**Chi-Square Test or the "Goodness of Fit" Test**

**© 1998 by Dr. Thomas W. MacFarland -- All Rights Reserved**

************ chi_squa.doc ************ Background: Parametric data have exact parameters, or boundaries, to the data. For example, scores on a final examination can range only from 000 to 100. Therefore, these scores are parametric data. As opposed to parametric data, nonparametric data are data that are instead typically counted and then put into groups or categories. Using test scores again as an example, nonparametric data could be viewed as the "number of pass test scores on a C++ programming final examination," with pass defined as all scores of 70 (out of 100) or greater. With this brief background on parametric and non- parametric data, the Chi-square test is perhaps the most frequently used (if not overused) non- paramteric statistical test. The Chi-square test, named for the Greek symbol "Chi," is used to test for differences in proportions between two or more groups. You may also see the Chi-square test called a "goodness of fit" test. That is to say, the Chi- square test is used to see if grouped data "fit" into declared groups, or if the data instead do not "fit" into the group. A Chi-square test typically involves: -- the assignment of frequency data (i.e., head counts) into a 2 by 2, 2 by 3, etc., table. The following figure represents a 2 by 2 (always in the order of ROWS by COLUMNS) table, with this table composed of four cells: N Scores >= 70 N Scores < 70 ------------------------------| Male | N = 45 | N = 38 | |-----------------------------| Female | N = 42 | N = 27 | ------------------------------ -- application of the actual Chi-square formula and subsequent decision-making as to whether differences between observed and expected counts in each cell are due to chance, or if the differences are instead due to true differences between the declared groups. A typical scenario for a Chi-square test would be the organization of data into discrete categories. Imagine a situation, such as the response among male and female workers to a question that allowed "yes," "no," or "undecided" as possible responses. The data would be organized into a 2 by 3 table (i.e., 2 rows by 3 columns, consisting of 2*3 = 6 cells): Response Yes No Undecided _______________________________ | | | | Male | N = 12 | N = 07 | N = 08 | | | | | Gender |------------------------------ | | | | Female | N = 09 | N = 11 | N = 14 | |_________|_________|_________| When using Chi-square, there are a few criteria that must be observed: -- Data must be presented as frequency (i.e., counted) data, such as the number of "yes" responses to a survey statement. Please recall, however, that parametric data can be organized into categories such as "The number of students with IQ > 100" or "The number of students with IQ <= 100." -- Ideally, the number of observed frequences for each cell should be five or more. Otherwise, it may be necessary to use Yates' correction formula to account for low cell counts. -- Regardless of the organization scheme, the data must be organized in a logical manner. Scenario: In this study, the Chi-square test will be used to determine if a passing grade on a high school mathematics mastery test, administered during the senior year of high school, can be used as a later indicator of the pass/fail rate in a C++ programming course among freshmen students in a community college. In this example, Dr. Dunbar teaches at Warren County Community College. She knows from local contacts that all 60 freshmen students in her most recent C++ programming class were required to sit for a mathematics mastery test during their senior year in high school. Dr. Dunbar decides to use the Chi-square test, comparing the proportion of pass/fail mathematics mastery test scores against the proportion of pass/fail C++ end-of-term grades. Correctly, Dr. Dunbar assumes that the Chi-square test is the most appropriate statistical test for this problem since it is used to test for differences in proportions between two or more groups (i.e., mathematics mastery test pass/fail rate and C++ end-of-term pass/fail rate). After a few phone calls to high school guidance counselors, Dr. Dunbar is able to assemble the mathematics mastery test scores (pass/fail status only) of these 60 students from her prior C++ programming class. Dr. Dunbar organizes the mathematics mastery test scores (pass/fail) and end-of-term grades (pass/ fail) in her C++ programming course into a 2 by 2 table: C Programming Class Pass Fail _______________________ | | Pass | N = 31 | N = 12 | Row 1 | | | Mathematics Test |---------------------| | | | Fail | N = 09 | N = 08 | Row 2 |_____________________| Column 1 Column 2 To summarize this 2 by 2 table: -- 31 students passed the mathematics mastery test and also passed the C++ programming course -- 12 students passed the mathematics mastery test but failed the C++ programming course -- 09 students failed the mathematics mastery test but passed the C++ programming course -- 08 students failed the mathematics mastery test and also failed the C++ programming course Then, Dr. Dunbar prepares a table (Table 1) that identifies student number, pass/fail status on the mathematics mastery test, and pass/fail status for end-of-term grade in the C++ programming class. Table 1 Pass/Fail Scores For Freshmen Students at Warren County Community College: Mathematics Mastery Test Score by End-of-Term Grade in a C++ Programming Course =================================================== Math C++ ================= Student Number Pass = 1 Fail = 2 --------------------------------------------------- 01 1 1 02 2 1 03 1 1 04 1 1 05 1 2 06 1 1 07 1 2 08 2 1 09 2 1 10 1 2 11 1 1 12 1 1 13 1 1 14 2 2 15 2 2 16 1 1 17 1 1 18 1 1 19 1 2 20 1 1 21 1 2 22 1 1 23 1 2 24 1 1 25 1 1 26 1 2 27 2 2 28 1 1 29 1 1 30 1 1 31 1 2 32 1 1 33 2 2 34 2 1 35 1 1 36 1 2 37 1 1 38 1 1 39 2 2 40 1 1 41 1 1 42 2 1 43 1 2 44 1 1 45 1 1 46 1 1 47 1 2 48 2 2 49 1 1 50 2 1 51 2 1 52 1 1 53 1 1 54 2 1 55 2 2 56 1 2 57 1 1 58 2 1 59 2 2 60 1 1 --------------------------------------------------- As you review this table, be sure to ask your advisor(s) for the proper form and style for the construction of a table. I prefer to place the title "flush left," but centered headings are also common. A new topic presented in this table is the use of numerical codes for pass/fail. In this template, I will use the following codes for pass/fail: -- pass = 1 -- fail = 2 You will also see this numerical coding scheme used in this template's SPSS run file. The exact code from the SPSS run file chi_squa.r01 follows: Value Labels Pass_M 1 'Passed' 2 'Failed' / Pass_Cpp 1 'Passed' 2 'Failed' Ho: Null Hypothesis: There is no difference between the pass/fail rate for senior high school students on a mathematics mastery test and their later pass/fail rate as freshmen college students in a C++ programming course (p <= .05). Notice how the Null Hypothesis (Ho) uses p <= .05. If this term is new to you, then you should know that the term p <= .05 is used to declare that there is a five percent or less probability that the final inference (i.e., decision that there is or is not a difference between the pass/fail rate on the mathematics mastery and the later pass/fail rate in the C++ programming course) is incorrect. That is to say, there is a five percent or less probability that any inference related to differences associated with this test will be incorrect. Most inferential analyses in the social sciences are conducted at p <= .05. However, you will see some problems set at the more restrictive p <= .01. I suggest that you consult with your advisor(s) for guidance on the most appropriate level of probability to use when you conduct your own analyses. Along with the use of "p," you will also see the term "alpha" to describe the level of probability. Personally, I prefer to use "alpha," but "p" is so common that I will use this term throughout this set of templates. Files: 1. chi_squa.doc 2. chi_squa.dat 3. chi_squa.r01 4. chi_squa.o01 5. chi_squa.con 6. chi_squa.lis Command: At the Unix prompt (%), key: %spss -m < chi_squa.r01 > chi_squa.o01 ************ chi_squa.dat ************ 01 1 1 02 2 1 03 1 1 04 1 1 05 1 2 06 1 1 07 1 2 08 2 1 09 2 1 10 1 2 11 1 1 12 1 1 13 1 1 14 2 2 15 2 2 16 1 1 17 1 1 18 1 1 19 1 2 20 1 1 21 1 2 22 1 1 23 1 2 24 1 1 25 1 1 26 1 2 27 2 2 28 1 1 29 1 1 30 1 1 31 1 2 32 1 1 33 2 2 34 2 1 35 1 1 36 1 2 37 1 1 38 1 1 39 2 2 40 1 1 41 1 1 42 2 1 43 1 2 44 1 1 45 1 1 46 1 1 47 1 2 48 2 2 49 1 1 50 2 1 51 2 1 52 1 1 53 1 1 54 2 1 55 2 2 56 1 2 57 1 1 58 2 1 59 2 2 60 1 1 ************ chi_squa.r01 ************ SET WIDTH = 80 SET LENGTH = NONE SET CASE = UPLOW SET HEADER = NO TITLE = Chi-Square COMMENT = This file examines pass/fail scores on a mathematics mastery test and the potential that students who pass this mastery test have a greater chance of passing a C++ programming class than students who did not pass this mastery test DATA LIST FILE = 'chi_squa.dat' FIXED / Stu_Code 20-21 Pass_M 39 Pass_Cpp 50 Variable Lables Stu_Code "Student Code" / Pass_M "Passed the Mathematics Competency Test" / Pass_Cpp "Passed the C++ Programming Course" Value Labels Pass_M 1 'Passed' 2 'Failed' / Pass_Cpp 1 'Passed' 2 'Failed' CROSSTABS TABLES = Pass_M by Pass_Cpp / STATISTICS = CHISQ ************ chi_squa.o01 ************ 1 SET WIDTH = 80 2 SET LENGTH = NONE 3 SET CASE = UPLOW 4 SET HEADER = NO 5 TITLE = Chi-Square 6 COMMENT = This file examines pass/fail scores on 7 a mathematics mastery test and the 8 potential that students who pass this 9 mastery test have a greater chance of 10 passing a C++ programming class than 11 students who did not pass this mastery 12 test 13 DATA LIST FILE = 'chi_squa.dat' FIXED 14 / Stu_Code 20-21 15 Pass_M 39 16 Pass_Cpp 50 17 This command will read 1 records from chi_squa.dat Variable Rec Start End Format STU_CODE 1 20 21 F2.0 PASS_M 1 39 39 F1.0 PASS_CPP 1 50 50 F1.0 18 Variable Lables 19 Stu_Code "Student Code" 20 / Pass_M "Passed the Mathematics Competency Test" 21 / Pass_Cpp "Passed the C++ Programming Course" 22 23 Value Labels 24 Pass_M 1 'Passed' 25 2 'Failed' 26 27 / Pass_Cpp 1 'Passed' 28 2 'Failed' 29 30 CROSSTABS TABLES = Pass_M by Pass_Cpp 31 / STATISTICS = CHISQ Memory allows for 11,915 cells with 2 dimensions for general CROSSTABS. PASS_M Passed the Mathematics Competency Test by PASS_CPP Passed the C++ Programming Course PASS_CPP Page 1 of 1 Count | |Passed Failed | Row | 1 | 2 | Total PASS_M --------+--------+--------+ 1 | 31 | 12 | 43 Passed | | | 71.7 +--------+--------+ 2 | 9 | 8 | 17 Failed | | | 28.3 +--------+--------+ Column 40 20 60 Total 66.7 33.3 100.0 Chi-Square Value DF Significance -------------------- ----------- ---- ------------ Pearson 2.01094 1 .15617 Continuity Correction 1.24145 1 .26519 Likelihood Ratio 1.95531 1 .16202 Mantel-Haenszel test for 1.97743 1 .15966 linear association Minimum Expected Frequency - 5.667 Number of Missing Observations: 0 ************ chi_squa.con ************ Outcome: In this example, the SPSS output file chi_squa.o01 has a great deal of information. However, for your interest in the use of the Chi-square test, you only need to concentrate on the part of the printout that shows that Pearson's Chi-square value = 2.01094: Computed Chi-square = 2.01 Knowing the computed Chi-square statistic, you should then compare it to the criterion Chi- square statistic: Criterion Chi-square = 3.84 (p <= .05, df = 1) The criterion Chi-square statistic of 3.84 was gained from the table of Chi-square statistics found in the appendix of nearly all leading statistics textbooks. (Be sure to notice in your own textbook how degrees of freedom are usually placed as row values and probability/alpha levels are placed as column headers.) Using this scheme, notice how the criterion Chi-square statistic equals 3.84 with the column header of p <= .05 and the row value of 1 degree of freedom. You may need to consult your statistics textbook for background information if this topic is totally new to you. After comparing the computed Chi-square statistic to the criterion Chi-square statistic, you will notice that: Computed Chi-square (2.01) < Criterion Chi- square (3.84) Because the computed Chi-square statistic (2.01) is less than the criterion Chi-square statistic (3.84), accept the Null Hypothesis (Ho). That is to say, there is no difference between the pass/ fail rate for senior high school students on a mathematics mastery test and their later pass/fail rate as freshmen college students in a C++ programming course (p <= .05). Another way to interpret this problem, without looking up criterion Chi-square statistics in tables, is to look at the significance value (i.e., probability of significant difference) presented as part of the SPSS output file: Chi-Square Value DF Significance ---------- ----------- -- ------------ Pearson 2.01094 1 .15617 By looking at this section of the printout, you can see that the probability of significance is .15617 (approximately 16 percent), which exceeds the previously declared value of p <= .05 (i.e., probability of significant difference, which was 5 percent). In this case, 16 percent probability of significant difference (the observed probability of significance) exceeds 5 percent or less probability of significance (the declared probability of significance), so you have another measure that confirms that there is no difference (at p <= .05) in the pass/fail rate for the mathematics mastery test and the later pass/fail rate in a C++ programming course. Conclusion: Dr. Dunbar, a college-level computer science teacher, had 60 