Chapter 11, Problem 11.5P

$\boxed{PS}$ Below are some of the tables you analyzed in problem 10.17. You already have the results of the chi square tests. Now, analyze these relationships in terms of association, using an appropriate measure of association. Write a paragraph summarizing your results.

a. Presidential preference and race/ethnicity

Preference	Race/Ethnicity
	White	Black	Latino	Totals
Romney	289	5	44	338
Obama Totals	$\frac{249}{538}$	$\frac{95}{100}$	$\frac{66}{110}$	$\frac{410}{748}$

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	$\frac{35}{65}$	$\frac{120}{300}$	$\frac{218}{336}$	$\frac{37}{47}$	$\frac{410}{748}$

c. Presidential preference by religion

Preference	Religion
	Protestant	Catholic	Jewish	None	Other	Totals
Romney	165	110	10	28	25	338
Obama Totals	$\frac{245}{410}$	$\frac{55}{165}$	$\frac{20}{30}$	$\frac{60}{88}$	$\frac{30}{55}$	$\frac{410}{748}$

10.15 $\boxed{PS}$ A random sample of 748 voters in a large city was asked how they voted in the presidential election of 2012. Calculate chi square and the column percentages for each table below and write a brief report describing the significance of the relationships as well as the patterns you observe.

a. Presidential preference and gender

	Gender
Preference	Male	Female	Totals
Romney	165	173	338
Obama Totals	$\frac{200}{365}$	$\frac{210}{383}$	$\frac{410}{748}$

b. Presidential preference and race/ethnicity

Preference	Race/Ethnicity
	White	Black	Latino	Totals
Romney	289	5	44	338
Obama Totals	$\frac{249}{538}$	$\frac{95}{100}$	$\frac{66}{110}$	$\frac{410}{748}$

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	$\frac{35}{65}$	$\frac{120}{300}$	$\frac{218}{336}$	$\frac{37}{47}$	$\frac{410}{748}$

d. Presidential preference by religion

Preference	Religion
	Protestant	Catholic	Jewish	None	Other	Totals
Romney	165	110	10	28	25	338
Obama Totals	$\frac{245}{410}$	$\frac{55}{165}$	$\frac{20}{30}$	$\frac{60}{88}$	$\frac{30}{55}$	$\frac{410}{748}$

To determine

(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 $f_e$ is given as,

$f_e=\frac{\text{Row}\text{marginal×Column}\text{marginal}}{N}$

Where N is the total of frequencies.

The chi square statistic is given by,

${\chi }^2\left(obtained\right)={\displaystyle \sum \frac{{\left(f_o-f_e\right)}^2}{f_e}}$

Where $f_o$ is the observed frequency,

And $f_e$ is the expected frequency.

The degrees of freedom for the bivariate table is given as,

$df=\left(r-1\right)\left(c-1\right)$

Where r is the number of rows and c is the number of columns.

For table larger than $2\times 2$, the Cramer’s V is given by the formula,

$V=\sqrt{\frac{{\chi }^2}{N\left(\min r-1,c-1\right)}}$

Where, ${\chi }^2$ is the chi square value,

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,

$f_e=\frac{\text{Row}\text{marginal×Column}\text{marginal}}{N}\to \left(1\right)$

Substitute 338 for row marginal, 538 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_1& =& \frac{338\times 538}{748}\\ & =& 243.11\end{array}$

Substitute 338 for row marginal, 100 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_2& =& \frac{338\times 100}{748}\\ & =& 45.19\end{array}$

Substitute 338 for row marginal, 110 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_3& =& \frac{338\times 110}{748}\\ & =& 49.71\end{array}$

Substitute 410 for row marginal, 538 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_4& =& \frac{410\times 538}{748}\\ & =& 294.89\end{array}$

Substitute 410 for row marginal, 100 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_5& =& \frac{410\times 100}{748}\\ & =& 54.81\end{array}$

Substitute 410 for row marginal, 110 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_6& =& \frac{410\times 110}{748}\\ & =& 60.29\end{array}$

Consider the following table,

	$f_o$	$f_e$	$f_o-f_e$	${\left(f_o-f_e\right)}^2$	${\left(f_o-f_e\right)}^2/f_e$
	289	243.11	45.89	2105.89	8.66
	5	45.19	$-40.19$	161.24	35.74
	44	49.71	$-5.71$	32.60	0.66
	249	294.89	$-45.89$	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		${\chi }^2=82.21$

The value $f_o-f_e$ is obtained as,

Substitute 289 for $f_o$ and 243.11 for $f_e$ in the above formula.

