How is the t-distribution a statistic considering it is dependent on parameter mu?

Question

Casella and Berger textbook teaches that a statistic is only the function of a random sample and never its unknown population parameter(s).

In the case of Student's t-distribution, while it utilizes the sample variance in place of the unknown variance population parameter, its distribution still relies on the mean mu population parameter.

Shouldn't this disqualify Student's t-distribution as being a statistic?

Thanks for any information

Answer 1

You seem to be confusing a number of concepts. A $t$-distribution is a probability distribution. If a random variable $T$ is Student's $t$-distributed with $\nu$ degrees of freedom, then it has density $$f_T(t) = \frac{\Gamma(\frac{\nu + 1}{2})}{\Gamma(\frac{\nu}{2})\Gamma(\frac{1}{2}) \sqrt{\nu}} \left(\frac{\nu}{t^2 + \nu}\right)^{(\nu + 1)/2}, \quad -\infty < t < \infty.$$ On the other hand, if we have a sample from a normally distributed population $\boldsymbol X = (X_1, X_2, \ldots, X_n)$ with $$X_i \sim \operatorname{Normal}(\mu, \sigma^2), \quad i = 1, 2, \ldots, n$$ independent and identically distributed, then the pivotal quantity $$T = \frac{\bar X - \mu}{S/\sqrt{n}}$$ is Student's $t$-distributed with $\nu = n-1$ degrees of freedom. When $\mu$ is unknown, then indeed $T$ is not a statistic. However, this quantity is used in the context of hypothesis testing; e.g., for a test of location of the form $$H_0 : \mu = \mu_0 \quad \text{vs.} \quad H_1 : \mu \ne \mu_0,$$ the test statistic $$T \mid H_0 = \frac{\bar X - \mu_0}{S/\sqrt{n}} \sim \operatorname{StudentT}(n-1)$$ is in fact a statistic and is Student's $t$-distributed. Note the conditional statement $T \mid H_0$ represents the assumption that the $X_i$ have mean $\mu = \mu_0$.

Answer 2

The t-distribution isn't a statistic. The t estimator, which is calculated using only the sample mean and standard deviation, is a statistic, and its distribution (under suitable assumptions and/or approximations) is the t-distribution with an appropriate number of degrees of freedom.

It's equivalent to the fact that the sample mean is (approximately) normally distributed, but the parameters of that distribution are based on (generally unknown) population parameters.

Answer 3

The statistic is $\dfrac{\frac1n\sum_i x_i }{\sqrt{\frac{1}{n(n-1)}\sum_j(x_j-\frac1n\sum_i x_i)^2}}$ where everything comes from the sample data.

If the $\{X_i\}$ are an i.i.d. sample from a normal distribution with mean $0$ and any positive variance, then this statistic has a Student $t$-distribution with $n−1$ degrees of freedom.