Chapter 13.2, Problem 7E

a.

To determine

To construct:a $95\%$ confidence interval for mean response when $x=2$

Answer to Problem 7E

  $7.89\pm 1.5431738$

Explanation of Solution

Given information: the sample size $25$ , $b_0=3.25,b_1=2.32,s_e=3.53,{\displaystyle \sum _{}^{}(x-\overline{x}}{)}^2=224.05,\overline{x}=0.98$

Formula Used:the formulas to be used when constructing a confidence interval for a mean response and when constructing a prediction interval for an individual response.

Let $x^{*}$ be a value of the explanatory variable $x$ , let $\stackrel{\wedge }{y}=b_0+b_1{x}^{*}$ be the predicted value corresponding to $x^{*}$ and let $n$ be the sample size .A level $100(1-\alpha )\%$ prediction interval for the mean response is

  $\stackrel{\wedge }{y}\pm t_{\alpha /2}.s_e\sqrt{\frac{1}{n}+\frac{{(x^{*}-\overline{x})}^2}{{\displaystyle \sum {(x-\overline{x})}^2}}}$

Here, the critical value $t_{\alpha /2}$ is based on $n-2$ degrees of freedom.

Let $x^{*}$ be a value of the explanatory variable $x$ , let $\stackrel{\wedge }{y}=b_0+b_1{x}^{*}$ be the predicted value corresponding to $x^{*}$ and let $n$ be the sample size .A level $100(1-\alpha )\%$ prediction interval for an individual response is

  $\stackrel{\wedge }{y}\pm t_{\alpha /2}.s_e\sqrt{1+\frac{1}{n}+\frac{{(x^{*}-\overline{x})}^2}{{\displaystyle \sum {(x-\overline{x})}^2}}}$

Here, the critical value $t_{\alpha /2}$ is based on $n-2$ degrees of freedom.

the predicted value corresponding to $x^{*}$ when $x=2$

  $\begin{array}{l}\stackrel{\wedge }{y}=b_0+b_1{x}^{*}\\ \stackrel{\wedge }{y}=3.25+2.32(2)\\ \stackrel{\wedge }{y}=3.25+4.64\\ \stackrel{\wedge }{y}=7.89\end{array}$

The t table corresponding t-value for the given confidence interval $95\%$ and degrees of freedom $23$ is not provided in the book. But below shows that the t-valueis $2.069$ .

  

Now to construct prediction interval for the mean response.

  $\begin{array}{l}\stackrel{\wedge }{y}\pm t_{\alpha /2}.s_e\sqrt{\frac{1}{n}+\frac{{(x^{*}-\overline{x})}^2}{{\displaystyle \sum {(x-\overline{x})}^2}}}\\ 7.89\pm t_{\alpha /2}3.53\sqrt{\frac{1}{25}+\frac{{(2-0.98)}^2}{224.05}}\\ 7.89\pm t_{\alpha /2}3.53\sqrt{\frac{1}{25}+\frac{{(1.02)}^2}{224.05}}\\ 7.89\pm t_{\alpha /2}3.53\sqrt{\frac{1}{25}+\frac{1.0404}{224.05}}\\ 7.89\pm t_{\alpha /2}3.53\sqrt{0.04+\frac{1.0404}{224.05}}\\ 7.89\pm t_{\alpha /2}3.53\sqrt{0.04+0.00464360633}\\ 7.89\pm t_{\alpha /2}3.53\sqrt{0.04464360633}\\ 7.89\pm t_{\alpha /2}3.53(0.21129033659)\\ 7.89\pm t_{\alpha /2}0.74585488817\end{array}$

  $\begin{array}{l}7.89\pm t_{\alpha /2}0.74585488817\\ 7.89\pm (2.069)(0.74585488817)\\ 7.89\pm 1.5431738\end{array}$

b.

To determine

To construct: a $95\%$ prediction interval for individual response when $x=2$

Answer to Problem 7E

  $7.89\pm 7.4648188$

Explanation of Solution

Given information: the sample size $25$ , $b_0=3.25,b_1=2.32,s_e=3.53,{\displaystyle \sum _{}^{}(x-\overline{x}}{)}^2=224.05,\overline{x}=0.98$

Formula Used:the formulas to be used when constructing a confidence interval for a mean response and when constructing a prediction interval for an individual response.

Let $x^{*}$ be a value of the explanatory variable $x$ , let $\stackrel{\wedge }{y}=b_0+b_1{x}^{*}$ be the predicted value corresponding to $x^{*}$ and let $n$ be the sample size .A level $100(1-\alpha )\%$ prediction interval for the mean response is

  $\stackrel{\wedge }{y}\pm t_{\alpha /2}.s_e\sqrt{\frac{1}{n}+\frac{{(x^{*}-\overline{x})}^2}{{\displaystyle \sum {(x-\overline{x})}^2}}}$

Here, the critical value $t_{\alpha /2}$ is based on $n-2$ degrees of freedom.

Let $x^{*}$ be a value of the explanatory variable $x$ , let $\stackrel{\wedge }{y}=b_0+b_1{x}^{*}$ be the predicted value corresponding to $x^{*}$ and let $n$ be the sample size .A level $100(1-\alpha )\%$ prediction interval for an individual response is

  $\stackrel{\wedge }{y}\pm t_{\alpha /2}.s_e\sqrt{1+\frac{1}{n}+\frac{{(x^{*}-\overline{x})}^2}{{\displaystyle \sum {(x-\overline{x})}^2}}}$

Here, the critical value $t_{\alpha /2}$ is based on $n-2$ degrees of freedom.

the predicted value corresponding to $x^{*}$ when $x=2$

  $\begin{array}{l}\stackrel{\wedge }{y}=b_0+b_1{x}^{*}\\ \stackrel{\wedge }{y}=3.25+2.32(2)\\ \stackrel{\wedge }{y}=3.25+4.64\\ \stackrel{\wedge }{y}=7.89\end{array}$

The t table corresponding t-value for the given confidence interval $95\%$ and degrees of freedom $23$ is not provided.But below shows that the t-value is $2.069$ .

  

Now to construct prediction interval for the individual response.

  $\begin{array}{l}\stackrel{\wedge }{y}\pm t_{\alpha /2}.s_e\sqrt{\frac{1}{n}+\frac{{(x^{*}-\overline{x})}^2}{{\displaystyle \sum {(x-\overline{x})}^2}}}\\ \stackrel{\wedge }{y}\pm t_{\alpha /2}.s_e\sqrt{1+\frac{1}{n}+\frac{{(x^{*}-\overline{x})}^2}{{\displaystyle \sum {(x-\overline{x})}^2}}}\\ 7.89\pm t_{\alpha /2}3.53\sqrt{1+\frac{1}{25}+\frac{{(2-0.98)}^2}{224.05}}\\ 7.89\pm t_{\alpha /2}3.53\sqrt{1+0.04+\frac{1.0404}{224.05}}\\ 7.89\pm t_{\alpha /2}3.53\sqrt{1+0.04+\frac{1.0404}{224.05}}\\ 7.89\pm t_{\alpha /2}3.53\sqrt{1.04464360634}\\ 7.89\pm t_{\alpha /2}3.53(1.02207808231)\\ 7.89\pm t_{\alpha /2}3.60793563055\end{array}$

  $\begin{array}{l}7.89\pm t_{\alpha /2}3.60793563055\\ 7.89\pm (2.069)(3.60793563055)\\ 7.89\pm 7.4648188\end{array}$