Pearson chi-square

The previous sections have dealt with the t-tests, methods applicable to continuous data. We will now consider tests for binary data, where the outcome at the subject level is a simple dichotomy success/failure. In a between-patient, parallel group trial our goal here is to compare two proportions or rates. 

In Section 3.2.2 we presented data from a clinical trial comparing trastuzumab to observation only after adjuvant chemotherapy in HER2-positive breast cancer. The incidence rates in the test treatment and control groups were respectively 7.0 per cent and 4.7 per cent. 

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 null and alternative hypotheses relate to the two true rates, 0j and 02  

This value captures the evidence in support of treatment differences. If what we have seen (Os) are close to what we should have seen with equal treatments ( s) then this statistic will have a value close to zero. If, however, the Os and the s are well separated then this statistic will have a larger positive value the more the Os and the s disagree, the larger it will be. For the moment this test statistic is not in the form of a signal-to-noise ratio, but we will see later that it can be formulated in that way. 

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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 ... 

The output data set pvalue includes a variable called P PCHI that contains the Pearson chi-square p-value you need. 

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 ... 

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. 

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

Pearson Chi-Square is significant, therefore a reiation exist between these variabies 2 Scaie form 1 to 3, where 3 is risk taker Scaie from 1 to 2, where 2 is risk taker... 

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... 

The generalization of Pearson s chi-square statistic for 2x2 contingency tables, which has been discussed in Section 16.2.3, can be written as ... 

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. 

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. 

ANACONDA then calculates the value of the Pearson s chi-squared statistic and the adjusted Pearson residual values. Pearson s statistic represents a global measure of the difference between observed and expected codon frequencies (20). 

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... 

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). 

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