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Chi-Square Predition Text
Chi-Square Test for Goodness of Fit - Chi-Square Goodness of Fit Test: Used when: Test method. Use the chi-square goodness of fit test to determine whether observed sample frequencies differ significantly from expected frequencies specified in the null hypothesis. -1 categorical variable -1- population -Appropriate when these conditions are met: -The sampling method is simple random sampling. -The population is at least 10 times as large as the sample. -The variable under study is categorical. -The expected value of the number of sample observations in each level of the variable is at least 5. (Q): Hypothesis Test (S): Sample Test (T): 1 categorical variable (N): 1 sample from 1 population Hypothesis: That is, if one is true, the other must be false; and vice versa.
H0: The data are consistent with a specified distribution. Ha: The data are not consistent with a specified distribution. Analyze Sample Data Using sample data, find the degrees of freedom, expected frequency counts, test statistic, and the P-value associated with the test statistic. Degrees of freedom: k- is equal to the number of levels (k) of the categorical variable minus Expected frequency counts. The expected frequency counts at each level of the categorical variable are equal to the sample size times the hypothesized proportion from the null hypothesis Ei = npi where Ei is the expected frequency count for the ith level of the categorical variable, n is the total sample size, and pi is the hypothesized proportion of observations in level i. Test statistic. The test statistic is a chi-square random variable (Χ2) defined by the following equation. Χ2 = Σ [ (Oi - Ei)2 / Ei ] where Oi is the observed frequency count for the ith level of the categorical variable, and Ei is the expected frequency count for the ith level of the categorical variable. P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a chi-square, use the Chi-Square Distribution Calculator to assess the probability associated with the test statistic. Use the degrees of freedom computed above.
The sampling method is simple random sampling. Each population is at least 10 times as large as its respective sample. The variables under study are each categorical. If sample data are displayed in a contingency table, the expected frequency count for each cell of the table is at least 5. This approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. State the Hypotheses Suppose that Variable A has r levels, and Variable B has c levels. The null hypothesis states that knowing the level of Variable A does not help you predict the level of Variable B. That is, the variables are independent. H0: Variable A and Variable B are independent. Ha: Variable A and Variable B are not independent. The alternative hypothesis is that knowing the level of Variable A can help you predict the level of Variable B. Note: Support for the alternative hypothesis suggests that the variables are related; but the relationship is not necessarily causal, in the sense that one variable "causes" the other. Formulate an Analysis Plan The analysis plan describes how to use sample data to accept or reject the null hypothesis. The plan should specify the following elements. Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.
Test method. Use the chi-square test for independence to determine whether there is a significant relationship between two categorical variables. Analyze Sample Data Using sample data, find the degrees of freedom, expected frequencies, test statistic, and the P-value associated with the test statistic. The approach described in this section is illustrated in the sample problem at the end of this lesson. Degrees of freedom. The degrees of freedom (DF) is equal to: DF = (r - 1) * (c - 1) where r is the number of levels for one catagorical variable, and c is the number of levels for the other categorical variable. Expected frequencies. The expected frequency counts are computed separately for each level of one categorical variable at each level of the other categorical variable. Compute r * c expected frequencies, according to the following formula. Er,c = (nr * nc) / n where Er,c is the expected frequency count for level r of Variable A and level c of Variable B, nr is the total number of sample observations at level r of Variable A, nc is the total number of sample observations at level c of Variable B, and n is the total sample size. Test statistic. The test statistic is a chi-square random variable (Χ2) defined by the following equation. Χ2 = Σ [ (Or,c - Er,c)2 / Er,c ] where Or,c is the observed frequency count at level r of Variable A and level c of Variable B, and Er,c is the expected frequency count at level r of Variable A and level c of Variable B.
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis. H0: Gender and voting preferences are independent. Ha: Gender and voting preferences are not independent. Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square test for independence. Analyze sample data. Applying the chi-square test for independence to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value. DF = (r - 1) * (c - 1) = (2 - 1) * (3 - 1) = 2 Er,c = (nr * nc) / n E1,1 = (400 * 450) / 1000 = 180000/1000 = 180 E1,2 = (400 * 450) / 1000 = 180000/1000 = 180 E1,3 = (400 * 100) / 1000 = 40000/1000 = 40 E2,1 = (600 * 450) / 1000 = 270000/1000 = 270 E2,2 = (600 * 450) / 1000 = 270000/1000 = 270 E2,3 = (600 * 100) / 1000 = 60000/1000 = 60 Χ2 = Σ [ (Or,c - Er,c)2 / Er,c ] Χ2 = (200 - 180)2/180 + (150 - 180)2/180 + (50 - 40)2/
- (250 - 270)2/270 + (300 - 270)2/270 + (50 - 60)2/ Χ2 = 400/180 + 900/180 + 100/40 + 400/270 + 900/270 + 100/ Χ2 = 2.22 + 5.00 + 2.50 + 1.48 + 3.33 + 1.67 = 16.
where DF is the degrees of freedom, r is the number of levels of gender, c is the number of levels of the voting preference, nr is the number of observations from level r of gender, nc is the number of observations from level c of voting preference, n is the number of observations in the sample, Er,c is the expected frequency count when gender is level r and voting preference is level c, and Or,c is the observed frequency count when gender is level r voting preference is level c. The P-value is the probability that a chi-square statistic having 2 degrees of freedom is more extreme than 16.2. We use the Chi-Square Distribution Calculator to find P(Χ2 > 16.2) = 0.0003. Interpret results. Since the P-value (0.0003) is less than the significance level (0.05), we cannot accept the null hypothesis. Thus, we conclude that there is a relationship between gender and voting preference. Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, each population was more than 10 times larger than its respective sample, the variables under study were categorical, and the expected frequency count was at least 5 in each cell of the contingency table. Chi-Square Test of Homogeneity - Chi-Square Test of Homogeneity -single categorical variable
- from two different populations. -It is used to determine whether frequency counts are distributed identically across different populations.
