Question 1202131: A trucking firm suspects that the average lifetime of 25,000 miles claimed for certained for certain tires is too high. To test the claim, the firm puts a random sample of 40 of these tires on its truck and later fins that their average lifetime is 24,421 miles and the standard deviation is 1,349. What can it conclude at 0.01 level of significance
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! sample mean is 24421
sample standard devition is 1349
sample size is 40
standard error is 1349 / sqrt(40) = 213.2956282.
t-score = (x-m)/s
x is the sample mean
m is the assumed population mean
s is the standard error.
formula becomes t = (24421 - 25000) / 213.2956282. = -2.714542276.
critical t-score with 39 degrees of freedom at with .01 two tailed alpha = -2.707913179.
the absolute value of the test t-score is greater than the absolute value of the critical t-score on the low end of the confidence interval.
this indicates that the test is significant.
this would lead to the conclusion that the average lifetime is probably lower than 25,000, although the evidence is not as stong as it could be if the test t-score was greater.
the test alpha was equal to .00491687336.
this was less than the critical alpha of .005, leading to the same conclusion.
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