Question

In​ soccer, serious fouls result in a penalty kick with one kicker and one defending goalkeeper. The accompanying table summarizes results from 295 kicks during games among top teams. In the​ table, jump direction indicates which way the goalkeeper​ jumped, where the kick direction is from the perspective of the goalkeeper. Use a 0.01 significance level to test the claim that the direction of the kick is independent of the direction of the goalkeeper jump. Do the results support the theory that because the kicks are so​ fast, goalkeepers have no time to​ react, so the directions of their jumps are independent of the directions of the​ kicks?

	left	center	right
Kick to left	51	5	44
Kick to center	41	12	32
kick to right	45	6	59

A. Determine the test statistic. x²

B. Use the​ chi-square test statistic to determine the​ P-value

C. Use the information from the previous steps to determine whether to reject or fail to reject the null hypothesis and interpret the results in the context of this problem

Answer 1

Solution:

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H₀: Two categorical variables are independent. (The direction of the kick is independent of the direction of the goalkeeper jump.)

Alternative hypothesis: H_a: Two categorical variables are dependent. (The direction of the kick is not independent of the direction of the goalkeeper jump.)

We are given level of significance = α = 0.01

A. Determine the test statistic. x²

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

We are given

Number of rows = r = 3

Number of columns = c = 3

Degrees of freedom = df = (r – 1)*(c – 1) = 2*2 = 4

α = 0.01

Critical value = 13.2767

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Column variable
Row variable	Left	Center	right	Total
Kick to left	51	5	44	100
Kick to center	41	12	32	85
Kick to right	45	6	59	110
Total	137	23	135	295

Expected Frequencies
	Column variable
Row variable	Left	Center	right	Total
Kick to left	46.44068	7.79661	45.76271	100
Kick to center	39.47458	6.627119	38.89831	85
Kick to right	51.08475	8.576271	50.33898	110
Total	137	23	135	295

Calculations
(O - E)
4.559322	-2.79661	-1.76271
1.525424	5.372881	-6.89831
-6.08475	-2.57627	8.661017

(O - E)^2/E
0.447612	1.003132	0.067897
0.058947	4.356019	1.22336
0.724759	0.7739	1.490162

Test Statistic = Chi square = ∑[(O – E)^2/E] = 10.14579

χ² statistic = 10.14579

B. Use the​ chi-square test statistic to determine the​ P-value

P-value = 0.038042

(By using Chi square table or excel)

C. Use the information from the previous steps to determine whether to reject or fail to reject the null hypothesis and interpret the results in the context of this problem

P-value > α = 0.01

So, we do not reject the null hypothesis H₀.

Fail to reject the null hypothesis H₀.

There is sufficient evidence to conclude that the direction of the kick is independent of the direction of the goalkeeper jump.