Question

Consider the set of ordered pairs shown below. Assuming that the regression equation is = 1.768 +0.712x and the SSE = 2.558, construct a 90% prediction interva for x= 1. 1 15 5 0 2 4 5 3 5 3 3 у 2 Click the icon to view a portion of the Student's t-distribution table. Calculate the upper and lower limits of the prediction interval. UPL = LPL = (Round to three decimal places as needed.)

Answer 1

Solution:

Given:

9 = 1.768+0.712 C

SSE = 2.558

Prediction level = 90%

x = 1

Formula:

(Yp - E , yp + E)

where

ýp = 1.768 +0.712 x

Yp = 1.768 +0.712 x 1

yp = 1.768 +0.712

ур = 2.480

and

$E = t_{c}\times S_{e}\times \sqrt{1+\frac{1}{n}+\frac{(x-\bar{x})^{2}}{SS_{xx}}}$

tc is t critical value for c = 90%  confidence level

Thus two tail area = 1 - c = 1 - 0.90 = 0.10

df = n - 2 =  6 - 2 = 4

Look in  t table for df = 4 and two tail area = 0.10 and find t critical value

t Table cum. prob t 50 5.76 t 90 0.50 0.25 t.20 0.20 0.40 t.85 0.15 0.30 0.10 0.20 1.00 0.50 0.05 0.10 one-tail two-tails df 1 21 3 4 0.000 0.000 0.000 1000 0.000 1.000 0.816 0.765 0.741 0.727 1.376 1.061 0.978 0.941 0.920 1.963 1.386 1.250 1190 1.156 3.078 1.886 1.638 1.592 1.476 6.314 2.92 2.353 2.132 2.015

t_c = 2.132

$\bar{x} = \frac{\sum x}{n}$

$SS_{xx}=\sum x^{2} \: \: -\: \: \left ( \sum x\times \sum x \: \: /\: \: n\: \: \right )$

Thus

x	x^2
5	25
0	0
4	16
3	9
3	9
1	1
$\sum x =16$	$\sum x^{2} =60$

Thus

$\bar{x} = \frac{\sum x}{n} = \frac{16}{6} = 2.6667$

$SS_{xx}=\sum x^{2} \: \: -\: \: \left ( \sum x\times \sum x \: \: /\: \: n\: \: \right )$

$SS_{xx}= 60 \: \: -\: \: \left ( 16 \times 16 \: \: /\: \: 6\: \: \right )$

$S_{e}= \sqrt{\frac{SSE}{n-2}}$

$S_{e}= \sqrt{\frac{2.558}{6-2}}$

$S_{e}= \sqrt{\frac{2.558}{4}}$

$S_{e}= \sqrt{ 0.6395 }$

thus

$E = t_{c}\times S_{e}\times \sqrt{1+\frac{1}{n}+\frac{(x-\bar{x})^{2}}{SS_{xx}}}$

$E =2.132 \times 0.7996874 \times \sqrt{1+\frac{1}{6}+\frac{(1- 2.6667 )^{2}}{17.3333 }}$

$E =2.132 \times 0.7996874 \times \sqrt{1+0.166667 +\frac{(-1.6667 )^{2}}{17.3333 }}$

$E =2.132 \times 0.7996874 \times \sqrt{1+0.166667 + 0.16025641 }$

$E =2.132 \times 0.7996874 \times \sqrt{ 1.326923 }$

$E =2.132 \times 0.7996874 \times 1.151921$

thus

(Yp - E , yp + E)

UPL = Upper Prediction Limit and LPL = Lower Prediction Limit

Thus

UPL = 4.444

LPL = 0.516