Question 448054: A local tire store suspects that the mean life of a new discount tire is less than 39,000 miles. The store selects randomly 18 of these new discount tires to test the claim. When the tires are tested, it is found that the mean life is 38,250 miles, with a sample standard deviation of 1,200 miles. Assume the distribution is normally distributed.
Using alpha () = 0.05, test the company's claim that the mean life is less than 39,000 miles.
(
H0:
Ha:
Test Statistic (t):
Critical t0:
Decision:
Interpretation:
Using alpha () = 0.01, test the company's claim that the mean life is less than 39,000 miles.
(References: example 1 though 5 pages 397 - 401, end of section exercises 23 – 28 pages 404 - 405) (6 points)
H0:
Ha:
Test Statistic (t):
Critical t0:
Decision:
Interpretation
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A local tire store suspects that the mean life of a new discount tire is less than 39,000 miles. The store selects randomly 18 of these new discount tires to test the claim. When the tires are tested, it is found that the mean life is 38,250 miles, with a sample standard deviation of 1,200 miles. Assume the distribution is normally distributed.
Using alpha () = 0.05, test the company's claim that the mean life is less than 39,000 miles
(
H0: u >= 39000
Ha: u < 39000 (claim)
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Test Statistic t(38250) = (38,250-39000)/[1200/sqrt(18)] = -2.6500
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Critical t0 = invT(0.05,17) = -1.7396
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Decision: Since the ts is in the reject interval, reject Ho
at the 5% level of significance.
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Interpretation:
Using alpha () = 0.01, test the company's claim that the mean life is less than 39,000 miles.
p-value = P(t < -2.6500 when df = 17) = 0.0084
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Since the p-value is less than 1%, reject Ho.
The test result supports the claim that the mean
life span of the tires is less than 39,000.
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Cheers,
Stan H.
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