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 = 63245 and s = 3035 kilometers, respectively. Assume that lifetimes of tires created from the new compound are distributed Normal with mean mu Test H_0: mu = 61000 against H_a: mu > 61000 at alpha = 0.05. Give the rejection region. Compute a 95% confidence lower bound for mu. 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

I.
Set Up Hypothesis
Null Hypothesis H0: U=61000
Alternate, It is more than Mean H1: U>61000
Test Statistic
Population Mean(U)=61000
Sample X(Mean)=63245
Standard Deviation(S.D)=3035
Number (n)=10
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =63245-61000/(3035/Sqrt(9))
to =2.1
| to | =2.1
Critical Value
The Value of |t | with n-1 = 9 d.f is 1.833
We got |to| =2.1 & | t | =1.833
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Right Tail - Ha : ( P > 2.1 ) = 0.03256
Hence Value of P0.05 > 0.03256,Here we Reject Ho

II.
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=63245
Standard deviation( sd )=3035
Sample Size(n)=10
Confidence Interval = [ 63245 ± t a/2 ( 3035/ Sqrt ( 10) ) ]
= [ 63245 - 2.26 * (959.751) , 63245 + 2.26 * (959.751) ]
= [ 61075.962,65414.038 ]

III.
We conclude that it is more tha 61000