Chi-square test of independence by hand

Performing the Chi-square test of independence manually

Introduction:

Chi-square tests of independence are used to determine whether there is a relationship between two categorical variables. This article provides a step-by-step guide on how to perform a Chi-square test of independence by hand, with an example involving smoking and being a professional athlete. The test compares observed frequencies to expected frequencies based on the assumption of independence. The test statistic, denoted (chi^2), is computed by summing the squared differences between observed and expected frequencies. The test statistic is then compared to a critical value from the Chi-square distribution to determine the significance of the relationship. This article also highlights the assumptions and conditions necessary for conducting the Chi-square test of independence.

Full Article: Performing the Chi-square test of independence manually

Chi-Square Tests of Independence: Testing Relationships Between Categorical Variables

A Chi-Square test of independence is a statistical test used to determine if there is a relationship between two categorical variables. This test is used to assess whether the values of one variable depend on the values of another variable. In simpler terms, it helps to establish if knowing the value of one variable provides any information about the value of another variable.

Performing a Chi-Square test of independence by hand and interpreting the results can be a bit intricate, so let’s dive into it using a concrete example. But before that, let’s understand the null and alternative hypotheses associated with this test.

The null hypothesis ((H_0)) states that the two variables are independent, meaning there is no relationship between them, and knowing the value of one variable does not assist in predicting the value of the other variable. On the other hand, the alternative hypothesis ((H_1)) suggests that the variables are dependent, implying that there is a relationship between them, and knowing the value of one variable helps predict the value of the other variable.

To examine this, we compare the observed frequencies in our sample to the expected frequencies under the assumption of independence. If the difference between the observed and expected frequencies is small, we cannot reject the null hypothesis. Conversely, if the difference is large, we can reject the null hypothesis and conclude that the variables are related. This threshold value, known as the critical value, depends on the significance level ((alpha)) and the degrees of freedom, which we will discuss in detail later.

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Now, let’s consider an example to demonstrate the Chi-Square test of independence. Suppose we want to determine if there is a statistically significant association between smoking and being a professional athlete. Both smoking and professional athlete categories have only two options, “yes” or “no.” We collected data on 28 individuals, and now we’ll analyze it using a contingency table.

The contingency table below summarizes the data, displaying the number of people in each subgroup and the totals by row, column, and overall.

Athlete: Yes No
Smoking: 14 4 18
Non-Smoking: 0 10 10
Total: 14 14 28

To conduct the Chi-Square test of independence, we need to calculate the expected frequencies assuming independence. We compute these expected frequencies for each subgroup using the formula

[ text{expected frequencies} = frac{text{total number of observations for the row} cdot text{total number of observations for the column}}{text{total number of observations}} ]

Based on the observed frequencies in the table above, we can compute the expected frequencies for each subgroup as follows:

Athlete: Yes No
Smoking: 9 9 18
Non-Smoking: 5 5 10
Total: 14 14 28

It’s important to note that the Chi-Square test of independence should only be performed when the expected frequencies in all groups are equal to or greater than 5. In our example, this assumption holds true as the minimum expected frequency is 5. If this condition is not met, it is recommended to use Fisher’s exact test instead.

Furthermore, this test assumes that the observations are independent. Although this assumption is not explicitly tested, it can be assumed if one observation does not have an impact on another. In the case of dependent observations, such as paired samples, different tests like McNemar’s or Cochran’s Q tests should be employed.

After calculating the observed and expected frequencies, we need to compare them to determine if they differ significantly. The difference between the observed and expected frequencies, known as the test statistic (or (t)-stat), is denoted as (chi^2) and is computed using the formula:

[ chi^2 = sum_{i, j} frac{(O_{ij} – E_{ij})^2}{E_{ij}} ]

Here, (O) represents the observed frequencies and (E) represents the expected frequencies. By squaring the differences, we ensure that negative differences don’t cancel out positive differences. We can apply this formula to our example:

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In the subgroup of athletes and non-smokers: (frac{(14-9)^2}{9} = 2.78)

In the subgroup of non-athletes and non-smokers: (frac{(0-5)^2}{5} = 5)

In the subgroup of athletes and smokers: (frac{(4-9)^2}{9} = 2.78)

In the subgroup of non-athletes and smokers: (frac{(10-5)^2}{5} = 5)

Finally, we sum up these values to obtain the test statistic:

[ chi^2 = 2.78 + 5 + 2.78 + 5 = 15.56 ]

The test statistic alone is insufficient to determine independence or dependence between the variables. To interpret the test statistic, it needs to be compared to the critical value. The critical value can be found in the statistical table of the Chi-Square distribution and depends on the significance level ((alpha)) and degrees of freedom. The calculation of degrees of freedom is beyond the scope of this article, but it is necessary for finding the critical value.

If the test statistic is greater than the critical value, it implies that the probability of observing such a difference between the observed and expected frequencies is low. Conversely, if the test statistic is smaller than the critical value, it suggests that the probability of observing such a difference is relatively high. In the latter case, we cannot reject the hypothesis of independence, while in the former case, we can conclude that a relationship exists between the variables.

In conclusion, the Chi-Square test of independence is a powerful tool for analyzing relationships between categorical variables. By performing this test and interpreting the results correctly, we can gain valuable insights into the association between different variables and make informed decisions based on these findings.

Note: To learn how to perform a Chi-Square test of independence using R, refer to the article “Chi-Square test of independence in R.”

Summary: Performing the Chi-square test of independence manually

The chi-square test of independence is used to determine whether two categorical variables are independent or dependent. This article explains how to perform the test by hand and interpret the results, using a specific example of smoking and being a professional athlete as the variables of interest. The test compares observed frequencies to expected frequencies under the assumption of independence. If the difference between the observed and expected frequencies is large, it suggests a relationship between the variables. The test statistic is then compared to a critical value to make a conclusion. The article also discusses assumptions and provides a formula for calculating the test statistic.

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