11.2 The chi-square distribution: test of independence  (Page 3/4)

The observed table and the question at the end of the problem, "Are the number of hours volunteered independent of the type of volunteer?" tell you this is a test ofindependence. The two factors are number of hours volunteered and type of volunteer . This test is always right-tailed.

$H_o$ : The number of hours volunteered is independent of the type of volunteer.

$H_a$ : The number of hours volunteered is dependent on the type of volunteer.

• $\frac{\mathrm{(225)(298)}}{\mathrm{(839)}}=90.57$ ; $\frac{\mathrm{(225)(379)}}{\mathrm{(839)}}=115.19$ ; $\frac{\mathrm{(225)(152)}}{\mathrm{(839)}}=49.24$ ;
• $\frac{\mathrm{(290)(298)}}{\mathrm{(839)}}=103.00$ ; $\frac{\mathrm{(290)(379)}}{\mathrm{(839)}}=131.00$ ; $\frac{\mathrm{(290)(162)}}{\mathrm{(839)}}=56.00$ ;
• $\frac{\mathrm{(194)(298)}}{\mathrm{(839)}}=104.42$ ; $\frac{\mathrm{(294)(379)}}{\mathrm{(839)}}=132.81$ ; $\frac{\mathrm{(294)(162)}}{\mathrm{(839)}}=56.77$ ;

The expected table is:

For example, the calculation for the expected frequency for the top left cell is

$E=\frac{\text{(row total)(column total)}}{\text{total number surveyed}}=\frac{255\cdot 298}{839}=90.57$

Calculate the test statistic: ${\chi }^2=12.99$

• ${\chi }^2=\underset{(i\cdot j)}{\Sigma }\frac{(\mathrm{Observed}-\mathrm{Expected}{)}^2}{\mathrm{Expected}}=\frac{(\mathrm{111}-\mathrm{90.57}{)}^2}{\mathrm{90.57}}+\frac{(\mathrm{96}-\mathrm{115.19}{)}^2}{\mathrm{115.19}}+\frac{(\mathrm{48}-\mathrm{49.24}{)}^2}{\mathrm{49.24}}+\frac{(\mathrm{96}-\mathrm{103.00}{)}^2}{\mathrm{103.00}}+$
• $\frac{(\mathrm{133}-\mathrm{131.00}{)}^2}{\mathrm{131.00}}+\frac{(\mathrm{61}-\mathrm{56.00}{)}^2}{\mathrm{56.00}}+\frac{(\mathrm{91}-\mathrm{104.42}{)}^2}{\mathrm{104.42}}+\frac{(\mathrm{150}-\mathrm{132.81}{)}^2}{\mathrm{132.81}}+\frac{(\mathrm{53}-\mathrm{56.77}{)}^2}{\mathrm{56.77}}$
• $=\frac{\mathrm{417.3849}}{\mathrm{90.57}}+\frac{\mathrm{368.2561}}{\mathrm{115.19}}+\frac{\mathrm{1.5376}}{\mathrm{49.24}}+\frac{4}{\mathrm{131.00}}+\frac{\mathrm{25}}{\mathrm{56.00}}+\frac{\mathrm{417.3849}}{\mathrm{90.57}}+\frac{\mathrm{180.0964}}{\mathrm{104.42}}+\frac{\mathrm{295.4961}}{\mathrm{132.81}}+\frac{\mathrm{14.2129}}{\mathrm{56.77}}$
• $=12.99$

Distribution for the test: ${\chi }_4^2$

$\text{df}=(\text{3 columns}-1)(\text{3 rows}-1)=\left(2\right)\left(2\right)=4$

Graph:

Probability statement: $\text{p-value}=P({\chi }^2>12.99)=0.0113$

Compare $\alpha$ and the $\text{p-value}$ : Since no $\alpha$ is given, assume $\alpha =0.05$ . $\text{p-value}=0.0113$ . $\alpha >\text{p-value}$ .

Make a decision: Since $\alpha >\text{p-value}$ , reject $H_o$ . This means that the factors are not independent. Because we reject the null hypothesis, we should look at the percentages. We are interested in whether there is a difference in the type of volunteers hours, so we are interested in the row percentages.

It appears that the community college students mainly volunteer for 1-3 hours on average (43.53% of them), while more 4-year college and non-students volunteer for 4-6 (45.86% and 51.02% respectively) hours on average.

Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the number of hours volunteered and the type of volunteer aredependent on one another.

For the above example, if there had been another type of volunteer, teenagers, what would the degrees of freedom be?