Chapter 10, Problem 10.1P

The following four $2\times 2$ tables, presented without labels or titles, provide simplified opportunities to practice your computational skills. Calculate chi square for each.

HINTS:

• Calculate the expected frequencies for each cell with Formula 10.2. Double check to make sure you are using the correct row and column marginals for each cell.

• It may he helpful to record the expected frequencies in table format—see Tables 10.4 and 10.8.

• Use a computational table to organize the calculation for Formula 10.1—see Tables 10.5 and 10.9.

• Follow the step-by-step instructions.

• Double-check to make sure you are using the correct values for each cell.

a.	$20$	$25$	$45$	c.	$25$	$15$	$40$
	$\frac{25}{45}$	$\frac{20}{45}$	$\frac{45}{90}$		$\frac{30}{55}$	$\frac{30}{45}$	$\frac{60}{100}$
b.	$10$	$15$	$25$	d.	$20$	$45$	$65$
	$\frac{20}{30}$	$\frac{30}{45}$	$\frac{50}{75}$		$\frac{15}{35}$	$\frac{20}{65}$	$\frac{35}{100}$

To determine

(a)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.12.

Explanation of Solution

Given:

The given table of information is,

a.	$20$	$25$	$45$
	$25$	$20$	$45$
	$45$	$45$	$90$

Description:

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.

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.

Calculation:

From the given information,

The row marginal is 45, the column marginal is 45 and N is 90.

The observed frequency is given as,

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

Substitute 45 for row marginal, 45 for column marginal and 90 for N in the above formula.

$\begin{array}{rll}{f}_e& =& \frac{\text{45}\times \text{45}}{90}\\ & =& 22.5\end{array}$

Consider the following table,

$6.25$	$f_o$	$f_e$	$f_o-f_e$	${\left(f_o-f_e\right)}^2$	${\left(f_o-f_e\right)}^2/f_e$
	20	22.5	$-2.5$	$6.25$	0.28
	25	22.5	$2.5$	$6.25$	0.28
	25	22.5	$2.5$	$6.25$	0.28
	20	22.5	$-2.5$	$6.25$	0.28
Total	90	90	0		${\chi }^2=1.12$

The value $f_o-f_e$ is obtained as,

Substitute 20 for $f_o$ and 22.5 for $f_e$ in the above formula.

$\begin{array}{rll}{f}_o-f_e& =& 20-22.5\\ & =& -2.5\end{array}$

Squaring the above obtained result,

$\begin{array}{rll}{\left(f_o-f_e\right)}^2& =& {\left(-2.5\right)}^2\\ & =& 6.25\end{array}$

Divide the above obtained result by $f_e$.

$\begin{array}{rll}\frac{{\left(f_o-f_e\right)}^2}{f_e}& =& \frac{6.25}{22.5}\\ & =& 0.28\end{array}$

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,

$\begin{array}{rll}{\chi }^2\left(obtained\right)& =& {\displaystyle \sum \frac{{\left(f_o-f_e\right)}^2}{f_e}}\\ & =& 0.28+0.28+0.28+0.28\\ & =& 1.12\end{array}$

Thus, the chi square value is 1.12.

Conclusion:

The chi square value is 1.12.

To determine

(b)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 0.0.

Explanation of Solution

Given:

The given table of information is,

b.	10	15	25
	20	30	50
	30	45	75

Description:

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.

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.

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 25 for row marginal, 30 for column marginal and 75 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_1& =& \frac{\text{25}\times \text{30}}{75}\\ & =& 10\end{array}$

Substitute 25 for row marginal, 45 for column marginal and 75 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_1& =& \frac{\text{25}\times 45}{75}\\ & =& 15\end{array}$

Substitute 50 for row marginal, 30 for column marginal and 75 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_1& =& \frac{\text{50}\times \text{30}}{75}\\ & =& 20\end{array}$

Substitute 50 for row marginal, 45 for column marginal and 75 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_1& =& \frac{\text{50}\times 45}{75}\\ & =& 30\end{array}$

Consider the following table,

$6.25$	$f_o$	$f_e$	$f_o-f_e$	${\left(f_o-f_e\right)}^2$	${\left(f_o-f_e\right)}^2/f_e$
	10	10	0	0	0
	15	15	0	0	0
	20	20	0	0	0
	30	30	0	0	0
Total	75	75	0	0	${\chi }^2=0.0$

The value $f_o-f_e$ is obtained as,

Substitute 10 for $f_o$ and 10 for $f_e$ in the above formula.

$\begin{array}{rll}{f}_o-f_e& =& 10-10\\ & =& 0\end{array}$

Squaring the above obtained result,

$\begin{array}{rll}{\left(f_o-f_e\right)}^2& =& {\left(0\right)}^2\\ & =& 0\end{array}$

Divide the above obtained result by $f_e$.

