## A sports statistician is interested in determining if there is a relationship between the number of home team and visiting team losses and d

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.

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2022-02-09T04:21:54+00:00
2022-02-09T04:21:54+00:00 1 Answer
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## Answers ( )

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

[tex]p_v = P(\chi^2_{3} >3.29)=0.349[/tex]

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 conceptsA

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

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

dependentof the sport.The level of significance assumed for this case is [tex]\alpha=0.01[/tex]

The statistic to check the hypothesis is given by:

[tex]\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]

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

And the calculations are given by:

[tex]E_{1} =\frac{70*303}{526}=40.32[/tex]

[tex]E_{2} =\frac{254*303}{526}=146.32[/tex]

[tex]E_{3} =\frac{44*303}{526}=25.35[/tex]

[tex]E_{4} =\frac{158*303}{526}=91.02[/tex]

[tex]E_{5} =\frac{70*223}{526}=29.68[/tex]

[tex]E_{6} =\frac{254*223}{526}=107.68[/tex]

[tex]E_{7} =\frac{44*223}{526}=18.65[/tex]

[tex]E_{8} =\frac{158*223}{526}=66.98[/tex]

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:

[tex]\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[/tex]Now we can calculate the degrees of freedom for the statistic given by:

[tex]df=(rows-1)(cols-1)=(4-1)(2-1)=3[/tex]

The critical value would be: [tex]chi^2_{crit}=11.345[/tex] 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:

[tex]p_v = P(\chi^2_{3} >3.29)=0.349[/tex]

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.