91Ó°ÊÓ

The table with observed values:

$\begin{array}{|cccccc|}\hline & \text{French Toast}& \text{Pancakes}& \text{Waffles}& \text{Omelettes}& \text{Total}\\ \text{Men}& 47& 35& 28& 53& 163\\ \text{Women}& 65& 59& 55& 60& 239\\ \text{Total}& 112& 94& 83& 113& 402\\ \hline\end{array}$

The table with expected values:

$\begin{array}{|cccccc|}\hline & \text{French Toast}& \text{Pancakes}& \text{Waffles}& \text{Omelettes}& \text{Total}\\ \text{Men}& 45.41& 38.11& 33.65& 45.82& 163\\ \text{Women}& 65& 59& 55& 60& 239\\ \text{Total}& 66.59& 55.89& 49.35& 67.18& 402\\ \hline\end{array}$

We want to test these hypothesis:

$H_0$: The distribution for breakfast preferences is the same for men and women

$H_1$: The distribution for breakfast preferences is not the same for men and women

Since there are two genders and 4 types of breakfast, the number of degrees of freedom is

$(2-1)(4-1)=3$

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

${\mathrm{χ}}^2=\underset{i=1}{\overset{2}{∑}}\underset{j=1}{\overset{4}{∑}}\frac{{\left(O{b}_{i,j}-E{x}_{i,j}\right)}^2}{E{x}_{i,j}}=4.012533$

Using the applet for ${\mathrm{χ}}^2$distribution, we get that $p$-value is $0.26011$:

Chi-Square Distribution

$X~{\mathrm{χ}}_{\left(df\right)}^2$

We are given $\mathrm{Î\pm }=0.05$and $p>\mathrm{Î\pm }$. Thus means that we cannot reject the null hypothesis.

There is insufficient evidence to conclude that the distribution of breakfast ordered is different for men and women.