Chapter 13, Problem 10CQ

To determine

To construct:a $95\%$ prediction intervalfor anindividual response when $x=20$

Answer to Problem 10CQ

  $34.199\pm t_{\alpha /2}6.07823093467$

Explanation of Solution

Given information: $x=20$ and $95\%$ prediction interval

$\begin{array}{lllllllll}x\hfill & 25\hfill & 13\hfill & 16\hfill & 19\hfill & 29\hfill & 19\hfill & 16\hfill & 30\hfill \\ y\hfill & 40\hfill & 20\hfill & 33\hfill & 30\hfill & 50\hfill & 37\hfill & 34\hfill & 37\hfill \end{array}$

Formula Used:the formula to be usedwhen 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 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.

  $s_e$ - Residual standard deviation $=\sqrt{\frac{{\displaystyle \sum (y-\stackrel{\wedge }{y{)}^2}}}{n-2}}$

Test statistic $t=\frac{(\overline{x_1}-\overline{x_2})-({\mu }_1-{\mu }_2)}{\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}}$

Calculation:sample means and the sample variances can be calculated as shown.

  $\begin{array}{llllllll}x\hfill & y\hfill & \overline{x}\hfill & \overline{y}\hfill & x-\overline{x}\hfill & y-\overline{y}\hfill & {\left(x-\overline{x}\right)}^2\hfill & {\left(y-\overline{y}\right)}^2\hfill \\ 25\hfill & 40\hfill & 20.875\hfill & 35.125\hfill & 4.125\hfill & 4.875\hfill & 17.015625\hfill & 23.765625\hfill \\ 13\hfill & 20\hfill & \hfill & \hfill & -7.875\hfill & -15.125\hfill & 62.015625\hfill & 228.76563\hfill \\ 16\hfill & 33\hfill & \hfill & \hfill & -4.875\hfill & -2.125\hfill & 23.765625\hfill & 4.515625\hfill \\ 19\hfill & 30\hfill & \hfill & \hfill & -1.875\hfill & -5.125\hfill & 3.515625\hfill & 26.265625\hfill \\ 29\hfill & 50\hfill & \hfill & \hfill & 8.125\hfill & 14.875\hfill & 66.015625\hfill & 221.26563\hfill \\ 19\hfill & 37\hfill & \hfill & \hfill & -1.875\hfill & 1.875\hfill & 3.515625\hfill & 3.515625\hfill \\ 16\hfill & 34\hfill & \hfill & \hfill & -4.875\hfill & -1.125\hfill & 23.765625\hfill & 1.265625\hfill \\ 30\hfill & 37\hfill & \hfill & \hfill & 9.125\hfill & 1.875\hfill & 83.265625\hfill & 3.515625\hfill \end{array}$

Sample means;

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ \overline{x}=\frac{(35+13+16+19+29+19+16+30)}{8}\\ \overline{x}=\frac{167}{8}\\ \overline{x}=20.875\end{array}$

  $\begin{array}{l}\overline{y}=\frac{{\displaystyle \sum y}}{n}\\ \overline{y}=\frac{(40+20+33+30+50+37+34+37)}{8}\\ \overline{y}=\frac{281}{8}\\ \overline{y}=35.125\end{array}$

Sample variances;

  $\begin{array}{l}{s}_x^2=\frac{{\displaystyle \sum {(x-\overline{x})}^2}}{n-1}\\ s_x^2=\frac{{(25-20.875)}^2+{(13-20.875)}^2+\mathrm{...........}+{(30-20.875)}^2}{8-1}\\ s_x^2=\frac{282.875}{7}\\ s_x^2=40.4107143\\ s_x=6.35694221\end{array}$

  $\begin{array}{l}{s}_y^2=\frac{{\displaystyle \sum {(y-\overline{y})}^2}}{n-1}\\ s_y^2=\frac{{(40-35.125)}^2+{(20-35.125)}^2+\mathrm{...........}+{(37-35.125)}^2}{8-1}\\ s_y^2=\frac{512.875}{7}\\ s_y^2=73.267857\\ s_y=8.5596645\end{array}$

The correlation coefficient is calculated as shown.

