Section 9: Chi-Square And ANOVA Tests

Section 9chi Square And Anova Testsrhonda Knehans Drakeassociate Profe

Section 9 Chi Square and ANOVA Tests Rhonda Knehans Drake Associate Professor, New York University Data Analytics, Interpretation and Reporting • Sometimes we want to evaluate the equality of more than 2 categories or groups. • For example is the average weight loss among three diet plans the same or are they different. • We cannot do this using the methods taught in Section 7. • In Section 7 we only learned how to evaluate one or two groups for significance. • To compare more than two groups we use the following tests: 1. Chi Square Tests 2. ANOVA Tests Introduction 3 • In Section 7 we learned how to compare the percentages of two groups for similarity. • When you wish to assess or compare the percentages of more than 2 groups for similarity you will employ the Chi-Square test.

There are two types of Chi-Square tests: 1. Goodness of Fit – does the distribution of the data fit a specific pattern? 2. Test of Independence – is there any difference in the distribution of data for two or more groups? Chi Square Tests 4 Example: You sample 100 people in NY arrested for drunk driving last year and note their age. Based on this data can you conclude the proportion of people arrested is different or the same for each age group? Our formal hypothesis we are testing is: • H0: Drunk drivers are distributed equally across all age categories • H1: Drunk drivers are not distributed equally across all age categories • To decide whether to accept H0 or reject H0 in favor of H1, we calculate the good of fit “Test Statistic” as shown on the next slide.

Age + Arrested Goodness of Fit I 5 Goodness of Fit II 6 Goodness of Fit III 1. 2. 3. 4. 5. 7 • Note that our goodness of fit test does not always have to be a test for an equal distribution across categories. • For example reconsidering our drunk driving example we could have also tested that the distribution is the same as last years or is the same as in another state. • Let’s consider another example here. Goodness of Fit IV 8 Example: In 2000, American Express asked a sample of 400 women over 45 who made financial decisions in the household. In 2010, they asked 400 women again. Are the distribution of responses in 2010 the same as in 2000 or have they shifted? Goodness of Fit V 9 Goodness of Fit VI 10 • If you are instead interested in determining if there is a difference in data distributions for different groups, then we conduct a test of independence • This is the second type of Chi-Square test Test of Independence I 11 Example: A sample of 300 adults were asked if they favor giving high school teachers more freedom to punish students for acting violent. Results are shown by gender: Do the opinions differ by gender is a natural question here. Our formal hypothesis we are testing is: • H0: Opinions do not differ by gender – gender and opinion are independent of each other • H1: Opinions do differ by gender – gender and opinion are dependent on one another To decide whether to accept H0 or reject H0 in favor of H1, we calculate the “Test Statistic” associated with the test of independence.

