In this section, we will study a number of important hypothesis tests that fall under the general term chi-square tests. These are named, as you might guess, because in each case the test statistics has (in the limit) a chi-square distribution. Although there are several different tests in this general category, they all share some common themes:

• In each test, there are one or more underlying multinomial samples, Of course, the multinomial model includes the Bernoulli model as a special case.
• Each test works by comparing the observed frequencies of the various outcomes with expected frequencies under the null hypothesis.
• If the model is incompletely specified, some of the expected frequencies must be estimated; this reduces the degrees of freedom in the limiting chi-square distribution.

We will start with the simplest case, where the derivation is the most straightforward; in fact this test is equivalent to a test we have already studied. We then move to successively more complicated models.

Topics

• One Sample Bernoulli Model
• Multi-Sample Bernoulli Model
• One Sample Multinomial Model
• Multi-Sample Multinomial Model
• Goodness of Fit Tests
• Tests of Independence
• Computational and Simulation Exercises

The One-Sample Bernoulli Model

Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample from the Bernoulli distribution with unknown success parameter \(p \in (0, 1)\). Thus, these are independent random variables taking the values 1 and 0 with probabilities \(p\) and \(1 - p\) respectively. We want to test \(H_0: p = p_0\) versus \(H_1: p \ne p_0\), where \(p_0 \in (0, 1)\) is specified. Of course, we have already studied such tests in the Bernoulli model. But keep in mind that our methods in this section will generalize to a variety of new models that we have not yet studied.

Let \(O_1 = \sum_{j=1}^n X_j\) and \(O_0 = n - O_1 = \sum_{j=1}^n (1 - X_j)\). These statistics give the number of times (frequency) that outcomes 1 and 0 occur, respectively. Moreover, we know that each has a binomial distribution; \(O_1\) has parameters \(n\) and \(p\), while \(O_0\) has parameters \(n\) and \(1 - p\). In particular, \(\E(O_1) = n p\), \(\E(O_0) = n (1 - p)\), and \(\var(O_1) = \var(O_0) = n p (1 - p)\). Moreover, recall that \(O_1\) is sufficient for \(p\). Thus, any good test statistic should be a function of \(O_1\). Next, recall that when \(n\) is large, the distribution of \(O_1\) is approximately normal, by the central limit theorem. Let \[ Z = \frac{O_1 - n p_0}{\sqrt{n p_0 (1 - p_0)}} \] Note that \(Z\) is the standard score of \(O_1\) under \(H_0\). Hence if \(n\) is large, \(Z\) has approximately the standard normal distribution under \(H_0\), and therefore \(V = Z^2\) has approximately the chi-square distribution with 1 degree of freedom under \(H_0\). As usual, let \(\chi_k^2\) denote the quantile function of the chi-square distribution with \(k\) degrees of freedom.

For purposes of generalization, the critical result in the next exercise is a special representation of \(V\). Let \(e_0 = n (1 - p_0)\) and \(e_1 = n p_0\). Note that these are the expected frequencies of the outcomes 0 and 1, respectively, under \(H_0\).

This representation shows that our test statistic \(V\) measures the discrepancy between the expected frequencies, under \(H_0\), and the observed frequencies. Of course, large values of \(V\) are evidence in favor of \(H_1\). Finally, note that although there are two terms in the expansion of \(V\) in Exercise 3, there is only one degree of freedom since \(O_0 + O_1 = n\). The observed and expected frequencies could be stored in a \(1 \times 2\) table.

The Multi-Sample Bernoulli Model

Suppose now that we have samples from several (possibly) different, independent Bernoulli trials processes. Specifically, suppose that \(\bs{X}_i = (X_{i,1}, X_{i,2}, \ldots, X_{i,n_i})\) is a random sample of size \(n_i\) from the Bernoulli distribution with unknown success parameter \(p_i \in (0, 1)\) for each \(i \in \{1, 2, \ldots, m\}\). Moreover, the samples \((\bs{X}_1, \bs{X}_2, \ldots, \bs{X}_m)\) are independent. We want to test hypotheses about the unknown parameter vector \(\bs{p} = (p_1, p_2, \ldots, p_m)\). There are two common cases that we consider below, but first let's set up the essential notation that we will need for both cases. For \(i \in \{1, 2, \ldots, m\}\) and \(j \in \{0, 1\}\), let \(O_{i,j}\) denote the number of times that outcome \(j\) occurs in sample \(\bs{X}_i\). The observed frequency \(O_{i,j}\) has a binomial distribution; \(O_{i,1}\) has parameters \(n_i\) and \(p_i\) while \(O_{i,0}\) has parameters \(n_i\) and \(1 - p_i\).

