15. Listed below are repair costs (in dollars) for different models of cars cras

Question

15. Listed below are repair costs (in dollars) for different models of cars crashed at 6 miles/hour in f crash tests and the same make and model of cars crashed at 6 miles/hour in full-rear cr ull-front 936 978 2252 1032 3 4312 3469 1480 1202 802 22 739 2767 Suppose we want to test the claim there is a linear correlation between the repair costs from full-front a) Use a significance level of -0.05, and find the critical value of the Correlation Coefficient, r, d Front,x Rear, y crashes and full-rear crashes. b) Find the value of the linear correlation coefficient,r (2 pes) Is there sufficient evidence of a linear correlation between the repair costs from full-front crashes and full-rear crashes? c) 2 pts) Circle the correct answer: Yes No Letting y represent the repair cost from a full-rear crash and x represent the repair cost from a full- front crash, find the regression equation. Write the equation in the form ya +bx. Round the values of a and b to 3 decimal places d) (4 pes) e) Find the best predicted repair costs for a full-rear crash for a particular model of car if the repair cost for a full-front crash is $3500. You must show your work or explain your answer to receive full credit (2 pt

Explanation / Answer

15.

b.

calculation procedure for correlation

sum of (x) = x = 16890

sum of (y) = y = 11303

sum of (x^2)= x^2 = 53892334

sum of (y^2)= y^2 = 23922183

sum of (x*y)= x*y = 24833485

to caluclate value of r( x,y) = covariance ( x,y ) / sd (x) * sd (y)

covariance ( x,y ) = [ x*y - N *(x/N) * (y/N) ]/n-1

= 24833485 - [ 7 * (16890/7) * (11303/7) ]/7- 1

= -348434.1837

and now to calculate r( x,y) = -348434.1837/ (SQRT(1/7*24833485-(1/7*16890)^2) ) * ( SQRT(1/7*24833485-(1/7*11303)^2)

=-348434.1837 / (1370.0457*900.0847)

=-0.2826

value of correlation is =-0.2826

coeffcient of determination = r^2 = 0.0798

properties of correlation

1. If r = 1 Corrlation is called Perfect Positive Corrlelation

2. If r = -1 Correlation is called Perfect Negative Correlation

3. If r = 0 Correlation is called Zero Correlation

& with above we conclude that correlation ( r ) is = -0.2826< 0, negative correlation

a.

Given that,
value of r =-0.2826
number (n)=7
null, Ho: =0
alternate, H1: !=0
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.571
since our test is two-tailed
reject Ho, if to < -2.571 OR if to > 2.571
we use test statistic (t) = r / sqrt(1-r^2/(n-2))
to=-0.2826/(sqrt( ( 1--0.2826^2 )/(7-2) )
to =-0.659
|to | =0.659
critical value
the value of |t | at los 0.05% is 2.571
we got |to| =0.659 & | t | =2.571
make decision
hence value of |to | < | t | and here we do not reject Ho
ANSWERS
---------------
null, Ho: =0
alternate, H1: !=0
test statistic: -0.659
critical value: -2.571 , 2.571

decision: do not reject Ho

c.

No,

there is evidence to support the claim that linear correlation between the repai cost for full front crash and full rear crash

d.

calculation procedure for regression

mean of X = X / n = 2412.8571

mean of Y = Y / n = 1614.7143

(Xi - Mean)^2 = 13139176.858

(Yi - Mean)^2 = 5671067.43

(Xi-Mean)*(Yi-Mean) = -2439039.285

b1 = (Xi-Mean)*(Yi-Mean) / (Xi - Mean)^2

= -2439039.285 / 13139176.858

= -0.186

bo = Y / n - b1 * X / n

bo = 1614.7143 - -0.186*2412.8571 = 2062.616

value of regression equation is, Y = bo + b1 X

Y'=2062.616-0.186* X

e.

Y'=2062.616-0.186* X

when X= 3500$ full front crash

then Y' = 2062.616 - (0.186*3500)

Y' =1411.616

( X) ( Y) X^2 Y^2 X*Y 936 1480 876096 2190400 1385280 978 1202 956484 1444804 1175556 2252 802 5071504 643204 1806104 1032 3191 1065024 10182481 3293112 3911 1122 15295921 1258884 4388142 4312 739 18593344 546121 3186568 3469 2767 12033961 7656289 9598723

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