Question

In soccer, serious fouls in the penalty box result in a penalty kick withone kicker and one defending goalkeeper. The table below summarizes results from 286 kicksduring games among top teams (based on data from “Action Bias Among Elite Soccer Goalkeepers:

The Case of Penalty Kicks,” by Bar-Eli et al., Journal of Economic Psychology,Vol.28, No. 5). In the table, jump direction indicates which way the goalkeeper jumped, where thekick direction is from the perspective of the goalkeeper. Use a 0.05 significance level to test theclaim that the direction of the kick is independent of the direction of the goalkeeper jump. Dothe results support the theory that because the kicks are so fast, goalkeepers have no time toreact, so the directions of their jumps are independent of the directions of the kicks?

	Goalkeeper Jump
	Left	Center	Right
Kick to Left	54	1	37
Kick to Center	41	10	31
Kick to Right	46	7	59

Short answer

The direction of the kick is dependent on the direction of the goalkeeper’s jump. Thus, the result is supportive of the theory.

Step-by-step solution

Given information

The data for the direction of kicks and the goalkeeper’s jump is recorded.

The level of significance is 0.05.

Compute the expected frequencies

Theexpected frequencyis computed as,

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

The table with row and column total is represented as,

	Left	Center	Right	Row Total
Kick to Left	54	1	37	92
Kick to Center	41	10	31	82
Kick to Right	46	7	59	112
Column Total	141	18	127	286

Theexpected frequency tableis represented as,

	Left	Center	Right
Kick to Left	45.3566	5.7902	40.8531
Kick to Center	40.4266	5.1608	36.4126
Kick to Right	55.2168	7.0490	49.7343

All the expected frequencies are above 5, and hence the requirement for the test is met assuming the sampling is done randomly.

State the null and alternate hypothesis

To test the independence of kick direction on goalkeeper’s jump, the hypothesis is formulated as follows:

\({H_0}:\)The direction of the kick is independent of the direction of the goalkeeper jump.

\({H_1}:\)The direction of the kick is dependent of the direction of the goalkeeper jump.

Compute the test statistic

The value of the test statisticis computed as,

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {54 - 45.3566} \right)}^2}}}{{45.3566}} + \frac{{{{\left( {1 - 5.7902} \right)}^2}}}{{5.7902}} + ... + \frac{{{{\left( {59 - 49.7343} \right)}^2}}}{{49.7343}}\\ = 14.5887\\ \approx 14.589\end{aligned}\]

Therefore, the value of the test statistic is 14.589.

Compute the degrees of freedom

The degrees of freedomare computed as,

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

Therefore, the degrees of freedom are 4.

Compute the critical value

From chi-square table, the critical value for row corresponding to 4 degrees of freedom and at 0.05 level of significance 9.488.

Therefore, the critical value is 9.488.

Also, the p-value is computed as 0.0056

State the decision

Since the critical (9.488) is less than the value of test statistic (14.589), in this case, the null hypothesis is rejected.

Therefore, the decision is to reject the null hypothesis.

State the conclusion

There isnot sufficient evidence to favour the claimthatthe direction of the kick is independent of the direction of the goalkeeper jump.

Therefore, the result is supportive of the theory that the direction of kicks is dependent to jumps of goalkeeper.