## Related questions with answers

The article “Human Lateralization from Head to Foot: Sex-Related Factors” (Science, 1978: 1291–1292) reports for both a sample of right-handed men and a sample of right-handed women the number of individuals whose feet were the same size, had a bigger left than right foot (a difference of half a shoe size or more), or had a bigger right than left foot.

$\begin{array}{|c|c|c|c|c|} \hline & & & & \text { Sample } \\ & \mathbf{L}>\mathbf{R} & \mathbf{L}=\mathbf{R} & \mathbf{L}<\mathbf{R} & \text { Size } \\ \hline \text { Men } & 2 & 10 & 28 & 40 \\ \hline \text { Women } & 55 & 18 & 14 & 87 \\ \hline \end{array}$

Does the data indicate that gender has a strong effect on the development of foot asymmetry? State the appropriate null and alternative hypotheses, compute the value of $\chi^{2}$, and obtain information about the P-value.

Solution

VerifiedTo test whether or not there is any effect of gender on the development of foot asymmetry, we will use the chi-square test of homogeneity (since the samples have been taken from two separate populations $-$ of men and women).

Let $\pi_{ij}$ be the probability of falling into the category described by the $i$-th row and $j$-th column of the given table (for instance, $\pi_{11}$ is the probability that a randomly selected man would have left foot bigger than right foot), for all $i \in \{1,2\}$ and $j \in \{1,2,3\}.$

We want to test:

$H_0: \pi_{1, j} = \pi_{2, j}, \text{ for all } j \in \{1,2,3\}$

$H_1: \text{Not all probabilities are as specified in } H_0$

at significance level $\alpha=0.05$ (our choice).

More informally, the above hypotheses can be stated as

$H_0':$ Gender and the development of foot asymmetry are not related

$H_1':$ Gender and the development of foot asymmetry are related.

First we need to make sure the assumptions required for carrying out the chi-square test are met, i.e. we need to make sure that at least $80\%$ of cells (i.e. at least $6\cdot 80\% = 4.8,$ i.e. at least 5 cells) have expected counts greater than 5.

Let $O_{ij}$ be the observed count in row $i$ and column $j,$ let $R_i$
be the row total and let $C_j$ be the column total, for every
$i \in \{1,2\}$ and $j \in \{1,2,3\}.$

Let also $N$ be the total sample size (i.e. the "total of totals").

The expected counts, denoted by $E_{ij}$ are calculated as

$E_{ij} = \frac{R_i \cdot C_j}{N}, \text{ for all } i \in \{1,2\} \text{ and } j \in \{1,2,3\} .$

So, for instance, the expected count in the top left cell is

$E_{11} = \frac{40\cdot (2+55)}{127} = \frac{40\cdot 57}{127} = 17.95.$

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