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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 鈥淎ction Bias Among Elite Soccer Goalkeepers:

The Case of Penalty Kicks,鈥 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

Expert verified

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

Step by step solution

01

Given information

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

The level of significance is 0.05.

02

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.

03

State the null and alternate hypothesis

To test the independence of kick direction on goalkeeper鈥檚 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.

04

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.

05

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.

06

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

07

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.

08

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.

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