Chapter 17, Problem 21CAP

1.

To determine

Find the Chi-square test for independence at 0.05 significance level.

Check whether to retain or reject the null hypothesis test by using the chi-square test for independence.

Answer to Problem 21CAP

The value of test statistic is 6.482.

The decision is to reject the null hypothesis.

The cravings for cocaine and relapse are related.

Explanation of Solution

Calculation:

The given information is that, a study was conducted by the tests of cravings for

cocaine and relapse. The summarized table is,

		Relapse		Totals
		Yes	No
Craving	Yes	20	10	30
	No	8	17	25
	Totals	28	27	$N=55$

Table 1

The formula for total participants is,

$N={\displaystyle \sum _{i=1}^n{f}_{o_i}}$

In the formula, $f_o$ denotes the observed frequencies for each category.

The formula for expected frequency is,

$f_e=\frac{\text{row total}\times \text{column total}}{N}$

In the formula, N denotes the total participants.

Test statistic:

The formula for chi-square test statistic is,

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

In the formula, $f_o$ denotes the observed frequency and $f_e$ denotes the expected frequency.

The formula for degrees of freedom for chi-square independence test is,

$df=\left(k_1-1\right)\left(k_2-1\right)$

In the formula, $k_1$ denotes the number of levels of first categorical variable and $k_2$ denotes the number of levels of second categorical variable.

Null hypothesis:

$H_0:$ The cravings for cocaine and relapse are independent.

Alternative hypothesis:

$H_1:$ The cravings for cocaine and relapse are related.

Expected frequency:

Substitute the corresponding values of roe totals, column totals and total frequencies for each level of categorical variable in the expected frequency formula.

Expected frequency		Relapse		Totals
		Yes	No
Craving	Yes	$\frac{30\times 28}{55}=15.3$	$\frac{30\times 27}{55}=14.7$	30
	No	$\frac{25\times 28}{55}=12.7$	$\frac{25\times 27}{55}=12.3$	25
	Totals	28	27	$N=55$

Table 2

Substitute, $k_1=2,k_2=2$ in degrees of freedom formula for chi-square independence test.

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

Critical value:

The given significance level is $\alpha =0.05$ and the degrees of freedom are 1.

From the Appendix C: Table C.7 Critical values for Chi-square:

• Locate the value 1 in the degrees of freedom (df) column.
• Locate the 0.05 in level of significance row.
• The intersecting value that corresponds to the 1 with level of significance 0.05 is 3.84.

The critical value for $df=1$ at a 0.05 level of significance is 3.84.

Test statistic value:

Substitute the corresponding values of observed and expected frequencies for each level of categorical variable in the test statistic formula.

$\begin{array}{c}{\chi }_{\text{obt}}^2=\frac{{\left(20-15.3\right)}^2}{15.3}+\frac{{\left(10-14.7\right)}^2}{14.7}+\frac{{\left(8-12.7\right)}^2}{12.7}+\frac{{\left(17-12.3\right)}^2}{12.3}\\ =\frac{22.09}{15.3}+\frac{22.09}{14.7}+\frac{22.09}{12.7}+\frac{22.09}{12.3}\\ =1.444+1.503+1.739+1.796\\ =6.482\end{array}$

Hence, the value of test statistic is 6.482.

Conclusion:

The test statistic value is 6.482.

The critical value is 3.84.

The test statistic value is greater than the critical value.

The null hypothesis is rejected.

The cravings for cocaine and relapse are related.

2.

To determine

Find the effect size using $\varphi$.

Find the effect size using Cramer’s V.

Answer to Problem 21CAP

The effect size using $\varphi$ is 0.34.

The effect size using Cramer’s V is 0.34.

Explanation of Solution

Calculation:

The test statistic is 6.482, and value of N is 55.

The formula for Phi coefficient is,

$\varphi =\sqrt{\frac{{\chi }^2}{N}}$

In the formula, ${\chi }^2$ denotes the calculated test statistic, N denotes the total participants.

The formula for effect size using Cramer’s V is,

$V=\sqrt{\frac{{\chi }^2}{N\times d{f}_{\text{Smaller}}}}$

In the formula, ${\chi }^2$ denotes the calculated test statistic, N denotes the total participants and $d{f}_{\text{Smaller}}$ denotes the smaller value for two sets of degrees of freedom, that is smaller of $\left(k_1-1\right)$ and $\left(k_2-1\right)$.

Substitute, $N=55$, and ${\chi }^2=6.482$ in effect size formula using Phi coefficient.

$\begin{array}{c}\varphi =\sqrt{\frac{6.482}{55}}\\ =\sqrt{0.1179}\\ =0.34\end{array}$

Hence, effect size using $\varphi$ is 0.34.

For $2\times 2$ chi-square, the degrees of freedom are $1\left(=2-1\right)$, and $1\left(=2-1\right)$.

The smaller degrees of freedom are 1.

Substitute, $N=55$, ${\chi }^2=6.482$, and $d{f}_{\text{smaller}}=1$ in effect size formula using Cramer’s V.

$\begin{array}{c}V=\sqrt{\frac{6.482}{55\times 1}}\\ =\sqrt{\frac{6.482}{55}}\\ =\sqrt{0.1179}\\ =0.34\end{array}$

Hence, effect size using Cramer’s V is 0.34.