freshmen students in her C++ programming course who sat for a mathematics mastery test during the senior year in high school. Based on the set of data in this one problem, Dr. Dunbar can expect that the pass/fail rate of these students in her C++ programming class will approximate the pass/fail rate these same students achieved on the mathematics mastery test. That is to say, approximately 72 percent of all students passed the mathematics mastery test and approximately 67 percent of all students passed the C++ programming course. With p <= .05, there is no difference in these two passing rates. Review: As you review this template, be sure to give attention to the following new concepts: -- Numerical codes were used to identify pass/fail (1 = pass and 2 = fail). -- Data are often organized in 2 by 2, 3 by 4, etc. tables. As data are organized into these tables, it is standard to state these organizational schemes as rows by columns. Further, a 2 by 3 table, with two rows and three columns, has six cells (2*3 = 6). You may sometimes see the term 2 x 3 instead of 2 by 3 when referencing these tables. I suggest that you avoid "2 x 3" and instead write "2 by 3" to avoid the confusion that may come about as some readers incorrectly see the "x" character as the name for a new variable. -- The computed Chi-square statistic (2.01), provided in the SPSS output file, was compared to a criterion Chi-square statistic (3.84). -- Consult your textbook for the table(s) that identifies these criterion statistics. Be sure to notice that the criterion value is dependent on the declared p level (probability of significance, or alpha) level. Most tests in the social sciences are conducted at the 5 percent level of significance (p <= .05). You may occasionally see p <= .10 (10 percent level of significance) and p <= .01 (1 percent level of significance) in the literature. -- The SPSS printout in the output file also includes the calculated significance value (.15617). By using this statistic, you may not need to review the table value of the criterion Chi-square statistic. -- When considering significance, an important point to mention here is that there is no basis to the incorrect statement "almost significant." The p level (whether .10, .05, or .01) is determined in advance and then included in the Null Hypothesis. Whichever test you use, differences are either significant at the declared p value or they are not significant. This is a discrete activity (i.e., The light bulb is either on or it is off. The light bulb is not "almost" on or "almost" off.) The MINITAB addendum for this analysis follows, as the file chi_squa.lis. This file represents a MINITAB-based analysis of this problem in interactive mode, as opposed to the previous analysis of this problem with SPSS in batch mode. ************ chi_squa.lis ************ % minitab MTB > outfile 'chi_squa.lis' Collecting Minitab session in file: chi_squa.lis MTB > # MINITAB addendum to chi_squa.dat MTB > read 'chi_squa.dat' c1 c2 c3 Entering data from file: chi_squa.dat 60 rows read. MTB > name c1 'Stu_Code' c2 'Pass_M' c3 'Pass_Cpp' MTB > print c1 Stu_Code 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 MTB > print c2 Pass_M 1 2 1 1 1 1 1 2 2 1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 2 2 1 1 1 1 2 1 1 2 1 1 1 1 1 2 1 2 2 1 1 2 2 1 1 2 2 1 MTB > print c3 Pass_Cpp 1 1 1 1 2 1 2 1 1 2 1 1 1 2 2 1 1 1 2 1 2 1 2 1 1 2 2 1 1 1 2 1 2 1 1 2 1 1 2 1 1 1 2 1 1 1 2 2 1 1 1 1 1 1 2 2 1 1 2 1 MTB > table 'Pass_M' by 'Pass_Cpp'; SUBC> chisquare 2. ROWS: Pass_M COLUMNS: Pass_Cpp 1 2 ALL 1 31 12 43 28.67 14.33 43.00 2 9 8 17 11.33 5.67 17.00 ALL 40 20 60 40.00 20.00 60.00 CHI-SQUARE = 2.011 WITH D.F. = 1 CELL CONTENTS -- COUNT EXP FREQ MTB > stop -------------------------- Disclaimer: All care was used to prepare the information in this tutorial. Even so, the author does not and cannot guarantee the accuracy of this information. The author disclaims any and all injury that may come about from the use of this tutorial. As always, students and all others should check with their advisor(s) and/or other appropriate professionals for any and all assistance on research design, analysis, selected levels of significance, and interpretation of output file(s). The author is entitled to exclusive distribution of this tutorial. Readers have permission to print this tutorial for individual use, provided that the copyright statement appears and that there is no redistribution of this tutorial without permission. Prepared 980316 Revised 980914 end-of-file 'chi_squa.ssi'