91Ó°ÊÓ

A sports club manager keeps track of the main sport that members participate in as well as their ages. $643$members of the sports club were randomly selected to see if there is a link between a member's age and his or her preferred sport.

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

Null hypothesis$H_0$is a zero. The age of a member and the main sport he or she participates in are independent.

Alternate hypothesis$H_a$is the age of a member and the main sport he or she participates in are also dependent.

Let's check the expected value table:

$\begin{array}{|lllll|}\hline \text{Sport}& 18-25& 26-30& 31-40& 41\text{and over}\\ \text{Racquetball}& 47.07& 53.10& 36.40& 39.42\\ \text{Tennis}& 60.72& 68.49& 46.95& 50.84\\ \text{swimming}& 61.52& 69.39& 47.57& 51.51\\ \hline\end{array}$

The expected value is $47.07$as follows,

=$\frac{172×176}{643}$

= $\mathbf{47}\mathbf{.}\mathbf{07}$

The test statistic is then calculated using equation given in step 3.

The test statistic$25.21$

The table provides the$P$value. The area of the probability Vs. variance curve after is the P-value. If the p value in the table is between $0$and $0.05$, the p value can be calculated using interpolation.

$0.0003$s the$P$value.

Alpha is$0.05$
Decision: null hypothesis rejected
Reason for decision = $(0.0003)<\text{alpha}(0.05)$
Conclusion:
The Conclusion is that Sport and age are dependent.