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Old 12-03-2003, 08:06 PM
BruceZ BruceZ is offline
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Join Date: Sep 2002
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Default sqrt(N/N-1) factor with the t-distribution

I'd forgotten I should use the t-test...I don't remember this sqrt(6/5) thing though, where does that come from?

That's because you used the maximum likelihood estimator for the variance, which divides by N, so you have to multiply the estimate of sigma by the factor sqrt(N/N-1) before you can use the t-distribution. Alternatively, you could have used the unbiased estimator for the variance, which divides by N-1, and then you can use the square root of this directly. If sigma were known, we would use the normal distribution, and the fact that
sqrt(N)*(Xn-u)/sigma is a normally distributed random variable, where u is the mean, and Xn are the samples. Since sigma is unknown, we are replacing sigma with an estimate of sigma, est(sigma), and then sqrt(N)*(Xn-u)/est(sigma) is distributed as a t-distribution with N-1 degrees of freedom instead of a normal distribution, but it is only a t-distribution if we use the variance estimate that divides by N-1, or equivalently, by the maximum likelihood estimate that divides by N, with sigma scaled by sqrt(N/N-1). This becomes a negligible factor as N becomes reasonably large.
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