Olympic figure skating For many people, the women’s figure skating competition is the highlight of the Olympic Winter Games. Scores in the short program x and scores in the free skate y were recorded for each of the 24 skaters who competed in both rounds during the 2010 Winter Olympics in Vancouver, Canada.28 Here is a scatterplot with least-squares regression line y^=−16.2+2.07x. For this model, s = 10.2 and r2 = 0.736.

The question explains the relationship between the short program and free skate scores. The regression line is as follows:

$\stackrel{⏜}{y}=-16.2+2.07x$

$Yu-Na$Kim received $78.50$in the short program and $150.06$in the free skate, which is impressive.

Thus,

we have,

$$\begin{gathered} x=-16.2 \\ y=2.07 \end{gathered}$$

Thus the predicted free skate score is as:

$$\begin{gathered} \stackrel{⏜}{y}=-16.2+2.07x \\ =-16.2+2.07(78.50) \\ =146.295 \end{gathered}$$

Thus, the residual will be calculated as:

$$\begin{gathered} Residual=y-\stackrel{⏜}{y} \\ =15.06-146.295 \\ =3.765 \end{gathered}$$

This means that while using the regression line to forecast $Yu-Na$ Kim's free skate score, we underestimated it by $3.765$

The relationship between the short program score and the free skate score is given in the question. And the regression line is as:

$\stackrel{⏜}{y}=-16.2+2.07x$

And also given that Yu-Na Kim scored $78.50$ in the short program and $150.06$

in the free skate.

Thus, we have,

$$\begin{gathered} x=-16.2 \\ y=2.07 \end{gathered}$$

As we know that the slope is the coefficient of $x$ in the least square regression equation and represents the average increase or decrease of $y$ per unit of $x$ Thus,

$b_1=2.07$

This means that the free skate score improved by $2.07$ points on average, while the short program score improved by one point.

The question explains the relationship between the short programme and free skate scores. The regression line is as follows:

$\stackrel{⏜}{y}=-16.2+2.07x$

And also given that Yu-Na Kim scored $78.50$in the short program and $150.06$in the free skate. Thus, we have,

$$\begin{gathered} x=-16.2 \\ y=2.07 \end{gathered}$$

And $s=10.2$is the value of $s$The standard error of the estimations, as we all know, is the average error of forecasts, and thus the average difference between actual and predicted values. As a result, the predicted free skate score using the least square regression line was$10.2$ points off from the actual free skate score.

The relationship between the short program score and the free skate score is given in the question. And the regression line is as:

$\stackrel{⏜}{y}=-16.2+2.07x$

And also given that $Yu-Na$Kim scored $78.50$in the short program and $150.06$in the free skate. Thus, we have,

$$\begin{gathered} x=-16.2 \\ y=2.07 \end{gathered}$$

And the value of $r^2$is,

$$\begin{gathered} r^2=0.736 \\ =73.6\% \end{gathered}$$

The coefficient of determination, as we know, is a measurement of how much variation in the answers y variable is explained by the least square regression model with the explanatory variable. As a result, the least square regression line utilizing the short program score as an explanatory variable may explain $73.6$ percent of the variation in the free skate score.