Question

An ice cream maker performs a nationwide survey about favorite flavors of ice cream in different geographic areas of the U.S. Based on the Table, do the numbers suggest that geographic location is independent of favorite ice cream flavors? Test at the $5\%$significance level.

$\begin{array}{|cccccccc|}\hline USRegion& \text{Stra}& \text{Choco}& \text{Van}& \text{RockyRoad}& \text{MintChocolateChip}& \text{Pistachio}& \text{Total}\\ \text{West}& 12& 21& 22& 19& 15& 8& 97\\ \text{Midwest}& 10& 32& 22& 11& 15& 6& 96\\ \text{East}& 8& 31& 27& 8& 15& 7& 96\\ \text{South}& 15& 28& 30& 8& 15& 6& 102\\ \text{Total}& 45& 112& 101& 46& 60& 27& 391\\ \hline\end{array}$

Short answer

Taking $\alpha =0.05$, we can see that $p>\alpha$. This means that we cannot reject the null hypothesis.

Step-by-step solution

Given Information

The table with observed values:

$\begin{array}{|cccccccc|}\hline & \text{Stra}& \text{Choco}& \text{Van}& \text{R R}& \text{M C C}& \text{Pist}& \text{Total}\\ \text{West}& 12& 21& 22& 19& 15& 8& 97\\ \text{Midwest}& 10& 32& 22& 11& 15& 6& 96\\ \text{East}& 8& 31& 27& 8& 15& 7& 96\\ \text{South}& 15& 28& 30& 8& 15& 6& 102\\ \text{Total}& 45& 112& 101& 46& 60& 27& 391\\ \hline\end{array}$

The table with expected values:

$\begin{array}{|cccccccc|}\hline & \text{Stra}& \text{Choco}& \text{Van}& \text{R R}& \text{M C C}& \text{Pist}& \text{Total}\\ \text{West}& 11.16& 27.79& 25.056& 11.41& 14.88& 6.7& 97\\ \text{Midwest}& 11.05& 27.5& 24.8& 11.29& 14.73& 6.63& 96\\ \text{East}& 11.05& 27.5& 24.8& 11.29& 14.73& 6.63& 96\\ \text{South}& 11.74& 29.22& 26.35& 12& 15.65& 7.04& 102\\ \text{Total}& 45& 112& 101& 46& 60& 27& 391\\ \hline\end{array}$

Explanation

We want to test the following hypotheses:

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

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

Since there are $4$regions and $6$flavors, the number of degrees of freedom is

$(4-1)(6-1)=15$

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

${\chi }^2=\sum _{i=1}^4\sum _{j=1}^6\frac{{\left(O{b}_{i,j}-E{x}_{i,j}\right)}^2}{E{x}_{i,j}}=14.06317$

Graph

Using the applet for ${\chi }^2$distribution, we get that $p$-value is $0.52074$:

Chi-Square Distribution

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