91影视

The data for different sports and home/Visiting team wins is provided.

The level of significance 0.10.

The formula forexpected frequencyis,

\(E = \frac{{\left( {row\;total} \right)\left( {column\;total} \right)}}{{\left( {grand\;total} \right)}}\)

The table with row and column total for observed counts is represented as,

Theexpected frequency tableis represented as,

The hypotheses are formulated as,

\({H_0}:\)Home/visitor wins are independent of the sport.

\({H_1}:\)Home/visitor wins are dependent on the sport.

The value of the test statisticis computed as,

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {127 - 115.9714} \right)}^2}}}{{115.9714}} + \frac{{{{\left( {53 - 58.5714} \right)}^2}}}{{58.5714}} + ... + \frac{{{{\left( {42 - 41.0143} \right)}^2}}}{{41.0143}}\\ = 4.7372\end{aligned}\]

Therefore, the value of the test statistic is 4.7372.

The degrees of freedomare computed as,

\(\begin{aligned}{c}\left( {r - 1} \right)\left( {c - 1} \right) = \left( {2 - 1} \right)\left( {4 - 1} \right)\\ = 3\end{aligned}\)

Therefore, the degrees of freedom are 3.

From the chi-square table, the critical value to 3 degrees of freedom and at 0.10 level of significance 6.2514.

Therefore, the critical value is 6.2514

Since the critical value (6.2514) is greater than the value of the test statistic (4.7372). In this case, the null hypothesis fails to be rejected.

Therefore, the decision is to fail to reject the null hypothesis.

There issufficient evidence to support the claim that Home/visitor wins are independent of the sport.

Thus, the wins are not dependent on the sport.

Thus, the advantage is not effective for the wins of baseball teams.