SOLUTION: 2. A local tire store suspects that the mean life of a new discount tire is less that 39,000 miles. To check the claim, the store selects randomly 18 of these new discount tires. W
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Question 310652: 2. A local tire store suspects that the mean life of a new discount tire is less that 39,000 miles. To check the claim, the store selects randomly 18 of these new discount tires. When they are tested, it is found that the mean life is 38,250 miles with a sample standard deviation s = 1200 miles. Assume the distribution is normally distributed.
a. Use the critical value t0 method from the normal distribution to test for the population mean . Test the company’s claim at the level of significance = 0.05.
(References: example 1 though 5 pages 397 - 401, end of section exercises 23 – 28 pages 404 - 405) (5 points)
1. H0 :
Ha :
2. =
3. Test statistics:
4. P-value or critical z0 or t0.
5. Rejection Region:
6. Decision:
7. Interpretation:
b. Use the critical value t0 method from the normal distribution to test for the population mean m. Test the company’s claim at the level of significance a = 0.01
(References: example 1 though 5 pages 397 - 401, end of section exercises 23 – 28 pages 404 - 405) (5 points)
1. H0 :
Ha :
2. a =
3. Test statistics:
4. P-value or critical z0 or t0.
5. Rejection Region:
6. Decision:
7. 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 that 39,000 miles. To check the claim, the store selects randomly 18 of these new discount tires. When they are tested, it is found that the mean life is 38,250 miles with a sample standard deviation s = 1200 miles. Assume the distribution is normally distributed.
a. Use the critical value t0 method from the normal distribution to test for the population mean . Test the company’s claim at the level of significance = 0.05.
---------------------------
1. H0 : u = 39000
Ha : u is not equal to 39000
-----------------
2. = 5%
3. Test statistics: t = (38250-39000)/[1200/sqrt(18)] = -2.6517
4. P-value or critical : p-value = 2*P(t < -2.6517 with df = 17) = 0.0168
5. Rejection Region: t< invT(0.025 with df=17)U t> invT(0.0975 with df=17)
6. Decision: Since the p-value is less than 5%, reject Ho.
7. Interpretation: The test results do not support the company claim.
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b. Use the critical value t0 method from the normal distribution to test for the population mean m. Test the company’s claim at the level of significance a = 0.01
(References: example 1 though 5 pages 397 - 401, end of section exercises 23 – 28 pages 404 - 405) (5 points)
1. H0 :
Ha :
2. a =
3. Test statistics:
4. P-value or critical z0 or t0.
5. Rejection Region:
6. Decision:
7. Interpretation:
----
Answers are the same as for the 1st problem except for the conclusion.
Since the p-value is greater than 1%, fail to reject Ho.
The test results support the company claim.
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Cheers,
Stan H.
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