The problem describes a situation where we want to test if the draft rankings of a fantasy football league have any predictive value for the final rankings of teams at the end of the season. We are given the draft projection and final rankings for 16 teams, along with $SS_M = 2.65$ and $SS_E = 337.35$. We need to determine if there is a statistically significant predictive relationship.

1. Problem Description

The problem describes a situation where we want to test if the draft rankings of a fantasy football league have any predictive value for the final rankings of teams at the end of the season. We are given the draft projection and final rankings for 16 teams, along with $SS_M = 2.65$ and $SS_E = 337.35$. We need to determine if there is a statistically significant predictive relationship.

2. Solution Steps

To test for a statistically significant predictive relationship, we will perform a regression analysis. We are given $SS_M$, which represents the Sum of Squares due to the model, and $SS_E$, which represents the Sum of Squares due to error. We will calculate the F-statistic and compare it to a critical value to determine significance.

First, we need to calculate the degrees of freedom.

The total number of teams is $n = 16$.

The degrees of freedom for the model ($df_M$) is $p-1$, where $p$ is the number of predictors. In this case, we only have one predictor (draft projection), so $df_M = 1 - 1 = 1$. This corresponds to using a single predictor variable, the draft ranking. However, since the problem states "assume $SS_M$ = 2.65", we will proceed, using $df_M = 1$.

The degrees of freedom for the error ($df_E$) is $n - p$, where $n$ is the number of observations (teams) and $p$ is the number of parameters in the model (including the intercept). In this case, $df_E = 16 - 2 = 14$. This is because we have 16 data points and we are estimating an intercept and a slope.

Next, we calculate the Mean Square for the Model ($MS_M$) and the Mean Square for Error ($MS_E$).

$MS_M = \frac{SS_M}{df_M} = \frac{2.65}{1} = 2.65$

$MS_E = \frac{SS_E}{df_E} = \frac{337.35}{14} = 24.0964$

Now, we calculate the F-statistic:

$F = \frac{MS_M}{MS_E} = \frac{2.65}{24.0964} = 0.11$

To determine statistical significance, we need to compare the calculated F-statistic to a critical F-value with $df_M = 1$ and $df_E = 14$. The F-critical value depends on the chosen alpha level (significance level). Let's assume an alpha level of 0.

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5. Using an F-distribution table or calculator, the F-critical value for $df_M = 1$ and $df_E = 14$ at $\alpha = 0.05$ is approximately 4.

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Since our calculated F-statistic (0.11) is less than the F-critical value (4.60), we fail to reject the null hypothesis. This means there is no statistically significant predictive relationship between the draft rankings and the final rankings at the 0.05 significance level.

3. Final Answer

There is no statistically significant predictive relationship between the draft rankings and the final rankings. The calculated F-statistic is 0.

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1. Assuming a significance level of 0.05, the critical F-value is approximately 4.

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0. Since 0.11 < 4.60, we fail to reject the null hypothesis.