Chapter 11, Problem 11.101LM

In fitting a least squares line ton= 15 data points, the following quantities were computed: SS_xx = 55, SS_yy = 198, SS_xy = − 88, $\bar{x}$ = 1.3, and $\bar{y}$ = 35.

a. Find the least squares line.

b. Graph the least squares line.

c. Calculate SSE.

d. Calculate s².

e. Find a 90% confidence interval for β₁. Interpret this estimate.

f. Find a 90% confidence interval for the mean value of y when x = 15.

g. Find a 90% prediction interval for y when x = 15.

To determine

To find: The least square line.

Answer to Problem 11.101LM

The fitted least square line is $\widehat{y}=37.08-1.6x$.

Explanation of Solution

Given info: The values are $S{S}_{xy}=-88$, $S{S}_{xx}=55$, $S{S}_{yy}=198$, $\overline{x}=1.3$, and $\overline{y}=35$.

Calculation:

The least square line is obtained below,

$y={\beta }_0+{\beta }_{1}x$

Where,

$\begin{array}{c}{\widehat{\beta }}_1=\frac{S{S}_{xy}}{S{S}_{xx}}\\ {\widehat{\beta }}_0=\overline{y}-{\widehat{\beta }}_1\overline{x}\end{array}$

Substitute the value $S{S}_{xy}=-88$, and $S{S}_{xx}=55$ in ${\widehat{\beta }}_1$,

The value of ${\widehat{\beta }}_1$ is,

$\begin{array}{c}{\widehat{\beta }}_1=\frac{S{S}_{xy}}{S{S}_{xx}}\\ =\frac{-88}{55}\\ =-1.6\end{array}$

Therefore, the value of ${\widehat{\beta }}_1$ is –1.6.

Substitute the values $\overline{y}=35$, $\overline{x}=1.3$, and ${\widehat{\beta }}_1=-1.6$ in ${\widehat{\beta }}_0$.

The value of ${\widehat{\beta }}_0$ is,

$\begin{array}{c}{\widehat{\beta }}_0=35-\left(-1.6\right)\left(1.3\right)\\ =37.08\end{array}$

The value of ${\widehat{\beta }}_0$ is 37.08.

The fitted regression line is,

$\begin{array}{c}\widehat{y}={\beta }_0+{\beta }_{1}x\\ =37.08-1.6x\end{array}$

Therefore, the fitted least square line is $\widehat{y}=37.08-1.6x$.

To determine

To plot: The line $\widehat{y}=37.08-1.6x$.

Answer to Problem 11.101LM

The plotted line is,

Explanation of Solution

Calculations:

To plot the line on the graph, first to obtain the at least two points on the line

From the information, the line is

$\widehat{y}=37.08-1.6x$ (1)

The two points are obtained below,

Let $x=1$.

Substitute the x value in equation (1) to get the value of y.

$\begin{array}{c}y=37.08-1.6\left(1\right)\\ =35.48\end{array}$

Therefore, $\left(1,35.48\right)$ is the one point on the line $\widehat{y}=37.08-1.6x$.

Let $x=5$.

Substitute the x value in equation (1) to get the value of y.

$\begin{array}{c}y=37.08-1.6\left(5\right)\\ =29.08\end{array}$

Therefore, $\left(5,29.08\right)$ is the point on the line $\widehat{y}=37.08-1.6x$.

Based on the points $\left(1,35.48\right)$ and $\left(5,29.08\right)$ the line $\widehat{y}=37.08-1.6x$ Plotted below,

1. 1. From the graph locate the value 1 on X-axis and 35.48 on Y-axis and mark the intersection point.
2. 2. Locate the value 5 on X-axis and 29.08 on Y-axis and mark the intersection point.
3. 3. Draw the line by combining the both intersection points.
4. 4. The Graph is shown below.

To determine

To find: The value of SSE.

Answer to Problem 11.101LM

The value of SSE is 57.2.

Explanation of Solution

Calculations:

The formula for SSE is obtained below:

$SSE=S{S}_{yy}-{\widehat{\beta }}_{1}S{S}_{xy}$

Substitute $S{S}_{xy}=-88$, $S{S}_{yy}=198$, and ${\widehat{\beta }}_1=-1.6$.

The value of SSE is,

$\begin{array}{c}SSE=S{S}_{yy}-{\widehat{\beta }}_{1}S{S}_{xy}\\ =\left(198\right)-\left(-1.6\right)\left(-88\right)\\ =198-140.8\\ =57.2\end{array}$

Therefore, the value of SSE is 57.2.

To determine

To find: The value of $s^2$.

Answer to Problem 11.101LM

The value of $s^2$ is 4.4.

Explanation of Solution

Calculation:

Formula for $s^2$ is obtained below:

$s^2=\frac{SSE}{n-2}$

Substitute 57.2 for SSE and 15 for n in the formula.

The value of $s^2$ is,

$\begin{array}{c}{s}^2=\frac{SSE}{n-2}\\ =\frac{57.2}{15-2}\\ =\frac{57.2}{13}\\ =4.4\end{array}$

Therefore, the value of $s^2$ is 4.4.

