Chapter 3.2, Problem 66E

(a)

To determine

To state the equation of the least squares regression line if each player had the same batting average the rest of the season as he did in the first month of the season.

Answer to Problem 66E

  $\widehat{y}=x$ .

Explanation of Solution

Let x be batting average during the first month of the season and y be the batting average during the season.

If each player has the same batting average in the first month as in the rest of the season then the data of the explanatory variable x will be the same as the data for the response variable y . If the data of the two variables are the same then the linear model will predict that the two variables are equal and thus the linear model will assume that the predicted y -variable is equal to the x -variable. Thus, we have the least square regression line as:

  $\widehat{y}=x$ .

(b)

To determine

To predict rest of the season batting average for a player who had a $0.200$ batting average the first month of the season and for a player who had a $0.400$ batting average the first month of the season.

Answer to Problem 66E

The predicted rest of the season batting average for a player who had a $0.200$ batting average the first month of the season is $0.2668$ .

The predicted rest of the season batting average for a player who had a $0.400$ batting average the first month of the season is $0.2886$ .

Explanation of Solution

It is given in the question the least squares regression line as:

  $\widehat{y}=0.245+0.109x$

Thus, we will predict rest of the season batting average for a player who had a $0.200$ batting average the first month of the season by:

  $\begin{array}{l}\widehat{y}=0.245+0.109x\\ =0.245+0.109(0.200)\\ =0.2668\end{array}$

Thus, the predicted rest of the season batting average for a player who had a $0.200$ batting average the first month of the season is $0.2668$ .

And we will predict rest of the season batting average for a player who had a $0.400$ batting average the first month of the season by:

  $\begin{array}{l}\widehat{y}=0.245+0.109x\\ =0.245+0.109(0.400)\\ =0.2886\end{array}$

Thus, the predicted rest of the season batting average for a player who had a $0.400$ batting average the first month of the season is $0.2886$ .

(c)

To determine

To explain how your answers to part (b) illustrate regression to the mean.

Explanation of Solution

From part (b), we have,

The predicted rest of the season batting average for a player who had a $0.200$ batting average the first month of the season is $0.2668$ .

The predicted rest of the season batting average for a player who had a $0.400$ batting average the first month of the season is $0.2886$ .

Thus, we estimate that the mean of the batting average will be somewhere between $0.200\text{ and 0}\text{.300}$ , as the batting average above $\text{0}\text{.300}$ is considered very good and we expect most of the player’s to be good while few are very good. The player with batting average during the first month of $0.200$ then has a batting average during the first month lower than the mean. And the player with a batting average during the first month of $0.400$ then has a batting average during the first month higher than the mean. Finally, we also note that player with batting average during the first month of $0.200$ has the higher predicted higher predicted batting average on the rest of the season, while the player with a batting average during the first month of $0.400$ has a lower predicted batting average on the rest of the season, which indicates that the prediction on the batting average appears to be closer to the mean.