# Test chi quadrato spss

The chi-square goodness-of-fit test is a single-sample nonparametric test, also referred to as the one-sample goodness-of-fit test or Pearson's chi-square goodness-of-fit test. It is used to determine whether the distribution of cases e. The proportion of cases expected in each group of the categorical variable can be equal or unequal e.

When you carry out a chi-square goodness-of-fit test, "hypothesising" whether you expect the proportion of cases in each group of your categorical variable to be "equal" or "unequal" is critical. Not only is it an important aspect of your research design, but from a practical perspective, it will determine how you carry out the chi-square goodness-of-fit test in SPSS Statistics, as well as how you interpret and write up your results.

In this "quick start" guide, we show you how to carry out a chi-square goodness-of-fit test using SPSS Statistics when you have "equal" expected proportions e.

In addition, we explain how to interpret the results from this test. However, if you have "unequal" expected proportions e. Therefore, assuming that you would like to know the SPSS Statistics procedure and interpretation of the chi-square goodness-of-fit test when you have equal expected proportions, you first need to understand the different assumptions that your data must meet in order for a chi-square goodness-of-fit to give you a valid result.

We discuss these assumptions next. When you choose to analyse your data using the chi-square goodness-of-fit test, part of the process involves checking to make sure that the data you want to analyse can actually be analysed using a chi-square goodness-of-fit test.

### How to Perform a Chi-Square Test of Independence in SPSS

You need to do this because it is only appropriate to use a chi-square goodness-of-fit test if your data meets four assumptions that are required for a chi-square goodness-of-fit test to give you a valid result.

In practice, checking for these assumptions is a relatively simple process, only requiring you to use SPSS Statistics. Therefore, before proceeding, check that your study design meets assumptions 1, 2 and 3.

Assuming they do, you will now need to check that your data meets assumption 4, which you can do using SPSS Statistics. We explain how to test for assumption 4 and how to interpret the SPSS Statistics output in our enhanced chi-square goodness-of-fit guide to help you get this right.

You can find out about our enhanced content as a whole on our Features: Overview page, or more specifically, learn how we help with testing assumptions on our Features: Assumptions page. In the section, Procedurewe illustrate the SPSS Statistics procedure required to perform a chi-square goodness-of-fit test assuming that no assumptions have been violated and when you have equal expected proportions. First, we set out the example we use to explain the chi-square goodness-of-fit procedure in SPSS Statistics.

A website owner, Christopher, wants to offer a free gift to people that purchase a subscription to his website. New subscribers can choose one of three gifts of equal value: a gift voucher, a cuddly toy or free cinema tickets. After people have signed up, Christopher wants to review the figures to see if the three gifts offered were equally popular. The people that have signed up reflect the "cases" i.

We have assigned codes of "1" for the gift certificate, which we labelled "Gift Certificate", "2" for the cuddly toy, which we labelled "Cuddly Toy", and "3" for the free cinema tickets, which we labelled "Cinema Tickets". If the frequency data has already been summated for the various categories, we need to create a second column that contains the respective frequency counts; we have called this variable frequency.

This type of data entry is shown below:. Note: If you have entered your data in this way, you cannot run the chi-square goodness-of-fit test without first "weighting" your cases.

It is required because it changes the way that SPSS Statistics deals with your data in order to run the chi-square goodness-of-fit test.Oddly, post hoc tests for the chi-square independence test are not widely used. This tutorial walks you through 2 options for obtaining and interpreting them in SPSS.

The data thus obtained are in edu-marit. All examples in this tutorial use this data file. So let's see if education level and marital status are associated in the first place: we'll run a chi-square independence test with the syntax below. This also creates a contingency table showing both frequencies and column percentages. Note that that SPSS wrongfully reports this 1-tailed significance as a 2-tailed significance.

But anyway, what we really want to know is precisely which percentages differ significantly from each other? These also apply to the percentages shown in the table: within each row, each possible pair of percentages is compared using a z-test. If they don't differ, they get a similar subscript. Reversely, within each row, percentages that don't share a subscript are significantly different. For example, the percentage of people with middle school who never married is This means that Which are all cells in this table row.

Now, a Bonferroni correction is applied for the number of tests within each row. This holds for all tests reported in this table. I'll verify these claims later on. The figure below suggests some basic steps.