To determine

To find: The 90% confidence interval for the slope ${\widehat{\beta }}_1$.

To interpret: The results.

Answer to Problem 11.101LM

The 90% confidence interval for the slope ${\widehat{\beta }}_1$ is $\left(-2.10,-1.10\right)$.

Explanation of Solution

Calculation:

The formula for the confidence interval for estimate of ${\beta }_1$ is obtained by,

${\widehat{\beta }}_1\pm t_{\frac{\alpha }{2}}\times s_{{\widehat{\beta }}_1}$

Where,

$s_{{\beta }_1}=\frac{s}{\sqrt{S{S}_{xx}}}$

For confidence coefficient is 0.90. So that level of significance $\alpha =0.10$.

$\begin{array}{c}{t}_{\frac{\alpha }{2}}=\frac{0.10}{2}\\ =0.05\end{array}$

The degrees of freedom is,

$\begin{array}{c}\text{Degrees of freedom}=n-2\\ =15-2\\ =13\end{array}$

From table III Appendix D:

1. 1. In the column locate the df as 13.
2. 2. In the row locate the level of significance at 0.05.
3. 3. The row and column of intersection point is 1.771 is the critical value.

Substitute ${\widehat{\beta }}_1=-1.6$, $s=2.0976$, $S{S}_{xx}=55$, and $t_{0.05}=2.132$ in the formula.

The 90% confidence interval for the slope ${\widehat{\beta }}_1$ is,

$\begin{array}{c}{\widehat{\beta }}_1\pm t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{S{S}_{xx}}}=-1.6\pm (1.771)\frac{2.0976}{\sqrt{55}}\\ =-1.6\pm 0.50\\ =\left(-2.10,-1.10\right)\end{array}$

The 90% confidence interval for the slope ${\widehat{\beta }}_1$ is $\left(-2.10,-1.10\right)$.

Interpretation:

It can be expected that the 90% confidence that each additional unit change in the mean value of y for each unit change in the x value is between –2.10 and –1.10.

To determine

To find: The 90% confidence interval for the mean value of y when $x=15$.

Answer to Problem 11.101LM

The 90% confidence interval is. $\left(6.15,20.01\right)$

Explanation of Solution

Calculations:

The formula for the 90% confidence interval is obtained below:

$\widehat{y}\pm t_{\alpha }\left(s\right)\sqrt{\frac{1}{n}+\frac{{\left(x-\overline{x}\right)}^2}{S{S}_{xx}}}$

The estimated value for $x_p=15$ is,

$\begin{array}{c}\widehat{y}=37.08-1.6x\\ =37.08-1.6\left(15\right)\\ =37.08-24\\ =13.08\end{array}$

The estimated value for $x_p=15$ is 13.08.

Substitute $\widehat{y}=13.08$, $s=2.0976$, $S{S}_{xx}=55$, $n=15$, $\overline{x}=1.3$ and $t_{0.05}=2.132$ in the formula.

The 90% confidence interval is,

$\begin{array}{c}\widehat{y}\pm t_{\alpha }\left(s\right)\sqrt{\frac{1}{n}+\frac{{\left(x-\overline{x}\right)}^2}{S{S}_{xx}}}=13.08\pm \left(1.771\right)\left(2.0976\right)\sqrt{\frac{1}{15}+\frac{{\left(15-1.3\right)}^2}{55}}\\ =13.08\pm \left(3.7148\right)\left(1.8655\right)\\ =13.08\pm 6.93\\ =\left(6.15,20.01\right)\end{array}$

Therefore, the 90% confidence interval is $\left(6.15,20.01\right)$.

To determine

To find: The 90% prediction interval for the mean value of y when $x=15$.

Answer to Problem 11.101LM

The 90% prediction interval is $\left(5.22,20.94\right)$.

Explanation of Solution

Calculations:

Formula for the 90% confidence interval is obtained below:

$\widehat{y}\pm t_{\alpha }\left(s\right)\sqrt{1+\frac{1}{n}+\frac{{\left(x-\overline{x}\right)}^2}{S{S}_{xx}}}$

The estimated value for $x_p=15$ is,

$\begin{array}{c}\widehat{y}=37.08-1.6x\\ =37.08-1.6\left(15\right)\\ =37.08-24\\ =13.08\end{array}$

The estimated value for $x_p=15$ is 13.08.

Substitute $\widehat{y}=13.08$, $s=2.0976$, $S{S}_{xx}=55$, $n=15$, $\overline{x}=1.3$ and $t_{0.05}=2.132$ in the formula.

The 90% prediction interval is,

$\begin{array}{c}\widehat{y}\pm t_{\alpha }\left(s\right)\sqrt{1+\frac{1}{n}+\frac{{\left(x-\overline{x}\right)}^2}{S{S}_{xx}}}=13.08\pm \left(1.771\right)\left(2.0976\right)\sqrt{1+\frac{1}{15}+\frac{{\left(15-1.3\right)}^2}{55}}\\ =13.08\pm \left(3.7148\right)\left(2.1158\right)\\ =13.08\pm 7.86\\ =\left(5.22,20.94\right)\end{array}$

Therefore, the 90% prediction interval is $\left(5.22,20.94\right)$.