$\frac{{\left(f_o-f_e\right)}^2}{f_e}=0$

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,

$\begin{array}{rll}{\chi }^2\left(obtained\right)& =& {\displaystyle \sum \frac{{\left(f_o-f_e\right)}^2}{f_e}}\\ & =& 0+0+0+0\\ & =& 0.0\end{array}$

Thus, the chi square value is 0.0.

Conclusion:

The chi square value is 0.0.

To determine

(c)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.51.

Explanation of Solution

Given:

The given table of information is,

c.	25	15	40
	30	30	60
	55	45	100

Description:

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.

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.

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 40 for row marginal, 55 for column marginal and 100 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_1& =& \frac{\text{40}\times 55}{100}\\ & =& 22\end{array}$

Substitute 40 for row marginal, 45 for column marginal and 100 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_1& =& \frac{\text{40}\times 45}{100}\\ & =& 18\end{array}$

Substitute 60 for row marginal, 55 for column marginal and 100 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_1& =& \frac{\text{60}\times \text{55}}{100}\\ & =& 33\end{array}$

Substitute 60 for row marginal, 45 for column marginal and 100 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_1& =& \frac{\text{60}\times 45}{100}\\ & =& 27\end{array}$

Consider the following table,

$6.25$	$f_o$	$f_e$	$f_o-f_e$	${\left(f_o-f_e\right)}^2$	${\left(f_o-f_e\right)}^2/f_e$
	25	22	3	9	0.41
	15	18	$-3$	9	0.50
	30	33	$-3$	9	0.27
	30	27	3	9	0.33
Total	100	100	0		${\chi }^2=1.51$

The value $f_o-f_e$ is obtained as,

Substitute 25 for $f_o$ and 22 for $f_e$ in the above formula.

$\begin{array}{rll}{f}_o-f_e& =& 25-22\\ & =& 3\end{array}$

Squaring the above obtained result,

$\begin{array}{rll}{\left(f_o-f_e\right)}^2& =& {\left(3\right)}^2\\ & =& 9\end{array}$

Divide the above obtained result by $f_e$.

$\begin{array}{rll}\frac{{\left(f_o-f_e\right)}^2}{f_e}& =& \frac{9}{22}\\ & =& 0.41\end{array}$

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,

$\begin{array}{rll}{\chi }^2\left(obtained\right)& =& {\displaystyle \sum \frac{{\left(f_o-f_e\right)}^2}{f_e}}\\ & =& 0.41+0.50+0.27+0.33\\ & =& 1.51\end{array}$

Thus, the chi square value is 1.51.

Conclusion:

The chi square value is 1.51.

To determine

(d)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.43.

Explanation of Solution

Given:

The given table of information is,

d.	20	45	65
	15	20	35
	35	65	100

Description:

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.

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.

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 65 for row marginal, 35 for column marginal and 100 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_1& =& \frac{\text{65}\times 35}{100}\\ & =& 22.75\end{array}$

Substitute 65 for row marginal, 65 for column marginal and 100 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_1& =& \frac{65\times 65}{100}\\ & =& 42.25\end{array}$

Substitute 35 for row marginal, 35 for column marginal and 100 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_1& =& \frac{\text{35}\times \text{35}}{100}\\ & =& 12.25\end{array}$

Substitute 35 for row marginal, 65 for column marginal and 100 for N in equation $\left(1\right)$.

$\begin{array}{rll}{f}_1& =& \frac{\text{35}\times 65}{100}\\ & =& 22.75\end{array}$

Consider the following table,

$6.25$	$f_o$	$f_e$	$f_o-f_e$	${\left(f_o-f_e\right)}^2$	${\left(f_o-f_e\right)}^2/f_e$
	20	22.75	-2.75	7.5625	0.332417582
	45	42.25	2.75	7.5625	0.178994083
	15	12.25	2.75	7.5625	0.617346939
	20	22.75	$-2.75$	7.5625	0.332417582
Total	100	100	0		${\chi }^2=1.46$

The value $f_o-f_e$ is obtained as,

Substitute 20 for $f_o$ and 22.75 for $f_e$ in the above formula.

$\begin{array}{rll}{f}_o-f_e& =& 20-22.75\\ & =& -2.75\end{array}$

Squaring the above obtained result,

$\begin{array}{rll}{\left(f_o-f_e\right)}^2& =& {\left(-2.75\right)}^2\\ & =& 7.56\end{array}$

Divide the above obtained result by $f_e$.

$\begin{array}{rll}\frac{{\left(f_o-f_e\right)}^2}{f_e}& =& \frac{7.56}{22.75}\\ & =& 0.33\end{array}$

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,

$\begin{array}{rll}{\chi }^2\left(obtained\right)& =& {\displaystyle \sum \frac{{\left(f_o-f_e\right)}^2}{f_e}}\\ & =& 0.33+0.18+0.62+0.33\\ & =& 1.46\end{array}$

Thus, the chi square value is 1.46.

Conclusion:

The chi square value is 1.46.