The Completely Specified Case

Consider a specified parameter vector \(\bs{p}_0 = (p_{0,1}, p_{0,2}, \ldots, p_{0,m}) \in (0, 1)^m\). We want to test the null hypothesis \(H_0: \bs{p} = \bs{p}_0\), versus \(H_1: \bs{p} \ne \bs{p}_0\). Since the null hypothesis specifies the value of \(p_i\) for each \(i\), this is called the completely specified case. Now let \(e_{i,0} = n_i (1 - p_{i,0})\) and let \(e_{i,1} = n_i p_{i,0}\). These are the expected frequencies of the outcomes 0 and 1, respectively, from sample \(\bs{X}_i\) under \(H_0\).

As a rule of thumb, large means that we need \(e_{i,j} \ge 5\) for each \(i \in \{1, 2, \ldots, m\}\) and \(j \in \{0, 1\}\). But of course, the larger these expected frequencies the better.

Once again, note that the test statistic \(V\) measures the discrepancy between the expected and observed frequencies, over all outcomes and all samples. There are \(2 \, m\) terms in the expansion of \(V\) in Exercise 4, but only \(m\) degrees of freedom, since \(O_{i,0} + O_{i,1} = n_i\) for each \(i \in \{1, 2, \ldots, m\}\). The observed and expected frequencies could be stored in an \(m \times 2\) table.

The Equal Probability Case

Suppose now that we want to test the null hypothesis \(H_0: p_1 = p_2 = \cdots = p_m\) that all of the success probabilities are the same, versus the complementary alternative hypothesis \(H_1\) that the probabilities are not all the same. Note, in contrast to the previous model, that the null hypothesis does not specify the value of the common success probability \(p\). But note also that under the null hypothesis, the \(m\) samples can be combined to form one large sample of Bernoulli trials with success probability \(p\). Thus, a natural approach is to estimate \(p\) and then define the test statistic that measures the discrepancy between the expected and observed frequencies, just as before. The challenge will be to find the distribution of the test statistic.

Let \(n = \sum_{i=1}^m n_i\) denote the total sample size when the samples are combined. Then the overall sample mean, which in this context is the overall sample proportion of successes, is \[ P = \frac{1}{n} \sum_{i=1}^m \sum_{j=1}^{n_i} X_{i,j} = \frac{1}{n} \sum_{i=1}^m O_{i,1} \] The sample proportion \(P\) is the best estimate of \(p\), in just about any sense of the word. Next, let \(E_{i,0} = n_i \, (1 - P)\) and \(E_{i,1} = n_i \, P\). These are the estimated expected frequencies of 0 and 1, respectively, from sample \(\bs{X}_i\) under \(H_0\). Of course these estimated frequencies are now statistics (and hence random) rather than parameters. Just as before, we define our test statistic \[ V = \sum_{i=1}^m \sum_{j=0}^1 \frac{(O_{i,j} - E_{i,j})^2}{E_{i,j}} \] It turns out that under \(H_0\), the distribution of \(V\) converges to the chi-square distribution with \(m - 1\) degrees of freedom as \(n \to \infty\).

Intuitively, we lost a degree of freedom over the completely specified case in because we had to estimate the unknown common success probability \(p\). Again, the observed and expected frequencies could be stored in an \(m \times 2\) table.

