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(adsbygoogle = window.adsbygoogle || []).push({}); _{1}empirical standard deviation.

,where N

- 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 studentĀ“s-t distribution with degree of freedom N_{2}-1=19"

The s_{2}is divided by root square of N_{1}however the studentĀ“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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Combining standard deviations

Tags:

**Physics Forums | Science Articles, Homework Help, Discussion**