Concept explainers
a. Presidential preference and race/ethnicity
Preference | Race/Ethnicity | |||||
White | Black | Latino | Totals | |||
Romney | 289 | 5 | 44 | 338 | ||
Obama Totals |
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b. Presidential preference by education
Preference | Education | ||||
Less than HS | HS Graduate | College Graduate | Post-Graduate Degree | Totals | |
Romney | 30 | 180 | 118 | 10 | 338 |
Obama Totals |
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c. Presidential preference by religion
Preference | Religion | |||||
Protestant | Catholic | Jewish | None | Other | Totals | |
Romney | 165 | 110 | 10 | 28 | 25 | 338 |
Obama Totals |
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10.15
a. Presidential preference and gender
Gender | |||
Preference | Male | Female | Totals |
Romney | 165 | 173 | 338 |
Obama Totals |
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b. Presidential preference and race/ethnicity
Preference | Race/Ethnicity | |||||
White | Black | Latino | Totals | |||
Romney | 289 | 5 | 44 | 338 | ||
Obama Totals |
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c. Presidential preference by education
Preference | Education | ||||
Less than HS | HS Graduate | College Graduate | Post-Graduate Degree | Totals | |
Romney | 30 | 180 | 118 | 10 | 338 |
Obama Totals |
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d. Presidential preference by religion
Preference | Religion | |||||
Protestant | Catholic | Jewish | None | Other | Totals | |
Romney | 165 | 110 | 10 | 28 | 25 | 338 |
Obama Totals |
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(a)
To find:
The column percentages, strength and pattern of relationship.
Answer to Problem 11.5P
Solution:
There is a strong positive association between Race/Ethnicity and preference of vote.
Explanation of Solution
Given:
The given statement is,
A random sample of 748 voters in a large city was asked how they voted in the presidential election of 2012.
The given table of information is,
Preference | Race/Ethnicity | |||
White | Black | Latino | Totals | |
Romney | 289 | 5 | 44 | 338 |
Obama | 249 | 95 | 66 | 410 |
Totals | 538 | 100 | 110 | 748 |
Approach:
If the Cramer’s V value is less than 0.10, it is said that there is a weak association between the categories.
If the Cramer’s V value lies between 0.11 and 0.30, it is said that there is a moderate association between the categories.
If the Cramer’s V value is greater than 0.30, it is said that there is a strong association between the categories.
Formula used:
For a chi square, the expected frequency
Where N is the total of frequencies.
The chi square statistic is given by,
Where
And
The degrees of freedom for the bivariate table is given as,
Where r is the number of rows and c is the number of columns.
For table larger than
Where,
N is the total number of elements,
r is the number of rows,
and c is the number of columns.
Calculation:
From the given information,
The observed frequency is given as,
Substitute 338 for row marginal, 538 for column marginal and 748 for N in equation
Substitute 338 for row marginal, 100 for column marginal and 748 for N in equation
Substitute 338 for row marginal, 110 for column marginal and 748 for N in equation
Substitute 410 for row marginal, 538 for column marginal and 748 for N in equation
Substitute 410 for row marginal, 100 for column marginal and 748 for N in equation
Substitute 410 for row marginal, 110 for column marginal and 748 for N in equation
Consider the following table,
289 | 243.11 | 45.89 | 2105.89 | 8.66 | |
5 | 45.19 | 161.24 | 35.74 | ||
44 | 49.71 | 32.60 | 0.66 | ||
249 | 294.89 | 2105.89 | 7.14 | ||
95 | 54.81 | 40.19 | 1615.24 | 29.47 | |
66 | 60.29 | 5.71 | 32.60 | 0.54 | |
Total | 748 | 748 | 0 |
The value
Substitute 289 for
Squaring the above obtained result,
Divide the above obtained result by
Proceed in a similar manner to obtain rest of the values of
The chi square value is given as,
Thus, the chi square value is 82.21.
The level of significance is
Number of rows is 2 and the number of columns of 3.
The degrees of freedom is given by,
And area of critical region is
The Cramer’s V is given by the formula,
Substitute 82.21 for
Since, 0.33 is greater than 0.30 it shows that the strength of association is strong.
Thus, there is a strong positive association between Race/Ethnicity and preference of vote.
Conclusion:
There is a strong positive association between Race/Ethnicity and preference of vote.
(b)
To find:
The column percentages, strength and pattern of relationship.
Answer to Problem 11.5P
Solution:
There is a moderate positive association between education and preference of vote.
Explanation of Solution
Given:
The given statement is,
A random sample of 748 voters in a large city was asked how they voted in the presidential election of 2012.
The given table of information is,
Preference | Education | ||||
Less than HS | HS Graduate | College Graduate | Post-Graduate Degree | Totals | |
Romney | 30 | 180 | 118 | 10 | 338 |
Obama | 35 | 120 | 218 | 37 | 410 |
Totals | 65 | 300 | 336 | 47 | 748 |
Approach:
If the Cramer’s V value is less than 0.10, it is said that there is a weak association between the categories.
If the Cramer’s V value lies between 0.11 and 0.30, it is said that there is a moderate association between the categories.
If the Cramer’s V value is greater than 0.30, it is said that there is a strong association between the categories.
Formula used:
For a chi square, the expected frequency
Where N is the total of frequencies.
The chi square statistic is given by,
Where
And
The degrees of freedom for the bivariate table is given as,
Where r is the number of rows and c is the number of columns.
