Chapter 10, Problem 10.15P

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 chi square value for the given information and test for its significance.

Answer to Problem 10.15P

Solution:

There is no significant relationship between the gender of the voter and the presidential preference.

The Percentage table suggests that though being not significant, both the genders prefer Obama over Romney.

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	Gender
	Male	Female	Totals
Romney	165	173	338
Obama	200	210	410
Totals	365	383	748

Approach:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

The five step model for hypothesis testing is,

Step 1. Making assumptions and meeting test requirements.

Step 2. Stating the null hypothesis.

Step 3. Selecting the sampling distribution and establishing the critical region.

Step 4. Computing test statistics.

Step 5. Making a decision and interpreting the results of the test.

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.

Calculation:

From the given information,

The observed frequency is given as,

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

Substitute 338 for row marginal, 365 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_1& =& \frac{338\times 365}{748}\\ & =& 164.93\end{array}$

Substitute 338 for row marginal, 383 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_2& =& \frac{338\times 383}{748}\\ & =& 173.07\end{array}$

Substitute 410 for row marginal, 365 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_3& =& \frac{410\times 365}{748}\\ & =& 200.07\end{array}$

Substitute 410 for row marginal, 383 for column marginal and 748 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_4& =& \frac{410\times 383}{748}\\ & =& 209.93\end{array}$

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	164.93	0.07	0.0049	0.00
	173	173.07	$-0.07$	0.0049	0.00
	200	200.07	$-0.07$	0.0049	0.00
	210	209.93	0.07	0.0049	0.00
Total	748	748	0		${\chi }^2=0.00$

The value $f_o-f_e$ is obtained as,

Substitute 165 for $f_o$ and 164.93 for $f_e$ in the above formula.

$\begin{array}{rll}{f}_o-f_e& =& 165-164.3\\ & =& 0.07\end{array}$

Squaring the above obtained result,

$\begin{array}{rll}{\left(f_o-f_e\right)}^2& =& {\left(0.07\right)}^2\\ & =& 0.0049\end{array}$

Divide the above obtained result by $f_e$.

$\begin{array}{rll}\frac{{\left(f_o-f_e\right)}^2}{f_e}& =& \frac{0.0049}{164.93}\\ & \approx & 0.00\end{array}$

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,

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

Thus, the chi square value is 0.00.

Follow the steps for chi square hypothesis testing.

Step 1. Making assumptions and meeting test requirements.

Model:

Independent random sampling.

Level of measurement is nominal.

Step 2. Stating the null hypothesis.

The statement of the null hypothesis is that there is no significant relationship between the gender of the voter and the presidential preference.

Thus, the null and the alternative hypotheses are,

$H_0:$ No significant relationship between the gender of the voter and the presidential preference.

$H_1:$ A significant relationship between the gender of the voter and the presidential preference.

Step 3. Selecting the sampling distribution and establishing the critical region.

The sampling distribution is chi square.

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

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

The degrees of freedom is given by,

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

And area of critical region is ${\chi }^2\left(critical\right)=3.841$.

Step 4. Computing test statistics.

The obtained chi square value is 0.00.

Step 5. Making a decision and interpreting the results of the test.

Since, $0.00<3.841$, the test statistics does not fall in the critical region. Thus, the null hypothesis is accepted.

This implies that, there is no significant relationship between the gender of the voter and the presidential preference.

The following table gives the column wise percentages of gender preferences..

Preference	Gender
	Male	Female
Romany	45.21%	45.17%
Obama	54.79%	54.83%
Totals	100%	100%

The columns of the table are computed as,

Substitute the values from the given table of information,

$\frac{165}{365}\times 100=45.21\mathrm{\%}$

Proceed in a similar manner to obtain rest of the values of the table.

The above table suggests that though being not significant, both the genders prefer Obama over Romney.

Conclusion:

There is no significant relationship between the gender of the voter and the presidential preference.

The Percentage table suggests that though being not significant, both the genders prefer Obama over Romney.

To determine

(b)

To find:

The chi square value for the given information and test for its significance.

Answer to Problem 10.15P

Solution:

There is a significant relationship between the race and ethnicity of the voter and the presidential preference.

The Percentage table suggests that mainly blacks prefer Obama with 95% majority.

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:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

The five step model for hypothesis testing is,

Step 1. Making assumptions and meeting test requirements.

Step 2. Stating the null hypothesis.

Step 3. Selecting the sampling distribution and establishing the critical region.

Step 4. Computing test statistics.

Step 5. Making a decision and interpreting the results of the test.

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.

Calculation:

From the given information,

The observed frequency is given as,

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

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	234.11	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.

Follow the steps for chi square hypothesis testing.

Step 1. Making assumptions and meeting test requirements.

Model:

Independent random sampling.

Level of measurement is nominal.

Step 2. Stating the null hypothesis.

The statement of the null hypothesis is that there is no significant relationship between the race and ethnicity of the voter and the presidential preference.

Thus, the null and the alternative hypotheses are,

$H_0:$ No significant relationship between the race and ethnicity of the voter and the presidential preference.

$H_1:$ A significant relationship between the race and ethnicity of the voter and the presidential preference.

Step 3. Selecting the sampling distribution and establishing the critical region.

The sampling distribution is chi square.

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 ${\chi }^2\left(critical\right)=5.991$.

Step 4. Computing test statistics.

The obtained chi square value is 82.21.

Step 5. Making a decision and interpreting the results of the test.

Since, $82.21>5.991$, the test statistics fall in the critical region. Thus, the null hypothesis is rejected.

This implies that, there is a significant relationship between the race and ethnicity of the voter and the presidential preference.

