Question 221501: : (a) State the null hypothesis and the alternate hypothesis. (b) State the decision rule. (c) Compute the value of the test statistic. (d) What is your decision regarding H0? (e) What is the p-value? Interpret it.
1. The manufacturer of the X-15 steel-belted radial truck tire claims that the mean mileage the tire can be driven before the tread wears out is 60,000 miles. The standard deviation of the mileage is 5,000 miles. The Crosset Truck Company bought 48 tires and found that the mean mileage for their trucks is 59,500 miles. Is Crosset's experience different from that claimed by the manufacturer at the .05 significance level?
How do I find the decision rule?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! (a) State the null hypothesis and the alternate hypothesis.
Ho: u = 60,000
H1: u < 60,000
(b) State the decision rule.
Reject Ho if p-value is greater than 5%.
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(c) Compute the value of the test statistic.
t(59,500) = (59,500-60,000)/[500/sqrt(48)] = -6.9282..
P-value = P(t<-6.9282) = 0.0000000052515
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(d) What is your decision regarding Ho?
Since the p-value is less than 5%, Fail to reject Ho.
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(e) What is the p-value? Interpret it.
p-value is the probability the test results could give stronger
proof for rejecting Ho.
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1. The manufacturer of the X-15 steel-belted radial truck tire claims that the mean mileage the tire can be driven before the tread wears out is 60,000 miles. The standard deviation of the mileage is 5,000 miles. The Crosset Truck Company bought 48 tires and found that the mean mileage for their trucks is 59,500 miles. Is Crosset's experience different from that claimed by the manufacturer at the .05 significance level?
How do I find the decision rule?
The decision rule depends on the significance level of the test
and the type of test (one or two-tail).
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Cheers,
Stan H.
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