The weights (in pounds) of 6 vehicles and the variability of their braking dista
ID: 3205884 • Letter: T
Question
The weights (in pounds) of 6 vehicles and the variability of their braking distances (in feet) when stopping on a dry surface are shown in the table. Can you conclude that there is a significant linear correlation between vehicle weight and variability in braking distance on a dry surface? Use alpha equals0.01 .
Weight, x 5920 5360 6500 5100 5820 4800
Variability in braking distance, y 1.71 1.97 1.92 1.56 1.64 1.50 Setup the hypothesis for the test. Upper H 0 : rho less than or equals less than equals greater than greater than or equals not equals 0 Upper H Subscript a : rho less than or equals less than equals greater than greater than or equals not equals 0 Identify the critical value(s). Select the correct choice below and fill in any answer boxes within your choice. (Round to three decimal places as needed.)
A. The critical value is ____ . B. The critical values are - t 0=____ and t 0 =____ .
Calculate the test statistic. t=____ (Round to three decimal places as needed.) What is your conclusion?
There is not is enough evidence at the 1 % level of significance to conclude that there is is not a significant linear correlation between vehicle weight and variability in braking distance on a dry surface.
Explanation / Answer
r( X,Y) =Co V ( X,Y) / S.D (X) * S.D (y)
r( X,Y) = Sum(XY) / N- Mean of (X) * Mean of (Y) / Sqrt( X^2/n - ( Mean of X)^2 ) Sqrt( Y^2/n - ( Mean of Y)^2 )
Co v ( X, Y ) = 1 /6 (57863.2) - [ 1/6 *33500 ] [ 1/6 *10.3] = 59.144
S. D ( X ) = Sqrt( 1/6*188948400-(1/6*33500)^2) = 563.728
S .D (Y) = Sqrt( 1/6*17.8646-(1/6*10.3)^2) = 0.175
r(x,y) = 59.144 / 563.728*0.175 = 0.5995
If r = 0.5995> 0 ,Perfect Positive Correlation
Hypothesis Procedure
value of r =0.5995
number (n)=6
null, Ho: =0
alternate, H1: !=0
level of significance, = 0.01
from standard normal table, two tailed t /2 =4.604
since our test is two-tailed
reject Ho, if to < -4.604 OR if to > 4.604
we use test statistic (t) = r / sqrt(1-r^2/(n-2))
to=0.5995/(sqrt( ( 1-0.5995^2 )/(6-2) )
to =1.5
|to | =1.5
critical value
the value of |t | at los 0.01% is 4.604
we got |to| =1.5 & | t | =4.604
make decision
hence value of |to | < | t | and here we do not reject Ho
ANSWERS
---------------
null, Ho: =0
alternate, H1: !=0
test statistic: 1.5
critical value: -4.604 , 4.604
decision: do not reject Ho
There is not is enough evidence at the 1 % level of significance to conclude that there is not a significant linear correlation between vehicle weight and variability in braking distance on a dry surface.
( X) ( Y) X^2 Y^2 X*Y 5920 1.71 35046400 2.9241 10123.2 5360 1.97 28729600 3.8809 10559.2 6500 1.92 42250000 3.6864 12480 5100 1.56 26010000 2.4336 7956 5820 1.64 33872400 2.6896 9544.8 4800 1.5 23040000 2.25 7200 33500 10.3 188948400 17.8646 57863.2 = TOTALSRelated Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.