Favor Against No Opinion Total Men Women Toal Test of Independence II 12 To calculate our test statistics we follow the steps outlined below Step 1/2: – First determine the numbers we would expect in each cell if they were independent. – If they were independent we would expect to see the same percent of men answering in favor as women, the same percent against, and so on. – We calculate this as shown below: Favor Against No Opinion Total Male If opinion is independent of gender then we would expect 60% of the males to be in favor or .60 x 175 = 105 If opinion is independent of gender then we would expect 34% of the males to be against or .34 x 175 = 59.5 If opinion is independent of gender then we would expect 6% of the males to have no opinion or .06 x 175 = 10. Toal Expected. If opinion is independent of gender then we would expect 60% of the females to be in favor or .60 x 125 = 75 If opinion is independent of gender then we would expect 34% of the females to be against or .34 x 125 = 42.5 If opinion is independent of gender then we would expect 6% of the females to have no opinion or .06 x 125 = 7. Total 180/300 = 60.0% 102/300 = 34.0% 18/300 = 6.0% 300 Test of Independence III 13 Step 2/2: We then calculate our test statistic as follows: Test of Independence IV 14 Test of Independence V 15 Let’s consider another example: A sample of students across the US were asked their GPA and if they binge drink regularly (defined as 5+ drinks at one time more than three times per week). Results are shown below by GPA. Binge Do Not Binge Total High GPA 1,260 3,588 4,848 Average GPA 2,157 4,186 6,343 Low GPA Total 3,858 8,271 12,129 Test of Independence VI 16 Let’s consider another example (continue….): • Do the percent that binge differ by quality of student? Our formal hypothesis we are testing is: – H0: Percent that binge or do not binge are the same by quality of student – showing independence – H1: Percent that binge or do not binge are not the same by quality of student – showing dependence To decide whether to accept H0 or reject H0 in favor of H1, we calculate the “Test Statistic” associated with the test of independence. Test of Independence VII 17 • To calculate our test statistics we first calculate the number of students that fall into each cell if there is no relationship between binge drinking and quality of student as shown below: Binge Do Not Binge Total High GPA If binge is independent of student quality then we would expect 31.81% of the best students to binge or .3181 x 4,848 = 1,542.05 Expected = 3,305.95 4,848 Average GPA Expected = 2,017.59 Expected = 4325.41 6,343 Low GPA Expected = 298.36 Expected = 639.64 938 Total 3,858/12129 = 31.81% 8,271/12,129 = 68.19% 12,129 Test of Independence VIII 18 Next we calculate our test statistic as follows: • We reject H0 and conclude H1 is true if the value of the excel function 1 – CHISDIST(TS Value, (R-1)(C-1)) is greater than 90%. • In this example 1-CHIDIST(189.78, 2) = 1-0 = 1 or 100% • Hence we reject H0 and conclude that the binge drinking is dependent on quality of student. Test of Independence IX 19 • In Section 7 we learned how to compare the averages or means of two groups for similarity. • When you wish to assess or compare the averages or means of more than 2 groups for similarity you will conduct an Analysis of Variance or ANOVA test. Analysis of Variance (ANOVA) I 20 Example: Fifteen fourth graders were selected and assigned to 3 groups in order to assess 3 different math teaching methods. At the end of the semester, each student was given a common math test. Results of the tests by group follow. Based on this data can you conclude any difference in the three teaching methods based on these test scores?

Our formal hypothesis we are testing is: – H0: Teaching methods do not differ – H1: There is a difference in teaching methods To decide whether to accept H0 or reject H0 in favor of H1, we calculate the “Test Statistic” associated with the ANOVA test. This entails six steps. Analysis of Variance (ANOVA) II 21 Our formal hypothesis we are testing is: – H0: Teaching methods do not differ – H1: There is a difference in teaching methods To decide whether to accept H0 or reject H0 in favor of H1, we calculate the “Test Statistic” associated with the ANOVA test. This entails six steps. Analysis of Variance (ANOVA) III 22 • Step 1: – Calculate the following for each teaching method: Analysis of Variance (ANOVA) IV 23 Analysis of Variance (ANOVA) V • Step 2: – 24 Analysis of Variance (ANOVA) VI • Step 2 (continue...) 25 Analysis of Variance (ANOVA) VII • Step 3: – 26 • Step 4: • Calculate MSB = SSB / (k – 1) where k = # of categories • Step 5: • Calculate MSW = SSW / (n – k) where n = total number of observations in study Analysis of Variance (ANOVA) VIII MSB = 492/2 MSB = 246 MSW = 2,384/12 MSW = 198. • Step 5: • Calculate the Test Statistics TS = MSB / MSW • Step 6: • Reject H0 and accept H1 that there is a difference between groups if 1 – FDIST (TS value, k – 1, n – k) is greater than 90%. Analysis of Variance (ANOVA) IX TS = 246/198.7 TS = 1. – FDIST (1.24 , 2 , 12) = 0. – 0.324 = 0.676 or 67.6% We don’t have any statistical prove to say they are different 28 First you input the data in excel Analysis of Variance (ANOVA) I • We can also do ANOVA in Excel: (Excel) 29 You then choose the ANOVA Single Factor feature within the data analysis area. Analysis of Variance (ANOVA) II (Excel) 30 You then highlight your data, check off the labels option and input where you wish to anchor the output. Analysis of Variance (ANOVA) III (Excel) 31 Your test statistic value One minus this value gives you your probability of 68% Analysis of Variance (ANOVA) IV (Excel) .1 Home Mail Corporation sells products by mail. The company’s management wants to find out if the number of orders received on each of the five days of the week is the same. The company took a sample of 400 orders received during a four-week period. The following table lists the frequency distribution for these orders by the day of the week. Conduct a goodness of fit test to determine if the distribution of orders is equally distributed by day of the week. 9.2 One hundred auto drivers, who were stopped by police for some violation, were also checked to see if they were wearing their seat belts. The following table shows the results of this survey. Conduct a test of independence to determine if seat belt usage differs by gender. The following table shows the results of a survey regarding seat belt usage by gender. Conduct a test of independence to determine if seat belt usage differs by gender.

