Chapter 8, Problem 51E

(a)

To determine

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

Answer to Problem 51E

The regression line is:

  $\text{<mover accent="true"><mo>H</mo><mo>^</mo></mover> igh Jump}=2.681-0.0067(\text{800-meter)}$ .

Explanation of Solution

In the question, the association between the high jump performance from the $800$ -meter results are examined. And the information is given in the table as:

800-m	High Jump
134.15	1.91
135.92	1.76
132.27	1.85
135.32	1.7
131.31	1.79
135.21	1.76
133.62	1.85
137.01	1.82
137.72	1.79
133.23	1.7
137.28	1.67
130.77	1.82
140.05	1.85
133.69	1.7
137.9	1.79
133.95	1.85
138.68	1.73
137.65	1.79
138.47	1.7
145.1	1.67
133.08	1.76
134.57	1.79
142.58	1.73
132.27	1.7
141.21	1.7
145.68	1.7

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:

  $\begin{array}{l}\text{<mover accent="true"><mo>H</mo><mo>^</mo></mover> igh Jump}=\alpha +\beta (\text{800-meter)}\\ =2.681-0.0067(\text{800-meter)}\end{array}$

Thus, the slope of the line interprets that high jump height is lower, on average, by $0.0067$ meters per additional second of $800$ -meter race time.

(b)

To determine

To find out what percent of the variability in high jumps can be accounted for by differences in $800$ m times.

Answer to Problem 51E

  $R^2=16.4\%$ .

Explanation of Solution

In the question, the association between the high jump performance from the $800$ -meter results are examined. And the regression line is:

  $\text{<mover accent="true"><mo>H</mo><mo>^</mo></mover> igh Jump}=2.681-0.0067(\text{800-meter)}$ .

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

  $R^2=16.4\%$

Thus, the value of $R^2$ explains the percentage of variation explained by the variables used. Thus, we can say that $16.4\%$ of the variation is explained by the high jump performance from the $800$ -meter race time.

(c)

To determine

To explain do good high jumpers tend to be fast runners.

Answer to Problem 51E

Yes, good jumpers tend to be fast runners.

Explanation of Solution

In the question, the association between the high jump performance from the $800$ -meter results are examined. And the regression line is:

  $\text{<mover accent="true"><mo>H</mo><mo>^</mo></mover> igh Jump}=2.681-0.0067(\text{800-meter)}$ .

Thus, we can say that good high jumpers tend to be fast runners because the slope is negative as calculated in part (a) and this implies faster runners tend to jump higher.

(d)

To determine

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

Explanation of Solution

In the question, the association between the high jump performance from the $800$ -meter results are examined. And the regression line is:

  $\text{<mover accent="true"><mo>H</mo><mo>^</mo></mover> igh Jump}=2.681-0.0067(\text{800-meter)}$ .

The residual plot is as:

  

From the above residual plot we can see that there is slight tendency for less variation in high-jump height among the slower runners than among that faster ones.

(e)

To determine

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

Answer to Problem 51E

No, this is not useful model and it does not be used to predict high-jump performance.

Explanation of Solution

In the question, the association between the high jump performance from the $800$ -meter results are examined. And the regression line is:

  $\text{<mover accent="true"><mo>H</mo><mo>^</mo></mover> igh Jump}=2.681-0.0067(\text{800-meter)}$ .

Thus, we think that this model is not especially useful model because the residual standard deviation is $0.06$ meters which is not much smaller than the standard deviation of all high jumps. The model does not appear to do a very good job of predicting as it does not give the accurate result.