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## Main Question or Discussion Point

We have two groups measuring the same resistors, the nominal value is unknown. Group 1 is slower and because of that they did not calcute the s

We have to give the confidence interval for the nominal value of the resistance at confidence level p=90% and prove that this is the right way to calculate it.

Somewhere i found an answer, but i dont really understand why is it correct and also i need to prove that its right.

The solution was this: "Even if we dont know s

The s

_{1}empirical standard deviation.- Group 1: N
_{1}=500 , R_{1}=6903 , s_{1}=unknown - Group 2: N
_{2}=20 , R_{2}=6880 , s_{2}=168.3

_{1},N_{2}is number of measurements, R_{1},R_{2}is average of the measured values and s_{1},s_{2}empirical standard deviationsWe have to give the confidence interval for the nominal value of the resistance at confidence level p=90% and prove that this is the right way to calculate it.

Somewhere i found an answer, but i dont really understand why is it correct and also i need to prove that its right.

The solution was this: "Even if we dont know s

_{1}, we can use estimator R_{1}, because its more accurate than R_{2}(because of the larger amount of information). However we can only use s_{2}empirical standard dev. and the confidence interval should be calculated by student´s-t distribution with degree of freedom N_{2}-1=19"The s

_{2}is divided by root square of N_{1}however the student´s distribution has degree of freedom N_{2}. Can someone explain me, why is this correct, why can we combine the two measurements this way or better show me the deduction?