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t-Distribution. Source: Pearson t-Distribution. Source: Pearson Braking Distance

ID: 3201633 • Letter: T

Question

t-Distribution. Source: Pearson t-Distribution. Source: Pearson Braking Distances: Wet Surface The weights (in pounds) of eight vehicles and the variabilities of their braking distances (in feet) when stopping on a wet surface are shown in the table. At = 0.05, is there enough evidence to conclude that there is a significant linear correlation between vehicle weight and variability in braking distance on a wet surface?

Weight, x 5890 5340 6500 4800 5940 5600 5100 5850

Variability, y 2.92 2.40 4.09 1.72 2.88 2.53 2.32 2.78

a)State the null and the alternative hypotheses.

b) Determine the critical values.

c) Find the standardized test statistic.

d) Make a decision and interpret your decision in the context of the original claim.

Question 1

State Ho and Ha.

Ho: 0

Ha: < 0

Ho: 0

Ha: > 0

Ho: = 0

Ha: 0

Find the critical value(s). (round your answer to 3 decimals)

to= ± _______

Find the standardized test statistic. (round to 3 decimal places)


t =______

Determine whether to reject or fail to reject the Ho and interpret your decision.

Reject Ho. There is not sufficient evidence at the 5% level of significance to conclude that there is significant correlation between vehicle weight and variability in braking distance on a wet surface.

Reject Ho. There is enough evidence at the 5% level of significance to conclude that there is significant linear correlation between vehicle weight and variability in braking distance on a wet surface.

Fail to reject Ho. There is enough evidence at the 5% level of significance to conclude that there is significant linear correlation between vehicle weight and variability in braking distance on a wet surface.

Fail to reject Ho. There is not sufficient evidence at the 5% level of significance to conclude that there is significant correlation between vehicle weight and variability in braking distance on a wet surface.

Ho: 0

Ha: < 0

Ho: 0

Ha: > 0

Ho: = 0

Ha: 0

Find the critical value(s). (round your answer to 3 decimals)

to= ± _______

Find the standardized test statistic. (round to 3 decimal places)


t =______

Determine whether to reject or fail to reject the Ho and interpret your decision.

Reject Ho. There is not sufficient evidence at the 5% level of significance to conclude that there is significant correlation between vehicle weight and variability in braking distance on a wet surface.

Reject Ho. There is enough evidence at the 5% level of significance to conclude that there is significant linear correlation between vehicle weight and variability in braking distance on a wet surface.

Fail to reject Ho. There is enough evidence at the 5% level of significance to conclude that there is significant linear correlation between vehicle weight and variability in braking distance on a wet surface.

Fail to reject Ho. There is not sufficient evidence at the 5% level of significance to conclude that there is significant correlation between vehicle weight and variability in braking distance on a wet 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 /8 (124226) - [ 1/8 *45020 ] [ 1/8 *21.64] = 305.862                              
S. D ( X ) = Sqrt( 1/8*255373800-(1/8*45020)^2) = 502.96                              
S .D (Y) = Sqrt( 1/8*61.779-(1/8*21.64)^2) = 0.637                              
r(x,y) = 305.862 / 502.96*0.637 = 0.9547      
                      
If r = 0.9547> 0 ,Perfect Positive Correlation  

                          
Given that,
value of r =0.9547
number (n)=8
null, Ho: =0
alternate, H1: !=0
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.447
since our test is two-tailed
reject Ho, if to < -2.447 OR if to > 2.447
we use test statistic (t) = r / sqrt(1-r^2/(n-2))
to=0.9547/(sqrt( ( 1-0.9547^2 )/(8-2) )
to =7.86
|to | =7.86
critical value
the value of |t | at los 0.05% is 2.447
we got |to| =7.86 & | t | =2.447
make decision
hence value of | to | > | t | and here we reject Ho

ANSWERS
---------------
null, Ho: =0
alternate, H1: !=0
test statistic: 7.86
critical value: -2.447 , 2.447
decision: reject Ho. rhere is enough evidence at the 5% level of significance
to conclude that there is significant linear correlation between vehicle weight and
variability in braking distance on a wet surface.

Correlation ( X) ( Y) X^2 Y^2 X*Y 5890 2.92 34692100 8.5264 17198.8 5340 2.4 28515600 5.76 12816 6500 4.09 42250000 16.7281 26585 4800 1.72 23040000 2.9584 8256 5940 2.88 35283600 8.2944 17107.2 5600 2.53 31360000 6.4009 14168 5100 2.32 26010000 5.3824 11832 5850 2.78 34222500 7.7284 16263 45020 21.64 255373800 61.779 124226 = TOTALS