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Chi-square with Ordinal Data

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.

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 Formula

The formula for the Chi-square statistic (χ2) is:

[Tex]\chi^2 = \sum \frac{(O_i – E_i)^2}{E_i}[/Tex]

where:

  • Oi is Observed Frequency
  • Ei is Expected frequency

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 Statistic

The steps to calculate Chi-Square Statistics are as follows:

Step 1: Define Hypotheses: Squash your null hypothesis (H0) and alternative hypotheses (H1).

Step 2: Create a Contingency Table: Tabulate observed frequencies.

Step 3: Calculate Expected Frequencies: For each cell, use [Tex]E_i = \frac{(\text{row total} \times \text{column total})}{\text{grand total}}[/Tex]

Step 4: Determine χ2: Using the above-mentioned formula ascertain the corresponding value of Chi-square for the categorical data.

Step 5: Determine Degrees of Freedom: The following should be applied as ((r–1)X(c–1)) where r is the number of rows & c is the number of columns.

Step 6: Compare with Critical Value: Here it is necessary to check the obtained Chi-square statistic against the Chi-square critical value from distribution table.

Expected vs. Observed Frequencies

Referring 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 Test

Various assumption for Chi-square test includes:

  • Independence: Observations must be independent of each other.
  • Expected Frequency: Each expected frequency should be at least 5.
  • Categorical Data: Data must be in categorical form.

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:

  • Customer satisfaction ratings (e.g., Excellent, Good, Fair, Poor)
  • Education levels (e.g., High School, Bachelor’s, Master’s).

Using Chi-Square Test with Ordinal Data

To use chi-square test with ordinal data follow the steps added below:

Step 1: Organize Data: Create a contingency table with ordinal categories.

Step 2: Compute Expected Frequencies: As before, calculate for each cell.

Step 3: Calculate Chi-Square Statistic: Use the same formula.

Step 4: Evaluate Results: Check the Chi-square value against the critical value.

Challenges with ordinal data include ensuring the data’s order is considered without assuming equal intervals between categories.

How to Interpret Chi-Square Test Results

The 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 Significance

The 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 Data

Let’s consider a hypothetical dataset of customer satisfaction and service quality ratings. The observed frequencies are as follows:


High

Medium

Low

Satisfied

30

20

10

Neutral

20

30

10

Unsatisfied

10

20

30

Solution:

Calculate Row Totals, Column Totals, and Grand Total:

[Tex]\begin{array}{|c|c|c|c|c|} \hline & \text{High} & \text{Medium} & \text{Low} & \text{Row Total} \\ \hline \text{Satisfied} & 30 & 20 & 10 & 60 \\ \hline \text{Neutral} & 20 & 30 & 10 & 60 \\ \hline \text{Unsatisfied} & 10 & 20 & 30 & 60 \\ \hline \text{Column Total} & 60 & 70 & 50 & 180 \\ \hline \end{array}[/Tex]

Calculate Expected Frequencies (E):

Expected frequency for each cell can be calculated using the formula:

[Tex]E = \frac{(\text{Row Total} \times \text{Column Total})}{\text{Grand Total}} [/Tex]

For example, the expected frequency for “Satisfied” and “High”:

[Tex]E_{\text{Satisfied, High}} = \frac{(60 \times 60)}{180} = 20 [/Tex]

Here are the expected frequencies for all cells:

[Tex]\begin{array}{|c|c|c|c|} \hline & \text{High} & \text{Medium} & \text{Low} \\ \hline \text{Satisfied} & 20 & 23.33 & 16.67 \\ \hline \text{Neutral} & 20 & 23.33 & 16.67 \\ \hline \text{Unsatisfied} & 20 & 23.33 & 16.67 \\ \hline \end{array}[/Tex]

Calculate Chi-Square Statistic (χ2):

The Chi-square statistic is calculated using the formula:

[Tex]\chi^2 = \sum \frac{(O_i – E_i)^2}{E_i} [/Tex]

Let’s calculate χ2:

