MATH 110 Module 10
Exam Page 1
Find the value of X2 for 20 degrees of freedom and an area of .010 in the right tail of the chi-
square distribution.
X2=37.566
Answer Key
Find the value of X2 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 X2 =37.566.
Exam Page 2
Find the value of X2 for 13 degrees of freedom and an area of .100 in the left tail of the chi-
square distribution.
(1-.100) = 0.900
X2 = 7.042
Answer Key
Find the value of X2 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 X2 =7.042.
Exam Page 3
, Find the value of X2 values that separate the middle 80 % from the rest of the distribution for 17
degrees of freedom.
Find the value of X2 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 X2 is .80 + .10 = .90. So we use this value and a DOF
of 17 to get X2 = 10.085.
The area to the right of the second X2 is .10. So we use this value and a DOF of 17 to get X2 =
24.764.
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
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.
Exam Page 5
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