$\begin{array}{rll}{f}_o-f_e& =& 289-243.11\\ & =& 45.89\end{array}$

Squaring the above obtained result,

$\begin{array}{rll}{\left(f_o-f_e\right)}^2& =& {\left(45.89\right)}^2\\ & =& 2105.89\end{array}$

Divide the above obtained result by $f_e$.

$\begin{array}{rll}\frac{{\left(f_o-f_e\right)}^2}{f_e}& =& \frac{2105.89}{45.98}\\ & =& 8.66\end{array}$

Proceed in a similar manner to obtain rest of the values of ${\left(f_o-f_e\right)}^2/f_e$ and refer above table for the rest of the values of ${\left(f_o-f_e\right)}^2/f_e$.

The chi square value is given as,

$\begin{array}{rll}{\chi }^2\left(obtained\right)& =& {\displaystyle \sum \frac{{\left(f_o-f_e\right)}^2}{f_e}}\\ & =& 8.66+35.74+0.66+7.14+29.47+0.54\\ & =& 82.21\end{array}$

Thus, the chi square value is 82.21.

The level of significance is $\alpha =0.05$.

Number of rows is 2 and the number of columns of 3.

The degrees of freedom is given by,

$\begin{array}{rll}df& =& \left(2-1\right)\left(3-1\right)\\ & =& 2\end{array}$

And area of critical region is ${\chi }^2\left(critical\right)=5.991$.

The Cramer’s V is given by the formula,

$V=\sqrt{\frac{{\chi }^2}{N\left(\min r-1,c-1\right)}}$

Substitute 82.21 for ${\chi }^2$, 2 for r, 3 for c and 748 for N in the above mentioned formula.

$\begin{array}{rll}V& =& \sqrt{\frac{82.21}{748\left(\min \left(2-1,3-1\right)\right)}}\\ & =& \sqrt{\frac{82.21}{748\left(\min \left(1,2\right)\right)}}\\ & =& \sqrt{\frac{82.21}{748\times 1}}\\ & =& 0.33\end{array}$

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.

To determine

(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 $f_e$ is given as,

$f_e=\frac{\text{Row}\text{marginal×Column}\text{marginal}}{N}$

Where N is the total of frequencies.

The chi square statistic is given by,

${\chi }^2\left(obtained\right)={\displaystyle \sum \frac{{\left(f_o-f_e\right)}^2}{f_e}}$

Where $f_o$ is the observed frequency,

And $f_e$ is the expected frequency.

The degrees of freedom for the bivariate table is given as,

$df=\left(r-1\right)\left(c-1\right)$

Where r is the number of rows and c is the number of columns.

For table larger than $2\times 2$, the Cramer’s V is given by the formula,

$V=\sqrt{\frac{{\chi }^2}{N\left(\min r-1,c-1\right)}}$

Where, ${\chi }^2$ is the chi square value,

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,

$f_e=\frac{\text{Row}\text{marginal×Column}\text{marginal}}{N}\to \left(1\right)$

Substitute 338 for row marginal, 65 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_1& =& \frac{338\times 65}{748}\\ & =& 29.37\end{array}$

Substitute 338 for row marginal, 300 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_2& =& \frac{338\times 300}{748}\\ & =& 135.56\end{array}$

Substitute 338 for row marginal, 336 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{c}{f}_3=\frac{338\times 336}{748}\\ =151.83\end{array}$

Substitute 338 for row marginal, 47 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_4& =& \frac{338\times 47}{748}\\ & =& 21.24\end{array}$