[Tex]\chi^2 = \frac{(30 – 20)^2}{20} + \frac{(20 – 23.33)^2}{23.33} + \frac{(10 – 16.67)^2}{16.67} \\ \quad + \frac{(20 – 20)^2}{20} + \frac{(30 – 23.33)^2}{23.33} + \frac{(10 – 16.67)^2}{16.67} \\ \quad + \frac{(10 – 20)^2}{20} + \frac{(20 – 23.33)^2}{23.33} + \frac{(30 – 16.67)^2}{16.67} \\ = \frac{(10)^2}{20} + \frac{(-3.33)^2}{23.33} + \frac{(-6.67)^2}{16.67} \\ \quad + \frac{(0)^2}{20} + \frac{(6.67)^2}{23.33} + \frac{(-6.67)^2}{16.67} \\ \quad + \frac{(-10)^2}{20} + \frac{(-3.33)^2}{23.33} + \frac{(13.33)^2}{16.67} \\ = \frac{100}{20} + \frac{11.09}{23.33} + \frac{44.49}{16.67} \\ \quad + \frac{0}{20} + \frac{44.49}{23.33} + \frac{44.49}{16.67} \\ \quad + \frac{100}{20} + \frac{11.09}{23.33} + \frac{177.69}{16.67} \\ = 5 + 0.48 + 2.67 \\ \quad + 0 + 1.91 + 2.67 \\ \quad + 5 + 0.48 + 10.66 \\ = 5 + 0.48 + 2.67 + 0 + 1.91 + 2.67 + 5 + 0.48 + 10.66 \\ = 28.37[/Tex]

Degrees of Freedom:

The degrees of freedom (df) are calculated using:

df = (r-1) × (c-1)

For our example:

df = (3-1) × (3-1) = 2 × 2 = 4

Compare with Critical Value:

We compare the calculated χ2 value with the critical value from the Chi-square distribution table at a specific significance level (e.g., 0.05). For df = 4, the critical value at 0.05 significance level is approximately 9.488.

Since χ2 = 28.37 is greater than 9.488, we reject the null hypothesis, suggesting a significant association between customer satisfaction and service quality.

Conclusion

Chi-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 Data

Problem 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:


High

Medium

Low

Satisfied

30

20

10

Neutral

20

30

10

Unsatisfied

10

20

30

Solution:

Calculate Row Totals and Column Totals:


High

Medium

Low

Row Total

Satisfied

30

20

10

60

Neutral

20

30

10

60

Unsatisfied

10

20

30

60

Column Total

60

70

50

180

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:


Low

Moderate

High

Never

25

15

10

Sometimes

20

30

15

Often

10

25

35

Solution:

Calculate Row Totals and Column Totals:


High

Medium

Low

Row Total

Never

25

15

10

50

Sometimes

20

30

15

65

Often

10

25

35

70

Column Total

55

70

60

185

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 Data

P1. 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:


High

Average

Low

Math

30

40

30

Science

20

30

50

English

10

20

70

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:


Low

Medium

High

Vegetarian

25

35

20

Non-Vegetarian

15

25

40

Vegan

20

30

10

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:


Good

Fair

Poor

Urban

30

20

50

Suburban

40

30

30

Rural

50

40

20

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 data

Can a chi-square test be used with ordinal data?

Yes, it is possible to conduct a chi-square test to check the connection with the ordinal level of data when testing two nominal variables. However, it appears to disregard the sequence in which the data is received.

What data type do you need for a chi-square test nominal ordinal categorical interval?

Whenever, performing a chi-square test, one is bound to use data which is in the form of categories though it could be nominal data, ordinal data or a dichotomous data.

What statistical test is used for ordinal data?

For ordinal data, the test could be the Mann-Whitney U test or the Kruskal-Wallis test since the arrangement of the data tends to be of importance.

Can chi-square be used for dichotomous variables?

Yes, the chi-square test can be used on dichotomous variables to find if there is an association between two binary variables.

What are the limitations of chi-square?

The chi-square test requires large sample sizes, assumes expected frequencies of at least 5, and does not consider the order of ordinal data.




Reffered: https://www.geeksforgeeks.org


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