2 out of 2 attempts An automobile manufacturer wishes to compare the lifetimes o

Question

2 out of 2 attempts An automobile manufacturer wishes to compare the lifetimes of two brands of tire. She obtains samples of 7 tires of each brand. On each of7 cars, she mounts one tire of each brand on each front wheel. The cars are driven until only 20% of the original tread remains. The distances, in miles, for each tire are presented in the following table. Car Brand 1 Brand 2 6,92534,934 45,30033,804 36,240 30,277 32,100 33,823 37,210 36,985 48,36035,383 6 738,200 47,717 Compute the value of the test statistic. Value of the test statistic

Explanation / Answer

Given that,
mean(x)=36190.71
standard deviation , s.d1=5633.565
number(n1)=7
y(mean)=36131.86
standard deviation, s.d2 =5507.088
number(n2)=7
null, Ho: u1 = u2
alternate, H1: u1 != u2
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) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =36190.71-36131.86/sqrt((31737054.60923/7)+(30328018.23974/7))
to =0.02
| to | =0.02
critical value
the value of |t | with min (n1-1, n2-1) i.e 6 d.f is 2.447
we got |to| = 0.01976 & | t | = 2.447
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.0198 ) = 0.985
hence value of p0.05 < 0.985,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.02
critical value: -2.447 , 2.447
decision: do not reject Ho
p-value: 0.985
we do not have enough evidence to support the claim

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.