Answer to Question #135359 in Statistics and Probability for Sana

Question #135359
You have to examine the relationship between the age and price for used cars
sold in the last year by a car dealership company. Fit a simple linear regression
model Price Sold on Car age.
(b) Find the estimated linear regression values of Y.
(c) Show that the sum of Deviations of estimated values from actual values is
zero.
(d) Show that the sum of squares of residuals is minimum.
(e) Find the standard error of estimate. Calculate the Correlation Co-efficient y on x.
(g) Find the Co-efficient of Determination.
Car Age (in years) Price (in dollars)
4 6300
4 5800
5 5700
5 4500
7 4500
7 4200
8 4100
9 3100
10 2100
11 2500
12 2200
1
Expert's answer
2020-09-29T18:16:47-0400

xyxyx2y2463002520016396900004580023200163364000055700285002532490000545002250025202500007450031500492025000074200294004917640000841003280064168100009310027900819610000102100210001004410000112500275001216250000122200264001444840000Sum=8245000265900690205880000\begin{matrix} & x & y & xy & x^2 & y^2 \\ & 4 & 6300 & 25200 & 16 & 39690000 \\ & 4 & 5800 & 23200 & 16 & 33640000 \\ & 5 & 5700 & 28500 & 25 & 32490000 \\ & 5 & 4500 & 22500 & 25 & 20250000 \\ & 7 & 4500 & 31500 & 49 & 20250000 \\ & 7 & 4200 & 29400 & 49 & 17640000 \\ & 8 & 4100 & 32800 & 64 & 16810000\\ & 9 & 3100 & 27900 & 81 & 9610000 \\ & 10 & 2100 & 21000 & 100 & 4410000 \\ & 11 & 2500 & 27500 & 121 & 6250000 \\ & 12 & 2200 & 26400 & 144 & 4840000 \\ Sum= & 82 & 45000 & 265900 & 690 & 205880000 \end{matrix}


xˉ=1ni=1nxi=82117.454545\bar{x}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nx_i=\dfrac{82}{11}\approx7.454545

yˉ=1ni=1nyi=450001140.909091\bar{y}=\dfrac{1}{n}\displaystyle\sum_{i=1}^ny_i=\dfrac{45000}{11}\approx40.909091

SSxx=i=1nxi21n(i=1nxi)2=SS_{xx}=\displaystyle\sum_{i=1}^nx_i^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nx_i)^2=

=6908221178.727273=690-\dfrac{82^2}{11}\approx78.727273

SSyy=i=1nyi21n(i=1nyi)2=SS_{yy}=\displaystyle\sum_{i=1}^ny_i^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^ny_i)^2=

=2058800004500021121789090.909091=205880000 -\dfrac{45000^2}{11}\approx21789090.909091


SSxy=i=1nxiyi1n(i=1nxi)(i=1nyi)=SS_{xy}=\displaystyle\sum_{i=1}^nx_iy_i-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nx_i)(\displaystyle\sum_{i=1}^ny_i)=

=26590082450001139554.545455=265900 -\dfrac{82\cdot45000}{11}\approx-39554.545455

Therefore, based on the above calculations, the regression coefficients (the slope m,m,

and the y-intercept nn) are obtained as follows:


m=SSxySSxx502.424942m=\dfrac{SS_{xy}}{SS_{xx}}\approx-502.424942

n=yˉmxˉn=\bar{y}-m\cdot\bar{x}\approx

40.909091(502.424942)7.4545457836.25866\approx40.909091-(-502.424942)\cdot7.454545\approx7836.25866

Therefore, we find that the regression equation is:


y=7836.25866502.424942xy=7836.25866-502.424942x

y(4)=7836.25866502.424942(4)=5826.56y(4)=7836.25866-502.424942(4)=5826.56

y(5)=7836.25866502.424942(5)=5324.13y(5)=7836.25866-502.424942(5)=5324.13

y(7)=7836.25866502.424942(7)=4319.28y(7)=7836.25866-502.424942(7)=4319.28

y(8)=7836.25866502.424942(8)=3816.86y(8)=7836.25866-502.424942(8)=3816.86

y(9)=7836.25866502.424942(9)=3314.43y(9)=7836.25866-502.424942(9)=3314.43

y(10)=7836.25866502.424942(10)=2812.01y(10)=7836.25866-502.424942(10)=2812.01

y(11)=7836.25866502.424942(11)=2309.58y(11)=7836.25866-502.424942(11)=2309.58

y(12)=7836.25866502.424942(12)=1807.16y(12)=7836.25866-502.424942(12)=1807.16


63005826.56+58005826.56+6300-5826.56+5800-5826.56++57005324.13+45005324.13++5700-5324.13+4500-5324.13++45004319.28+42004319.28++4500-4319.28+4200-4319.28++41003816.86+31003314.43++4100-3816.86+3100-3314.43++21002812.01+25002309.58++2100-2812.01+2500-2309.58++22001807.16=0.020+2200-1807.16=0.02\approx0

RSS=i=1n(yiobsyipred)2=SSyymSSxy=RSS=\displaystyle\sum_{i=1}^n(y_{iobs}-y_{ipred})^2=SS_{yy}-mSS_{xy}=


=SSyy(1SSxy2SSxxSSyy)==SS_{yy}(1-\dfrac{SS_{xy}^2}{SS_{xx}SS_{yy}})=

=21789090.909091(1(39554.545455)278.727273(21789090.909091))==21789090.909091(1-\dfrac{(-39554.545455)^2}{78.727273(21789090.909091)})=

=1915896.5844=1915896.5844

The sum of square of residuals is minimum for points lying on the regression line and so cannot be less than 1915896.58441915896.5844 for any other line.


sest=RSSn21915896.5844112461.39s_{est}=\sqrt{\dfrac{RSS}{n-2}}\approx\sqrt{\dfrac{1915896.5844}{11-2}}\approx461.39

r=SSxySSxxSSyy=39554.54545578.72727321789090.909091r=\dfrac{SS_{xy}}{\sqrt{SS_{xx}}\sqrt{SS_{yy}}}=\dfrac{-39554.545455}{\sqrt{78.727273}\sqrt{21789090.909091}}\approx

0.955024\approx-0.955024

Strong negarive correlation.


r20.912071r^2\approx0.912071

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Assignment Expert
28.04.21, 08:52

Dear Jeco, please use the panel for submitting a new question.

Jeco
27.04.21, 16:37

2. You have to examine the relationship between the age and price for used cars sold in the last year by a car dealership company with the given table of data. Car Age (in years) Price (in dollars) 4 6300 4 5800 5 5700 5 4500 7 4500 7 4200 8 4100 9 3100 10 2100 11 2500 12 2200 What is the estimated cost value of car ages 2 and 15?

Leave a comment