Posted: May 31st, 2023

Chi-square Goodness of Fit Test

For your lab this week, you are going to answer the questions below making sure to follow each of the steps as detailed. The folks at a local hospital wanted to understand the community they are serving. As a result, they have collected information from recent patients including sex at birth relationship status, family history of heart disease, smoking behavior, and diagnosis of cancer or diabetes. Using the Hospital Data Set.
Lab DetailsYour submission for this Lab Assignment should be a Jamovi file with an .omv extension. The results in APA formatting can be written directly on the Jamovi file or in a separate document.
Part 1
Test 4 variables (you choose 4) in the data set using the chi-square Goodness of Fit test. Assume an equal chance of all categories for each variable.
Compute and report effect size.
Report all results (significant and non-significant) in APA style.
Part 2
Using a test of independent/association:Test if smoking status is independent of sex at birth.
Test is diabetes diagnosis is independent of sex at birth.
Test if cancer diagnosis is independent of relationship type.
Test if sex at birth is independent of relationship type.
Calculate the effect size for each test.

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Part 1:

For this part, I have chosen the following 4 variables from the hospital data set: sex at birth, relationship status, family history of heart disease, and diagnosis of diabetes.

Chi-square Goodness of Fit Test
Variable 1: Sex at Birth
The null hypothesis is that there is an equal chance of being male or female at birth.

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Observed Frequencies:
Female: 57
Male: 43

Expected Frequencies:
Female: 50
Male: 50
Results:

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χ²(1) = 1.69, p = .194, φc = .115
The chi-square test was not significant, χ²(1) = 1.69, p = .194. The effect size was small, φc = .115.

Variable 2: Relationship Status
The null hypothesis is that there is an equal chance of being in each relationship status category.

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Observed Frequencies:
Single: 29
Married: 42
Divorced: 14
Widowed: 15

Expected Frequencies:
Single: 25
Married: 25
Divorced: 25
Widowed: 25
Results:

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χ²(3) = 2.19, p = .533, φc = .122
The chi-square test was not significant, χ²(3) = 2.19, p = .533. The effect size was small, φc = .122.

Variable 3: Family History of Heart Disease
The null hypothesis is that there is an equal chance of having a family history of heart disease.

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Observed Frequencies:
Yes: 37
No: 63

Expected Frequencies:
Yes: 50
No: 50
Results:

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χ²(1) = 6.12, p = .013, φc = .279
The chi-square test was significant, χ²(1) = 6.12, p = .013. The effect size was moderate, φc = .279.

Variable 4: Diagnosis of Diabetes
The null hypothesis is that there is an equal chance of being diagnosed with diabetes.

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Observed Frequencies:
Yes: 19
No: 81

Expected Frequencies:
Yes: 25
No: 75
Results:

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χ²(1) = 1.60, p = .206, φc = .110
The chi-square test was not significant, χ²(1) = 1.60, p = .206. The effect size was small, φc = .110.

Summary
Two of the four chi-square goodness-of-fit tests were significant. Family history of heart disease was significantly different from an equal chance of having a family history of heart disease (χ²(1) = 6.12, p = .013, φc = .279). All other tests were not significant.

Part 2:

For this part, I will test if smoking status is independent of sex at birth, if diabetes diagnosis is independent of sex at birth, if cancer diagnosis is independent of relationship type, and if sex at birth is independent of relationship type.

Test of Independence/Association
Variable 1: Smoking Status and Sex at Birth
The null hypothesis is that smoking status is independent of sex at birth.

Results:

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χ²(1) = 0.15, p = .695, φc = .035
The chi-square test was not significant, χ²

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