I have asked this question without success earlier, perhaps it was not really clear what I was looking for. Here is a clearer explanation, hopefully clear enough to get some help!

My data has 378 entries where each row represents the shared score for two individuals playing a game. For example, on row one, the two players made 10 points total. Row two, they made 7 points, and so on.

I would like to calculate mean points per player for the entire sample and calculate the standard error. The idea is probably to calculate the total mean of every row's mean. But I have no clue then, how to calculate the error: this is my question!

Can anybody help me? I'd greatly appreciate some imput!

Thank you very much in advance!


Each row is for a different team of two players, and contains only a score, from which we dont know who made the points. I want to know how many points a player scored, which I obtain by calculating the mean for each row (score/2), then the global mean of all rows.

Team - Score
1 - 10
2 - 7
3 - 8
4 - 8
378 - 10

  • $\begingroup$ Could you provide a few rows of sample data (with headers) to give an idea of what the data looks like and what it means? $\endgroup$ Aug 1 '17 at 0:22
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    $\begingroup$ In particular, this would help clarify whether you have one number (a shared score) or two numbers (each players' score) in each row. Also, does the entire dataset represent the same two players, or are there different players in different games? And what do you mean by "mean points per player for the entire sample"? $\endgroup$ Aug 1 '17 at 0:29
  • $\begingroup$ I just edited the post with more details. I hope this helps. The data contains more lines, but the relevant ones are just team and score. $\endgroup$ Aug 1 '17 at 2:06
  • $\begingroup$ Ok, so are there 378*2 = 756 different players then, i.e., each player only plays once? Or do players play on multiple teams? $\endgroup$ Aug 1 '17 at 2:36
  • $\begingroup$ Each row is a team of two players that play only once, for one team. $\endgroup$ Aug 1 '17 at 17:15

If I understand your scenario correctly, there are latent individual scores $X_1,\dots,X_{2n}$, which we can model as being iid random variables, with $n=378$. The data that we observe are team scores $Y_1,\dots,Y_n$, defined as sums over pairs of individual scores: $$Y_i = X_{2i-1} + X_{2i}$$ And again, if I understand correctly, you are wanting to perform inference on the distribution of the latent individual scores; specifically, you would like to estimate the expected value of $X_1$, (which is equal to the expected value of $X_i$ for any $i$, since we are assuming the $X_i$ are iid). This is fairly straightforward. Taking the expected value on both sides of the equation $Y_i = X_{2i-1} + X_{2i}$ gives us the relationship $E(Y_1) = 2E(X_1)$, so you can simply perform inference on $E(Y_1)$ and then translate the results into a statement about $E(X_1)$. To make this concrete, you could construct the t confidence interval for $E(Y_1)$ as follows: $$E(Y_1) \approx \overline Y \pm \frac{t^*S}{\sqrt n}$$ where \begin{align*} \overline Y &= \frac1n\sum_{i=1}^N Y_i\\ S &= \sqrt{\frac1{n-1}\sum_{i=1}^N (Y_i-\overline Y)^2}\\ t^* &= \text{critical value from t distribution with $\text{df} = n-1$} \\ &\quad\text{(e.g., $t^* \approx 1.966$ for a 95% confidence interval when $n=378$)} \end{align*} This then translates into a confidence interval for $E(X_1)$ as follows $$E(X_1) \approx \frac{\overline Y}2 \pm \frac{t^*S}{2\sqrt n}$$ So your estimate for $E(X_1)$ is $\overline Y/2$, with a margin of error of $\frac{t^*S}{2\sqrt n}$. If you are specifically looking for the standard error (i.e., an approximation of the standard deviation of the estimate $\frac{\overline Y}2$), then you would remove the factor of $t^*$ from this last expression; i.e., it would be $\frac{S}{2\sqrt n}$.

  • $\begingroup$ This demonstration is very clear, thank you. One question though : what does the margin of error correspond to? Is it the SE for a 95% C.I.? I'm trying to make my mind on how to state clearly my mean values. $\endgroup$ Aug 1 '17 at 19:00
  • $\begingroup$ The margin of error is the distance between our estimate and either endpoint of the C.I.; in other words, it is half the width of the C.I. This is not the same as the standard error. $\endgroup$ Aug 1 '17 at 20:48
  • $\begingroup$ Ok, I understand, thank you. I am wondering what would be the correct measure of variability to use (SD, SE, CI), say in a scientific publication, to better describe the mean. I am planning to compare this mean with another sample of players..? $\endgroup$ Aug 1 '17 at 21:27
  • $\begingroup$ It might depend on what the norm is for the particular publication, but I think that a confidence interval is easiest to interpret, so I would go with that by default. $\endgroup$ Aug 2 '17 at 17:37

If we assume that the scores are drawn iid (and I don't see why we wouldn't, since it's different players of different teams), then the 378 observations of average points per player (score/2 for each row) allow you to characterize the distribution of the statistic you're trying to estimate (in your case the "mean of means"). The standard deviation of this sampling distribution is the standard error you are trying to calculate!

  • $\begingroup$ I am not sure if I understood you well Yannis. Normally the SE = SD/sqrt(n). Are you simply suggesting that I can calculate the mean of means and its SD/SE the same way as I would if these were already rows of "score per player"? $\endgroup$ Aug 1 '17 at 17:27
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    $\begingroup$ A standard error is defined as the standard deviation of a sampling distribution. You have the sampling distribution of the statistic you're interested in, so the standard deviation of that is the standard error. And yes, you divide each row by 2, then the SD of that vector is the standard error of the mean of that vector. $\endgroup$ Aug 1 '17 at 17:30

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