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Introductory statistics classes often present various hypothesis tests — t-test, ANOVA, etc. — in a disconnected and *ad hoc* way. However, it’s often more useful to think about statistical hypothesis tests under a unifying framework, and then derive specific tests as special cases. This can make it easier to reason about each test’s assumptions and guarantees.

In this post, we review the fact that some of the most common statistical tests can be placed in the framework of a linear model. This is largely based on Jonas Kristoffer Lindeløv’s excellent post on the same topic. While much of this post is a review of his, I try to add more detail around the specific assumptions under each test.

Consider a sample of $n$ univariate random variables, $\mathbf{y} = (y_1, \dots, y_n)$ whose true, unobserved mean is $\mu.$ We denote the empirical mean as $\bar{y} = \frac1n \sum_{i=1}^n y_i.$ Suppose we would like to test whether the true mean is equal to zero or not. We denote the null and alternative hypotheses as

\[H_0: \mu = 0,~~~H_1: \mu \neq 0.\]Recall that a t-test assumes that the data is Gaussian distributed. We can thus model the data with the following linear model:

\[y_i = \mu + \epsilon_i,~~\epsilon_i \sim N(0, \sigma^2),\]where the $y_i$’s are modeled with a shared intercept $\mu$ plus noise.

Thus, testing whether $\mu = 0$ is equivalent to testing whether the intercept is equal to zero in this linear model.

In a two-sample t-test, we wish to test whether the means of two groups are equal. Consider data from two groups with sample sizes $n$ and $m$. Denote the data as

\[\mathbf{y} = (y_1, \dots, y_{n + m}),~~\mathbf{g} = (\underbrace{0, \dots, 0}_{\text{$n$ times}}, \underbrace{1, \dots, 1}_{\text{$m$ times}}),\]where $g_i \in \{0, 1\}$ is the group label for sample $y_i.$ Suppose the true, unobserved means for the two groups are $\mu_1$ and $\mu_2,$ respectively. The null and alternative hypotheses in a two-sample t-test are given by

\[H_0: \mu_1 = \mu_2,~~H_1: \mu_1 \neq \mu_2.\]Analogous to the one-sample t-test, we could now consider incorporating group-specific intercept terms in the linear model,

\[y_i = 1_{\{g_i = 0\}} \mu_1 + 1_{\{g_i = 1\}} \mu_2 + \epsilon_i,~~\epsilon_i \sim N(0, \sigma^2),~~~\textbf{(Incorrect)}\]where $1_{\{\cdot\}}$ is the indicator function. However, under this model, $\mu_1$ and $\mu_2$ will not be identifiable. In other words, in the matrix form of this linear model, notice that the columns of the design matrix will be linearly dependent.

Instead, we can model the data as a sum of a global intercept $\mu$ and one group-specific intercept $\mu_1:$

\[y_i = \mu + 1_{\{g_i = 0\}} \beta_1 + \epsilon_i,~~\epsilon_i \sim N(0, \sigma^2)\]In this case, the second group’s intercept will be represented by $\mu,$ and the first group’s intercept, $\mu_1,$ can be retrieved as $\beta_1 - \mu.$

Under this model, testing whether $\mu_1 = \mu_2$ is equivalent to testing whether $\beta_1 = 0.$

In a paired-sample t-test, we again have two groups of data, but now we also assume that the sample sizes are the same and that there is a natural pairing between the samples of each group. Specifically, let $\mathbf{y}^{(1)} = (y_1^{(1)}, \cdots, y_n^{(1)})$ and $\mathbf{y}^{(2)} = (y_1^{(2)}, \cdots, y_n^{(2)})$ be the two groups’ data, where $y_i^{(1)}$ and $y_i^{(2)}$ are paired for each $i.$

\[y_i^{(1)} - y_i^{(2)} = \mu + \epsilon,~~\epsilon \sim N(0, \sigma^2).\]Under this model, testing whether $\mu_1 = \mu_2$ is equivalent to testing whether $\mu = 0.$

Analysis of variance (ANOVA) tests generalize t-tests in order to test whether multiple ($> 2$) groups have the same means or not. Continuing our notation from above, let $g_i$ denote the group label of sample $i$ where there are $K$ distinct groups. Then we have the following model for an ANOVA test:

\[y_i = \mu + \epsilon_i + \sum\limits_{k = 2}^K 1_{\{g_i = k\}} \beta_k,~~\epsilon_i \sim N(0, \sigma^2).\]Under this model, testing whether $\mu_1 = \cdots = \mu_K$ is equivalent to testing whether $\beta_2 = \cdots = \beta_K = 0.$

Chi-squared ($\chi^2$) tests are used to test whether two or more proportions are the same. Suppose we have $K$ groups where the $k$th group contains $n_k$ samples. Denote the total sample size by $n = \sum_k n_k.$ The null hypothesis for a $\chi^2$ test is

\[H_0: \frac{n_1}{n} = \cdots \frac{n_K}{n},\]and the alternative hypothesis is that at least one of these proportions is unequal to the rest. Notice that the null hypothesis is equivalent to the hypothesis that the counts are equal, $n_1 = \cdots = n_K.$

We can model the count of each group with a Poisson GLM:

\[n_k \sim \text{Po}(\lambda_k),~~\lambda_k = \exp\left\{ \mu + \beta_k \right\},\]where we have used a $\log$ link function, and the linear predictor is given by $\mu + \beta_k.$

Testing whether the proportions are the same in these groups is equivalent to testing whether $\beta_2 = \cdots = \beta_K = 0.$

- Jonas Kristoffer Lindeløv’s blog post