Question

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.

Short answer

Part (a) The residual is $3.765$

Part (b) The free skate score improved by $2.07$points on average, while the short program score improved by one point.

Part (c) The predicted free skate score using the least square regression line was $10.2$points off from the actual free skate score.

Part (d) 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.

Step-by-step solution

Part (a) Step 1: Given information

Part (a) Step 2: Calculation

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 $3.765$

Part (b) Step 1: Calculation

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.

Part (c) Step 1: Explanation

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.

Part (d) Step 1: Explanation

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.