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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.
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. X^2 = 7.
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.
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) =. (.20/2) =. X^2 (.90) = 10. X^2 (.10) = 24.
Answer Key
Exam Page 4
Find the critical value of F for DOF=(4,17) and area in the right tail of.
DOF(4,17) F = 2.
Answer Key
Exam Page 5
Exam Page 6
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.
Passed Failed Row Totals Drove 4 hours or less 450 130 580 Drove 8 hours or more 325 145 470
Is there is a relationship between hours of driving and alertness? (Do a test for independence.) Test at the .5 % level of significance.
H0: drivers are still alert H1: drivers are not still alert
H 0 : The mayor’s distribution is correct. H 1 : The mayor’s distribution is not correct.
This is a multinomial experiment, for multinomial experiments, we use the chi-square distribution.
Calculate the degrees of freedom for three possible outcomes: DOF=3-1=2. Our level of significance is 5 % (.05). So look up DOF of 2 and .05 on the Chi-square distribution table to get 5.991.
For the expected frequencies, we will use Ei = npi. So, the expected frequencies (of the n=285 in the sample) based on the mayor’s distribution:
This is greater than the critical value of 5.991. Therefore, we reject the null hypothesis.
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.
H 0 : Driving hours and alertness are independent events.
H 1 : 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. PASS FAIL Totals 4 hours 450 130 580 8 hours 325 145 470 Totals 775 275 1050
Eij = (TRiTCi)/total number in sample*
E for 4 hrs/passed = (580775)/1050 = 428. E for 4 hrs/failed = (580275)/1050 = 151. E for 8 hrs/passed = (470775)/1050 = 346. E for 8 hrs/failed = (470275)/1050 = 123.
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.
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
