MATH 110 Module 10 Exam Questions and Answers- Portage Learning

Stati sti cs - Portage Online Summer 2018




Module 10 Problem Set 1-5 Answers

Problem Set 10.1 Solutions


1. Solution. Look across the top of the chi-square distribution table for .01, then look down the left
column for 14. These two meet at X2 =29.141.


2. Solution. Look across the top of the chi-square distribution table for .025, then look down the left
column for 11. These two meet at X2 =21.920.


3. Solution. Since the chi-square distribution table gives the area in the right tail, we must use 1 - .
05 = .95. Look across the top of the chi-square distribution table for .95, then look down the left
column for 8. These two meet at X2 =2.733.


4. Solution. 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
20 to get X2 = 12.443.


The area to the right of the second X2 is .10. So we use this value and a DOF of 20 to get X2 =
28.412.


Problem Set 10.2 Solutions


1. Solution. In order to solve this, we turn to the F distribution table that an area of .01. DOF=(5,11)
indicates that degrees of freedom for the numerator is 5 and degrees of freedom for the denominator
is 11. So, we look up these in the table and find that F=5.32.


2. Solution. In order to solve this, we turn to the F distribution table that an area of .10. DOF=(3,10)
indicates that degrees of freedom for the numerator is 3 and degrees of freedom for the denominator
is 10. So, we look up these in the table and find that F=2.73.


3. Solution. In order to solve this, we turn to the F distribution table that an area of .05. DOF=(10,20)
indicates that degrees of freedom for the numerator is 10 and degrees of freedom for the
denominator is 20. So, we look up these in the table and find that F=2.35.




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, Stati sti cs - Portage Online Summer 2018



4. Solution. In order to solve this, we turn to the F distribution table that an area of .025. DOF=(7,13)
indicates that degrees of freedom for the numerator is 7 and degrees of freedom for the denominator
is 13. So, we look up these in the table and find that F=2.83.


Problem Set 10.3 Solutions


1. Solution.


We will set the null and alternate hypothesis:


H0: Your grandfather’s estimated distribution is correct.


H1: Your grandfather’s estimated distribution is not correct.


This is a multinomial experiment because it has more than two possible outcomes (there are four
possible outcomes). For multinomial experiments, we use the chi-square distribution.


Calculate the degrees of freedom for four possible outcomes: DOF=4-1=3. Our level of significance
is 5 % (.05). So look up DOF of 3 and .05 on the Chi-square distribution table to get 7.815.


For the expected frequencies, we will use Ei = npi. So, the expected frequencies (of the n=220 in the
sample) based on your grandfather’s estimate:


1900 and before E1 = 220(.20) = 44


1901 to 1940 E2 = 220(.25) = 55


1941 to 1970 E3 = 220(.40) = 88


After 1970 E4 = 220(.15) = 33


The observed frequencies:


1900 and before 46


1901 to 1940 53


1941 to 1970 79



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