Suppose that data were sampled from r populations, and assume that the categorical variable had c levels. At any specified level of the categorical variable, the null hypothesis states that each population has the same proportion of observations. Thus, H0: Plevel 1 of population 1 = Plevel 1 of population 2 =... = Plevel 1 of population r H0: Plevel 2 of population 1 = Plevel 2 of population 2 =... = Plevel 2 of population r
... H0: Plevel c of population 1 = Plevel c of population 2 =... = Plevel c of population r The alternative hypothesis (Ha) is that at least one of the null hypothesis statements is false. Formulate an Analysis Plan The analysis plan describes how to use sample data to accept or reject the null hypothesis. The plan should specify the following elements. Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used. Test method. Use the chi-square test for homogeneity to determine whether observed sample frequencies differ significantly from expected frequencies specified in the null hypothesis. The chi-square test for homogeneity is described in the next section. Analyze Sample Data Using sample data from the contingency tables, find the degrees of freedom, expected frequency counts, test statistic, and the P- value associated with the test statistic. The analysis described in
this section is illustrated in the sample problem at the end of this lesson. Degrees of freedom. The degrees of freedom (DF) is equal to: DF = (r - 1) * (c - 1) where r is the number of populations, and c is the number of levels for the categorical variable. Expected frequency counts. The expected frequency counts are computed separately for each population at each level of the categorical variable, according to the following formula. Er,c = (nr * nc) / n where Er,c is the expected frequency count for population r at level c of the categorical variable, nr is the total number of observations from population r, nc is the total number of observations at treatment level c, and n is the total sample size. Test statistic. The test statistic is a chi-square random variable (Χ2) defined by the following equation. Χ2 = Σ [ (Or,c - Er,c)2 / Er,c ] where Or,c is the observed frequency count in population r for level c of the categorical variable, and Er,c is the expected frequency count in population r for level c of the categorical variable. P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a chi-square, use the Chi-Square Distribution Calculator to assess the probability associated with the test statistic. Use the degrees of freedom computed above. Interpret Results
Null hypothesis: The null hypothesis states that the proportion of boys who prefer the Lone Ranger is identical to the proportion of girls. Similarly, for the other programs. Thus, H0: Pboys who prefer Lone Ranger = Pgirls who prefer Lone Ranger H0: Pboys who prefer Sesame Street = Pgirls who prefer Sesame Street H0: Pboys who prefer The Simpsons = Pgirls who prefer The Simpsons Alternative hypothesis: At least one of the null hypothesis statements is false. Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square test for homogeneity. Analyze sample data. Applying the chi-square test for homogeneity to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value. DF = (r - 1) * (c - 1) = (2 - 1) * (3 - 1) = 2 Er,c = (nr * nc) / n E1,1 = (100 * 100) / 300 = 10000/300 = 33. E1,2 = (100 * 110) / 300 = 11000/300 = 36. E1,3 = (100 * 90) / 300 = 9000/300 = 30. E2,1 = (200 * 100) / 300 = 20000/300 = 66. E2,2 = (200 * 110) / 300 = 22000/300 = 73. E2,3 = (200 * 90) / 300 = 18000/300 = 60.
Χ2 = Σ [ (Or,c - Er,c)2 / Er,c ] Χ2 = (50 - 33.3)2/33.3 + (30 - 36.7)2/36.7 + (20 - 30)2/
- (50 - 66.7)2/66.7 + (80 - 73.3)2/73.3 + (70 - 60)2/ Χ2 = (16.7)2/33.3 + (-6.7)2/36.7 + (-10.0)2/30 + (-16.7)2/66.7 + (3.3)2/73.3 + (10)2/ Χ2 = 8.38 + 1.22 + 3.33 + 4.18 + 0.61 + 1.67 = 19. where DF is the degrees of freedom, r is the number of populations, c is the number of levels of the categorical variable, nr is the number of observations from population r, nc is the number of observations from level c of the categorical variable, n is the number of observations in the sample, Er,c is the expected frequency count in population r for level c, and Or,c is the observed frequency count in population r for level c. The P-value is the probability that a chi-square statistic having 2 degrees of freedom is more extreme than 19.39. We use the Chi-Square Distribution Calculator to find P(Χ2 > 19.39) = 0.0000. (The actual P-value, of course, is not exactly zero. If the Chi-Square Distribution Calculator reported more than four decimal places, we would find that the actual P-value is a very small number that is less than 0.00005 and greater than zero.) Interpret results. Since the P-value (0.0000) is less than the significance level (0.05), we reject the null hypothesis. Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, each population was more than 10 times larger than its respective sample, the variable under study
