Pearson chi-square test

If the normal approximation to the binomial distribution is valid (that is, not more than 20% of expected cell counts are less than 5) for drug therapy and symptom of headache, then you can use the Pearson chi-square test to test for a difference in proportions. To get the Pearson chi-square / -value for the preceding 2x2 table, you run SAS code like the following ... 

Here you can still use the Pearson chi-square test as shown in the 2x2 table example as long as your response variable is nominal and merely descriptive. If your response variable is ordinal, meaning that it is an ordered sequence, and you can use a parametric test, then you should use the Mantel-Haenszel test statistic for parametric tests of association. For instance, if in our previous example the variable called headache was coded as a 2 when the patient experienced extreme headache, a 1 if mild headache, and a 0 if no headache, then headache would be an ordinal variable. You can get the Mantel-Haenszel /pvalue by running the following SAS code ... 

The chi-square test for comparing two proportions or rates was developed by Karl Pearson around 1900 and pre-dates the development of the t-tests. The steps involved in the Pearson chi-square test can be set down as follows ... 

The Pearson chi-square test extends in a straightforward way when there are more than two outcome categories. 

As we shall see later the data type to a large extent determines the class of statistical tests that we undertake. Commonly for continuous data we use the t-tests and their extensions analysis of variance and analysis of covariance. For binary, categorical and ordinal data we use the class of chi-square tests (Pearson chi-square for categorical data and the Mantel-Haenszel chi-square for ordinal data) and their extension, logistic regression. 

Pearson s chi-square test is what we refer to as a large sample test this means that provided the sample sizes are fairly large then it works well. Unfortunately when the sample sizes in the treatment groups are not large there can be problems. Under these circumstances we have an alternative test, Fisher s exact test. 

In such a contingency table we would typically be interested in whether there were differences in the probabilities or rates of failures which result in the observed numbers of failures rii j,i,j =1,2,3 between the different locations or systems. Thereare two main statistical tests which could be used to do this the Chi-Squared Test and Fisher s Exact Test (Pearson 1900, Fisher 1922). 

A Pearson chi square analysis [42] was performed to evaluate the answers. The postulated null hypothesis was The perceived approaching speed is not influenced by adding additional motion cues . The answers were classified in three categories. The perceived relative speed of the following vehicle was underestimated, correctly estimated, or overestimated. The statistic test for significance (see Table 1) is performed by comparing the sum over all residues = with the value for a... 

P-values Age = Wilcoxon rank-sum, Gender = Pearson s chi-square, Race = Fisher s exact test. 

Pearson calculated the probabilities associated with values of this test statistic when the treatments are the same, to produce the null distribution. This distribution is called the chi-square distribution on one degree of freedom, denoted x i> and is displayed in Figure 4.2. Note that values close to zero have the highest probability. Values close to zero for the test statistic would only result when the Os and the s agree closely, whereas large values are unlikely when the treatments are the same. 

The test that is used more frequently than any other to check the goodness of data is the criterion (chi square), or Pearson s x test. The x test is based on the quantity... 

---