The weights (in pounds) of 6 vehicles and the variability of their braking dista
ID: 3205888 • 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
A. The critical value is -4.604, +4.604
B. The critical values are - t 0=-4.604 and t 0 =+4.604
Calculate the test statistic. t=1.5(Round to three decimal places as needed.) What is your conclusion? There is not is 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.
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