Chapter 17, Problem 23CAP

(a)

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 23CAP

The value of test statistic is 4.855.

The decision is to retain the null hypothesis.

The noise level and exam grades during an exam are independent.

Explanation of Solution

Calculation:

The given information is that, a study was conducted by the tests of loudness of noise during an exam (low, medium, high) and exam grades (pass, fail). The summarized table is,

		Noise level			Totals
		Low	Medium	High
Exam	Pass	20	18	8	46
	Fail	8	6	10	24
	Totals	28	24	18	$N=70$

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 noise level and exam grades during an exam are independent.

Alternative hypothesis:

$H_1:$ The noise level and exam grades during an exam 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.

		Noise level			Totals
		Low	Medium	High
Exam	Pass	$\frac{46\times 28}{70}=18.4$	$\frac{46\times 24}{70}=15.8$	$\frac{46\times 18}{70}=11.8$	46
	Fail	$\frac{24\times 28}{70}=9.6$	$\frac{24\times 24}{70}=8.2$	$\frac{24\times 18}{70}=6.2$	24
	Totals	28	24	18	$N=70$

Table 2

Substitute, $k_1=2,k_2=3$ in degrees of freedom formula for chi-square independence test.

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

Critical value:

The given significance level is $\alpha =0.05$ and the degrees of freedom are 2.

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

• Locate the value 2 in the degrees of freedom (df) column.
• Locate the 0.05 in level of significance row.
• The intersecting value that corresponds to the 2 with level of significance 0.05 is 5.99.

The critical value for $df=2$ at a 0.05 level of significance is 5.99.

Test statistic value:

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

$\begin{array}{c}{\chi }_{\text{obt}}^2=\frac{{\left(20-18.4\right)}^2}{18.4}+\frac{{\left(18-15.8\right)}^2}{15.8}+\frac{{\left(8-11.8\right)}^2}{11.8}+\frac{{\left(8-9.6\right)}^2}{9.6}+\frac{{\left(6-8.2\right)}^2}{8.2}+\frac{{\left(10-6.2\right)}^2}{6.2}\\ =\frac{2.56}{18.4}+\frac{4.84}{15.8}+\frac{14.44}{11.8}+\frac{2.56}{9.6}+\frac{4.84}{8.2}+\frac{14.4}{6.2}\\ =0.139+0.306+1.224+0.267+0.590+2.329\\ =4.855\end{array}$

Hence, the value of test statistic is 4.855.

Conclusion:

The test statistic value is 4.855.

The critical value is 5.99.

The test statistic value is less than the critical value.

The null hypothesis is retained.

The noise level and exam grades during an exam are independent.

(b)

To determine

Find the effect size using Cramer’s V.

Answer to Problem 23CAP

The effect size using Cramer’s V is 0.26.

Explanation of Solution

Calculation:

The test statistic is 4.855, and value of N is 70.

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 $\left(k_2-1\right)$.

For $2\times 3$ chi-square, the degrees of freedom are $1\left(=2-1\right)$, and $2\left(=3-1\right)$.

The smaller degrees of freedom are 1.

Substitute, $N=70$, ${\chi }^2=4.855$, and $d{f}_{\text{smaller}}=1$ in effect size formula using Cramer’s V.

$\begin{array}{c}V=\sqrt{\frac{4.855}{70\times 1}}\\ =\sqrt{\frac{4.855}{70}}\\ =\sqrt{0.0694}\\ =0.26\end{array}$

Hence, effect size using Cramer’s V is 0.26.