A sports researcher is interested in determining if there is a relationship between the number of home team and visiting team wins and different sports. A random sample of 526 games is selected and the results are given below. Calculate the chi-square test statistic $\chi ^ { 2 }$ used to test the claim that the number of home team and visiting team wins is independent of the sport. Use $\alpha = 0.01$.
$\begin{array}{l|cccc}
& \text { Football } & \text { Basketball } & \text { Soccer } & \text { Baseball } \
\hline \text { Home team wins } & 39 & 156 & 25 & 83 \
\text { Visiting team wins } & 31 & 98 & 19 & 75
\end{array}$

A) 3.290
B) 2.919
C) 5.391
D) 4.192

Question

A sports researcher is interested in determining if there is a relationship between the number of home team and visiting team wins and different sports. A random sample of 526 games is selected and the results are given below. Calculate the chi-square test statistic $\chi ^ { 2 }$ used to test the claim that the number of home team and visiting team wins is independent of the sport. Use $\alpha = 0.01$.
$\begin{array}{l|cccc} 
& \text { Football } & \text { Basketball } & \text { Soccer } & \text { Baseball } \
\hline \text { Home team wins } & 39 & 156 & 25 & 83 \
\text { Visiting team wins } & 31 & 98 & 19 & 75
\end{array}$

A) 3.290
B) 2.919
C) 5.391
D) 4.192

Quizplus · Accepted Answer

The chi-square test statistic is calculated as:$\chi ^ { 2 } = \sum \frac{(O - E) ^ { 2 }}{E}$where O is the observed frequency and E is the expected frequency.The expected frequency is calculated as:$E = \frac{(\text { row total })(\text { column total })}{\text { grand total }}$The grand total is 526.The row totals are:Football: 70Basketball: 254Soccer: 44Baseball: 158The column totals are:Home team wins: 303Visiting team wins: 223The expected frequencies are:Football:Home team wins: $\frac{(70)(303)}{526} = 40.57$Visiting team wins: $\frac{(70)(223)}{526} = 29.43$Basketball:Home team wins: $\frac{(254)(303)}{526} = 145.64$Visiting team wins: $\frac{(254)(223)}{526} = 108.36$Soccer:Home team wins: $\frac{(44)(303)}{526} = 25.19$Visiting team wins: $\frac{(44)(223)}{526} = 18.81$Baseball:Home team wins: $\frac{(158)(303)}{526} = 91.59$Visiting team wins: $\frac{(158)(223)}{526} = 66.41$The chi-square test statistic is:$\chi ^ { 2 } = \frac{(39 - 40.57) ^ { 2 }}{40.57} + \frac{(31 - 29.43) ^ { 2 }}{29.43} + \frac{(156 - 145.64) ^ { 2 }}{145.64} + \frac{(98 - 108.36) ^ { 2 }}{108.36} + \frac{(25 - 25.19) ^ { 2 }}{25.19} + \frac{(19 - 18.81) ^ { 2 }}{18.81} + \frac{(83 - 91.59) ^ { 2 }}{91.59} + \frac{(75 - 66.41) ^ { 2 }}{66.41} = 3.290$The degrees of freedom are:$df = (r - 1)(c - 1) = (4 - 1)(2 - 1) = 3$The critical value of chi-square with 3 degrees of freedom and $\alpha = 0.01$ is 11.345.Since the chi-square test statistic (3.290) is less than the critical value (11.345), we fail to reject the null hypothesis and conclude that the number of home team and visiting team wins is independent of the sport.

A Sports Researcher Is Interested in Determining If There Is χ2\chi ^ { 2 }χ2

A Sports Researcher Is Interested in Determining If There Is $\chi ^ { 2 }$