MATH 110 Module 10 Exam (New, 2023-2024) / MATH110 Module 10 Exam: Portage Learning, Exams of Mathematics

Download MATH 110 Module 10 Exam (New, 2023-2024) / MATH110 Module 10 Exam: Portage Learning and more Exams Mathematics in PDF only on Docsity! Find the value of X 2 for 20 degrees of freedom and an area of .010 in the right tail of the chi- square distribution. Look across the top of the chi-square distribution table for .010, then look down the left column for 20. These two meet at X 2 =37.566. Find the value of X 2 for 13 degrees of freedom and an area of .100 in the left tail of the chi- square distribution. Since the chi-square distribution table gives the area in the right tail, we must use 1 - .10 = .90. Look across the top of the chi-square distribution table for .90, then look down the left column for 13. These two meet at X 2 =7.042. MATH 110 Module 10 Exam Page 1 Find the value of X 2 for 20 degrees of freedom and an area of .010 in the right tail of the chi- square distribution. X 2 =37.566 Answer Key Exam Page 2 Find the value of X 2 for 13 degrees of freedom and an area of .100 in the left tail of the chi- square distribution. (1-.100) = 0.900 X 2 = 7.042 Answer Key Exam Page 3 Find the value of X 2 values that separate the middle 80 % from the rest of the distribution for 17 degrees of freedom. In this case, we have 1-.80=.20 outside of the middle or .20/2 = .1 in each of the tails. Notice that the area to the right of the first X 2 is .80 + .10 = .90. So we use this value and a DOF of 17 to get X 2 = 10.085. The area to the right of the second X 2 is .10. So we use this value and a DOF of 17 to get X 2 = 24.764. Find the critical value of F for DOF=(4,17) and area in the right tail of .05 In order to solve this, we turn to the F distribution table that an area of .05. DOF=(4,17) indicates that degrees of freedom for the numerator is 4 and degrees of freedom for the denominator is 17. So, we look up these in the table and find that F=2.96. Find the value of X 2 values that separate the middle 80 % from the rest of the distribution for 17 degrees of freedom. (1-.80) = .20 (.20/2) = .10 X 2 (.90) = 10.085 X 2 (.10) = 24.769 Answer Key Exam Page 4 Find the critical value of F for DOF=(4,17) and area in the right tail of .05 DOF(4,17) F = 2.96 Answer Key Exam Page 5 A trucking company wants to find out if their drivers are still alert after driving long hours. So, they give a test for alertness to two groups of drivers. They give the test to 580 drivers who have just finished driving 4 hours or less and they give the test to 470 drivers who have just finished driving 8 hours or more. The results of the tests are given below. Is there is a relationship between hours of driving and alertness? (Do a test for independence.) Test at the .5 % level of significance. H0: Driving hours and alertness are independent events. H1: Driving hours and alertness are not independent events. We have two rows and three columns, so # of Rows =2 and # of Columns=2. The degrees of freedom are given by: DOF = (# of Rows-1)(# of Columns-1)=(2-1)(2-1)=1. Using this, along with .005 (for the .5 % level of significance) we find in the chi-square table a critical value of 7.879. DOF = (2-1)(2-1) = 1 Critical Value = 7.879 PASS FAIL Totals 4 hours 450 130 580 8 hours 325 145 470 Totals 775 275 1050 Eij = (TRi*TCi)/total number in sample E for 4 hrs/passed = (580*775)/1050 = 428.0952 E for 4 hrs/failed = (580*275)/1050 = 151.9048 E for 8 hrs/passed = (470*775)/1050 = 346.9048 E for 8 hrs/failed = (470*275)/1050 = 123.0952 X 2 = (450-428.0952) 2 /428.0952 +(130-151.9048) 2 /151.9048 +(325-346.9048) 2 /396.9048 +(145- 123.0952) 2 /123.0952 = 9.56 9.56>7.879 , therefore, we will reject the null hypothesis. Answer Key Passed Failed Row Totals Drove 4 hours or less 450 130 580 Drove 8 hours or more 325 145 470 This value is more than the critical value of 7.879. So, we reject the null hypothesis.