The Chi-Square Test of Independence determines whether there is an association between categorical variables (i.e., whether the variables are independent or related). It is a nonparametric test. This test is also known as:. Chi-Square Test of Association. This test utilizes a contingency table to analyze the data. A contingency table (also known as a cross-tabulation, crosstab, or two-way table) is an arrangement in which data is classified according to two categorical variables. The categories for one variable appear in the rows, and the categories for the other variable appear in columns.
Dividing a chi-square-distributed random variable by its degrees of freedom is merely rescaling; it doesn't change the shape parameter in the gamma distribution. The expected value does become the same as that of a $ chi^2_1$ distribution, but the shape of the density function is quite different. Entrance to the Seattle Underground Tour in Pioneer Square, Seattle. 'Not on par with the best deep dish in the Chi, but a respectable. Find this Pin and more on Seattle Love by FlowVella. A true look at Seattle's history can be found at the Museum of History & Industry (MOHAI).
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Each variable must have two or more categories. Each cell reflects the total count of cases for a specific pair of categories. There are several tests that go by the name 'chi-square test' in addition to the Chi-Square Test of Independence. Look for context clues in the data and research question to make sure what form of the chi-square test is being used. The Chi-Square Test of Independence is commonly used to test the following:. Statistical independence or association between two or more categorical variables. The Chi-Square Test of Independence can only compare categorical variables.
It cannot make comparisons between continuous variables or between categorical and continuous variables. Additionally, the Chi-Square Test of Independence only assesses associations between categorical variables, and can not provide any inferences about causation. If your categorical variables represent 'pre-test' and 'post-test' observations, then the chi-square test of independence is not appropriate. This is because the assumption of the independence of observations is violated. In this situation, McNemar's Test is appropriate. There are two different ways in which your data may be set up initially. The format of the data will determine how to proceed with running the Chi-Square Test of Independence.
At minimum, your data should include two categorical variables (represented in columns) that will be used in the analysis. The categorical variables must include at least two groups. Your data may be formatted in either of the following ways: If you have the raw data (each row is a subject):. Cases represent subjects, and each subject appears once in the dataset.
That is, each row represents an observation from a unique subject. The dataset contains at least two nominal categorical variables (string or numeric).
The categorical variables used in the test must have two or more categories. If you have frequencies (each row is a combination of factors): An example of using the chi-square test for this type of data can be found in the. Cases represent the combinations of categories for the variables.
Each row in the dataset represents a distinct combination of the categories. The value in the 'frequency' column for a given row is the number of unique subjects with that combination of categories. You should have three variables: one representing each category, and a third representing the number of occurrences of that particular combination of factors.
Before running the test, you must activate Weight Cases, and set the frequency variable as the weight. In SPSS, the Chi-Square Test of Independence is an option within the Crosstabs procedure. Recall that the Crosstabs procedure creates a contingency table or two-way table, which summarizes the distribution of two categorical variables. To create a crosstab and perform a chi-square test of independence, click Analyze Descriptive Statistics Crosstabs.
A Row(s): One or more variables to use in the rows of the crosstab(s). You must enter at least one Row variable. B Column(s): One or more variables to use in the columns of the crosstab(s). You must enter at least one Column variable. Also note that if you specify one row variable and two or more column variables, SPSS will print crosstabs for each pairing of the row variable with the column variables.
The same is true if you have one column variable and two or more row variables, or if you have multiple row and column variables. A chi-square test will be produced for each table.
Additionally, if you include a layer variable, chi-square tests will be run for each pair of row and column variables within each level of the layer variable. Not sure which variable should be the 'row' and which should be the 'column'? You can 'exchange' the row and column variables without affecting the results of the chi-square test of independence - the test statistic and p-value will be identical. C Statistics: Opens the Crosstabs: Statistics window, which contains fifteen different inferential statistics for comparing categorical variables.
To run the Chi-Square Test of Independence, make sure that the Chi-square box is checked off. D Cells: Opens the Crosstabs: Cell Display window, which controls which output is displayed in each cell of the crosstab. (Note: in a crosstab, the cells are the inner sections of the table. They show the number of observations for a given combination of the row and column categories.) There are three options in this window that are useful (but optional) when performing a Chi-Square Test of Independence: 1 Observed: The actual number of observations for a given cell.
