# 10 Chi-Square Test

The *Chi-Square Test* applies to *nominal/categorical* variables. It has two forms:

- A one-way classification to estimate how close an observed distribution is to an expected distribution. This type of test is called a
*goodness-of-fit test*. - A two-way classification to estimate whether two variables from the same population are related or not. This type of test is called a
*contingency test*or*test of independence*.

The dataset described below will be used to exemplify both types of analyses. It contains all hospital discharges in New York State in 1993 (12,844 records) for patients admitted with an Acute Myocardial Infarction (heart attack) who did not have surgery. The data is defined as shown in Table 10.1.

Table 10.1: Variables in the heart attack data set

Variable | Explanation |
---|---|

age | Patient age in years |

gender | Patient’s gender, coded M for males and F for females |

diagnosis | Code based on ICD classification |

drg | Diagnosis Related Group (121 - with complications who did not die; 122 - without complications who did not die; 123 - patients who died) |

los | Hospital length of stay |

died | 1 if the patient died in the hospital; 0 if not |

charges | $ amount of hospital charges |

Let’s take a look at how the data looks like by listing the first few rows of the dataset (Table 10.2).

Table 10.2: First few rows of the heart attack data set

diagnosis | gender | drg | died | charges | los | age |
---|---|---|---|---|---|---|

41041 | F | 122 | 0 | 4752 | 10 | 79 |

41041 | F | 122 | 0 | 3941 | 6 | 34 |

41091 | F | 122 | 0 | 3657 | 5 | 76 |

41081 | F | 122 | 0 | 1481 | 2 | 80 |

41091 | M | 122 | 0 | 1681 | 1 | 55 |

41091 | M | 121 | 0 | 6379 | 9 | 84 |

## 10.1 Goodness-of-Fit Test

Looks at a single categorical variable from a population and attempts to assess how close to or consistent with a hypothesized distribution the actual distribution of that variable is.

*Null hypothesis (H _{0})*: The data follows the

*theoretical*distribution.

Before running the analysis, let’s look at a basic descriptive statistic, frequency analysis.

Table 10.3: Gender variable frequencies

Gender | Freq |
---|---|

F | 5065 |

M | 7779 |

Running the chi-square goodness-of-fit test for the *gender* variable will attempt to test for equal counts in every cell of the design.

```
Chi-squared test for given probabilities
data: table(my.ha$gender)
X-squared = 570, df = 1, p-value <2e-16
```

Given the small *p-value*, the null hypothesis can be rejected and it can be considered that the alternative is true.

Example of how to write it up:

*The null hypothesis stating that patients are equally distributed across genders, \(\chi^{2}(df = 1) = 573.48, p < 0.001\) is rejected. Therefore, there are significant differences between the expected number of patients for each gender and the observed number.*

Observed frequencies can help explain the direction of the relationship for the possible underlying explanation or reason. This explanation can be added to the text if relevant for the study.

## 10.2 Test of Independence

In this case the *Chi-Square Test* is used to determine if a significant relationship exists between two categorical variables. The test compares the frequency of each category of one of the variables to the frequency for each of the categories of the other variable.

In this example we will try to find if there number of patients deaths is in some way related to their gender. Therefore, we will be testing the null hypothesis:

*The null hypothesis (H _{0})*: There is no relationship between

*gender*and the

*patient dying*in the hospital.

Let’s look at some cell counts first. An idea about how these look like will be useful later when compared with expected cell counts computed from the statistical test.

Table 10.4: Gender count and Died (0 = no, 1 = yes) by gender variable counts

0 | 1 | |
---|---|---|

F | 4298 | 767 |

M | 7136 | 643 |

Now let’s run the test:

```
Pearson's Chi-squared test with Yates' continuity correction
data: table(my.ha$gender, my.ha$died)
X-squared = 150, df = 1, p-value <2e-16
```

The obtained *p-value* < .05 indicates that the null hypothesis (indicating independence) can be safely rejected and the alternative hypothesis (indicating the existence of a relationship between *gender* and *died*) should be accepted.

If the null hypothesis were true, the expected counts are shown in Table 10.5, compared to the observed frequencies shown in Table 10.4.

Table 10.5: Expected frequencies

0 | 1 | |
---|---|---|

F | 4509 | 556 |

M | 6925 | 854 |

A mosaic plot (Figure 10.1) can help visualize the relative cell sizes.

Now let’s look at a chi-square analysis when at least one of the variables has more than one level. In this case we will be using the *drg* variable89 The *drg* (Diagnosis Related Group) has three levels: 121 - with complications who did not die; 122 - without complications who did not die; 123 - patients who died..

The cell counts in this case are shown in Table 10.6.

Table 10.6: Gender count and Diagnostic Related Group variable counts

121 | 122 | 123 | |
---|---|---|---|

F | 2328 | 1970 | 767 |

M | 3059 | 4077 | 643 |

Running the test returns the results shown below.

```
Pearson's Chi-squared test
data: table(my.ha$gender, my.ha$drg)
X-squared = 280, df = 2, p-value <2e-16
```

And the mosaic plot is presented in Figure 10.2.

Table 10.7 shows the expected frequencies when the null hypothesis (H_{0}) is true. Compare them with the observed frequencies shown in Table 10.6.

Table 10.7: Expected frequencies

121 | 122 | 123 | |
---|---|---|---|

F | 2124 | 2385 | 556 |

M | 3263 | 3662 | 854 |

Example of how to write it up:

*The null hypothesis stating that there is no relationship between gender and diagnosis related group classification, \(\chi^{2}(df = 2) = 283.43, p < 0.001\) is rejected. Therefore, the differences between the number of males and females in the different diagnosis related group are significant.*