  $\begin{array}{l}r=\frac{1}{(n-1)}\frac{{\displaystyle \sum (x-\overline{x})(y-\overline{y})}}{s_x{s}_y}\\ r=\frac{1}{(8-1)}\frac{(4.125)(4.875)+\mathrm{........}+(9.125)(1.875)}{(6.35694221)(8.5596645)}\\ r=\frac{1}{(8-1)}\frac{299.125}{(6.35694221)(8.5596645)}\\ r=0.785325\end{array}$

Also, the least square regression line can be calculated as shown.

  $\stackrel{\wedge }{y}=b_0+b_{1}x$

Here;

  $\begin{array}{l}{b}_1=r\frac{s_y}{s_x}\\ b_1=0.785325\left(\frac{6.35694221}{8.5596645}\right)\\ b_1=1.0574\end{array}$

Here;

  $\begin{array}{l}{b}_0=\overline{y}-b_1\overline{x}\\ b_0=35.125-(1.0574)(20.875)\\ b_0=13.051\end{array}$

Therefore,

  $\begin{array}{l}\stackrel{\wedge }{y}=b_0+b_{1}x\\ \stackrel{\wedge }{y}=13.051+1.0574x\end{array}$

Now we need to calculate the residuals to find the $s_e$ - Residual standard deviation.

  $\begin{array}{lllll}x\hfill & y\hfill & \stackrel{\wedge }{y}=13.051+1.0574x\hfill & y-\stackrel{\wedge }{y}\hfill & {\left(y-\stackrel{\wedge }{y}\right)}^2\hfill \\ 25\hfill & 40\hfill & 39.486\hfill & 0.514\hfill & 0.264196\hfill \\ 13\hfill & 20\hfill & 26.7972\hfill & -6.7972\hfill & 46.20193\hfill \\ 16\hfill & 33\hfill & 29.9694\hfill & 3.0306\hfill & 9.184536\hfill \\ 19\hfill & 30\hfill & 33.1416\hfill & -3.1416\hfill & 9.869651\hfill \\ 29\hfill & 50\hfill & 43.7156\hfill & 6.2844\hfill & 39.49368\hfill \\ 19\hfill & 37\hfill & 33.1416\hfill & 3.8584\hfill & 14.88725\hfill \\ 16\hfill & 34\hfill & 29.9694\hfill & 4.0306\hfill & 16.24574\hfill \\ 30\hfill & 37\hfill & 44.773\hfill & -7.773\hfill & 60.41953\hfill \end{array}$

  $\begin{array}{l}{s}_e=\sqrt{\frac{{\displaystyle \sum (y-\stackrel{\wedge }{y{)}^2}}}{n-2}}\\ s_e=\sqrt{\frac{0.264196+\mathrm{........}+60.41953}{8-2}}\\ s_e=\sqrt{\frac{196.5665}{6}}\\ s_e=\sqrt{32.76109}\\ s_e=5.72373\end{array}$

Finally we can now calculate the prediction interval for an individual response.

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

Here; when $x=20$ ;

  $\begin{array}{l}\stackrel{\wedge }{y}=13.051+1.057x\\ \stackrel{\wedge }{y}=13.051+1.057(20)\\ \stackrel{\wedge }{y}=34.199\end{array}$

And;

  $\begin{array}{l}$ {(x^{*}-\overline{x})}^2$$ ={(20-20.875)}^2$=0.765625\end{array}$

And;

  $\begin{array}{l}{\displaystyle \sum {(x-\overline{x})}^2}\\ =282.875\end{array}$

Therefore;

  $\begin{array}{l}\stackrel{\wedge }{y}\pm t_{\alpha /2}.s_e\sqrt{1+\frac{1}{n}+\frac{{(x^{*}-\overline{x})}^2}{{\displaystyle \sum {(x-\overline{x})}^2}}}\\ 34.199\pm t_{\alpha /2}(5.72373)\sqrt{1+\frac{1}{8}+\frac{0.765625}{282.875}}\\ 34.199\pm t_{\alpha /2}(5.72373)(1.0619353)\\ 34.199\pm t_{\alpha /2}6.07823093467\end{array}$

Please note there wasn’t a table given in the book to find corresponding t-value for the given confidence interval $95\%$ and degrees of freedom $6$ as shown below.

  

The t table below statescorresponding t-value for the given confidence interval $95\%$ and degrees of freedom $6$ can be read as $1.94$ .