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Chi-square test is a primordial technique employed by statisticians to evaluate the hypothesis concerned with an association between two variables. This article will take you through an understanding of the Chi-square test especially when used with ordinal data. In this article, we will learn the general concept of the test, the assumption placed on it and how and in what manner all these results may be understood and analyzed. Table of Content What is Chi-Square Test?Chi-square test can be described as the statistical method that is used in testing the variation between the categories’ explanations of the collected data. It is mainly applied to examine hypotheses relating to the correlation between two variables. It is used comprehensively in numerous areas of studies including marketing, political science, and biology to determine if the distribution of categorical variables deviates from what is expected by chance. Chi-Square Test FormulaThe formula for the Chi-square statistic (χ2) is:
where:
Each component represents the observed and expected frequencies for each category. It is a mathematical formula used when arriving at the test statistic which is the sum of the squared difference between the frequency that occurred, and that which was expected, all divided by the expected frequency. How to Calculate the Chi-Square StatisticThe steps to calculate Chi-Square Statistics are as follows:
Expected vs. Observed FrequenciesReferring to the Chi-square test, observed frequencies ‘O’ represent true data values gathered using a study or an experiment. On the other hand, expected frequencies ‘E’ are believed frequencies which would exist if the two variables under investigation are independent of each other. Assumptions for Chi-Square TestVarious assumption for Chi-square test includes:
Meeting these assumptions ensures the validity of the Chi-square test results. What is Ordinal Data?Ordinal data are quantitative data that can be ranked in order but the intervals between them are not fixed. The main difference between nominal data and ordinal data is that the former is characterized by ordering while the intervals between terms are irregular. Examples include:
Using Chi-Square Test with Ordinal DataTo use chi-square test with ordinal data follow the steps added below:
Challenges with ordinal data include ensuring the data’s order is considered without assuming equal intervals between categories. How to Interpret Chi-Square Test ResultsThe p-value shows the likelihood of these associations existing by mere coincidence. If the p-value is less than the specifically set significance level for instance 0.05 then the null hypothesis is rejected, implying a relationship between the two variables. Understanding Statistical SignificanceThe term statistical significance refers to the results which cannot be attributed to random occurrence. Communicate these results clearly, emphasizing their practical implications and limitations. Example of Chi-Square with Ordinal DataLet’s consider a hypothetical dataset of customer satisfaction and service quality ratings. The observed frequencies are as follows:
Solution:
ConclusionChi-square test is one of the several relevant statistic tests one can use and it is meant for dealing with categorical data hence makes it relevant with ordinal data. Therefore, after elaborating on the assumptions, calculations, and the interpretation process, you can apply this test to your data. This guide should help you grasp the concepts and perform Chi-square tests confidently. Problems on Chi-Square with Ordinal DataProblem 1: A teacher wants to determine if there is a significant association between students’ satisfaction levels and their grades in a class. The satisfaction levels are “Satisfied,” “Neutral,” and “Unsatisfied,” while the grades are “High,” “Medium,” and “Low.” The data collected is as follows:
Solution: Calculate Row Totals and Column Totals:
