Chi-Square Distribution: Testing Goodness of Fit, Normality, and Contingency Tables, Lecture notes of Mathematics

Download Chi-Square Distribution: Testing Goodness of Fit, Normality, and Contingency Tables and more Lecture notes Mathematics in PDF only on Docsity! 10.1 10.2 10.3 10.4 Chapter 10 CHI —- SQUARE DISTRIBUTION Testing Goodness of Fit Testing Normality Contingency Tables Exercises Chi - Square Distribution Chi — square distribution was discovered by Karl Pearson. The distribution was introduced to determine whether or not discrepancies between observed theoretical counts were significant. The test used to find out how well an observed frequency distribution conforms to or fits some theoretical frequency distribution is referred to asa “goodness of fit test”. Also, chi — square distribution can be used to test the normality of any distribution. Testing a hypothesis made about several population proportions are sometimes considered. In this section, a discussion for testing thee normality with the use of chi — square is being emphasized. On the other hand, tables representing rows and columns are often called contingency tables. This particular topic is equally important. It helps us determine whether the two classifications of variables are independent. The value of chi — square varies for each number of degrees of freedom, one of the assumptions that apply fora contingency table is to have at least 5 expected frequencies for everyone of the X categories. 10.1 Testing Goodness of Fit Testing goodness of fit can be used to test how well an observed frequency distribution fits to some theoretical frequency distribution. Suppose, for example, we want to test the claim that the fatal accidents occur at the different widths of the road. Example 1: Width of the Road 4.0m —- 4.5m 4.6m — 5.0m 5.1m-—5.5m 5.6m — 6.0m Numbers of 95 90 83 73 Accidents =(0-E)? xX? = —— E Where: 0 - observed frequency x2 x2 df x2 (123 - 174)? (72-75)? (55-36)? (50-15)? = + + + 174 75 36 15 = 14.95 + 0.12 + 10.03 + 81.67 = 106.99 = 3 = 11.345 Computed X? is very much larger than the tabular X? value; thus, the claim of the students is correct and should be rejected. This means that there is at least one the proportions that is not equal to value claimed. 10.2 Testing Normally Many statistical tests require normally in the distribution. Chi — square analysis is one of the tests that can be used to determine if the distribution is normal. Summary of steps on how to apply chi — square for testing normality are listed below. Step 1: Use the mean and the standard deviation of the sample to estimate the mean and the standard deviation of the population if not know or assumed. Step 2: Group the sample data into class intervals or categories. Step 3: Calculate for the Z values for the class boundaries. Step 4: Determine the area under the standard normal curve between Z values to obtain the hypothesis proportion of the sample in each class. Step 5: Multiply each proportion by the total number of observations to obtain fe. Step 6: Compute for the X?. Remarks: 1) The hypothesis being tested is that the sample came from a population that has a normal distribution. 2) The degrees of freedom for the chi — square test is k — 1 —m, where k is the number of classes and m is the number of population parameters estimated. But if the sample mean and standard deviation have been used to estimate the population mean and the standard deviation, then m = 2; thus, the degrees of freedom (df) = k — 3. 10.3 Contingency Tables Teenagers and young adults have their own styles of studying. Some prefer to study with music; other don't. a group of psychologists conducted a study to determine the particular age of the students who like music. At the .01 level of significance, test the claim that style of studying in independent of the listed age groups. T he table below summarizes the information. Age Group Study Habit 9-12 13-16 17-20 21-24 With Music 89 75 63 52 Without Music 28 20 34 39 In contingency tables, we intend to test that the row variable is independent of the column variable. Computation for expected frequency for the contingency table is different from the one in the goodness of fit. The expected frequency E can be computed with the use of this formula. (row total) (column total) E = grand total Age Group 9-12 | 13-16 | 17-20 | 21-26 Total Study Habit With music Observed 89 75 63 52 279 frequency 81.61 66.26 67.66 63.47 Expected frequency 28 20 34 39 121 Without music 35.39 | 28.74 29.35 27.53 Observed frequency Expected frequency Total 117 95 97 91 400 (279) (117) (121) (117) E, = —— =81.61 E, = — =385.39 400 400