9. An automobile manufacturer wishes to compare the lifetimes of two brands of t
ID: 3311899 • Letter: 9
Question
9. An automobile manufacturer wishes to compare the lifetimes of two brands of tire. She obtains samples of six tires of each brand. On each of six cars, she mounts one tire of each brand on each front wheel, The cars are driven until only 20% ofthe original tread remains The distances, in miles, for each tire are presented in the following table. Can you conclude that there is a difference between the mean lifetimes of the two brands of tire? Car Brand 1 Brand 2 36,925 45,300 36,240 32,100 37,210 48,360 38,200 34,318 42,280 35,500 31,950 38,015 47,800 33,215 4 a. State the appropriate null and alternate hypotheses b. Compute the value of the test statistic. c. Find the P-value and state your conclusionExplanation / Answer
Given that,
mean(x)=39190.7143
standard deviation , s.d1=5633.5649
number(n1)=7
y(mean)=37582.5714
standard deviation, s.d2 =5663.4022
number(n2)=7
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.1
from standard normal table, two tailed t /2 =1.943
since our test is two-tailed
reject Ho, if to < -1.943 OR if to > 1.943
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =39190.7143-37582.5714/sqrt((31737053.48251/7)+(32074124.47896/7))
to =0.5326
| to | =0.5326
critical value
the value of |t | with min (n1-1, n2-1) i.e 6 d.f is 1.943
we got |to| = 0.53263 & | t | = 1.943
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.5326 ) = 0.613
hence value of p0.1 < 0.613,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.5326
critical value: -1.943 , 1.943
decision: do not reject Ho
p-value: 0.613
conclude that no difference between the mean lifetimes of the two
brands of tire
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