Chapter 13.2, Problem 17E

An experiment to study the relationship between x = Time spent exercising (minutes) and y = Amount of oxygen consumed during the exercise period resulted in the following summary statistics.

$\begin{array}{l}n=20\sum x=50\sum y=16,705\sum x^2=150\\ \sum y^2=14,194,231\sum xy=44,194\end{array}$

1. a. Estimate the slope and y intercept of the population regression line.
2. b. One sample observation on oxygen usage was 757 for a 2-minute exercise period. What amount of oxygen consumption is predicted for this exercise period, and what is the corresponding residual?
3. c. Calculate a 99% confidence interval for the average change in oxygen consumption associated with a 1-minute increase in exercise time.

To determine

Estimate the slope.

Find the y intercept of the population regression equation.

Answer to Problem 17E

The slope is 97.26.

The y intercept of the population regression equation is 592.1.

Explanation of Solution

Calculation:

The given information is that, the variable x denotes the time spent exercising (minutes) and y denotes amount of oxygen consumed during the exercise period. The summary statistics are,

$n=20,{\displaystyle \sum x}=50,{\displaystyle \sum y}=16,705,{\displaystyle \sum x^2}=150,$${\displaystyle \sum y^2}=14,194,231,{\displaystyle \sum xy}=44,194$

Formula for Slope:

$b=\frac{S_{xy}}{S_{xx}}$

In the formula, $S_{xy}={\displaystyle \sum xy}-\frac{\left({\displaystyle \sum x}\right)\left({\displaystyle \sum y}\right)}{n}$ and $S_{xx}={\displaystyle \sum x^2}-\frac{{\left({\displaystyle \sum x}\right)}^2}{n}$.

Formula for intercept:

$a=\overline{y}-b\overline{x}$

In the formula $\overline{y}=\frac{{\displaystyle \sum y}}{n},\overline{x}=\frac{{\displaystyle \sum x}}{n}$

$S_{xy}$:

Substitute ${\displaystyle \sum xy}=44,194,{\displaystyle \sum x}=50,{\displaystyle \sum y}=16,705,n=20$

$\begin{array}{c}{S}_{xy}=44,194-\frac{\left(50\right)\left(16,705\right)}{20}\\ =44,194-\frac{835,250}{20}\\ =44,194-41,762.5\\ =2,431.5\end{array}$

$S_{xx}$:

Substitute $n=20,{\displaystyle \sum x}=50,{\displaystyle \sum x^2}=150$

$\begin{array}{c}{S}_{xx}={\displaystyle \sum x^2}-\frac{{\left({\displaystyle \sum x}\right)}^2}{n}\\ =150-\frac{{\left(50\right)}^2}{20}\\ =150-\frac{2,500}{20}\\ =150-125\\ =25\end{array}$

Slope:

Substitute, $S_{xy}=2,431.5,S_{xx}=25$

$\begin{array}{c}b=\frac{2,431.5}{25}\\ =97.26\end{array}$

Hence, the slope is 97.26.

$\overline{y}$:

Substitute $n=20,{\displaystyle \sum y}=16,705$ in the formula.

$\begin{array}{c}\overline{y}=\frac{16,705}{20}\\ =835.25\end{array}$

$\overline{x}$:

Substitute $n=20,{\displaystyle \sum x}=50$ in the formula.

$\begin{array}{c}\overline{x}=\frac{50}{20}\\ =2.5\end{array}$

Intercept:

Substitute $\overline{y}=835.25,b=97.26,\overline{x}=2.5$ in the formula

$\begin{array}{c}a=835.25-\left(97.26\right)2.5\\ =835.25-243.15\\ =592.1\end{array}$

Hence, the y intercept of the population regression equation is 592.1.

To determine

Find the amount of oxygen consumption that is predicted for 2-minute exercise period.

Find the corresponding residual.

Answer to Problem 17E

The amount of oxygen consumption that is predicted for 2-minute exercise period is 786.62.

The corresponding residual is –29.62.

Explanation of Solution

Calculation:

The given information is that for a 2 minute period the oxygen intake was 757. From part (a), the value of slope is $b=97.26$ and intercept is $a=592.1$.

The estimated regression line is given by, $y=a+bx$

Substitute, $b=97.26$ and $\alpha =592.1$ in the regression equation

$y=592.1+97.26x$

Predicted value:

Substitute $x=2$ in the estimated regression equation.

