Chapter 11.4, Problem 7E

a.

To determine

Construct the 95% confidence interval for the mean response when $x=2$.

Answer to Problem 7E

The 95% confidence interval for the mean response when $x=2$ is $\left(6.34,9.44\right)$.

Explanation of Solution

Calculation:

The given information is that, the sample size is 25 and parameter estimates are given below:

$b_0=3.25,b_1=2.32,s_e=3.53,{{\displaystyle \sum \left(x-\overline{x}\right)}}^2=224.05\text{ and }\overline{x}=0.98$.

Construction of confidence interval:

$\widehat{y}\pm t_{\frac{\alpha }{2}}\cdot s_e\sqrt{\frac{1}{n}+\frac{{\left(x^{\ast }-\overline{x}\right)}^2}{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}}$

Where,

A particular value of $x^{\ast }$ is substituted to get the confidence interval.

$s_e$ represents the residual standard deviation.

$\overline{x}$ represents the mean of predictor variable.

$t_{\frac{\alpha }{2}}$ represents the table value of t distribution for a given level of significance.

From the given information the predicted value is calculated as follows:

$\widehat{y}=b_0+b_1{x}^{*}$

Substitute $b_0$ as 3.25, $b_1$ as 2.32 and $x^{*}$ as 2.

$\begin{array}{c}\widehat{y}=b_0+b_1{x}^{*}\\ =3.25+\left(2.32\right)\left(2\right)\\ =3.25+4.64\\ =7.89\end{array}$

Thus, the predicted value $\widehat{y}$ is 7.89.

Critical value:

Software procedure:

Step-by-step procedure to find the critical value using MINITAB is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• In Degrees of freedom, enter 23.
• Click the Shaded Area tab.
• Choose Probability and Two tail for the region of the curve to shade.
• Enter the Probability value as 0.05.
• Click OK.

Output obtained from MINITAB is given below:

95% confidence interval is calculated as follows:

$\widehat{y}\pm t_{\frac{\alpha }{2}}\cdot s_e\sqrt{\frac{1}{n}+\frac{{\left(x^{\ast }-\overline{x}\right)}^2}{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}}$

Substitute $\widehat{y}$ as 7.89, $x^{*}$ as 2, n as 25, $s_e$ as 3.53, ${{\displaystyle \sum \left(x-\overline{x}\right)}}^2\text{as }224.05\text{, }\overline{x}\text{ as }0.98$ and $t_{\frac{\alpha }{2}}$ as 2.069 in the above formula,

$\begin{array}{c}\widehat{y}\pm t_{\frac{\alpha }{2}}\cdot s_e\sqrt{\frac{1}{n}+\frac{{\left(x^{\ast }-\overline{x}\right)}^2}{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}}=7.89\pm \left(2.069\right)\cdot \left(3.53\right)\sqrt{\frac{1}{25}+\frac{{\left(2-0.98\right)}^2}{224.05}}\\ =7.89\pm \left(2.069\right)\cdot \left(3.53\right)\sqrt{0.04+0.00464}\\ =7.89\pm \left(2.069\right)\cdot \left(3.53\right)\sqrt{0.045}\\ =7.89\pm 1.55\end{array}$

                                           $=\left(6.34,9.44\right)$

Thus, the 95% confidence interval is $\left(6.34,9.44\right)$.

b.

To determine

Construct the 95% prediction interval for the individual response when $x=2$.

Answer to Problem 7E

The 95% prediction interval for the individual response when $x=2$ is $\left(0.42,15.36\right)$.

Explanation of Solution

Calculation:

The given information is that the sample size is 25 and parameter estimates are given below:

$b_0=3.25,b_1=2.32,s_e=3.53,{{\displaystyle \sum \left(x-\overline{x}\right)}}^2=224.05\text{ and }\overline{x}=0.98$.

Construction of prediction interval:

$\widehat{y}\pm t_{\frac{\alpha }{2}}\cdot s_e\sqrt{1+\frac{1}{n}+\frac{{\left(x^{\ast }-\overline{x}\right)}^2}{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}}$

Where,

A particular value of $x^{\ast }$ is substituted to get the confidence interval.

$s_e$ represents the residual standard deviation.

$\overline{x}$ represents the mean of predictor variable.

$t_{\frac{\alpha }{2}}$ represents the table value of t distribution for a given level of significance.

From the given information the predicted value is calculated as follows:

$\widehat{y}=b_0+b_1{x}^{*}$

Substitute $b_0$ as 3.25, $b_1$ as 2.32 and $x^{*}$ as 2.

$\begin{array}{c}\widehat{y}=b_0+b_1{x}^{*}\\ =3.25+\left(2.32\right)\left(2\right)\\ =3.25+4.64\\ =7.89\end{array}$

Thus, the predicted value $\widehat{y}$ is 7.89.

95% prediction interval is calculated as follows:

$\widehat{y}\pm t_{\frac{\alpha }{2}}\cdot s_e\sqrt{1+\frac{1}{n}+\frac{{\left(x^{\ast }-\overline{x}\right)}^2}{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}}$

Substitute $\widehat{y}$ as 7.89, $x^{*}$ as 2, n as 25, $s_e$ as 3.53, ${{\displaystyle \sum \left(x-\overline{x}\right)}}^2\text{as }224.05\text{, }\overline{x}\text{ as }0.98$ and $t_{\frac{\alpha }{2}}$ as 2.069 in the above formula,

$\begin{array}{c}\widehat{y}\pm t_{\frac{\alpha }{2}}\cdot s_e\sqrt{1+\frac{1}{n}+\frac{{\left(x^{\ast }-\overline{x}\right)}^2}{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}}=7.89\pm \left(2.069\right)\cdot \left(3.53\right)\sqrt{1+\frac{1}{25}+\frac{{\left(2-1.95\right)}^2}{360.26}}\\ =7.89\pm \left(2.069\right)\cdot \left(3.53\right)\sqrt{1+0.04+0.00464}\\ =7.89\pm \left(2.069\right)\cdot \left(3.53\right)\sqrt{1.045}\\ =7.89\pm 7.47\end{array}$

                                               $=\left(0.42,15.36\right)$

Thus, the 95% prediction interval is $\left(0.42,15.36\right)$.