The One-Sample Multinomial Model

Our next model generalizes the one-sample Bernoulli model in a different direction. Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a sequence of multinomial trials. Thus, these are independent, identically distributed random variables, each taking values in a set \(S\) with \(k\) elements. If we want, we can assume that \(S = \{0, 1, \ldots, k - 1\}\); the one-sample Bernoulli model then corresponds to \(k = 2\). Let \(f\) denote the common probability density function of the sample variables on \(S\), so that \(f(j) = \P(X_i = j)\) for \(i \in \{1, 2, \ldots, n\}\) and \(j \in S\). The values of \(f\) are assumed unknown, but of course we must have \(\sum_{j \in S} f(j) = 1\), so there are really only \(k - 1\) unknown parameters. For a given probability density function \(f_0\) on \(S\) we want to test \(H_0: f = f_0\) versus \(H_1: f \ne f_0\).

By this time, our general approach should be clear. We let \(O_j\) denote the number of times that outcome \(j \in S\) occurs in sample \(\bs{X}\): \[ O_j = \sum_{i=1}^n \bs{1}(X_i = j) \] Note that \(O_j\) has the binomial distribution with parameters \(n\) and \(f(j)\). Thus, \(e_j = n \, f_0(j)\) is the expected number of times that outcome \(j\) occurs, under \(H_0\). Out test statistic, of course, is \[ V = \sum_{j \in S} \frac{(O_j - e_j)^2}{e^j} \] It turns out that under \(H_0\), the distribution of \(V\) converges to the chi-square distribution with \(k - 1\) degrees of freedom as \(n \to \infty\). Note that there are \(k\) terms in the expansion of \(V\), but only \(k - 1\) degrees of freedom since \(\sum_{j \in S} O_j = n\).

Again, as a rule of thumb, we need \(e_j \ge 5\) for each \(j \in S\), but the larger the expected frequencies the better.

The Multi-Sample Multinomial Model

As you might guess, our final generalization is to the multi-sample multinomial model. Specifically, suppose that \(\bs{X}_i = (X_{i,1}, X_{i,2}, \ldots, X_{i,n_i})\) is a random sample of size \(n_i\) from a distribution on a set \(S\) with \(k\) elements, for each \(i \in \{1, 2, \ldots, m\}\). Moreover, we assume that the samples \((\bs{X}_1, \bs{X}_2, \ldots, \bs{X}_m)\) are independent. Again there is no loss in generality if we take \(S = \{0, 1, \ldots, k - 1\}\). Then \(k = 2\) reduces to the multi-sample Bernoulli model, and \(m = 1\) corresponds to the one-sample multinomial model.

Let \(f_i\) denote the common probability density function of the variables in sample \(\bs{X}_i\), so that \(f_i(j) = \P(X_{i,l} = j)\) for \(i \in \{1, 2, \ldots, m\}\), \(l \in \{1, 2, \ldots, n_i\}\), and \(j \in S\). These are generally unknown, so that our vector of parameters is the vector of probability density functions: \(\bs{f} = (f_1, f_2, \ldots, f_m)\). Of course, \(\sum_{j \in S} f_i(j) = 1\) for \(i \in \{1, 2, \ldots, m\}\), so there are actually \(m \, (k - 1)\) unknown parameters. We are interested in testing hypotheses about \(\bs{f}\). As in the multi-sample Bernoulli model, there are two common cases that we consider below, but first let's set up the essential notation that we will need for both cases. For \(i \in \{1, 2, \ldots, m\}\) and \(j \in S\), let \(O_{i,j}\) denote the number of times that outcome \(j\) occurs in sample \(\bs{X}_i\). The observed frequency \(O_{i,j}\) has a binomial distribution with parameters \(n_i\) and \(f_i(j)\).

The Completely Specified Case

Consider a given vector of probability density functions on \(S\), denoted \(\bs{f}_0 = (f_{0,1}, f_{0,2}, \ldots, f_{0,m})\). We want to test the null hypothesis \(H_0: \bs{f} = \bs{f}_0\), versus \(H_1: \bs{f} \ne \bs{f}_0\). Since the null hypothesis specifies the value of \(f_i(j)\) for each \(i\) and \(j\), this is called the completely specified case. Let \(e_{i,j} = n_i \, f_{0,i}(j)\). This is the expected frequency of outcome \(j\) in sample \(\bs{X}_i\) under \(H_0\).

