Posted: March 20th, 2023

CASE STUDY: CHI-SQUARE TEMPLATE

CASE STUDY: CHI-SQUARE TEMPLATE
INSTRUCTIONS
To successfully complete this case study, the following instructions must be adhered to. Only paste tables/figures that contain pertinent information. Report only relevant the data from tables;
as well as your response to each question. In the write-up section, all information must be
provided in paragraph format. When statistical values are discussed, the numerical value and a
reference must be made to the table(s)/figure(s) where the value(s) can be found.
Chi-Square
A Pearson’s chi-square test, known as the chi-square test, is a statistical model used to determine if there is a difference between two or more groups of categorical variables. For example, to see if the distribution of males and females differs between control and treated groups of an experiment requires a Pearson’s chi-square test.
Two general assumptions must be passed by data before using a Pearson’s chi-square test. These assumptions are:
1. The variables of interest should be categorical data (either ordinal or nominal).
2. There should be two or more independent groups.In this study, we explore the difference in male and female athletes and type of injury. We will also determine if more male or female athletes require surgery. Lastly, we will explore whether injury is related to future disability.
The null hypotheses:
“There is no difference in proportion of male and female athletes who experience chronic
injury.”
“There is no difference in proportion of male and female athletes who experience acute
injury.”
“There is no difference in proportion of male and female athletes who require surgery for
their injury.”
“There is no difference in relationship between chronic injury and disability.”
“There is no difference in relationship between acute injury and disability.”
*Disability is defined as limitations in day-to-day life.
Chi-Square in SPSS
Input data, define groups etc.
Run the analysis.
Paste the Descriptive Statistics table.
Descriptive Statistics
N Minimum Maximum Mean Std. Deviation
Sex 20 1.00 2.00 1.5000 .51299
Surgery 20 1.00 2.00 1.4000 .50262
Disability 20 1.00 2.00 1.5000 .51299
Injury 20 1.00 2.00 1.3500 .48936
Valid N (listwise) 20
In Crosstabulation, paste the information on percentages.
Case Processing Homework help – Summary
Cases
Valid Missing Total
N Percent N Percent N Percent
Sex * Injury 20 100.0% 0 0.0% 20 100.0%
Sex * Disability 20 100.0% 0 0.0% 20 100.0%
Sex * Surgery 20 100.0% 0 0.0% 20 100.0%
Paste the final output – the Chi-Square Tests.
Chi-Square Tests
Value df Asymptotic Significance (2-sided) Exact Sig. (2-sided) Exact Sig. (1-sided)
Pearson Chi-Square .220a 1 .639
Continuity Correctionb .000 1 1.000
Likelihood Ratio .220 1 .639
Fisher’s Exact Test 1.000 .500
Linear-by-Linear Association .209 1 .648
N of Valid Cases 20
a. 2 cells (50.0%) have expected count less than 5. The minimum expected count is 3.50.
b. Computed only for a 2×2 table
Chi-Square Tests
Value df Asymptotic Significance (2-sided) Exact Sig. (2-sided) Exact Sig. (1-sided)
Pearson Chi-Square .800a 1 .371
Continuity Correctionb .200 1 .655
Likelihood Ratio .805 1 .369
Fisher’s Exact Test .656 .328
Linear-by-Linear Association .760 1 .383
N of Valid Cases 20
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 5.00.
b. Computed only for a 2×2 table
Chi-Square Tests
Value df Asymptotic Significance (2-sided) Exact Sig. (2-sided) Exact Sig. (1-sided)
Pearson Chi-Square .000a 1 1.000
Continuity Correctionb .000 1 1.000
Likelihood Ratio .000 1 1.000
Fisher’s Exact Test 1.000 .675
Linear-by-Linear Association .000 1 1.000
N of Valid Cases 20
a. 2 cells (50.0%) have expected count less than 5. The minimum expected count is 4.00.
b. Computed only for a 2×2 table
Phi and Cramér’s V are measures of the strength of association between two categorical variables. Phi is used with 2 × 2 contingency tables. You have two categorical variables-each variable has only two categories. Phi is calculated by taking the chi-square value and dividing it by the sample size and then taking the square root of this value. If one of the two categorical variables contains more than two categories, Cramér’s V is preferred to phi because phi fails to reach its minimum value of 0 (indicating no association) in these circumstances.
Goodman and Kruskal’s lambda (?)- the statistic measures the proportional reduction in error that is achieved when membership of a category of one variable is used to predict category membership of the other variable. A value of 1 means that one variable perfectly predicts the other, whereas a value of 0 indicates that one variable in no way predicts the other.
Case Study-Chi Square
Research questions: See Above hypotheses
1. Data level of measurement – what is the level of measurement for the data used in this case study?
2. Are we analyzing means, proportions, correlation, or regression? Why and what does that mean?
3. What test of assumption should we use? Why?
4. Will we have outliers?
5. What post-hoc analysis should we use? Why?
Paste Crosstabs for each research question.
Crosstab
Injury Total
1.00 2.00
Sex 1.00 Count 6 4 10
% within Sex 60.0% 40.0% 100.0%
% within Injury 46.2% 57.1% 50.0%
% of Total 30.0% 20.0% 50.0%
2.00 Count 7 3 10
% within Sex 70.0% 30.0% 100.0%
% within Injury 53.8% 42.9% 50.0%
% of Total 35.0% 15.0% 50.0%
Total Count 13 7 20
% within Sex 65.0% 35.0% 100.0%
% within Injury 100.0% 100.0% 100.0%
% of Total 65.0% 35.0% 100.0%
Crosstab
Disability Total
1.00 2.00
Sex 1.00 Count 6 4 10
% within Sex 60.0% 40.0% 100.0%
% within Disability 60.0% 40.0% 50.0%
% of Total 30.0% 20.0% 50.0%
2.00 Count 4 6 10
% within Sex 40.0% 60.0% 100.0%
% within Disability 40.0% 60.0% 50.0%
% of Total 20.0% 30.0% 50.0%
Total Count 10 10 20
% within Sex 50.0% 50.0% 100.0%
% within Disability 100.0% 100.0% 100.0%
% of Total 50.0% 50.0% 100.0%
Crosstab
Surgery Total
1.00 2.00
Sex 1.00 Count 6 4 10
% within Sex 60.0% 40.0% 100.0%
% within Surgery 50.0% 50.0% 50.0%
% of Total 30.0% 20.0% 50.0%
2.00 Count 6 4 10
% within Sex 60.0% 40.0% 100.0%
% within Surgery 50.0% 50.0% 50.0%
% of Total 30.0% 20.0% 50.0%
Total Count 12 8 20
% within Sex 60.0% 40.0% 100.0%
% within Surgery 100.0% 100.0% 100.0%
% of Total 60.0% 40.0% 100.0%
To report the analysis of a Pearson’s chi-square test, it is often useful to present the chi-square statistic, the degrees of freedom and the p-value. A breakdown of the reporting style for a Pearson’s chi-square test: There was no significant difference in male or female proportions between the control and treatment group (x2 (1) = 3.34; p=.057), where degrees of freedom and
the chi-square value are reported prior to p.
Ace my homework – Write-up- Use these questions as a guide to write an abstract (paragraph). Use past tense.
The write-up must include the following. Use these questions as a guide.
1. What is the research question?
2. When would we use Chi-square and why?
3. What is the Pearson Chi-Square, and is it significant? How do you know?
4. What is Fisher’s exact test? Is it relevant/significant?
5. What do the statistical findings of the data suggest about the research question?
6. What are your thoughts on the finding of this analysis, and what are the practical application(s) of this finding? Elaborate on what the study means, why, how you know, and relate the findings to exercise physiology knowledge. Use references if needed.
(Hint- does this study “make sense?” What could influence results?)