Substitute 410 for row marginal, 65 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_5& =& \frac{410\times 65}{748}\\ & =& 35.63\end{array}$

Substitute 410 for row marginal, 300 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{c}{f}_6=\frac{410\times 300}{748}\\ =164.44\end{array}$

Substitute 410 for row marginal, 336 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_7& =& \frac{410\times 336}{748}\\ & =& 184.17\end{array}$

Substitute 410 for row marginal, 47 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_8& =& \frac{410\times 47}{748}\\ & =& 25.76\end{array}$

Consider the following table,

	$f_o$	$f_e$	$f_o-f_e$	${\left(f_o-f_e\right)}^2$	${\left(f_o-f_e\right)}^2/f_e$
	30	29.37	0.63	0.40	0.01
	180	135.56	44.44	1974.91	14.57
	118	151.83	$-33.83$	1144.47	7.54
	10	21.24	$-11.24$	126.34	5.95
	35	35.63	$-0.63$	0.40	0.01
	120	164.44	$-44.44$	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		${\chi }^2=51.19$

The value $f_o-f_e$ is obtained as,

Substitute 30 for $f_o$ and 29.37 for $f_e$ in the above formula.

$\begin{array}{rll}{f}_o-f_e& =& 30-29.37\\ & =& 0.63\end{array}$

Squaring the above obtained result,

$\begin{array}{rll}{\left(f_o-f_e\right)}^2& =& {\left(0.63\right)}^2\\ & =& 0.40\end{array}$

Divide the above obtained result by $f_e$.

$\begin{array}{rll}\frac{{\left(f_o-f_e\right)}^2}{f_e}& =& \frac{0.40}{29.37}\\ & =& 0.01\end{array}$

Proceed in a similar manner to obtain rest of the values of ${\left(f_o-f_e\right)}^2/f_e$ and refer above table for the rest of the values of ${\left(f_o-f_e\right)}^2/f_e$.

The chi square value is given as,

$\begin{array}{rll}{\chi }^2\left(obtained\right)& =& {\displaystyle \sum \frac{{\left(f_o-f_e\right)}^2}{f_e}}\\ & =& 0.01+14.57+7.54+\mathrm{...}+4.90\\ & =& 51.19\end{array}$

Thus, the chi square value is 51.19.

The level of significance is $\alpha =0.05$.

Number of rows is 2 and the number of columns of 4.

The degrees of freedom is given by,

$\begin{array}{rll}df& =& \left(2-1\right)\left(4-1\right)\\ & =& 3\end{array}$

And area of critical region is ${\chi }^2\left(critical\right)=7.815$.

The Cramer’s V is given by the formula,

$V=\sqrt{\frac{{\chi }^2}{N\left(\min r-1,c-1\right)}}$

Substitute 51.19 for ${\chi }^2$, 2 for r, 4 for c and 748 for N in the above mentioned formula.

$\begin{array}{rll}V& =& \sqrt{\frac{51.19}{748\left(\min \left(2-1,4-1\right)\right)}}\\ & =& \sqrt{\frac{51.19}{748\left(\min \left(1,3\right)\right)}}\\ & =& \sqrt{\frac{51.19}{748\times 1}}\\ & =& 0.26\end{array}$

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.

To determine

(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 $f_e$ is given as,

$f_e=\frac{\text{Row}\text{marginal×Column}\text{marginal}}{N}$

Where N is the total of frequencies.

The chi square statistic is given by,

${\chi }^2\left(obtained\right)={\displaystyle \sum \frac{{\left(f_o-f_e\right)}^2}{f_e}}$

Where $f_o$ is the observed frequency,

And $f_e$ is the expected frequency.

The degrees of freedom for the bivariate table is given as,

$df=\left(r-1\right)\left(c-1\right)$

Where r is the number of rows and c is the number of columns.

