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I have experimenatal data from medical industry. In the experiment 4 observations were weighted, each one 2 times for more precise measurement.

As a final weight, a mean from each pair is reported. So I have 8 values, a pair (2) for each observation and 4 means.

We assume a normal distrubution with unknown mean and standard deviation.

I need to calculate 95% CI for mean weight. I use this formula: $$ X¯±t∗S/\sqrt{n}, $$ where where $S$ is the sample standard deviation and where $t$ cuts probability 0.025 from the upper tail of Student's t distribution with $n−1$

degrees of freedom.

Should I do this from 8 values (7 df, n=8) or from calculated mean values from each pair(3 df, n=4)?

Maybe a should use another formula? Why?

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Well, I think since you do your measurements twice (and take the mean values) then your data becomes the mean-valued-observations. Based on this you should use $t_{3}$ (student's distribution with $3$ degrees of freedom). This approach is more interpretable since by taking the mean values you, in a way, take more reliable and less noisy data.

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