A research engineer for a tire manufacturer is investigating

A research engineer for a tire manufacturer is investigating tire life for a new rubber
compound. She has built n = 10 tires and tested them to end-of-life in a road test. The
sample mean and standard deviation are x bar = 63245 and s = 3025 and kilometers,
respectively. Assume that lifetimes of tires created from the new compound are
distributed Normal with mean .
(a) Test Ho : mu = 61000 against Ha : mu > 61000 at alpha = 0.05 . Give the rejection region.
(b) Compute a 95% confidence lower bound for mu .
(c) Use the bound found in part (b) to test the hypotheses given in part (a). (You should
have the same conclusion as in (a), but this time justify your answer based only on the
confidence lower bound.)

Solution

the test statistic is given by T=(Xbar-mu(H₀))/(s/sqrt(n)) which under H₀ follows a t distribution with n-1 degrees of freedom.

here H₁:mu>61000 so we use a right tailed test.

the rejection region is given as

we reject H₀ at alpha=0.05 iff t>t_alpha,n-1    where t is the observed value of T and t_alpha,n-1 is the upper alpha point of t distribution with n-1 degrees of freedom.   [answer]

here Xbar=63245   mu(H0)=61000   s=3025 n=10 alpha=0.05

now t=(63245-61000)/(3025/sqrt(10))=2.34688

and t_alpha,9=1.83311

hence t> t_alpha,9      so on the basis of the given data at hand we reject H0 at 5% level of significance and conclude that mu is not 61000

b) the 95% confidence lower bound for mu is Xbar-s/sqrt(n)* t_alpha/2,n-1=63245-3025/sqrt(10)*2.26216=61081.042 [answer]

c) here we reject the test if mu(H0)<95% confidence lower bound.

now mu(h0)=61000<61081.042=95% confidence lower bound.

hence we reject H0 and conclude the same as in part a) that mu is not equal to 61000 [answer]