Chapter 8, Problem 52E

(a)

To determine

To find out what is the regression equation and what does the slope mean.

Answer to Problem 52E

The regression line is:

  $\text{<mover accent="true"><mo>L</mo><mo>^</mo></mover> ong Jump}=4.2+1.11(\text{High jump)}$ .

Explanation of Solution

In the question, the association between the long jump performance on the high-jump results are examined. And the information is given in the table as:

High Jump	Long Jump
1.91	6.51
1.76	6.3
1.85	6.51
1.7	6.25
1.79	6.21
1.76	6.42
1.85	6.19
1.82	6.23
1.79	6.02
1.7	5.84
1.67	6.36
1.82	6.35
1.85	6.1
1.7	6.02
1.79	5.97
1.85	5.9
1.73	6.03
1.79	6.36
1.7	6.21
1.67	6.15
1.76	5.98
1.79	6.16
1.73	6.02
1.7	5.92
1.7	6.22
1.7	5.7

Thus, we will create a regression line by using excel as:

We will first select the data given in the table and then go to the insert tab. In the tab we will use the scatterplot option from the charts options and then the scatterplot will appear on the screen. Now, we will go to the design tab from the chart tools. We will then select the quick layout option from it. Then in it we will select the layout $9$ from it and the scatterplot with the model and regression line will appear as:

  

Thus, the regression line for this context is:

  $\begin{array}{l}\text{<mover accent="true"><mo>L</mo><mo>^</mo></mover> ong Jump}=\alpha +\beta (\text{High jump)}\\ =4.2+1.11(\text{High jump)}\end{array}$

Thus, the slope of the line interprets that long jump height is higher, on average, by $1.11$ meters per additional second of high jump results.

(b)

To determine

To find out what percentage of the variability in long jumps can be accounted for by high-jumps performances.

Answer to Problem 52E

  $12.64\%$ .

Explanation of Solution

In the question, the association between the long jump performance on the high-jump results are examined. And the regression line is:

  $\text{<mover accent="true"><mo>L</mo><mo>^</mo></mover> ong Jump}=4.2+1.11(\text{High jump)}$ .

Thus, from the above scatterplot in part (a) we can see that the coefficient of determination is also given, that is:

  $R^2=12.64\%$

Thus, the value of $R^2$ explains the percentage of variation explained by the variables used. Thus, we can say that $12.64\%$ of the variation is explained by the long jump performance on high jump results.

(c)

To determine

To explain do good high jumpers tend to be good long jumpers.

Answer to Problem 52E

Yes, good high jumpers tend to be good long jumpers.

Explanation of Solution

In the question, the association between the long jump performance on the high-jump results are examined. And the regression line is:

  $\text{<mover accent="true"><mo>L</mo><mo>^</mo></mover> ong Jump}=4.2+1.11(\text{High jump)}$ .

Thus, we can say that good high jumpers tend to be good long jumpers because the slope is positive as calculated in part (a) and this implies that higher values are better for both variables.

(d)

To determine

To explain what does the residuals plot reveal about the model.

Explanation of Solution

In the question, the association between the long jump performance on the high-jump results are examined. And the regression line is:

  $\text{<mover accent="true"><mo>L</mo><mo>^</mo></mover> ong Jump}=4.2+1.11(\text{High jump)}$ .

The residual plot is as:

  

From the above residual plot we can see that the average high runners tend to have closer to average long jumps than the really low of high high-jumps.

(e)

To determine

To explain do you think this is a useful model and would you use it to predict long-jump performance.

Answer to Problem 52E

Yes, this is useful model and it is used to predict long-jump performance.

Explanation of Solution

In the question, the association between the long jump performance on the high-jump results are examined. And the regression line is:

  $\text{<mover accent="true"><mo>L</mo><mo>^</mo></mover> ong Jump}=4.2+1.11(\text{High jump)}$ .

Thus, we think that this model is especially useful model because the residual standard deviation is a lot smaller than the standard deviation of all long jumps. The model does appear to do a very good job of predicting as it does give the accurate result.