Question

A $2013$poll in California surveyed people about taxing sugar-sweetened beverages. The results are presented in the Table and are classified by ethnic group and response type. Are the poll responses independent of the participants’ ethnic group? Conduct a test of independence at the $5\%$significance level.

$\begin{array}{|llllll|}\hline \begin{array}{l}\text{Opinionl}\\ \text{Ethnicity}\end{array}& \begin{array}{l}\text{Asian-}\\ \text{American}\end{array}& \begin{array}{l}\text{White/Non-}\\ \text{Hispanic}\end{array}& \begin{array}{l}\text{African-}\\ \text{American}\end{array}& \text{Latino}& \begin{array}{l}\text{Row}\\ \text{Total}\end{array}\\ \text{Against tax}& 48& 433& 41& 160& 682\\ \text{In Favor of tax}& 54& 234& 24& 147& 459\\ \text{No opinion}& 16& 43& 16& 19& 94\\ \text{Column Total}& 118& 710& 81& 326& 1235\\ \hline\end{array}$

Short answer

Taking alpha $=0.05$, we can see that $p<\alpha$. This means that we reject null hypothesis.

Step-by-step solution

Given Information

The table with observed values:

$\begin{array}{|cccccc|}\hline & \text{Asian Am}& \text{White/NH}& \text{Af Am}& \text{Latino}& \text{Total}\\ \text{Against tax}& 48& 433& 41& 160& 628\\ \text{In Favor of tax}& 54& 234& 24& 147& 459\\ \text{No opinion}& 16& 43& 16& 19& 84\\ \text{Total}& 118& 710& 71& 272& 1171\\ \hline\end{array}$

The table with expected values:

$\begin{array}{|cccccc|}\hline & \text{Asian Am}& \text{White/NH}& \text{Af Am}& \text{Latino}& \text{Total}\\ \text{Against tax}& 63.28& 380.77& 38.07686& 145.87& 628\\ \text{In Favor of tax}& 46.25& 278.3& 27.8& 106.62& 459\\ \text{No opinion}& 8.46& 50.93& 5.09& 19.51& 84\\ \text{Total}& 118& 710& 71& 272& 1171\\ \hline\end{array}$

Hypotheses test

We want to test these hypothesis:

$H_0$: Geographic location is independent of favorite ice cream flavors

$H_1$: Geographic location depends on favorite ice cream flavors

Since there are $3$regions and $4$flavors, the number of degrees of freedom is

$(3-1)(4-1)=6$

We are using ${\chi }^2$distribution. Test statistic is given by:

${\chi }^2=\sum _{i=1}^3\sum _{j=1}^4\frac{{\left(O{b}_{i,j}-E{x}_{i,j}\right)}^2}{E{x}_{i,j}}=67.93506$

Graph

Using the applet for ${\chi }^2$distribution, we get that $p$-value is $0$:

Chi-Square Distribution

$X~{\chi }_{\left(df\right)}^2$

Graphical explanation

Taking alpha $=0.05$, we can see that $p<\alpha$. This means that we reject null hypothesis.

There is evidence to conclude that the poll responses are independent of the participants' ethnic group.