You probably want to select both frequencies and column percentages for education level. We recommend you add totals for education levels as well. Next, our z-tests are found in the Test Statistics tab shown below. For each significant pair, the key of the category with the smaller column proportion appears in the category with the larger column proportion.

### How To Perform A Pearson’s Chi-Square Test In SPSS

Significance level for upper case letters A, B, C :. Tests are adjusted for all pairwise comparisons within a row of each innermost subtable using the Bonferroni correction.

Within each row, each possible pair of column proportions is compared using a z-test. If 2 proportions differ significantly, then the higher is flagged with the column letter of the lower. Somewhat confusingly, SPSS flags the frequencies instead of the percentages.Our tutorials reference a dataset called "sample" in many examples.

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If you'd like to download the sample dataset to work through the examples, choose one of the files below:. The Chi-Square Test of Independence determines whether there is an association between categorical variables i. It is a nonparametric test. This test utilizes a contingency table to analyze the data. A contingency table also known as a cross-tabulationcrosstabor 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. 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 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. The null hypothesis H 0 and alternative hypothesis H 1 of the Chi-Square Test of Independence can be expressed in two different but equivalent ways:.

H 0 : "[ Variable 1 ] is independent of [ Variable 2 ]" H 1 : "[ Variable 1 ] is not independent of [ Variable 2 ]". 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:. An example of using the chi-square test for this type of data can be found in the Weighting Cases tutorial.

Recall that the Crosstabs procedure creates a contingency table or two-way tablewhich summarizes the distribution of two categorical variables. 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.

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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. C Layer: An optional "stratification" variable.A chi-square independence test evaluates if two categorical variables are associated in some population. We'll therefore try to refute the null hypothesis that two categorical variables are perfectly independent in some population.

If this is true and we draw a sample from this population, then we may see some association between these variables in our sample. This is because samples tend to differ somewhat from the populations from which they're drawn. However, a strong association between variables is unlikely to occur in a sample if the variables are independent in the entire population.

If we do observe this anyway, we'll conclude that the variables probably aren't independent in our population after all. That is, we'll reject the null hypothesis of independence.

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A sample of students evaluated some course. Apart from their evaluations, we also have their genders and study majors. We'd now like to know: is study major associated with gender?

And -if so- how? Since study major and gender are nominal variableswe'll run a chi-square test to find out. In the main dialog, we'll enter one variable into the R ow s box and the other into C olumn s. Since sex has only 2 categories male or femaleusing it as our column variable results in a table that's rather narrow and high.

It will fit more easily into our final report than a wider table resulting from using major as our column variable.

Anyway, both options yield identical test results. Under S tastistics we'll just select C hi-Square. Clicking P aste results in the syntax below. You can use this syntax if you like but I personally prefer a shorter version shown below. I simply type it into the Syntax Editor window, which for me is much faster than clicking through the menu. Both versions yield identical results.

First off, we take a quick look at the Case Processing Summary to see if any cases have been excluded due to missing values. That's not the case here. With other data, if many cases are excluded, we'd like to know why and if it makes sense. Next, we inspect our contingency table. Note that its marginal frequencies -the frequencies reported in the margins of our table- show the frequency distributions of either variable separately. Since this holds, we can rely on our significance test for which we use Pearson Chi-Square.

This probability is 0.

Conclusion: we reject the null hypothesis that our variables are independent in the entire population.The binomial test is useful for determining if the proportion of people in one of two categories is different from a specified amount. For example, if we asked people to select one of two pets, either a cat or a dog, we could determine if the proportion of people who selected a cat is different from.

That is, is the proportion of people who selected a cat different from the proportion of people who selected a dog. SPSS assumes that the variable that specifies the category is numeric. In the sample data set, the PET variable corresponds to the question described above, but it is a string variable.

So we will have to recode the variable before we can perform the binomial test. If you don't remember how to automatically recode a variable, see the tutorial on transforming variables. Determine if the hypotheses are one- or two-tailed. These hypotheses are two-tailed as the null is written with an equal sign. If you do not already have View Value Labels turned on, do so if there is a check next to Value Labels when you pull down the View menu, the labels are turned on, otherwise you should click on Value Labels to turn it on.

