Question

Consider the relationship between the number of bids an item on eBay received and the item's selling price. The following is a sample of 5 items sold through an auction. Price in Dollars 29 32 40 45 49 Number of Bids 7 89 Table Copy Data Step 3 of 5: Calculate the estimated variance of slope, s. Round your answer to three decimal places. Answer(How to Enter) 2 Points Keypad BTables Next

Answer 1

Solution:

We are given the data on the number of bids an items on eBay received and the items selling price.

Part 3) We have to find the estimated variance of slope, that is: S²_b1

$S_{b1}=\frac{S_{e}}{\sqrt{SS_{xx}}}$

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

$SSE = SS_{yy}-\frac{SS_{xy}^{2}}{SS_{xx}}$

$SS_{xy}=\sum xy \: \: -\: \: \left ( \sum x\times \sum y \: \: /\: \: n\: \: \right )$

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

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

Thus we need to make following table:

x: Price	y: Number of Bids	x^2	y^2	xy
29	2	841	4	58
32	3	1024	9	96
40	7	1600	49	280
45	8	2025	64	360
49	9	2401	81	441
$\dpi{80} \sum x = 195$	$\dpi{80} \sum y = 29$	$\dpi{80} \sum x^{2} = 7891$	$\dpi{80} \sum y^{2} = 207$	$\dpi{80} \sum xy =1235$

Thus

$SS_{xy}=\sum xy \: \: -\: \: \left ( \sum x\times \sum y \: \: /\: \: n\: \: \right )$

$SS_{xy}=1235 \: \: -\: \: \left ( 195 \times 29 \: \: /\: \: 5\: \: \right )$

$SS_{xy}=1235 \: \: -\: \: 1131$

$SS_{xy}=104$

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

$SS_{xx}=7891 \: \: -\: \: \left ( 195 \times 195 \: \: /\: \: 5\: \: \right )$

$SS_{xx}=7891 \: \: -\: \: 7605$

$SS_{xx}= 286$

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

$SS_{yy}=207 \: \: -\: \: \left ( 29 \times 29 \: \: /\: \: 5\: \: \right )$

$SS_{yy}=207 \: \: -\: \: 168.2$

$SS_{yy}=38.8$

Thus

$SSE = SS_{yy}-\frac{SS_{xy}^{2}}{SS_{xx}}$

$SSE = 38.8 -\frac{104 ^{2}}{286 }$

$SSE = 38.8 -37.818182$

$SSE = 0.981818$

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

$S_{e}=\sqrt{\frac{0.981818 }{5-2}}$

$S_{e}=\sqrt{\frac{0.981818 }{3}}$

$S_{e}=\sqrt{0.327273 }$

$S_{e}=0.572078$

$S_{b1}=\frac{S_{e}}{\sqrt{SS_{xx}}}$

$S_{b1}=\frac{0.572078 }{\sqrt{286 }}$

$S_{b1}=\frac{0.572078 }{16.911535 }$

$S_{b1}=0.03383$

Thus estimated variance is given by:

$S^{2}_{b1}=0.03383^{2}$

$S^{2}_{b1}=0.00114$

$S^{2}_{b1}=0.001$

Part 4) 95% confidence interval for slope:

Formula:

$(b_{1}-E\: \: ,\: \: b_{1}+E)$

where

$b_{1}=\frac{SS_{xy}}{SS_{xx}}$

$b_{1}=\frac{104 }{286}$

$b_{1}=0.36364$

and

$E = t_{c}\times S_{b_{1}}$

Thus we need to find t critical value for c = 95% confidence level.

Thus two tailed area = 1 - c = 1 - 0.95 = 0.05

and df = n - 2 = 5 - 2 = 3

Thus look in t table for df = 3 and two tail area = 0.05 and find t critical value.

Thus from t table , we get:

t_c = 3.182

Thus

$E = t_{c}\times S_{b_{1}}$

$E = 3.182 \times 0.03383$

$E = 0.10764$

Thus

Lower end point = $b_{1}-E=0.36364-0.10764=0.25600 = 0.256$

Upper end point = $b_{1}+E=0.36364+0.10764=0.47128 = 0.471$

Lower end point = 0.256

Upper end point = 0.471

Part 5) 90% confidence interval for slope:

Formula:

$(b_{1}-E\: \: ,\: \: b_{1}+E)$

We need to find t critical value for c = 90% confidence level.

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

and df = n - 2 = 5 - 2 = 3

Thus look in t table for df = 3 and two tail area = 0.10 and find t critical value.

Thus from t table , we get:

t_c = 2.353

Thus

$E = t_{c}\times S_{b_{1}}$

$E = 2.353 \times 0.03383$

$E = 0.07960$

Thus

Lower end point = $b_{1}-E=0.36364-0.07960 =0.28404 = 0.284$

Upper end point = $b_{1}+E=0.36364+0.07960 =0.44323 = 0.443$

Thus

Lower end point =0.284

Upper end point = 0.443