Question

A sports statistician is interested in determining if there is a relationship between the number of home team and visiting team losses and different sports. A random sample of 526 games is selected and the results are given below. Find the critical value chi Subscript alpha Superscript 2 to test the claim that the number of home team and visiting team losses is independent of the sport. Use alphaequals0.01. Round to three decimal places.

Answer 1

Answer:

The critical value would be: $\chi^2_{crit}=11.345$ and we use the following excel code to find it: "=CHISQ.INV(1-0.01,3)"

$p_v = P(\chi^2_{3} >3.29)=0.349$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(3.29,3,TRUE)"

Since the p value is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance that the two variables are independent.

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

                      F          B         S      Bs     Total

home wins    39      156      25      83      303

Visitor wins   31       98        19      75       223

Total              70      254     44      158      526

We need to conduct a chi square test in order to check the following hypothesis:

H0: The number of home team and visiting team losses is independent of the sport.

H1: The number of home team and visiting team losses is dependent of the sport.

The level of significance assumed for this case is $\alpha=0.01$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{70*303}{526}=40.32$

$E_{2} =\frac{254*303}{526}=146.32$

$E_{3} =\frac{44*303}{526}=25.35$

$E_{4} =\frac{158*303}{526}=91.02$

$E_{5} =\frac{70*223}{526}=29.68$

$E_{6} =\frac{254*223}{526}=107.68$

$E_{7} =\frac{44*223}{526}=18.65$

$E_{8} =\frac{158*223}{526}=66.98$

And the expected values are given by:

                        F           B            S          Bs     Total

home wins    40.32   146.32    25.35    91.02    303

Visitor wins   29.68   107.68    18.65    66.98    223

Total                70         254         44        158      526

And now we can calculate the statistic:

$\chi^2 = \frac{(39-40.32)^2}{40.32}+\frac{(156-146.32)^2}{146.32}+\frac{(25-25.35)^2}{25.35}+\frac{(83-91)^2}{91}+\frac{(31-29.68)^2}{29.68}+\frac{(98-107.68)^2}{107.68}+\frac{(19-18.65)^2}{18.65}+\frac{(75-66.98)^2}{66.98} =3.29$Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(4-1)(2-1)=3$

The critical value would be: $chi^2_{crit}=11.345$ and we use the following excel code to find it: "=CHISQ.INV(1-0.01,3)"

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >3.29)=0.349$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(3.29,3,TRUE)"

Since the p value is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance that the two variables are independent.