I created linear model to give prediction for a team member (individual). Can I use this model to give average (individual) prediction in a team by providing average values of features among team members, or by averaging predictions by each team member?


Training Data

Independent Variables are Task Count and Average Points while Points are y-value.

$$\begin{array}{c|c|c|} \text{Team Id} & \text{Sprint Id} & \text{Team Member} & \text{Task Count} & \text{Avg Points} & \text{Points} \\ \hline \text{1} & 123 & Mike & 4 & 5 & 20 \\ \hline \text{2} & 273 & Chris & 5 & 3 & 15 \\ \hline \text{3} & 403 & James & 7 & 2 & 14 \\ \hline \text{4} & 298 & Paul & 2 & 6 & 12 \\ \hline \end{array}$$

Actual Data that linear model will use to predict

I average independent values by category within a team. The averaged values for each category will be given to predict average story points for an individual within a team. $$\begin{array}{c|c|c|} \text{Team Id} & \text{Sprint Id} & \text{Avg. Task Count} & \text{Avg. Avg Points} & \text{Avg. Points} \\ \hline \text{1} & 342 & 4 & 4 & 5 \\ \hline \text{2} & 713 & 5 & 5 & 3 \\ \hline \text{3} & 663 & 6 & 7 & 2 \\ \hline \text{4} & 188 & 7 & 2 & 6 \\ \hline \end{array}$$

Not sure if I can use the model with average values by category to predict average points.

  • $\begingroup$ Post an example in your question. $\endgroup$ Aug 27, 2019 at 6:00
  • $\begingroup$ @user2974951 I just added the example in my question above. $\endgroup$ Aug 27, 2019 at 13:06
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    $\begingroup$ This can be done, but one is inclined to suspect responses within groups may be strongly correlated, so you should be reluctant to assume independence of responses: consider using a model that accommodates within-group dependence. $\endgroup$
    – whuber
    Aug 27, 2019 at 13:20
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    $\begingroup$ There are many. Two common ones are mixed models (in which variation of individuals within the group is modeled with a single Normally-distributed random variable) and generalized least squares models (in which some explicit form of correlation among within-group individuals is hypothesized). BTW, in at least one case you seem to use the word "dependent" in your question where you mean independent. Also, the connection between the two tables in the question is truly obscure--they aren't clarifying what you're trying to do. $\endgroup$
    – whuber
    Aug 27, 2019 at 14:30
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    $\begingroup$ That's not much of a question--yes, you can use the model (as you could use any model) in that way, because the rules of mathematics permit it. The questions I think you want to ask are (1) how to estimate the uncertainty of that prediction and (2) what models can give reliable predictions with least uncertainty. $\endgroup$
    – whuber
    Aug 27, 2019 at 14:51

1 Answer 1


If I understand your question correctly, what you are asking is:

  • Given a linear model (e.g. y=a1*x1 + a2*x2+...+b),
  • applied to a group of data points, (e.g. each data point is a player, with their own features (e.g. x1_i, x2_i,... for each player i),
  • would averaging the features across players (e.g. using x1_avg, ...), and using them in the model, give the same result as using the features of each player and averaging the results?

If so, the answer is - yes, but only for a linear model (i.e. not a quadratic, polynomial, neural network or whatever).

This can be verified quite simply: for example, if there are only two players and two features: enter image description here

  • $\begingroup$ This partially correct. The issue isn't whether averaging is appropriate for estimating the mean response: even in a nonlinear model, one would estimate the mean response of some arbitrary set of individuals by averaging the individual estimates. The hard part is constructing a prediction interval for that response, especially for groups of individuals who are somehow related and not just randomly selected. $\endgroup$
    – whuber
    Aug 27, 2019 at 13:36
  • $\begingroup$ Thanks. I am asking in general to see if I can use linear model (which is trained with non-average data) with average data. $\endgroup$ Aug 27, 2019 at 14:13

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