Question 1026627: A trucking firm suspects that the mean lifetime of a certain tire it uses is less than 36,000 miles. To check the claim, the firm randomly selects 54 of these tires and gets a mean lifetime of 35,630 miles with a population standards deviation of 1200 miles. At alpha=0.05, test the trucking firms claim.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! mean to test against is 36,000.
population standard deviation is 1200.
sample mean is 35,630.
alpha = .05.
sample size = 54.
z = (x-m)/se
z is the z-score.
x is the sample mean.
m is the mean to test against.
se is the standard error.
standard error = standard deviation divided by square root of sample size.
se = 1200 / sqrt(54) = 163.3.
z = (35,630 - 36,000) / 163.3 = -370 / 163.3 = -2.27.
look up z-score in the z-score table to find that the area to the left of that z-score is equal to .0116.
your threshold alpha is .05.
the sample alpha is less than that.
this is an indication that the average life of the tire is actually less than 36,000 miles.
if the mean life of the tire was 36,000 miles, as claimed, then the likelihood of a sample having a mean life of 35,630 would only happen 1% of the time, on average.
to be sure, they would look at all the samples they took, and if more than 5% of those samples had a mean life less than the threshold, they would conclude that the tires do indeed have a mean life that is less than 36,000 miles.
the threshold of 5% can be found by looking at the z-score table and finding the z-score that has an area to the left of it that is .05 or less.
looking at the z-score table, i see that a z-score of -1.65 is just under the threshold of .05.
a z-score of -1.65 has an area to the left of it that is equal to .0495.
to find the raw score that relates to that, use the z-score formula again.
let x = the raw score.
z = (x-m) / se
z = -1.65
m = 36,000
se = 163.3
formula becomes:
-1.65 = (x - 36,000) / 163.3
solve for x to get:
x = -1.65 * 163.3 + 36,000.
you get x = 35,730.555
based on an alpha of .05, they would expect to get a sample with a mean less than 35,730 miles in less than 5% of the samples that they take.
since they take regular samples, they should be able to look at all the samples that they took and see how many of them had a mean of 35,730 or less.
if the percentage of those samples was greater than 5%, then they could reasonably conclude that the average life of the tires was actually less than 36,000 which is contrary to the manufacturer's claim.
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