Anonymous
Nov 29, 2023
Using three solved examples, explain the concept of chi-square?
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Nov 29, 2023
Example 1: Testing for Independence
Suppose we want to determine if there is a relationship between gender and smoking habits. We collect data from a sample of 200 individuals and categorize them into four groups: male smokers, male non-smokers, female smokers, and female non-smokers. The observed frequencies are as follows:
Smokers Non-smokers
Males 40 60
Females 30 70
To test for independence, we use the chi-square test. The null hypothesis is that there is no association between gender and smoking habits. The expected frequencies under the null hypothesis can be calculated by assuming independence between the two variables. After performing the calculations, we obtain a chi-square statistic of 4.17 with 1 degree of freedom. Comparing this value to the critical chi-square value, we can determine if there is a significant relationship between gender and smoking habits.
Example 2: Goodness of Fit
Suppose we want to test if the observed frequencies of different eye colors in a population match the expected frequencies based on a theoretical distribution. We collect data from a sample of 500 individuals and categorize them into four groups: blue, brown, green, and other. The observed frequencies are as follows:
Blue Brown Green Other
Observed 150 250 80 20
To test for goodness of fit, we use the chi-square test. The null hypothesis is that the observed frequencies match the expected frequencies. The expected frequencies can be calculated based on the theoretical distribution. After performing the calculations, we obtain a chi-square statistic of 12.5 with 3 degrees of freedom. Comparing this value to the critical chi-square value, we can determine if the observed frequencies significantly deviate from the expected frequencies.
Example 3: Homogeneity
Suppose we want to determine if there is a difference in the distribution of political party preferences among three different age groups: young, middle-aged, and elderly. We collect data from a sample of 300 individuals and categorize them into six groups: young Democrats, young Republicans, middle-aged Democrats, middle-aged Republicans, elderly Democrats, and elderly Republicans. The observed frequencies are as follows:
Democrats Republicans
Young 40 60
Middle-aged 50 50
Elderly 30 70
To test for homogeneity, we use the chi-square test. The null hypothesis is that the distribution of political party preferences is the same across all age groups. The expected frequencies under the null hypothesis can be calculated assuming the same distribution. After performing the calculations, we obtain a chi-square statistic of 6.67 with 2 degrees of freedom. Comparing this value to the critical chi-square value, we can determine if there is a significant difference in the distribution of political party preferences among the age groups.
Suppose we want to determine if there is a relationship between gender and smoking habits. We collect data from a sample of 200 individuals and categorize them into four groups: male smokers, male non-smokers, female smokers, and female non-smokers. The observed frequencies are as follows:
Smokers Non-smokers
Males 40 60
Females 30 70
To test for independence, we use the chi-square test. The null hypothesis is that there is no association between gender and smoking habits. The expected frequencies under the null hypothesis can be calculated by assuming independence between the two variables. After performing the calculations, we obtain a chi-square statistic of 4.17 with 1 degree of freedom. Comparing this value to the critical chi-square value, we can determine if there is a significant relationship between gender and smoking habits.
Example 2: Goodness of Fit
Suppose we want to test if the observed frequencies of different eye colors in a population match the expected frequencies based on a theoretical distribution. We collect data from a sample of 500 individuals and categorize them into four groups: blue, brown, green, and other. The observed frequencies are as follows:
Blue Brown Green Other
Observed 150 250 80 20
To test for goodness of fit, we use the chi-square test. The null hypothesis is that the observed frequencies match the expected frequencies. The expected frequencies can be calculated based on the theoretical distribution. After performing the calculations, we obtain a chi-square statistic of 12.5 with 3 degrees of freedom. Comparing this value to the critical chi-square value, we can determine if the observed frequencies significantly deviate from the expected frequencies.
Example 3: Homogeneity
Suppose we want to determine if there is a difference in the distribution of political party preferences among three different age groups: young, middle-aged, and elderly. We collect data from a sample of 300 individuals and categorize them into six groups: young Democrats, young Republicans, middle-aged Democrats, middle-aged Republicans, elderly Democrats, and elderly Republicans. The observed frequencies are as follows:
Democrats Republicans
Young 40 60
Middle-aged 50 50
Elderly 30 70
To test for homogeneity, we use the chi-square test. The null hypothesis is that the distribution of political party preferences is the same across all age groups. The expected frequencies under the null hypothesis can be calculated assuming the same distribution. After performing the calculations, we obtain a chi-square statistic of 6.67 with 2 degrees of freedom. Comparing this value to the critical chi-square value, we can determine if there is a significant difference in the distribution of political party preferences among the age groups.
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