$\begin{array}{c}y=592.1+97.26\left(2\right)\\ =592.1+194.52\\ =786.62\end{array}$

Hence, the amount of oxygen consumption that is predicted for 2-minute exercise period is 786.62.

Residual:

The formula for residual is,

$\text{Residual}=y-\widehat{y}$

Substitute, $y=757,\widehat{y}=786.62$ in the formula.

$\begin{array}{c}y=757-786.62\\ =-29.62\end{array}$

Hence, the corresponding residual is –29.62.

To determine

Find 99% confidence interval for the average change in oxygen consumption associated with a 1-minute increase in exercise time.

Answer to Problem 17E

The 99% confidence interval for the average change in oxygen consumption associated with a 1-minute increase in exercise time is $\left(87.7641,106.7559\right)$.

Explanation of Solution

Calculation:

The given information is that, the variable x denotes the time spent exercising (minutes) and y denotes amount of oxygen consumed during the exercise period. The level of significance is 0.01. The summary statistics are,

$n=20,{\displaystyle \sum x}=50,{\displaystyle \sum y}=16,705,{\displaystyle \sum x^2}=150,$${\displaystyle \sum y^2}=14,194,231,{\displaystyle \sum xy}=44,194$

Formula for Degrees of freedom:

The formula for degrees of freedom is,

$df=n-2$

In the formula, n is the total number of observations.

Formula for estimated standard deviation of statistic b:

$s_b=\frac{s_e}{\sqrt{S_{xx}}}$

In the formula, $s_e=\sqrt{\frac{S{S}_{\text{Resid}}}{n-2}}$, $S{S}_{\text{Resid}}={\displaystyle \sum y^2}-a{\displaystyle \sum y}-b{\displaystyle \sum xy}$, n denotes the total number of observations and $S_{xx}={\displaystyle \sum x^2}-\frac{{\left({\displaystyle \sum x}\right)}^2}{n}$

Formula for Confidence interval for $\beta$:

$\text{Confidence interva}l=b\pm \left(t\text{critical}\text{value}\right)s_b$

In the formula b denotes the estimated slope and $s_b$ denotes the estimated standard deviation of statistic b.

Degrees of freedom:

Substitute $n=20$ in the degrees of freedom formula.

$\begin{array}{c}df=20-2\\ =18\end{array}$

From the Appendix: Table 3 the t Critical Values:

• Locate the value 18 in the degrees of freedom (df) column.
• Locate the 0.99 in the row of central area captured.
• The intersecting value that corresponds to the 18 with confidence level 0.99 is $\pm 2.88$

Thus, the critical value for $df=18$ with two-tailed test is $\pm 2.88$.

$S{S}_{\text{Resid}}$:

Substitute, ${\displaystyle \sum y^2}=14,194,231,{\displaystyle \sum y}=16,705,$$a=592.1,b=97.26,{\displaystyle \sum xy}=44,194$ in the formula.

$\begin{array}{c}S{S}_{\text{Resid}}=14,194,231-592.1\left(16,705\right)-97.26\left(44,194\right)\\ =14,194,231-9,891,030.5-4,298,308.44\\ =4,892.06\end{array}$

$s_e$:

Substitute, $S{S}_{\text{Resid}}=4,892.06,n=20$ in the formula.

$\begin{array}{c}{s}_e=\sqrt{\frac{4,892.06}{20-2}}\\ =\sqrt{\frac{4,892.06}{18}}\\ =\sqrt{271.7811}\\ =16.4858\end{array}$

$s_b$:

Substitute, $s_e=16.4858,S_{xx}=25$ in the formula.

$\begin{array}{c}{s}_b=\frac{16.4858}{\sqrt{25}}\\ =\frac{16.4858}{5}\\ =3.2972\end{array}$

Confidence interval:

Substitute, $b=97.26,s_b=3.2972,t\text{critical value}=2.88$ in the confidence interval formula.

$\begin{array}{c}\text{Confidence interval}=97.26\pm \left(2.88\right)3.2972\\ =97.26\pm 9.4959\\ =\left(87.7641,106.7559\right)\end{array}$

Hence, the 99% confidence interval for $\beta$ is $\left(87.7641,106.7559\right)$.

Interpretation:

The 99% confidence interval is interpreted as, there is 99% confidence that the slope for the population regression line or the average change in oxygen consumption associated with a 1-minute increase in exercise time lies between 87.7641 and 106.7559.