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!
EDIT :
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
Best Answer
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 $\frac{S}{2\sqrt n}$.