The method described in this section is commonly used in two circumstances: - Given a sample of cases that can be classified into several groups, determine if the sample is representative of the general population - Evaluate whether data resemble a particular distribution, such as a normal distribution or a geometric distribution

Example¶

Are jurors racially representative of the overall population? \(H_0\): The jurors are a random sample, i.e. there is no racial bias, and the observed counts reflect natural sampling fluctuation. \(H_A\): The jurors are not randomly sampled, i.e. there is racial bias in juror selection.

In previous hypothesis tests, we constructed a test statistic of the following form: (point estimate - null value) / SE of point estimate. In this scenario, we must construct a test statistic for each group: - White - \(Z_1 = \frac{205-198}{\sqrt{198}} = 0.50\) - Black - \(Z_2=\frac{26-19.25}{\sqrt{19.25}}=1.54\) - Hispanic - \(Z_3=\frac{25-33}{\sqrt{33}}=-1.39\) - Other - \(Z_4=\frac{19-24.75}{\sqrt{24.75}}=-1.16\) We take the squared value of each of these Z's and sum them. Using the squared value has two effects: - Any negative values are now positive - Unusual differences will become much larger and easier to notice \(Z^2_1 + Z^2_2 + Z^2_3 + Z^2_4 = 5.89 = X^2\) \(X^2\) summarizes how strongly the observed counts tend to deviate from the null counts. If the null hypothesis is true, then \(X^2\) follows a chi-square distribution.

The chi-square distribution is sometimes used to characterize data sets and statistics that are always positive and typically right skewed.

While a normal distribution has two parameters - mean and standard deviation - chi-square only has one, called degrees of freedom (df). As df gets larger, - the center becomes larger - the variability increases - the distribution is more symmetric

Finding a p-value for a chi-square distribution¶

A large \(X^2\) value suggests strong evidence favouring the alternative hypothesis (that there was racial bias). However, we cannot quantify what the chance was of observing such a large test statistic if the null hypothesis actually was true. This is where the chi-square distribution comes in. If the null hypothesis is true and there is no racial bias, \(X^2\) would follow a chi-square distribution with three degrees of freedom. Under certain conditions, \(X^2\) follows a chi-square distribution with \(k-1\) degrees of freedom, where \(k\) is the number of bins.

Definitions¶

Chi-square test for one-way table¶

Suppose we are to evaluate whether there is convincing evidence that a set of observed counts \(O_1, O_2,...O_k\) in \(k\) categories are unusually different from what might be expected under a null hypothesis. Call the expected counts that are based on the null hypothesis \(E_1,E_2,...E_k\). If each expected count is at least 5 and the null hypothesis is true, then the test statistic below follows a chi-square distribution with \(k-1\) degrees of freedom: \(\(X^2=\frac{(O_1-E_1)^2}{E_1}+\frac{(O_2-E_2)^2}{E_2}+...+\frac{(O_k-E_k)^2}{E_k}+\)\) The p-value for this test statistic is found by looking at the upper tail of this chi-square distribution. We consider the upper tail because large values of \(X^2\) would provide greater evidence against the null hypothesis.

Conditions for Chi-square test¶

There are two conditions that must be checked before performing a chi-square test: - Independence - Each case must be independent of all others in the table - Sample size/distribution - each scenario must have at least 5 expected cases