Paper For Above instruction

Assessing Group Differences Through Chi-Square and ANOVA Tests

Statistical analysis plays a fundamental role in evaluating differences among groups in various contexts, such as understanding demographic distributions or comparing treatment effects. The Chi-Square test and Analysis of Variance (ANOVA) are essential tools used when comparing more than two categories or groups. While Chi-Square tests are primarily employed to assess relationships between categorical variables, ANOVA is utilized to compare the means of three or more groups to determine if significant differences exist.

The Chi-Square test of goodness of fit evaluates whether the observed data distribution matches a specific expected pattern. For example, if a transportation company wants to check whether the number of orders received on different days is evenly distributed, a goodness of fit test can be employed. Conversely, the Chi-Square test of independence examines whether two categorical variables, such as gender and seat belt usage, are related or independent. For example, it can determine whether seat belt usage varies significantly between males and females in a given sample.

In contrast, ANOVA compares the averages or means across three or more groups. For instance, to evaluate whether multiple teaching methods produce different student test scores, an ANOVA test can determine if at least one method results in significantly different outcomes. This test involves calculating the variability within groups and between groups and comparing these variances to determine if the observed differences are statistically significant.

Let's explore each test in detail with relevant examples.

Chi Square Goodness of Fit

This test assesses whether the distribution of observed data aligns with a predetermined distribution. For example, a survey of 100 drunk drivers' ages can be analyzed to determine if the age distribution across categories (e.g., 20-29, 30-39) fits a uniform distribution. The null hypothesis posits that the observed frequencies match the expected frequencies, while the alternative suggests they do not.

Another example involves comparing the distribution of responses from women over 45 in 2000 and 2010 to identify shifts over the decade. This type of analysis helps determine if the distribution has changed significantly over time or due to other factors.

Chi Square Test of Independence

This test examines whether two categorical variables are independent. For example, a study surveying whether opinions about punishing students are related to gender utilizes the Chi-Square test of independence. The null hypothesis states that gender and opinion are independent; the alternative states they are dependent.

The test involves calculating the expected frequencies assuming independence and comparing them with the observed frequencies. If differences are statistically significant, it suggests a relationship exists between the variables.

Analysis of Variance (ANOVA)

ANOVA tests whether the means of three or more groups differ significantly. For example, an educational researcher might compare test scores across three different teaching methods. The hypotheses are that the methods do not differ (null) and that at least one method creates a different outcome (alternative).

The process involves calculating the variance within groups and between groups to generate an F-statistic. If this statistic exceeds a critical value, it indicates significant differences among the group means.

Application and Interpretation

These statistical tests are invaluable in research and decision-making processes. For example, companies analyzing customer preferences can use Chi-Square tests to see if purchasing behaviors vary across regions or demographics. Educational institutions can use ANOVA to assess the effectiveness of various teaching strategies or curricula.

It is also important to note that proper data collection, adherence to assumptions, and accurate calculations are crucial for valid results. Software tools like Excel facilitate these analyses by providing built-in functions for Chi-Square and ANOVA tests, enhancing efficiency and accuracy.

Conclusion

In summary, understanding and properly applying Chi-Square tests and ANOVA are essential skills in statistical analysis. Each has specific use cases—Chi-Square for analyzing categorical data distributions and relationships, and ANOVA for comparing group means. Mastery of these methods enables researchers and analysts to draw meaningful conclusions, inform strategies, and support data-driven decisions in various domains.

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