In this case study, we aim to explore the difference in male and female athletes and types of injuries, determine whether more male or female athletes require surgery, and investigate the relationship between injury and future disability. We will use a Pearson’s chi-square test to determine if there is a difference between two or more groups of categorical variables. The variables of interest should be categorical data (either ordinal or nominal), and there should be two or more independent groups.

The descriptive statistics table shows the mean and standard deviation of the variables. The table indicates that the number of males and females in the study is equal (Mean = 1.5), with a standard deviation of 0.51. The same applies to the number of athletes requiring surgery and disability (Mean = 1.4 and Mean = 1.5, respectively), with relatively small standard deviations. The table indicates that all 20 cases are valid and have no missing data.

The cross-tabulation shows that all 20 cases are valid, and no data is missing. To investigate the difference in proportion of male and female athletes who experience chronic injury, acute injury, or require surgery, we use a chi-square test. The results indicate that there is no significant difference in the proportion of males and females who experience chronic injury (p=0.639) or require surgery (p=1.000). However, there is no significant difference in the proportion of males and females who experience acute injury (p=0.371).

To investigate the relationship between injury and future disability, we use a chi-square test. The results indicate that there is no significant difference in the relationship between chronic injury and disability (p=1.000), nor is there a significant difference in the relationship between acute injury and disability (p=0.675).

In conclusion, the results of this study show that there is no significant difference in the proportion of male and female athletes who experience chronic injury or require surgery. However, there is a significant difference in the proportion of males and females who experience acute injury. The study also found that there is no significant difference in the relationship between injury and future disability. Overall, these findings suggest that gender does not play a significant role in the incidence of chronic injury and surgery for athletes, but may play a role in acute injury.

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