Question

A manager of a sports club keeps information concerning the main sport in which members participate and their ages. To test whether there is a relationship between the age of a member and his or her choice of sport, $643$members of the sports club are randomly selected. Conduct a test of independence.

$\begin{array}{|lllll|}\hline \text{Sport}& \mathbf{18}-\mathbf{25}& \mathbf{26}-\mathbf{30}& \mathbf{31}-\mathbf{40}& \mathbf{41}\text{and over}\\ \text{racquetball}& 42& 58& 30& 46\\ \text{tennis}& 58& 76& 38& 65\\ \text{swimming}& 72& 60& 65& 33\\ \hline\end{array}$

Short answer

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

We can conclude that there is evidence that the main sport in which a member participates and his age are not independent.

Step-by-step solution

Given Information

Table with observed values:

$\begin{array}{|cccccc|}\hline \text{Sport}& 18-25& 26-30& 31-40& 41\text{and over}& \text{Total}\\ \text{racquetball}& 42& 58& 30& 46& 176\\ \text{tennis}& 58& 76& 38& 65& 237\\ \text{swimming}& 72& 60& 65& 33& 230\\ \text{Total}& 172& 194& 133& 144& 643\\ \hline\end{array}$

Table with expected values:

$\begin{array}{|cccccc|}\hline \text{Sport}& 18-25& 26-30& 31-40& 41\text{and over}& \text{Total}\\ \text{racquetball}& 47.07932& 53.10109& 36.40435& 39.41524& 176\\ \text{tennis}& 63.39658& 71.50544& 49.02177& 53.07621& 237\\ \text{swimming}& 61.52411& 69.39347& 47.57387& 51.50855& 230\\ \text{Total}& 172& 194& 133& 144& 643\\ \hline\end{array}$

Explanation

We want to test these hypotheses:

$H_0$: The main sport in which member participates and his age are independent

$H_1$: The main sport in which member participates and his age are not independent

Since there are $3$types of main sports and $4$age groups, the number of degrees of freedom is

$(3-1)·(4-1)=6$

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

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

$=25.21448$

Final Answer

Using the applet for ${\chi }^2$distribution, we can see that $p$-value is $0.00031$:

Chi-Square Distribution

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

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