This option is enabled by default. 2 Expected: The expected number of observations for that cell (see the test statistic formula).
3 Unstandardized Residuals: The 'residual' value, computed as observed minus expected. Problem Statement In the sample dataset, respondents were asked their gender and whether or not they were a cigarette smoker. There were three answer choices: Nonsmoker, Past smoker, and Current smoker. Suppose we want to test for an association between smoking behavior (nonsmoker, current smoker, or past smoker) and gender (make or female) using a Chi-Square Test of Independence (we'll use α = 0.05).
Before the Test Before we test for 'association', it is helpful to understand what an 'association' and a 'lack of association' between two categorical variables looks like. One way to visualize this is using clustered bar charts. Let's look at the clustered bar chart produced by the Crosstabs procedure. This is the chart that is produced if you use Smoking as the row variable and Gender as the column variable (running the syntax later in this example): The 'clusters' in a clustered bar chart are determined by the row variable (in this case, the smoking categories). The color of the bars is determined by the column variable (in this case, gender).
The height of each bar represents the total number of observations in that particular combination of categories. This type of chart emphasizes the differences within the categories of the row variable. Notice how within each smoking category, the heights of the bars (i.e., the number of males and females) are very similar. That is, there are an approximately equal number of male and female nonsmokers; approximately equal number of male and female past smokers; approximately equal number of male and female current smokers. If there were an association between gender and smoking, we would expect these counts to differ between groups in some way.
Running the Test. Open the Crosstabs dialog ( Analyze Descriptive Statistics Crosstabs). Select Smoking as the row variable, and Gender as the column variable. Click Statistics. Check Chi-square, then click Continue. (Optional) Check the box for Display clustered bar charts. Syntax CROSSTABS /TABLES=Smoking BY Gender /FORMAT=AVALUE TABLES /STATISTICS=CHISQ /CELLS=COUNT /COUNT ROUND CELL /BARCHART.
Output Tables The first table is the Case Processing summary, which tells us the number of valid cases used for analysis. Only cases with nonmissing values for both smoking behavior and gender can be used in the test. The next tables are the crosstabulation and chi-square test results. The key result in the Chi-Square Tests table is the Pearson Chi-Square. The value of the test statistic is 3.171. The footnote for this statistic pertains to the expected cell count assumption (i.e., expected cell counts are all greater than 5): no cells had an expected count less than 5, so this assumption was met.
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Because the test statistic is based on a 3x2 crosstabulation table, the degrees of freedom (df) for the test statistic is $$ df = (R - 1).(C - 1) = (3 - 1).(2 - 1) = 2.1 = 2 $$. The corresponding p-value of the test statistic is p = 0.205. Decision and Conclusions Since the p-value is greater than our chosen significance level ( α = 0.05), we do not reject the null hypothesis. Rather, we conclude that there is not enough evidence to suggest an association between gender and smoking. Based on the results, we can state the following:. No association was found between gender and smoking behavior ( Χ 2(2) = 3.171, p = 0.205).
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Definition 1: The chi-square distribution with k degrees of freedom, abbreviated χ 2( k), has probability density function k does not have to be an integer and can be any positive real number. For more technical details about the chi-square distribution, including proofs of some of the propositions described below. Except for the proof of Corollary 2 knowledge of calculus will be required. Observation: The chi-square distribution is the where α = k/2 and β = 2.: The χ 2( k) distribution has mean k and variance 2 k Observation: The key statistical properties of the chi-square distribution are:. Mean = k.
Median = k -2⁄3 for large k. Mode = k – 1 for k 2. Range = 0.∞). Variance = 2 k.
Skewness =. Kurtosis = 12/ k The following are the graphs of the pdf with degrees of freedom df = 5 and 10. As df grows larger the fat part of the curve shifts to the right and becomes more like the graph of a normal distribution. Figure 1 – Chart of chi-square distributions: Suppose x has standard normal distribution N(0, 1) and let x 1, x kbe k independent sample values of x, then the random variable has a chi-square distribution χ 2( k).