The national institute of standards and technology (nist) supplies "standard materials" whose physical properties are supposed to be known. for example, you can buy from nist an iron rod whose electrical conductivity is supposed to be 10.1 at 293 kelvin. (the units for conductivity are microsiemens per centimeter. distilled water has conductivity 0.5.) of course, no measurement is exactly correct. nist knows the variability of its measurements very well, so it is quite realistic to assume that the population of all measurements of the same rod has the normal distribution with mean μ equal to the true conductivity and standard deviation σ = 0.1. here are six measurements on the same standard iron rod, which is supposed to have conductivity 10.1.

Question

Adison

Mathematics

31 January, 14:20

The national institute of standards and technology (nist) supplies "standard materials" whose physical properties are supposed to be known. for example, you can buy from nist an iron rod whose electrical conductivity is supposed to be 10.1 at 293 kelvin. (the units for conductivity are microsiemens per centimeter. distilled water has conductivity 0.5.) of course, no measurement is exactly correct. nist knows the variability of its measurements very well, so it is quite realistic to assume that the population of all measurements of the same rod has the normal distribution with mean μ equal to the true conductivity and standard deviation σ = 0.1. here are six measurements on the same standard iron rod, which is supposed to have conductivity 10.1.

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Answers (2)

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Atticus · Answer 1

I assume that your problem that these are the 6 measurements: 10.1, 10.02, 9.89, 10.05, 10.16, 10.21, and 10.11

Mean of 10.8333

Sample size is 6

Standard dev is 0.1

Alpha is 1 - 0.90 Z (0.1) = 1.645

So the computation would be:

Mean ± Z * s/sqrt (n)

= 10.0833 ± 1.645*0.1/sqrt (6)

= 10.0833± 1.645*0.0408

= 10.0833± 0.0672

=10.0161, 10.1505 is the interval.

Interpretation: We are 90% confident that the iron rod’s true conductivity is between 10.0161 and 10.1504 microsiemens per centimeter.

Koen Bradshaw · Answer 2

Answer: Let be X: liquid whose electrical conductivity X âĽ n (5; 0.2) n = 6 Xmean = (1/6) * (5.32+4.88+5.1+4.73+5.15+4.75) = 4.9883 H0: ÎĽx = 5 H1: ÎĽx â‰ 5, Î± = 0.01 Decision Rule: If abs (Xmean - ÎĽx) * sqrt (n) / Ďx) > z (1 - Î±/2) reject H0. If (abs (4.9883 - 5) * 2.45/0.2 > 2.575, Reject H0. If 0.14 â‰Ż 2.575, we don't Reject H0. Conclusion: The electrical conductivity is 5.