- #1

- 19

- 0

_{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 deviations

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

_{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?