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I want to estimate the variance of a normally distributed population. I can take N samples and calculate the sample mean and sample variance, which would normally suffice; however, the sampling method itself has inherent variability that is known. Is there a way to "subtract out" the known test method variance the get a better estimate of the population variance?

To provide some context, I'm measuring density of an asphalt road after construction. Random locations are selected for measuring density for quality assurance. There are two methods, A and B, for measuring density and they have different known variance. The method variance was found by repeated measurements on known references. If I use Method A, with high variance, to and get N samples of the road density, I'll end up with a sample variance that is higher than if I used the more precise Method B. And in both cases, my variance will still be higher than the true variance of my sample. If I know the actual test method variance, isn't there some way to leverage that information when making an estimate of the sample and population variance?

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  • $\begingroup$ Repeated Measures ANOVA? $\endgroup$ – olooney Jun 28 at 14:05
  • $\begingroup$ Can you explain how that would apply to my situation? I'm not sure I follow. $\endgroup$ – Bryan Jun 28 at 20:47
  • $\begingroup$ "The population is normally distributed with an unknown population variance." This sentence makes no sense, if you're doing a classical (fequentist) test. If you context is Bayesian, then you need to say more. $\endgroup$ – BruceET Jun 28 at 21:20
  • $\begingroup$ The sentence makes sense to me. What's the problem you see with it? $\endgroup$ – Glen_b Jun 29 at 7:17
  • $\begingroup$ The best estimate of the population mean is going to be the sample mean. I am not sure what you are asking. $\endgroup$ – Peter Flom Jun 29 at 12:37

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