For table larger than $2\times 2$, the Cramer’s V is given by the formula,

$V=\sqrt{\frac{{\chi }^2}{N\left(\min r-1,c-1\right)}}$

Where, ${\chi }^2$ is the chi square value,

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,

$f_e=\frac{\text{Row}\text{marginal×Column}\text{marginal}}{N}\to \left(1\right)$

Substitute 338 for row marginal, 410 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_1& =& \frac{338\times 410}{748}\\ & =& 185.27\end{array}$

Substitute 338 for row marginal, 165 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_2& =& \frac{338\times 165}{748}\\ & =& 74.56\end{array}$

Substitute 338 for row marginal, 30 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_3& =& \frac{338\times 30}{748}\\ & =& 13.56\end{array}$

Substitute 338 for row marginal, 88 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_4& =& \frac{338\times 88}{748}\\ & =& 39.76\end{array}$

Substitute 338 for row marginal, 55 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_5& =& \frac{338\times 55}{748}\\ & =& 24.85\end{array}$

Substitute 410 for row marginal, 410 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_6& =& \frac{410\times 410}{748}\\ & =& 224.73\end{array}$

Substitute 410 for row marginal, 165 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{c}{f}_7=\frac{410\times 165}{748}\\ =90.44\end{array}$

Substitute 410 for row marginal, 30 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_8& =& \frac{410\times 30}{748}\\ & =& 16.44\end{array}$

Substitute 410 for row marginal, 88 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_9& =& \frac{410\times 88}{748}\\ & =& 48.24\end{array}$

Substitute 410 for row marginal, 55 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_{10}& =& \frac{410\times 55}{748}\\ & =& 30.15\end{array}$

Consider the following table,

	$f_o$	$f_e$	$f_o-f_e$	${\left(f_o-f_e\right)}^2$	${\left(f_o-f_e\right)}^2/f_e$
	165	185.27	$-20.27$	410.87	2.22
	110	74.56	35.44	1255.99	16.85
	10	13.56	$-3.56$	12.67	0.93
	28	39.76	$-11.76$	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	$-35.44$	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	$-0.15$	0.02	0.00
Total	748	748	0		${\chi }^2=42.84$

The value $f_o-f_e$ is obtained as,

Substitute 165 for $f_o$ and 185.27 for $f_e$ in the above formula.

$\begin{array}{rll}{f}_o-f_e& =& 165-185.27\\ & =& -20.27\end{array}$

Squaring the above obtained result,

$\begin{array}{rll}{\left(f_o-f_e\right)}^2& =& {\left(-20.27\right)}^2\\ & =& 410.87\end{array}$

Divide the above obtained result by $f_e$.

$\begin{array}{rll}\frac{{\left(f_o-f_e\right)}^2}{f_e}& =& \frac{410.87}{185.27}\\ & =& 2.22\end{array}$

Proceed in a similar manner to obtain rest of the values of ${\left(f_o-f_e\right)}^2/f_e$ and refer above table for the rest of the values of ${\left(f_o-f_e\right)}^2/f_e$.

The chi square value is given as,

$\begin{array}{rll}{\chi }^2\left(obtained\right)& =& {\displaystyle \sum \frac{{\left(f_o-f_e\right)}^2}{f_e}}\\ & =& 2.22+16.85+0.93+\mathrm{...}+0.00\\ & =& 42.84\end{array}$

Thus, the chi square value is 42.84.

The level of significance is $\alpha =0.05$.

Number of rows is 2 and the number of columns of 5.

The degrees of freedom is given by,

$\begin{array}{rll}df& =& \left(2-1\right)\left(5-1\right)\\ & =& 4\end{array}$

And area of critical region is ${\chi }^2\left(critical\right)=9.488$.

The Cramer’s V is given by the formula,

$V=\sqrt{\frac{{\chi }^2}{N\left(\min r-1,c-1\right)}}$

Substitute 42.84 for ${\chi }^2$, 2 for r, 5 for c and 748 for N in the above mentioned formula.

$\begin{array}{rll}V& =& \sqrt{\frac{42.84}{748\left(\min \left(2-1,5-1\right)\right)}}\\ & =& \sqrt{\frac{42.84}{748\left(\min \left(1,4\right)\right)}}\\ & =& \sqrt{\frac{42.84}{748\times 1}}\\ & =& 0.24\end{array}$

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.