As usual, our rule of thumb is that we need \(e_{i,j} \ge 5\) for each \(i \in \{1, 2, \ldots, m\}\) and \(j \in S\). But of course, the larger these expected frequencies the better.

As always, the test statistic \(V\) measures the discrepancy between the expected and observed frequencies, over all outcomes and all samples. There are \(m k\) terms in the expansion of \(V\) in Exercise 8, but we lose \(m\) degrees of freedom, since \(\sum_{j \in S} O_{i,j} = n_i\) for each \(i \in \{1, 2, \ldots, m\}\).

The Equal PDF Case

Suppose now that we want to test the null hypothesis \(H_0: f_1 = f_2 = \cdots = f_m\) that all of the probability density functions are the same, versus the complementary alternative hypothesis \(H_1\) that the probability density functions are not all the same. Note, in contrast to the previous model, that the null hypothesis does not specify the value of the common success probability density function \(f\). But note also that under the null hypothesis, the \(m\) samples can be combined to form one large sample of multinomial trials with probability density function \(f\). Thus, a natural approach is to estimate the values of \(f\) and then define the test statistic that measures the discrepancy between the expected and observed frequencies, just as before.

Let \(n = \sum_{i=1}^m n_i\) denote the total sample size when the samples are combined. Under \(H_0\), our best estimate of \(f(j)\) is \[ P_j = \frac{1}{n} \sum_{i=1}^m O_{i,j} \] Hence our estimate of the expected frequency of outcome \(j\) in sample \(\bs{X}_i\) under \(H_0\) is \(E_{i,j} = n_i P_j\). Again, this estimated frequency is now a statistic (and hence random) rather than a parameter. Just as before, we define our test statistic \[ V = \sum_{i=1}^m \sum_{j \in S} \frac{(O_{i,j} - E_{i,j})^2}{E_{i,j}} \] As you no doubt expect by now, it turns out that under \(H_0\), the distribution of \(V\) converges to a chi-square distribution as \(n \to \infty\). But let's see if we can determine the degrees of freedom heuristically.

A Goodness of Fit Test

A goodness of fit test is an hypothesis test that an unknown sampling distribution is a particular, specified distribution or belongs to a parametric family of distributions. Such tests are clearly fundamental and important. The one-sample multinomial model leads to a quite general goodness of fit test.

To set the stage, suppose that we have an observable random variable \(X\) for an experiment, taking values in a general set \(S\). Random variable \(X\) might have a continuous or discrete distribution, and might be single-variable or multi-variable. We want to test the null hypothesis that \(X\) has a given, completely specified distribution, or that the distribution of \(X\) belongs to a particular parametric family.

Our first step, in either case, is to sample from the distribution of \(X\) to obtain a sequence of independent, identically distributed variables \(\bs{X} = (X_1, X_2, \ldots, X_n)\). Next, we select \(k \in \N_+\) and partition \(S\) into \(k\) (disjoint) subsets. We will denote the partition by \(\{A_j: j \in J\}\) where \(\#(J) = k\). Next, we define the sequence of random variables \(\bs{Y} = (Y_1, Y_2, \ldots, Y_n)\) by \(Y_i = j\) if and only if \(X_i \in A_j\) for \(i \in \{1, 2, \ldots, n\}\) and \(j \in J\).

The Completely Specified Case

Let \(H\) denote the statement that \(X\) has a given, completely specified distribution. Let \(f_0\) denote the probability density function on \(J\) defined by \(f_0(j) = \P(X \in A_j \mid H)\) for \(j \in J\). To test hypothesis \(H\), we can formally test \(H_0: f = f_0\) versus \(H_1: f \ne f_0\), which of course, is precisely the problem we solved in the one-sample multinomial model.

Generally, we would partition the space \(S\) into as many subsets as possible, subject to the restriction that the expected frequencies all be at least 5.

