A sports researcher aims to determine whether there is a relationship between the number of home and visiting team wins across various sports. A random sample of 526 games was selected, and the results are as follows. Calculate the critical value of Chi-square to test the hypothesis that the number of home and visiting team wins is independent of the sport, with a significance level of α = 0.01. The results are: - Football: Home wins 39, Visitor wins 31 - Basketball: Home wins 156, Visitor wins 98 - Soccer: Home wins 25, Visitor wins 19 - Baseball: Home wins 83, Visitor wins 75

To determine if there is a relationship between the number of home and visiting team wins across various sports, we'll use a chi-square test of independence. Let's begin by setting up a contingency table with the given data:

| | Home Wins | Visitor Wins | Total | |------------|-----------|--------------|-------| | Football | 39 | 31 | 70 | | Basketball | 156 | 98 | 254 | | Soccer | 25 | 19 | 44 | | Baseball | 83 | 75 | 158 | | Total | 303 | 223 | 526 |

Now, we can calculate the expected frequencies, which are based on the assumption that home and visitor wins are independent of sports. The expected frequency for each cell is calculated using the formula: E = (row total × column total) / grand total. Let's calculate this for each cell:

For football, Expected Home Wins = (70 * 303) / 526 ≈ 40.39 Expected Visitor Wins = (70 * 223) / 526 ≈ 29.61

For basketball, Expected Home Wins = (254 * 303) / 526 ≈ 145.95 Expected Visitor Wins = (254 * 223) / 526 ≈ 108.05

For soccer, Expected Home Wins = (44 * 303) / 526 ≈ 25.3 Expected Visitor Wins = (44 * 223) / 526 ≈ 18.7

For baseball, Expected Home Wins = (158 * 303) / 526 ≈ 91.36 Expected Visitor Wins = (158 * 223) / 526 ≈ 66.64

With the expected frequencies, we can now calculate the chi-square statistic using the formula: χ² = Σ((O - E)² / E) where O represents the observed frequency and E represents the expected frequency.

Let's do these calculations:

For football, χ² (Home) = (39 - 40.39)² / 40.39 ≈ 0.048 χ² (Visitor) = (31 - 29.61)² / 29.61 ≈ 0.067

For basketball, χ² (Home) = (156 - 145.95)² / 145.95 ≈ 0.702 χ² (Visitor) = (98 - 108.05)² / 108.05 ≈ 0.937

For soccer, χ² (Home) = (25 - 25.3)² / 25.3 ≈ 0.004 χ² (Visitor) = (19 - 18.7)² / 18.7 ≈ 0.005

For baseball, χ² (Home) = (83 - 91.36)² / 91.36 ≈ 0.789 χ² (Visitor) = (75 - 66.64)² / 66.64 ≈ 1.053

Sum all these values to get the total chi-square statistic: χ² = 0.048 + 0.067 + 0.702 + 0.937 + 0.004 + 0.005 + 0.789 + 1.053 ≈ 3.605

With our chi-square statistic calculated, we can now find the critical value from the chi-square distribution table. To determine the degrees of freedom for the test, use the formula (rows - 1) * (columns - 1). We have 4 sports (rows) and 2 categories of wins (columns), so our degrees of freedom are:

DF = (4 - 1) * (2 - 1) = 3 * 1 = 3

Consulting a chi-square distribution table with 3 degrees of freedom and a significance level of α = 0.01, we find the critical value. For α = 0.01 and 3 degrees of freedom, the critical chi-square value is approximately 11.345.

Since our calculated chi-square statistic of 3.605 is less than the critical value of 11.345, we would fail to reject the null hypothesis, concluding that there is not enough evidence at the 0.01 level of significance to suggest that the number of home and visitor wins is dependent on the sport.

Extra: A chi-square test is used to determine if there is a significant association between two categorical variables. In this case, we're looking at the association between the types of wins (home or visitor) and types of sports. The χ² statistic measures the discrepancy between the observed and expected frequencies. If the value of χ² is large, it suggests that the observed frequencies deviate significantly from the expected frequencies, and hence, there is an association between the variables.

The 'critical value' is a threshold against which the calculated χ² value is compared. If the calculated χ² is higher than the critical value, this indicates that the observed frequency distribution is significantly different from what would be expected under the null hypothesis, leading to its rejection. Conversely, if the calculated χ² is lower, we fail to reject the null hypothesis, suggesting that the observed distribution is not significantly different from what would be expected under the hypothesis of independence. The significance level (α) sets the threshold for how extreme the observed results must be to reject the null hypothesis. Common levels of significance are 0.05, 0.01, and 0.001.