Look at the first observation for the recoded variable: In the sample data set, the first value corresponds to a person who would select a dog as a pet. Make a note of this value as we will need it later. To perform the binomial test, select Analyze Nonparametric Tests Binomial: The Binomial dialog box appears: Select the variable of interest from the list at the left by clicking on it, and then move it into the Test Variable List by clicking on the arrow button.

In this example, I selected the variable that I automatically recoded previously PETNUM and moved it into the Test Variable List box: If the value of the first observation determined above is the same as the value in your hypothesis, then you should enter the hypothesis proportion into the Test Proportion box if it does not already contain it.

In this example, the first observation is DOG, and the hypothesis is stated in terms of CAT, so we will not perform this step. If the value of the first observation DOG in this example is not the same as the value in your hypothesis CAT in this examplethen you should enter 1 - the hypothesis proportion into the Test Proportion box if it does not already contain it.

Note: when the test proportion is. We will enter 1. The column labeled N tells us that there were 8 people who reported that they would select a cat and 38 people who reported that they would select a dog.

The Observed Prop. The next column, Test Prop. The last column, Asymp. Decide whether to reject H 0. The p value in this example is.

## Calculate and Interpret Chi Square in SPSS

Thus, we reject H 0 that the mean proportion of people who would select a cat as a pet is equal to. Chi-Squared, One-Variable Test. The chi-squared one-variable test serves a purpose similar to the binomial test, except that it can be used when there are more than two categories to the variable. Thus, if you want to determine if the number of people in each of several categories differ from some predicted values, the chi-squared one-variable test is appropriate.

For example, we could test to see the number of people primarily interested in five different areas of psychology is equal. This corresponds to the AREA variable in the sample data set. SPSS assumes that the variable that specifies the categories is numeric. In the sample data set, the AREA variable corresponds to the question described above, but it is a string variable.Suppose we want to know whether or not gender is associated with political party preference.

We take a simple random sample of voters and survey them on their political party preference. In order for the test to work correctly, we need to tell SPSS that the variables Party and Gender should be weighted by the variable Count. The first table displays the number of missing cases in the dataset. We can see that there are 0 missing cases in this example. The second table displays a crosstab of the total number of individuals by gender and political party preference.

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The third table shows the results of the Chi-Square Test of Independence. The test statistic is. The null hypothesis for the Chi-Square Test of Independence is that the two variables are independent. In this case, our null hypothesis is that gender and political party preference are independent. Since the p-value. This means we do not have sufficient evidence to say that there is an association between gender and political party preference. Your email address will not be published. Skip to content Menu.

Posted on June 4, by Zach. Step 1: Enter the data. First, enter the data in the following format: Step 2: Use weighted cases. Step 4: Interpret the results. Published by Zach. View all posts by Zach. Leave a Reply Cancel reply Your email address will not be published.These assumptions are:. I will apply the above example to explore the difference in male and female numbers between two groups control and treated.

The first variable Sex contains the information regarding the sex of the individual. In total, there are 40 individuals, 20 in each group. A new window will now appear. In it, move one of the variables into the Row s window and the other variable into the Column s window.

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Next, click the Statistics Now click the Continue button. Next, click the Cells In the new window, tick the options for RowColumn and Total under the Percentages header. This will give the percentages within each subgroup in the results output.

Click the Continue button. The first contains information regarding the number of cases involved in the test. In the Crosstabulation window, there are further descriptive information regarding the numbers and proportions in percentages of males or females, in this example, for each group.

The statistical output we are interested in can be found in the final window: Chi-Square Tests.

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There are a few figures quoted in each column, these are:. Thus, we accept the null hypothesis and reject the alternative hypothesis. Featured image credit: Quinn Dombrowski via Flickr. Save my name, email, and website in this browser for the next time I comment.

Sign in. Log into your account. Privacy Policy. Password recovery. Forgot your password? Get help. Top Tip Bio. These assumptions are: The variables of interest should be categorical data either ordinal or nominal.

There should be two or more independent groups of interest.

Below is a snapshot of some of the data within SPSS. Finally, perform the test by clicking on the OK button. There are a few figures quoted in each column, these are: Value — This is the chi-square x 2 statistic. Asymptotic Significance 2-sided — The P value for a 2-sided analysis. Exact Sig.

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