The Partially Specified Case

Often we don't really want to test whether \(X\) has a completely specified distribution (such as the normal distribution with mean 5 and variance 9), but rather whether the distribution of \(X\) belongs to a specified parametric family (such as the normal). A natural course of action in this case would be to estimate the unknown parameters and then proceed just as above. As we have seen before, the expected frequencies would be statistics \(E_j\) because they would be based on the estimated parameters. As a rule of thumb, we lose a degree of freedom in the chi-square statistic \(V\) for each parameter that we estimate, although the precise mathematics can be complicated.

A Test of Independence

Suppose that we have observable random variables \(X\) and \(Y\) for an experiment, where \(X\) takes values in a set \(S\) with \(k\) elements, and \(Y\) takes values in a set \(T\) with \(m\) elements. Let \(f\) denote the joint probability density function of \((X, Y)\), so that \(f(i, j) = \P(X = i, Y = j)\) for \(i \in S\) and \(j \in T\). Recall that the marginal probability density functions of \(X\) and \(Y\) are the functions \(g\) and \(h\) respectively, where \begin{align} g(i) = & \sum_{j \in T} f(i, j), \quad i \in S \\ h(j) = & \sum_{i \in S} f(i, j), \quad j \in T \end{align} Usually, of course, \(f\), \(g\), and \(h\) are unknown. In this section, we are interested in testing whether \(X\) and \(Y\) are independent, a basic and important test. Formally then we want to test the null hypothesis \[ H_0: f(i, j) = g(i) \, h(j), \quad (i, j) \in S \times T \] versus the complementary alternative \(H_1\).

Our first step, of course, is to draw a random sample \((\bs{X}, \bs{Y}) = ((X_1, Y_1), (X_2, Y_2), \ldots, (X_n, Y_n))\) from the distribution of \((X, Y)\). Since the state spaces are finite, this sample forms a sequence of multinomial trials. Thus, with our usual notation, let \(O_{i,j}\) denote the number of times that \((i, j)\) occurs in the sample, for each \((i, j) \in S \times T\). This statistic has the binomial distribution with trial parameter \(n\) and success parameter \(f(i, j)\). Under \(H_0\), the success parameter is \(g(i) \, h(j)\). However, since we don't know the success parameters, we must estimate them in order to compute the expected frequencies. Our best estimate of \(f(i, j)\) is the sample proportion \(\frac{1}{n} O_{i,j}\). Thus, our best estimates of \(g(i)\) and \(h(j)\) are \(\frac{1}{n} N_i\) and \(\frac{1}{n} M_j\), respectively, where \(N_i\) is the number of times that \(i\) occurs in sample \(\bs{X}\) and \(M_j\) is the number of times that \(j\) occurs in sample \(\bs{Y}\): \begin{align} N_i & = \sum_{j \in T} O_{i,j} \\ M_j & = \sum_{i \in S} O_{i,j} \end{align} Thus, our estimate of the expected frequency of \((i, j)\) under \(H_0\) is \[ E_{i,j} = n \, \frac{1}{n} \, N_i \frac{1}{n} \, M_j = \frac{1}{n} \, N_i \, M_j \] Of course, we define our test statistic by \[ V = \sum_{i \in J} \sum_{j \in T} \frac{(O_{i,j} - E_{i,j})^2}{E_{i,j}} \] As you now expect, the distribution of \(V\) converges to a chi-square distribution as \(n \to \infty\). But let's see if we can determine the appropriate degrees of freedom on heuristic grounds.

The observed frequencies are often recorded in a \(k \times m\) table, known as a contingency table, so that \(O_{i,j}\) is the number in row \(i\) and column \(j\). In this setting, note that \(N_i\) is the sum of the frequencies in the \(i\)th row and \(M_j\) is the sum of the frequencies in the \(j\)th column. Also, for historical reasons, the random variables \(X\) and \(Y\) are sometimes called factors and the possible values of the variables categories.

Computational and Simulation Exercises

Computational Exercises

In each of the following exercises, specify the number of degrees of freedom of the chi-square statistic, give the value of the statistic and compute the \(P\)-value of the test.

Data Analysis Exercises

Simulation Exercises

In the simulation exercises below, you will be able to explore the goodness of fit test empirically.