Calculate Expected Frequencies (E): [Tex]E_i = \frac{R_i \times C_j}{N}[/Tex] [Tex]E_{11} = \frac{(40 \times 30)}{100} = 12[/Tex] [Tex]E_{12} = \frac{(40 \times 70)}{100} = 28[/Tex] [Tex]E_{21} = \frac{(60 \times 30)}{100} = 18[/Tex] [Tex]E_{22} = \frac{(60 \times 70)}{100} = 42[/Tex] Calculate the Chi-Square Statistic (χ2): [Tex]\chi^2 = \sum \frac{(O_{ij} – E_{ij})^2}{E_{ij}}[/Tex] [Tex]\chi^2 = \frac{(10-12)^2}{12} + \frac{(30-28)^2}{28} + \frac{(20-18)^2}{18} + \frac{(40-42)^2}{42}[/Tex] [Tex]\chi^2 = \frac{4}{12} + \frac{4}{28} + \frac{4}{18} + \frac{4}{42}[/Tex] [Tex]\chi^2 = 0.33 + 0.14 + 0.22 + 0.10[/Tex] [Tex]\chi^2 = 0.79[/Tex] Degrees of Freedom: [Tex]\text{df} = (r – 1) \times (c – 1)[/Tex] [Tex]\text{df} = (2-1) \times (2-1) = 1[/Tex] Compare with Critical Value: At α=0.05 and df=4, the critical value from the Chi-square table is 9.488. Since χ2=29.119 is greater than 9.488, we reject the null hypothesis and conclude that there is a significant association between satisfaction levels and grades. Problem 2: A researcher wants to study the relationship between exercise frequency and stress levels among adults. The exercise frequency is categorized as “Never,” “Sometimes,” and “Often,” while stress levels are categorized as “Low,” “Moderate,” and “High.” The data collected is as follows:
Solution: Calculate Row Totals and Column Totals:
Calculate Expected Frequencies (E): [Tex]E_{ij} = \frac{(R_i \times C_j)}{N}[/Tex] [Tex]E_{11} = \frac{(50 \times 40)}{120} = 16.67[/Tex] [Tex]E_{12} = \frac{(50 \times 80)}{120} = 33.33[/Tex] [Tex]E_{21} = \frac{(70 \times 40)}{120} = 23.33[/Tex] [Tex]E_{22} = \frac{(70 \times 80)}{120} = 46.67[/Tex] Calculate the Chi-Square Statistic (χ2): [Tex]\chi^2 = \sum \frac{(O_{ij} – E_{ij})^2}{E_{ij}}[/Tex] [Tex]\chi^2 = \frac{(20-16.67)^2}{16.67} + \frac{(30-33.33)^2}{33.33} + \frac{(20-23.33)^2}{23.33} + \frac{(50-46.67)^2}{46.67}[/Tex] [Tex]\chi^2 = \frac{11.11}{16.67} + \frac{11.11}{33.33} + \frac{11.11}{23.33} + \frac{11.11}{46.67}[/Tex] [Tex]\chi^2 = 0.67 + 0.33 + 0.48 + 0.24[/Tex] [Tex]\chi^2 = 1.72[/Tex] Degrees of Freedom: [Tex]\text{df} = (r – 1) \times (c – 1)[/Tex] [Tex]\text{df} = (2-1) \times (2-1) = 1[/Tex] Compare with Critical Value: At α=0.05 and df=4, the critical value from the Chi-square table is 9.488. Since χ2=25.80 is greater than 9.488, we reject the null hypothesis and conclude that there is a significant association between exercise frequency and stress levels. Practice Problems on Chi-Square with Ordinal DataP1. A survey was conducted to study the association between students’ interest in different subjects (Math, Science, English) and their performance levels (High, Average, Low). The data collected is as follows:
Determine if there is a significant association between interest in subjects and performance levels using the Chi-square test. P2. A health study aims to explore the relationship between diet type (Vegetarian, Non-Vegetarian, Vegan) and cholesterol levels (Low, Medium, High). The data collected is as follows:
Use the Chi-square test to determine if there is a significant association between diet type and cholesterol levels. P3. An environmental scientist is studying the relationship between pollution levels in different areas (Urban, Suburban, Rural) and the health outcomes of residents (Good, Fair, Poor). The data collected is:
Determine if there is a significant association between area type and health outcomes using the Chi-square test. FAQs on Chi-Square Test with ordinal dataCan a chi-square test be used with ordinal data?
What data type do you need for a chi-square test nominal ordinal categorical interval?
What statistical test is used for ordinal data?
Can chi-square be used for dichotomous variables?
What are the limitations of chi-square?
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Reffered: https://www.geeksforgeeks.org
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Type: | Geek |
Category: | Coding |
Sub Category: | Tutorial |
Uploaded by: | Admin |
Views: | 26 |