the variability by a factor of $1/\sqrt{n}$. As for which one of these is 'right', I don't know how to answer that. They are answers to different questions.

With more information, I may update this answer, but this may be what's going on.

Now that I understand your data I would certainly use the original set in your analysis rather than the four averages. The "noise" that you describe is actually a crucial part of your data. The higher $R^2$ in the model with fewer data points is because you have taken out the variation around the means.

First, which of your two plots and regressions to use?

By using a technique like linear regression on your original data (the 120 points) you can simultaneously look at the average perceived velocity for each of your four levels of actual velocity, and conduct inference based on a full understanding of the level of variation in the data.

If you conduct it on just the four means, you have split your analysis into two phases. The first phase finds the average for each level of actual velocity; then the second (where you conduct regression) tries to do inference to the general population. You can't really do what you want to in this second phase because you have lost all the information about the randomness in your first phase.

You might want to come back to your picture of the four averages as part of a result-reporting summary of your data, but all inference should be based on a model fitted to the original set.

Second, what analysis to do

Because you have repeated observations with characteristics in common (eg different trials; and each observer with different goes) you can't just fit a regression to all 120 points. You need to somehow control for observer effects. I would fit a model like

perceived ~ actual * observer


while hoping that can be reduced to

perceived ~ actual + observer


For this to work properly actual needs to be an ordered factor. R will then automatically apply the correct contrasts to it in fitting a model (sorry I don't have time to explain that better but there would be material on the web about what that means).

However, there are lots of potential fishhooks. Can perceived velocity really be treated as continuous? Does it have an approximately normal distribution (obviously not really, but it may be close enough)? Is the variance independent of the mean? Evading these will be tricky and you should use R's graphic capabilities fully in exploring your data. If the assumptions behind linear regression do not hold, you may be able to use the polr() function from library(MASS) to fit an ordinal response regression instead.