Super Bowl and\({R^2}\)Let x represent years coded as 1, 2, 3, . . . for years starting in 1980, and let y represent the numbers of points scored in each Super Bowl from 1980. Using the data from 1980 to the last Super Bowl at the time of this writing, we obtain the following values of\({R^2}\)for the different models: linear: 0.147; quadratic: 0.255; logarithmic: 0.176; exponential: 0.175; power: 0.203. Based on these results, which model is best? Is the best model a good model? What do the results suggest about predicting the number of points scored in a future Super Bowl game?

The relation between y (number of points scored in each Super Bowl from 1980) and x (years starting from 1980 and coded as 1,2,3…..) is modeled by 5 different models. The \({R^2}\) value of each model is provided.

To determine which of the 5 models is the best fit for the given two variables, the model with the highest value of\({R^2}\)is considered the most appropriate.

The following are the\({R^2}\)values corresponding to the 5 models:

• Linear:\({R^2} = 0.147\)
• Quadratic:\({R^2} = 0.255\)
• Logarithmic:\({R^2} = 0.176\)
• Exponential:\({R^2} = 0.175\)
• Power:\({R^2} = 0.203\)

The highest value of\({R^2}\)is equal to 0.255 and corresponds to the quadratic model.

Thus, it can be concluded that the quadratic model is the best.

For a fitted model to be good, the value of\({R^2}\)should be substantially large.

Here, the value of\({R^2}\)corresponding to the quadratic model is very low, equal to 0.255.

Thus, the fitted quadratic model is not a good model.

Since the model is not a good model, the predicted values obtained from this model are not accurate.

Thus, it is not advisable to use the quadratic model to predict the number of points scored in a future Super Bowl game.