



Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
MATH 110 Module 10 Exam (New, 2023-2024) / MATH110 Module 10 Exam/ MATH 110 Statistics Module 10 Exam/ MATH110 Statistics Module 10 Exam: Portage Learning
Typology: Exams
1 / 6
This page cannot be seen from the preview
Don't miss anything!
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.
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.
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.
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.
Find the critical value of F for DOF=(4,17) and area in the right tail of.
DOF(4,17) F = 2.
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.
Passed Failed Row Totals Drove 4 hours or less 450 130 580 Drove 8 hours or more 325 145 470