The Chi-Square Distribution: Goodness-of-Fit Test 1.6 2008/07/05 14:16:12 GMT-5 2008/10/27 20:41:11.182 GMT-5 Susan Dean deansusan@deanza.edu Barbara Illowsky illowskybarbara@deanza.edu Susan Dean deansusan@deanza.edu Barbara Illowsky illowskybarbara@deanza.edu Connexions cnx@cnx.org chi elementary fit good square statistics test This module describes how the chi-square distribution is used to conduct goodness-of-fit test. In this type of hypothesis test, you determine whether the data "fit" a particular distribution or not. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test (meaning the distribution for the hypothesis test is chi-square) to determine if there is a fit or not. The null and the alternate hypotheses for this test may be written in sentences or may be stated as equations or inequalities. The test statistic for a goodness-of-fit test is: Σ n ( O − E ) 2 E where: O = observed values (data) E = expected values (from theory) n = the number of different data cells or categories The observed values are the data values and the expected values are the values you would expect to get if the null hypothesis were true. There are n terms of the form ( O − E ) 2 E .The degrees of freedom are df = (number of columns - 1)(number of rows - 1).The goodness-of-fit test is almost always right tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve. Absenteeism of college students from math classes is a major concern to math instructors because missing class appears to increase the drop rate. Three statistics instructors wondered whether the absentee rate was the same for every day of the school week. They took a sample of absent students from three of their statistics classes during one week of the term. The results of the survey appear in the table. Monday Tuesday Wednesday Thursday Friday # of students absent 28 22 18 20 32 Determine the null and alternate hypotheses needed to run a goodness-of-fit test. Since the instructors wonder whether the absentee rate is the same for every school day, we could say in the null hypothesis that the data "fit" a uniform distribution. H o : The rate at which college students are absent from their statistics class fits a uniform distribution.The alternate hypothesis is the opposite of the null hypothesis. H a : The rate at which college students are absent from their statistics class does not fit a uniform distribution. How many students do you expect to be absent on any given school day? The total number of students in the sample is 120. If the null hypothesis were true, you would divide 120 by 5 to get 24 absences expected per day. The expected number is based on a true null hypothesis. What are the degrees of freedom (df)? There are 5 days of the week or 5 "cells" or categories. df = no. cells - 1 = 5 - 1 = 4 Employers particularly want to know which days of the week employees are absent in a five day work week. Most employers would like to believe that employees are absent equally during the week. That is, the average number of times an employee is absent is the same on Monday, Tuesday, Wednesday, Thursday, or Friday. Suppose a sample of 20 absent days was taken and the days absent were distributed as follows: Day of the Week Absent Monday Tuesday Wednesday Thursday Friday Number of Absences 5 4 2 3 6 For the population of employees, do the absent days occur with equal frequencies during a five day work week? Test at a 5% significance level. The null and alternate hypotheses are: H o : The absent days occur with equal frequencies, that is, they fit a uniform distribution. H a : The absent days occur with unequal frequencies, that is, they do not fit a uniform distribution. If the absent days occur with equal frequencies, then, out of 20 absent days, there would be 4 absences on Monday, 4 on Tuesday, 4 on Wednesday, 4 on Thursday, and 4 on Friday. These numbers are the expected (E) values. The values in the table are the observed (O) values or data.This time, calculate the χ 2 test statistic by hand. Make a chart with the following headings:Expected (E) values Observed (O) values ( O - E) ( O - E)2 ( O - E)2 E Now add (sum) the last column. Verify that the sum is 2.5. This is the χ 2 test statistic.To find the p-value, calculate P ( χ 2 > 2.5 ) . This test is right-tailed.The dfs are the number of cells- 1 = 4 .Next, complete a graph like the one below with the proper labeling and shading. (You should shade the right tail. It will be a "large" right tail for this example because the p-value is "large.") Use a computer or calculator to find the p-value. You should get p-value=0.6446.The decision is to not reject the null hypothesis.Conclusion: At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies.TI-83+ and TI-84: Press 2nd DISTR. Arrow down to χ2cdf. Press ENTER. Enter (2.5,1E99,4). Rounded to 4 places, you should see 0.6446 which is the p-value.TI-83+ and some TI-84 calculators do not have a special program for the test statistic for the goodness-of-fit test. The next example (Example 11-3) has the calculator instructions. