The mean tread life of a manufacturer's best selling tire is known to be 22,560 miles....

The mean tread life of a manufacturer's best selling tire is known to be 22,560 miles. Because the distribution of individual tire life is positively skewed, it cannot be assumed to be normal. R&D has come up with a new rubber additive that they suspect will increase mean tire life without affecting variability of tire life. A sample of 36 tires with the new additive has a sample mean tread life of 24,470 miles; the sample based estimate of the standard deviation is 1200 miles. At .05 level of significance, test whether the additive is working. Clearly state your hypothesis and conclusion.

Solutions

Expert Solution

Given that,
population mean(u)=22560
sample mean, x =24470
standard deviation, s =1200
number (n)=36
null, Ho: μ=22560
alternate, H1: μ>22560
level of significance, alpha = 0.05
from standard normal table,right tailed t alpha/2 =1.69
since our test is right-tailed
reject Ho, if to > 1.69
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =24470-22560/(1200/sqrt(36))
to =9.55
| to | =9.55
critical value
the value of |t alpha| with n-1 = 35 d.f is 1.69
we got |to| =9.55 & | t alpha | =1.69
make decision
hence value of | to | > | t alpha| and here we reject Ho
p-value :right tail - Ha : ( p > 9.55 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: μ=22560
alternate, H1: μ>22560
test statistic: 9.55
critical value: 1.69
decision: reject Ho
p-value: 0
we have evidence that new rubber additive will increase mean tire life without affecting variability of tire life

orchestra answered 1 year ago

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