The following table gives the column wise percentages of gender preferences..

Preference	Race/Ethnicity
	White	Black	Latino
Romany	53.72%	5%	40%
Obama	46.28%	95%	60%
Totals	100%	100%	100%

The columns of the table are computed as,

Substitute the values from the given table of information,

$\frac{298}{538}\times 100=53.72\mathrm{\%}$

Proceed in a similar manner to obtain rest of the values of the table.

The above table suggests that mainly blacks prefer Obama with 95% majority.

Conclusion:

There is a significant relationship between the race and ethnicity of the voter and the presidential preference.

The Percentage table suggests that mainly blacks prefer Obama with 95% majority.

To determine

(c)

To find:

The chi square value for the given information and test for its significance.

Answer to Problem 10.15P

Solution:

There is a significant relationship between the education level of the voter and the presidential preference.

The Percentage table suggests that, except the high school graduates, rest all the others prefer Obama over Romney.

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

Apprach:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

The five step model for hypothesis testing is,

Step 1. Making assumptions and meeting test requirements.

Step 2. Stating the null hypothesis.

Step 3. Selecting the sampling distribution and establishing the critical region.

Step 4. Computing test statistics.

Step 5. Making a decision and interpreting the results of the test.

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.

Calculation:

From the given information,

The observed frequency is given as,

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

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 $\left(1\right)$.

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

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 $\left(1\right)$.

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

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.

Follow the steps for chi square hypothesis testing.

Step 1. Making assumptions and meeting test requirements.

Model:

Independent random sampling.

Level of measurement is nominal.

Step 2. Stating the null hypothesis.

The statement of the null hypothesis is that there is no significant relationship between the education level of the voter and the presidential preference.

Thus, the null and the alternative hypotheses are,

$H_0:$ No significant relationship between the education level of the voter and the presidential preference.

$H_1:$ A significant relationship between the education level of the voter and the presidential preference.

Step 3. Selecting the sampling distribution and establishing the critical region.

The sampling distribution is chi square.

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 ${\chi }^2\left(critical\right)=7.815$.

Step 4. Computing test statistics.

The obtained chi square value is 82.21.

Step 5. Making a decision and interpreting the results of the test.

Since, $51.19>7.815$, the test statistics fall in the critical region. Thus, the null hypothesis is rejected.

This implies that, there is a significant relationship between the education level of the voter and the presidential preference.

The following table gives the column wise percentages of gender preferences..

Preference	Education
	Less than HS	HS Graduate	College Graduate	Post-Graduate Degree
Romany	46.15%	60%	35.12%	21.28%
Obama	53.85%	40%	64.88%	78.72%
Totals	100%	100%	100%	100%

The columns of the table are computed as,

Substitute the values from the given table of information,

$\frac{298}{538}\times 100=53.72\mathrm{\%}$

Proceed in a similar manner to obtain rest of the values of the table.

The above table suggests that, except the high school graduates, rest all the others prefer Obama over Romney.

Conclusion:

There is a significant relationship between the education level of the voter and the presidential preference.

The Percentage table suggests that, except the high school graduates, rest all the others prefer Obama over Romney.

To determine

(d)

To find:

The chi square value for the given information and test for its significance.

Answer to Problem 10.15P

Solution:

There is a significant relationship between the religion of the voter and the presidential preference.

The Percentage table suggests that, except the voters from the Catholic religion, rest all the other religion voters prefer Obama over Romney.

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:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

The five step model for hypothesis testing is,

Step 1. Making assumptions and meeting test requirements.

Step 2. Stating the null hypothesis.

Step 3. Selecting the sampling distribution and establishing the critical region.

Step 4. Computing test statistics.

Step 5. Making a decision and interpreting the results of the test.

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.

Calculation:

From the given information,

The observed frequency is given as,

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

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 $\left(1\right)$.

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

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.

Follow the steps for chi square hypothesis testing.

Step 1. Making assumptions and meeting test requirements.

Model:

Independent random sampling.

Level of measurement is nominal.

Step 2. Stating the null hypothesis.

The statement of the null hypothesis is that there is no significant relationship between the religion of the voter and the presidential preference.

Thus, the null and the alternative hypotheses are,

$H_0:$ No significant relationship between the religion of the voter and the presidential preference.

$H_1:$ A significant relationship between the religion of the voter and the presidential preference.

Step 3. Selecting the sampling distribution and establishing the critical region.

The sampling distribution is chi square.

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,

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

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

Step 4. Computing test statistics.

The obtained chi square value is 42.84.

Step 5. Making a decision and interpreting the results of the test.

Since, $42.84>9.488$, the test statistics fall in the critical region. Thus, the null hypothesis is rejected.

This implies that, there is a significant relationship between the religion of the voter and the presidential preference.

The following table gives the column wise percentages of gender preferences..

Preference	Religion
	Protestant	Catholic	Jewish	None	Other
Romany	40.24%	66.67%	33.33%	31.82%	45.45%
Obama	59.75%	33.33%	66.67%	68.18%	54.55%
Totals	100%	100%	100%	100%	100%

The columns of the table are computed as,

Substitute the values from the given table of information,

$\frac{165}{410}\times 100=40.24\mathrm{\%}$

Proceed in a similar manner to obtain rest of the values of the table.

The above table suggests that, except the voters from the Catholic religion, rest all the other religion voters prefer Obama over Romney.

Conclusion:

There is a significant relationship between the religion of the voter and the presidential preference.

The Percentage table suggests that, except the voters from the Catholic religion, rest all the other religion voters prefer Obama over Romney.