For table larger than
Where,
N is the total number of elements,
r is the number of rows,
and c is the number of columns.
Calculation:
From the given information,
The observed frequency is given as,
Substitute 338 for row marginal, 65 for column marginal and 748 for N in equation
Substitute 338 for row marginal, 300 for column marginal and 748 for N in equation
Substitute 338 for row marginal, 336 for column marginal and 748 for N in equation
Substitute 338 for row marginal, 47 for column marginal and 748 for N in equation
Substitute 410 for row marginal, 65 for column marginal and 748 for N in equation
Substitute 410 for row marginal, 300 for column marginal and 748 for N in equation
Substitute 410 for row marginal, 336 for column marginal and 748 for N in equation
Substitute 410 for row marginal, 47 for column marginal and 748 for N in equation
Consider the following table,
30 | 29.37 | 0.63 | 0.40 | 0.01 | |
180 | 135.56 | 44.44 | 1974.91 | 14.57 | |
118 | 151.83 | 1144.47 | 7.54 | ||
10 | 21.24 | 126.34 | 5.95 | ||
35 | 35.63 | 0.40 | 0.01 | ||
120 | 164.44 | 1974.91 | 12.00 | ||
218 | 184.17 | 33.83 | 1144.47 | 6.21 | |
37 | 25.76 | 11.24 | 126.34 | 4.90 | |
Total | 748 | 748 | 0 |
The value
Substitute 30 for
Squaring the above obtained result,
Divide the above obtained result by
Proceed in a similar manner to obtain rest of the values of
The chi square value is given as,
Thus, the chi square value is 51.19.
The level of significance is
Number of rows is 2 and the number of columns of 4.
The degrees of freedom is given by,
And area of critical region is
The Cramer’s V is given by the formula,
Substitute 51.19 for
Since, 0.26 lies between 0.11 and 0.30 it shows that the strength of association is moderate.
Thus, there is a moderate positive association between education and preference of vote.
Conclusion:
There is a moderate positive association between education and preference of vote.
(c)
To find:
The column percentages, strength and pattern of relationship.
Answer to Problem 11.5P
Solution:
There is a moderate positive association between Religion and preference of vote.
Explanation of Solution
Given:
The given statement is,
A random sample of 748 voters in a large city was asked how they voted in the presidential election of 2012.
The given table of information is,
Preference | Religion | |||||
Protestant | Catholic | Jewish | None | Other | Totals | |
Romney | 165 | 110 | 10 | 28 | 25 | 338 |
Obama | 245 | 55 | 20 | 60 | 30 | 410 |
Totals | 410 | 165 | 30 | 88 | 55 | 748 |
Approach:
If the Cramer’s V value is less than 0.10, it is said that there is a weak association between the categories.
If the Cramer’s V value lies between 0.11 and 0.30, it is said that there is a moderate association between the categories.
If the Cramer’s V value is greater than 0.30, it is said that there is a strong association between the categories.
Formula used:
For a chi square, the expected frequency
Where N is the total of frequencies.
The chi square statistic is given by,
Where
And
The degrees of freedom for the bivariate table is given as,
Where r is the number of rows and c is the number of columns.
For table larger than
Where,
N is the total number of elements,
r is the number of rows,
and c is the number of columns.
Calculation:
From the given information,
The observed frequency is given as,
Substitute 338 for row marginal, 410 for column marginal and 748 for N in equation
Substitute 338 for row marginal, 165 for column marginal and 748 for N in equation
Substitute 338 for row marginal, 30 for column marginal and 748 for N in equation
Substitute 338 for row marginal, 88 for column marginal and 748 for N in equation
Substitute 338 for row marginal, 55 for column marginal and 748 for N in equation
Substitute 410 for row marginal, 410 for column marginal and 748 for N in equation
Substitute 410 for row marginal, 165 for column marginal and 748 for N in equation
Substitute 410 for row marginal, 30 for column marginal and 748 for N in equation
Substitute 410 for row marginal, 88 for column marginal and 748 for N in equation
Substitute 410 for row marginal, 55 for column marginal and 748 for N in equation
Consider the following table,
165 | 185.27 | 410.87 | 2.22 | ||
110 | 74.56 | 35.44 | 1255.99 | 16.85 | |
10 | 13.56 | 12.67 | 0.93 | ||
28 | 39.76 | 138.30 | 3.48 | ||
25 | 24.85 | 0.15 | 0.02 | 0.00 | |
245 | 224.73 | 20.27 | 410.87 | 1.83 | |
55 | 90.44 | 1255.99 | 13.89 | ||
20 | 16.44 | 3.56 | 12.67 | 0.77 | |
60 | 48.24 | 11.76 | 138.30 | 2.87 | |
30 | 30.15 | 0.02 | 0.00 | ||
Total | 748 | 748 | 0 |
The value
Substitute 165 for
Squaring the above obtained result,
Divide the above obtained result by
Proceed in a similar manner to obtain rest of the values of
The chi square value is given as,
Thus, the chi square value is 42.84.
The level of significance is
Number of rows is 2 and the number of columns of 5.
The degrees of freedom is given by,
And area of critical region is
The Cramer’s V is given by the formula,
Substitute 42.84 for
Since, 0.24 lies between 0.11 and 0.30 it shows that the strength of association is moderate.
Thus, there is a moderate positive association between Religion and preference of vote.
Conclusion:
There is a moderate positive association between Religion and preference of vote.
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Chapter 11 Solutions
Essentials Of Statistics
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