The western United States has a number of four-lane interstate highways that cut

Question

The western United States has a number of four-lane interstate highways that cut through long tracts of wilderness. To prevent car accidents with wild animals, the highways are bordered on both sides with 12-foot-high woven wire fences. Although the fences prevent accidents, they also disturb the winter migration pattern of many animals. To compensate for this disturbance, the highways have frequent wilderness underpasses designed for exclusive use by deer, elk, and other animals. In Colorado, there is a large group of deer that spend their summer months in a region on one side of a highway and survive the winter months in a lower region on the other side. To determine if the highway has disturbed deer migration to the winter feeding area, the following data were gathered on a random sample of 10 wilderness districts in the winter feeding area. Row B represents the average January deer count for a 5-year period before the highway was built, and row A represents the average January deer count for a 5-year period after the highway was built. The highway department claims that the January population has not changed. Test this claim against the claim that the January population has dropped. Use a 5% level of significance. Units used in the table are hundreds of deer. (Let d = B A.)

Wilderness District B: Before highway A: After highway 10 10.3 7.412.95.8 17.49.9 20.5 16.2 18.9 11.6 9.18.2 10.0 4.3 4.07.115.2 8.3 12.2 7.3 1 2 3 4 5 6 7 8 9

Explanation / Answer

paired T test

Given that,
null, H0: Ud = 0
alternate, H1: Ud > 0
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.833
since our test is right-tailed
reject Ho, if to > 1.833
we use Test Statistic  
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 4.52
We have d = 4.52
pooled variance = calculate value of Sd= S^2 = sqrt [ 354.02-(45.2^2/10 ] / 9 = 4.079
to = d/ (S/n) = 3.504
critical Value
the value of |t | with n-1 = 9 d.f is 1.833
we got |t o| = 3.504 & |t | =1.833
make Decision
hence Value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 3.5045 ) = 0.00334
hence value of p0.05 > 0.00334,here we reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud > 0
test statistic: 3.504
critical value: reject Ho, if to > 1.833
decision: Reject Ho
p-value: 0.00334
we have enough evidence to support the claim

X Y X-Y (X-Y)^2 10.3 9.1 1.2 1.44 7.4 8.2 -0.8 0.64 12.9 10 2.9 8.41 5.8 4.3 1.5 2.25 17.4 4 13.4 179.56 9.9 7.1 2.8 7.84 20.5 15.2 5.3 28.09 16.2 8.3 7.9 62.41 18.9 12.2 6.7 44.89 11.6 7.3 4.3 18.49 45.2 354.02

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