SOLUTION: a trucking firm suspects that the mean lifetime of a certain tire it uses is less than 34,000 miles. To check the claim, the firm randomly selects and tests 54 of these tires and g
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-> SOLUTION: a trucking firm suspects that the mean lifetime of a certain tire it uses is less than 34,000 miles. To check the claim, the firm randomly selects and tests 54 of these tires and g
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Question 1203080: a trucking firm suspects that the mean lifetime of a certain tire it uses is less than 34,000 miles. To check the claim, the firm randomly selects and tests 54 of these tires and gets a mean lifetime of 33,390 miles with a standard deviation of 1200 miles. At α = 0.05
You can put this solution on YOUR website! population mean = 34000 = m
sample size = 54 = n
sample mean = 33390 = x
sample standard deviation = 1200 = d
one tailed confidence interval is .95 with .05 tail on the left end of it.
it's a one tail confidence interval because the test is to determine if the mean lifetime of the sample is less than the population mean.
since the standad deviation is taken from the sample, the t-test is used.
the standard error of the test is equal to the standard deviation of the sample divided by the square root of the sample size = d / sqrt(n) = 1200 / sqrt(54) = 163.2993162.
the t-score formula is t = (x-m)/s which is equal to (33390 - 34000) / 163.2993162 = -3.73547.
the area to the left of that t-score, with 53 degrees of freedom, is equal to .00029828.
this is considerably less than the critical p-value of .05.
also, the critical t-score at 95% left side one tailed confidence interval is equal to t = -1.674116.
the test t-score of -3.73547 is greater than this, supporting the conclusion that the results of the test are significant.
this means that the mean lifetime of those certain tires can be assumed to actually be less than the 34000 originally stated.
this is what the t-test results look like on a graph.
