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 5 = 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

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   61000  
Ha:    u   >   61000  
              
As we can see, this is a    right   tailed test.      
              
              
Getting the test statistic, as              
              
X = sample mean =    63245          
uo = hypothesized mean =    61000          
n = sample size =    10          
s = standard deviation =    3035          
              
Thus, t = (X - uo) * sqrt(n) / s =    2.339147726          
              
Thus, getting the critical t,              
df = n - 1 =    9          
tcrit =    +   1.833112933      

Thus, we reject Ho if t > 1.8331.
              
Comparing t > 1.8331, we   REJECT THE NULL HYPOTHESIS.          

Thus, there is significant evidence that the mean tire life is greater than 61000 km. [CONCLUSION]

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b)

Lower Bound = X - t(alpha) * s / sqrt(n)              
      
              
where              
alpha = (1 - confidence level)=    0.05          
X = sample mean =    63245          
t(alpha) = critical t for the confidence interval =    1.833112933          
s = sample standard deviation =    3035          
n = sample size =    10          
df = n - 1 =    9          
Thus,              
              
Lower bound =    61485.66754          

              
Thus, the confidence interval is u > 61485.66754. [ANSWER]

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c)

As the lower bound of the interval above is still greater than 61000, then there is significant evidence at 0.05 level that the mean tire life is greater than 61000 km.   [CONCLUSION]