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A sample mean has a standard normal distribution $$z = \sqrt{n} \frac{\bar{x} - \mu}{\sigma}$$ . This statement seems to assume that true population mean and variance is known. If that is the case, why is t-score then defined as $$t = \sqrt{n} \frac{\bar{x} - \mu}{s}$$ where $$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}$$ and $\mu$ is still known presumably? Sample and population statistics are mixed here which is making it hard for me to understand why Bessel's correction is applied to variance. If it's to compensate for an unknown population distribution, how come we still use true population mean ($\mu$) in our formula?

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2 Answers 2

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With known variance, your $z$, which could be written as $\frac{\bar{x} - \mu}{\sqrt{\sigma^2/n}}$, has a standard normal distribution with mean $0$ and variance $1$. You can use it to test whether the observations are significantly different from a particular possible value for the mean or to construct confidence intervals for the population mean from the observations

With unknown variance, your $t$, which could be written as $\frac{\bar{x} - \mu}{\sqrt{\sum_i (x_i-\bar{x})^2/(n(n-1))}}$, has a Student's $t$-distribution with $n-1$ degrees of freedom. Again, you can use it to test whether the observations are significantly different from a particular possible value for the mean or to construct confidence intervals for the population mean from the observations

You need the $n(n-1)$ divisor to get the Student's $t$-distribution. Otherwise you have a scaled $t$-distribution, which is less convenient. You can regard this as first using Bessel's correction to get $s^2$ to be an unbiased estimate of unknown $\sigma^2$ and then divide by $n$; or you can use the uncorrected sample variance and then divide by $n-1$. It makes no difference which you use, but you must use one to get a $t$ distribution on the correct scale

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  • $\begingroup$ Thank you. To dive deeper, Jensen's inequality would state that Bessel's correction can only remove bias from variance (linear property), but only reduce bias for the deviation, which is sufficient for most approximations. $\endgroup$
    – GViz
    Commented Oct 8, 2019 at 22:03
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No, neither $\mu$ nor $\sigma$ is assumed to be known. We assume the $x_i$ are Gaussian IIDs, so $\bar{x}$ is Gaussian too. But we have to esimtate $\mu$, without bias, as the sample mean $\bar{x}$ of the $x_i\sim N(\mu,\,\sigma^2)$ with $\sigma>0$. Then $s^2$ is an unbiased estimator of $\sigma^2$, and $t$ has a $\nu=N-1$ Student's $t$-distribution. That last part can be proven with the theory of ratio distributions, allowing us to test a hypothesis about $\mu$.

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  • $\begingroup$ Thank you, makes a bit more sense. Just one followup, if we estimate $\mu$ as $\bar{x}$, the numerator seems to go to zero. What am I missing here? $\endgroup$
    – GViz
    Commented Oct 8, 2019 at 20:54
  • $\begingroup$ @GViz. What you're missing is that, although the definition of $s$ in terms of $\bar{x}$ predicts the distribution of $t$, we test a null hypothesis claiming a specific value for $\mu$ by using $t$ to quantify how far $\bar{x}$ deviates from the claimed $\mu$ in practice. $\endgroup$
    – J.G.
    Commented Oct 8, 2019 at 21:10
  • $\begingroup$ Great, so true mean would be a tested claim or more typically derived from the t-distribution tables. $\endgroup$
    – GViz
    Commented Oct 8, 2019 at 22:00
  • $\begingroup$ @GViz Definitely the former. You can't get a realistic $\mu$ for your dataset from standard tables. $\endgroup$
    – J.G.
    Commented Oct 8, 2019 at 22:04
  • $\begingroup$ Apologies, I meant lay out confidence intervals with the latter. $\endgroup$
    – GViz
    Commented Oct 8, 2019 at 22:08

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