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MATH 110 Module 10 Exam (Latest-2023) / MATH110 Module 10 Exam/ MATH 110 Statistics Module 10 Exam/ MATH110 Statistics Module 10 Exam: Portage Learning |100% Verified and Correct Q & A|

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MATH 110 Module 10 Exam (Latest-2023) / MATH110 Module 10 Exam/ MATH 110 Statistics Module 10 Exam/ MATH110 Statistics Module 10 Exam: Portage Learning |100% Verified and Correct Q & A|

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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.

(1-.80) = .20
(.20/2) = .10
X2(.90) = 10.085
X2(.10) = 24.769




Answer Key

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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