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter whatever else is asked and press calculate or draw. Make sure you clear any lists before you start. See below.To Clear Lists in the calculators: Go into STAT EDIT and arrow up to the list name area of the particular list. Press CLEAR and then arrow down. The list will be cleared. Or, you can press STAT and press 4 (for ClrList). Enter the list name and press ENTER. One study indicates that the number of televisions that American families have is distributed (this is the given distribution for the American population) as follows: Number of Televisions Percent 0 10 1 16 2 55 3 11 over 3 8 The table contains expected (E) percents. A random sample of 600 families in the far western United States resulted in the following data: Number of Televisions Frequency Total = 600 0 66 1 119 2 340 3 60 over 3 15 The table contains observed (O) frequency values. At the 1% significance level, does it appear that the distribution "number of televisions" of far western United States families is different from the distribution for the American population as a whole? This problem asks you to test whether the far western United States families distribution fits the distribution of the American families. This test is always right-tailed. The first table contains expected percentages. To get expected (E) frequencies, multiply the percentage by 600. The expected frequencies are: Number of Televisions Percent Expected Frequency 0 10 ( 0.10 ) ⋅ ( 600 ) = 60 1 16 ( 0.16 ) ⋅ ( 600 ) = 96 2 55 ( 0.55 ) ⋅ ( 600 ) = 330 3 11 ( 0.11 ) ⋅ ( 600 ) = 66 over 3 8 ( 0.08 ) ⋅ ( 600 ) = 48 Therefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI calculators, you can let the calculator do the math. For example, instead of 60, enter .10*600. H o : The "number of televisions" distribution of far western United States families is the same as the "number of televisions" distribution of the American population. H a : The "number of televisions" distribution of far western United States families is different from the "number of televisions" distribution of the American population.Distribution for the test: χ 4 2 where df = (the number of cells) - 1 = 5 - 1 = 4 . df ≠ 600 − 1 Calculate the test statistic: χ 2 = 29.65 Graph: Probability statement: p-value = P ( χ 2 > 29.65 ) = 0.000006 .Compare α and the p-value: α = 0.01 p-value = 0.000006 So, α > p-value .Make a decision: Since α > p-value , reject H o .This means you reject the belief that the distribution for the far western states is the same as that of the American population as a whole.Conclusion: At the 1% significance level, from the data, there is sufficient evidence to conclude that the "number of televisions" distribution for the far western United States is different from the "number of televisions" distribution for the American population as a whole.TI-83+ and some TI-84 calculators: Press STAT and ENTER. Make sure to clear lists L1, L2, and L3 if they have data in them (see the note at the end of Example 11-2). Into L1, put the observed frequencies 66, 119, 349, 60, 15. Into L2, put the expected frequencies .10*600, .16*600, .55*600, .11*600, .08*600. Arrow over to list L3 and up to the name area "L3". Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see "sum" (Enter L3). Rounded to 2 decimal places, you should see 29.65. Press 2nd DISTR. Press 7 or Arrow down to 7:χ2cdf and press ENTER. Enter (29.65,1E99,4). Rounded to 4 places, you should see 5.77E-6 = .000006 (rounded to 6 decimal places) which is the p-value. Suppose you flip two coins 100 times. The results are 20 HH, 27 HT, 30 TH, and 23 TT. Are the coins fair? Test at a 5% significance level. This problem can be set up as a goodness-of-fit problem. The sample space for flipping two fair coins is {HH, HT, TH, TT}. Out of 100 flips, you would expect 25 HH, 25 HT, 25 TH, and 25 TT. This is the expected distribution. The question, "Are the coins fair?" is the same as saying, "Does the distribution of the coins (20 HH, 27 HT, 30 TH, 23 TT) fit the expected distribution?" Random Variable: Let X = the number of heads in one flip of the two coins. X takes on the value 0, 1, 2. (There are 0, 1, or 2 heads in the flip of 2 coins.) Therefore, the number of cells is 3. Since X = the number of heads, the observed frequencies are 20 (for 2 heads), 57 (for 1 head), and 23 (for 0 heads or both tails). The expected frequencies are 25 (for 2 heads), 50 (for 1 head), and 25 (for 0 heads or both tails). This test is right-tailed. H o : The coins are fair. H a : The coins are not fair.Distribution for the test: χ 2 2 where df = 3 - 1 = 2 .Calculate the test statistic: χ 2 = 2.14 Graph: Probability statement: p-value = P ( χ 2 > 2.14 ) = 0.3430 Compare α and the p-value: α = 0.05 p-value = 0.3430 So, α < p-value . Make a decision: Since α < p-value , do not reject H o . Conclusion: The coins are fair.TI-83+ and some TI- 84 calculators: Press STAT and ENTER. Make sure you clear lists L1, L2, and L3 if they have data in them. Into L1, put the observed frequencies 20, 57, 23. Into L2, put the expected frequencies 25, 50, 25. Arrow over to list L3 and up to the name area "L3". Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see "sum".Enter L3. Rounded to 2 decimal places, you should see 2.14. Press 2nd DISTR. Arrow down to 7:χ2cdf (or press 7). Press ENTER. Enter 2.14,1E99,2). Rounded to 4 places, you should see .3430 which is the p-value.For the newer TI-84 calculators, check STAT TESTS to see if you have Chi2 GOF. If you do, see the calculator instructions (a NOTE) before Example 11-3